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Learning Objectives
• Assess the relative strengths of acids and bases according to their ionization constants
• Rationalize trends in acid–base strength in relation to molecular structure
• Carry out equilibrium calculations for weak acid–base systems
We can rank the strengths of acids by the extent to which they ionize in aqueous solution. The reaction of an acid with water is given by the general expression:
$\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq) \nonumber$
Water is the base that reacts with the acid $\ce{HA}$, $\ce{A^{−}}$ is the conjugate base of the acid $\ce{HA}$, and the hydronium ion is the conjugate acid of water. A strong acid yields 100% (or very nearly so) of $\ce{H3O+}$ and $\ce{A^{−}}$ when the acid ionizes in water; Figure $1$ lists several strong acids. A weak acid gives small amounts of $\ce{H3O+}$ and $\ce{A^{−}}$.
Figure $1$: Some of the common strong acids and bases are listed here.
Six Strong Acids Six Strong Bases
$\ce{HClO4}$ perchloric acid $\ce{LiOH}$ lithium hydroxide
$\ce{HCl}$ hydrochloric acid $\ce{NaOH}$ sodium hydroxide
$\ce{HBr}$ hydrobromic acid $\ce{KOH}$ potassium hydroxide
$\ce{HI}$ hydroiodic acid $\ce{Ca(OH)2}$ calcium hydroxide
$\ce{HNO3}$ nitric acid $\ce{Sr(OH)2}$ strontium hydroxide
$\ce{H2SO4}$ sulfuric acid $\ce{Ba(OH)2}$ barium hydroxide
The relative strengths of acids may be determined by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the acid-ionization constant, Ka. For the reaction of an acid $\ce{HA}$:
$\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq) \nonumber$
we write the equation for the ionization constant as:
$K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}} \nonumber$
where the concentrations are those at equilibrium. As noted in the section on equilibrium constants, although water is a reactant in the reaction, it is the solvent as well, so its activity has a value of 1, which does not change the value of $K_a$.
Note
It is a common error to claim that the molar concentration of the solvent is in some way involved in the equilibrium law. This error is a result of a misunderstanding of solution thermodynamics. For example, it is often claimed that Ka = Keq[H2O] for aqueous solutions. This equation is incorrect because it is an erroneous interpretation of the correct equation Ka = Keq($\textit{a}_{H_2O}$). Because $\textit{a}_{H_2O}$ = 1 for a dilute solution, Ka = Keq(1), or Ka = Keq.
The larger the $K_a$ of an acid, the larger the concentration of $\ce{H3O+}$ and $\ce{A^{−}}$ relative to the concentration of the nonionized acid, $\ce{HA}$. Thus a stronger acid has a larger ionization constant than does a weaker acid. The ionization constants increase as the strengths of the acids increase.
The following data on acid-ionization constants indicate the order of acid strength: $\ce{CH3CO2H} < \ce{HNO2} < \ce{HSO4-}$
\begin{aligned} \ce{CH3CO2H}(aq) + \ce{H2O}(l) &⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \quad &K_\ce{a}=1.8×10^{−5} \[4pt] \ce{HNO2}(aq)+\ce{H2O}(l) &⇌\ce{H3O+}(aq)+\ce{NO2-}(aq) &K_\ce{a}=4.6×10^{-4} \[4pt] \ce{HSO4-}(aq)+\ce{H2O}(l) &⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) & K_\ce{a}=1.2×10^{−2} \end{aligned} \nonumber
Another measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is the ratio of the concentration of the ionized acid to the initial acid concentration, times 100:
$\% \:\ce{ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}×100\% \label{PercentIon}$
Because the ratio includes the initial concentration, the percent ionization for a solution of a given weak acid varies depending on the original concentration of the acid, and actually decreases with increasing acid concentration.
Example $1$: Calculation of Percent Ionization from pH
Calculate the percent ionization of a 0.125-M solution of nitrous acid (a weak acid), with a pH of 2.09.
Solution
The percent ionization for an acid is:
$\ce{\dfrac{[H3O+]_{eq}}{[HNO2]_0}}×100 \nonumber$
The chemical equation for the dissociation of the nitrous acid is:
$\ce{HNO2}(aq)+\ce{H2O}(l)⇌\ce{NO2-}(aq)+\ce{H3O+}(aq). \nonumber$
Since $10^{−pH} = \ce{[H3O+]}$, we find that $10^{−2.09} = 8.1 \times 10^{−3}\, M$, so that percent ionization (Equation \ref{PercentIon}) is:
$\dfrac{8.1×10^{−3}}{0.125}×100=6.5\% \nonumber$
Remember, the logarithm 2.09 indicates a hydronium ion concentration with only two significant figures.
Exercise $1$
Calculate the percent ionization of a 0.10 M solution of acetic acid with a pH of 2.89.
Answer
1.3% ionized
We can rank the strengths of bases by their tendency to form hydroxide ions in aqueous solution. The reaction of a Brønsted-Lowry base with water is given by:
$\ce{B}(aq)+\ce{H2O}(l)⇌\ce{HB+}(aq)+\ce{OH-}(aq) \nonumber$
Water is the acid that reacts with the base, $\ce{HB^{+}}$ is the conjugate acid of the base $\ce{B}$, and the hydroxide ion is the conjugate base of water. A strong base yields 100% (or very nearly so) of OH and HB+ when it reacts with water; Figure $1$ lists several strong bases. A weak base yields a small proportion of hydroxide ions. Soluble ionic hydroxides such as NaOH are considered strong bases because they dissociate completely when dissolved in water.
As we did with acids, we can measure the relative strengths of bases by measuring their base-ionization constant (Kb) in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, $\ce{B}$:
$\ce{B}(aq)+\ce{H2O}(l)⇌\ce{HB+}(aq)+\ce{OH-}(aq), \nonumber$
we write the equation for the ionization constant as:
$K_\ce{b}=\ce{\dfrac{[HB+][OH- ]}{[B]}} \nonumber$
where the concentrations are those at equilibrium. Again, we do not see water in the equation because water is the solvent and has an activity of 1. The chemical reactions and ionization constants of the three bases shown are:
\begin{aligned} \ce{NO2-}(aq)+\ce{H2O}(l) &⇌\ce{HNO2}(aq)+\ce{OH-}(aq) \quad &K_\ce{b}=2.17×10^{−11} \[4pt] \ce{CH3CO2-}(aq)+\ce{H2O}(l) &⇌\ce{CH3CO2H}(aq)+\ce{OH-}(aq) &K_\ce{b}=5.6×10^{−10} \[4pt] \ce{NH3}(aq)+\ce{H2O}(l) &⇌\ce{NH4+}(aq)+\ce{OH-}(aq) &K_\ce{b}=1.8×10^{−5} \end{aligned} \nonumber
A table of ionization constants of weak bases appears in Table E2. As with acids, percent ionization can be measured for basic solutions, but will vary depending on the base ionization constant and the initial concentration of the solution.
Consider the ionization reactions for a conjugate acid-base pair, $\ce{HA − A^{−}}$:
$\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq) \nonumber$
with $K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}$.
$\ce{A-}(aq)+\ce{H2O}(l)⇌\ce{OH-}(aq)+\ce{HA}(aq) \nonumber$
with $K_\ce{b}=\ce{\dfrac{[HA][OH]}{[A- ]}}$.
Adding these two chemical equations yields the equation for the autoionization for water:
\begin{align*} \cancel{\ce{HA}(aq)}+\ce{H2O}(l)+\cancel{\ce{A-}(aq)}+\ce{H2O}(l) &⇌ \ce{H3O+}(aq)+\cancel{\ce{A-}(aq)}+\ce{OH-}(aq)+\cancel{\ce{HA}(aq)} \[4pt] \ce{2H2O}(l) &⇌\ce{H3O+}(aq)+\ce{OH-}(aq) \end{align*} \nonumber
As shown in the previous chapter on equilibrium, the $K$ expression for a chemical equation derived from adding two or more other equations is the mathematical product of the other equations’ $K$ expressions. Multiplying the mass-action expressions together and cancelling common terms, we see that:
$K_\ce{a}×K_\ce{b}=\ce{\dfrac{[H3O+][A- ]}{[HA]}×\dfrac{[HA][OH- ]}{[A- ]}}=\ce{[H3O+][OH- ]}=K_\ce{w} \nonumber$
For example, the acid ionization constant of acetic acid (CH3COOH) is 1.8 × 10−5, and the base ionization constant of its conjugate base, acetate ion ($\ce{CH3COO-}$), is 5.6 × 10−10. The product of these two constants is indeed equal to $K_w$:
$K_\ce{a}×K_\ce{b}=(1.8×10^{−5})×(5.6×10^{−10})=1.0×10^{−14}=K_\ce{w} \nonumber$
The extent to which an acid, $\ce{HA}$, donates protons to water molecules depends on the strength of the conjugate base, $\ce{A^{−}}$, of the acid. If $\ce{A^{−}}$ is a strong base, any protons that are donated to water molecules are recaptured by $\ce{A^{−}}$. Thus there is relatively little $\ce{A^{−}}$ and $\ce{H3O+}$ in solution, and the acid, $\ce{HA}$, is weak. If $\ce{A^{−}}$ is a weak base, water binds the protons more strongly, and the solution contains primarily $\ce{A^{−}}$ and $\ce{H3O^{+}}$—the acid is strong. Strong acids form very weak conjugate bases, and weak acids form stronger conjugate bases (Figure $2$).
Figure $3$ lists a series of acids and bases in order of the decreasing strengths of the acids and the corresponding increasing strengths of the bases. The acid and base in a given row are conjugate to each other.
The first six acids in Figure $3$ are the most common strong acids. These acids are completely dissociated in aqueous solution. The conjugate bases of these acids are weaker bases than water. When one of these acids dissolves in water, their protons are completely transferred to water, the stronger base.
Those acids that lie between the hydronium ion and water in Figure $3$ form conjugate bases that can compete with water for possession of a proton. Both hydronium ions and nonionized acid molecules are present in equilibrium in a solution of one of these acids. Compounds that are weaker acids than water (those found below water in the column of acids) in Figure $3$ exhibit no observable acidic behavior when dissolved in water. Their conjugate bases are stronger than the hydroxide ion, and if any conjugate base were formed, it would react with water to re-form the acid.
The extent to which a base forms hydroxide ion in aqueous solution depends on the strength of the base relative to that of the hydroxide ion, as shown in the last column in Figure $3$. A strong base, such as one of those lying below hydroxide ion, accepts protons from water to yield 100% of the conjugate acid and hydroxide ion. Those bases lying between water and hydroxide ion accept protons from water, but a mixture of the hydroxide ion and the base results. Bases that are weaker than water (those that lie above water in the column of bases) show no observable basic behavior in aqueous solution.
Example $2$: The Product Ka × Kb = Kw
Use the $K_b$ for the nitrite ion, $\ce{NO2-}$, to calculate the $K_a$ for its conjugate acid.
Solution
Kb for $\ce{NO2-}$ is given in this section as 2.17 × 10−11. The conjugate acid of $\ce{NO2-}$ is HNO2; Ka for HNO2 can be calculated using the relationship:
$K_\ce{a}×K_\ce{b}=1.0×10^{−14}=K_\ce{w} \nonumber$
Solving for Ka, we get:
\begin{align*} K_\ce{a} &=\dfrac{K_\ce{w}}{K_\ce{b}} \[4pt] &=\dfrac{1.0×10^{−14}}{2.17×10^{−11}} \[4pt] &=4.6×10^{−4} \end{align*} \nonumber
This answer can be verified by finding the Ka for HNO2 in Table E1
Exercise $2$
We can determine the relative acid strengths of $\ce{NH4+}$ and $\ce{HCN}$ by comparing their ionization constants. The ionization constant of $\ce{HCN}$ is given in Table E1 as 4.9 × 10−10. The ionization constant of $\ce{NH4+}$ is not listed, but the ionization constant of its conjugate base, $\ce{NH3}$, is listed as 1.8 × 10−5. Determine the ionization constant of $\ce{NH4+}$, and decide which is the stronger acid, $\ce{HCN}$ or $\ce{NH4+}$.
Answer
$\ce{NH4+}$ is the slightly stronger acid (Ka for $\ce{NH4+}$ = 5.6 × 10−10).
The Ionization of Weak Acids and Weak Bases
Many acids and bases are weak; that is, they do not ionize fully in aqueous solution. A solution of a weak acid in water is a mixture of the nonionized acid, hydronium ion, and the conjugate base of the acid, with the nonionized acid present in the greatest concentration. Thus, a weak acid increases the hydronium ion concentration in an aqueous solution (but not as much as the same amount of a strong acid).
Acetic acid ($\ce{CH3CO2H}$) is a weak acid. When we add acetic acid to water, it ionizes to a small extent according to the equation:
$\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber$
giving an equilibrium mixture with most of the acid present in the nonionized (molecular) form. This equilibrium, like other equilibria, is dynamic; acetic acid molecules donate hydrogen ions to water molecules and form hydronium ions and acetate ions at the same rate that hydronium ions donate hydrogen ions to acetate ions to reform acetic acid molecules and water molecules. We can tell by measuring the pH of an aqueous solution of known concentration that only a fraction of the weak acid is ionized at any moment (Figure $4$). The remaining weak acid is present in the nonionized form.
For acetic acid, at equilibrium:
$K_\ce{a}=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}}=1.8 \times 10^{−5} \nonumber$
Table $1$: Ionization Constants of Some Weak Acids
Ionization Reaction Ka at 25 °C
$\ce{HSO4- + H2O ⇌ H3O+ + SO4^2-}$ 1.2 × 10−2
$\ce{HF + H2O ⇌ H3O+ + F-}$ 3.5 × 10−4
$\ce{HNO2 + H2O ⇌ H3O+ + NO2-}$ 4.6 × 10−4
$\ce{HNCO + H2O ⇌ H3O+ + NCO-}$ 2 × 10−4
$\ce{HCO2H + H2O ⇌ H3O+ + HCO2-}$ 1.8 × 10−4
$\ce{CH3CO2H + H2O ⇌ H3O+ + CH3CO2-}$ 1.8 × 10−5
$\ce{HCIO + H2O ⇌ H3O+ + CIO-}$ 2.9 × 10−8
$\ce{HBrO + H2O ⇌ H3O+ + BrO-}$ 2.8 × 10−9
$\ce{HCN + H2O ⇌ H3O+ + CN-}$ 4.9 × 10−10
Table $1$ gives the ionization constants for several weak acids; additional ionization constants can be found in Table E1.
At equilibrium, a solution of a weak base in water is a mixture of the nonionized base, the conjugate acid of the weak base, and hydroxide ion with the nonionized base present in the greatest concentration. Thus, a weak base increases the hydroxide ion concentration in an aqueous solution (but not as much as the same amount of a strong base).
For example, a solution of the weak base trimethylamine, (CH3)3N, in water reacts according to the equation:
$\ce{(CH3)3N}(aq)+\ce{H2O}(l)⇌\ce{(CH3)3NH+}(aq)+\ce{OH-}(aq) \nonumber$
This gives an equilibrium mixture with most of the base present as the nonionized amine. This equilibrium is analogous to that described for weak acids.
We can confirm by measuring the pH of an aqueous solution of a weak base of known concentration that only a fraction of the base reacts with water (Figure 14.4.5). The remaining weak base is present as the unreacted form. The equilibrium constant for the ionization of a weak base, $K_b$, is called the ionization constant of the weak base, and is equal to the reaction quotient when the reaction is at equilibrium. For trimethylamine, at equilibrium:
$K_\ce{b}=\ce{\dfrac{[(CH3)3NH+][OH- ]}{[(CH3)3N]}} \nonumber$
The ionization constants of several weak bases are given in Table $2$ and Table E2.
Table $2$: Ionization Constants of Some Weak Bases
Ionization Reaction Kb at 25 °C
$\ce{(CH3)2NH + H2O ⇌ (CH3)2NH2+ + OH-}$ 5.9 × 10−4
$\ce{CH3NH2 + H2O ⇌ CH3NH3+ + OH-}$ 4.4 × 10−4
$\ce{(CH3)3N + H2O ⇌ (CH3)3NH+ + OH-}$ 6.3 × 10−5
$\ce{NH3 + H2O ⇌ NH4+ + OH-}$ 1.8 × 10−5
$\ce{C6H5NH2 + H2O ⇌ C6N5NH3+ + OH-}$ 4.3 × 10−10
Example $3$: Determination of Ka from Equilibrium Concentrations
Acetic acid is the principal ingredient in vinegar; that's why it tastes sour. At equilibrium, a solution contains [CH3CO2H] = 0.0787 M and $\ce{[H3O+]}=\ce{[CH3CO2- ]}=0.00118\:M$. What is the value of $K_a$ for acetic acid?
Solution
We are asked to calculate an equilibrium constant from equilibrium concentrations. At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:
$\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber$
\begin{align*} K_\ce{a} &=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}} \[4pt] &=\dfrac{(0.00118)(0.00118)}{0.0787} \[4pt] &=1.77×10^{−5} \end{align*} \nonumber
Exercise $3$
What is the equilibrium constant for the ionization of the $\ce{HSO4-}$ ion, the weak acid used in some household cleansers:
$\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \nonumber$
In one mixture of NaHSO4 and Na2SO4 at equilibrium, $\ce{[H3O+]}$ = 0.027 M; $\ce{[HSO4- ]}=0.29\:M$; and $\ce{[SO4^2- ]}=0.13\:M$.
Answer
$K_a$ for $\ce{HSO_4^-}= 1.2 ×\times 10^{−2}$
Example $4$: Determination of Kb from Equilibrium Concentrations
Caffeine, C8H10N4O2 is a weak base. What is the value of Kb for caffeine if a solution at equilibrium has [C8H10N4O2] = 0.050 M, $\ce{[C8H10N4O2H+]}$ = 5.0 × 10−3 M, and [OH] = 2.5 × 10−3 M?
Solution
At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:
$\ce{C8H10N4O2}(aq)+\ce{H2O}(l)⇌\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq) \nonumber$
so
$K_\ce{b}=\ce{\dfrac{[C8H10N4O2H+][OH- ]}{[C8H10N4O2]}}=\dfrac{(5.0×10^{−3})(2.5×10^{−3})}{0.050}=2.5×10^{−4} \nonumber$
Exercise $4$
What is the equilibrium constant for the ionization of the $\ce{HPO4^2-}$ ion, a weak base:
$\ce{HPO4^2-}(aq)+\ce{H2O}(l)⇌\ce{H2PO4-}(aq)+\ce{OH-}(aq) \nonumber$
In a solution containing a mixture of $\ce{NaH2PO4}$ and $\ce{Na2HPO4}$ at equilibrium with:
• $[\ce{OH^{−}}] = 1.3 × 10^{−6} M$
• $\ce{[H2PO4^{-}]=0.042\:M}$ and
• $\ce{[HPO4^{2-}]=0.341\:M}$.
Answer
Kb for $\ce{HPO4^2-}=1.6×10^{−7}$
Example $5$: Determination of Ka or Kb from pH
The pH of a 0.0516-M solution of nitrous acid, $\ce{HNO2}$, is 2.34. What is its $K_a$?
$\ce{HNO2}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{NO2-}(aq) \nonumber$
Solution
We determine an equilibrium constant starting with the initial concentrations of HNO2, $\ce{H3O+}$, and $\ce{NO2-}$ as well as one of the final concentrations, the concentration of hydronium ion at equilibrium. (Remember that pH is simply another way to express the concentration of hydronium ion.)
We can solve this problem with the following steps in which x is a change in concentration of a species in the reaction:
We can summarize the various concentrations and changes as shown here. Because water is the solvent, it has a fixed activity equal to 1. Any small amount of water produced or used up during the reaction will not change water's role as the solvent, so the value of its activity remains equal to 1 throughout the reaction.
To get the various values in the ICE (Initial, Change, Equilibrium) table, we first calculate $\ce{[H3O+]}$, the equilibrium concentration of $\ce{H3O+}$, from the pH:
$\ce{[H3O+]}=10^{−2.34}=0.0046\:M \nonumber$
The change in concentration of $\ce{H3O+}$, $x_{\ce{[H3O+]}}$, is the difference between the equilibrium concentration of H3O+, which we determined from the pH, and the initial concentration, $\mathrm{[H_3O^+]_i}$. The initial concentration of $\ce{H3O+}$ is its concentration in pure water, which is so much less than the final concentration that we approximate it as zero (~0).
The change in concentration of $\ce{NO2-}$ is equal to the change in concentration of $\ce{[H3O+]}$. For each 1 mol of $\ce{H3O+}$ that forms, 1 mol of $\ce{NO2-}$ forms. The equilibrium concentration of HNO2 is equal to its initial concentration plus the change in its concentration.
Now we can fill in the ICE table with the concentrations at equilibrium, as shown here:
Finally, we calculate the value of the equilibrium constant using the data in the table:
$K_\ce{a}=\ce{\dfrac{[H3O+][NO2- ]}{[HNO2]}}=\dfrac{(0.0046)(0.0046)}{(0.0470)}=4.5×10^{−4} \nonumber$
Exercise $5$
The pH of a solution of household ammonia, a 0.950-M solution of NH3, is 11.612. What is Kb for NH3.
Answer
$K_b = 1.8 × 10^{−5}$
Example $6$: Equilibrium Concentrations in a Solution of a Weak Acid
Formic acid, HCO2H, is the irritant that causes the body’s reaction to ant stings.
What is the concentration of hydronium ion and the pH in a 0.534-M solution of formic acid?
$\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−4} \nonumber$
Solution
1. Determine x and equilibrium concentrations. The equilibrium expression is:
$\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \nonumber$
Because water is the solvent, it has a fixed activity equal to 1. Any small amount of water produced or used up during the reaction will not change water's role as the solvent, so the value of its activity remains equal to 1 throughout the reaction so we do not need to consider it when setting up the ICE table.
The table shows initial concentrations (concentrations before the acid ionizes), changes in concentration, and equilibrium concentrations follows (the data given in the problem appear in color):
2. Solve for $x$ and the equilibrium concentrations. At equilibrium:
\begin{align*} K_\ce{a} &=1.8×10^{−4}=\ce{\dfrac{[H3O+][HCO2- ]}{[HCO2H]}} \[4pt] &=\dfrac{(x)(x)}{0.534−x}=1.8×10^{−4} \end{align*} \nonumber
Now solve for $x$. Because the initial concentration of acid is reasonably large and $K_a$ is very small, we assume that $x << 0.534$, which permits us to simplify the denominator term as $(0.534 − x) = 0.534$. This gives:
$K_\ce{a}=1.8×10^{−4}=\dfrac{x^{2}}{0.534} \nonumber$
Solve for $x$ as follows:
\begin{align*} x^2 &=0.534×(1.8×10^{−4}) \[4pt] &=9.6×10^{−5} \[4pt] x &=\sqrt{9.6×10^{−5}} \[4pt] &=9.8×10^{−3} \end{align*} \nonumber
To check the assumption that $x$ is small compared to 0.534, we calculate:
\begin{align*} \dfrac{x}{0.534} &=\dfrac{9.8×10^{−3}}{0.534} \[4pt] &=1.8×10^{−2} \, \textrm{(1.8% of 0.534)} \end{align*} \nonumber
$x$ is less than 5% of the initial concentration; the assumption is valid.
We find the equilibrium concentration of hydronium ion in this formic acid solution from its initial concentration and the change in that concentration as indicated in the last line of the table:
\begin{align*} \ce{[H3O+]} &=~0+x=0+9.8×10^{−3}\:M. \[4pt] &=9.8×10^{−3}\:M \end{align*} \nonumber
The pH of the solution can be found by taking the negative log of the $\ce{[H3O+]}$, so:
$pH = −\log(9.8×10^{−3})=2.01 \nonumber$
Exercise $6$: acetic acid
Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a 0.100-M solution of acetic acid, CH3CO2H?
$\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−5} \nonumber$
Hint
Determine $\ce{[CH3CO2- ]}$ at equilibrium.) Recall that the percent ionization is the fraction of acetic acid that is ionized × 100, or $\ce{\dfrac{[CH3CO2- ]}{[CH3CO2H]_{initial}}}×100$.
Answer
percent ionization = 1.3%
The following example shows that the concentration of products produced by the ionization of a weak base can be determined by the same series of steps used with a weak acid.
Example $7$: Equilibrium Concentrations in a Solution of a Weak Base
Find the concentration of hydroxide ion in a 0.25-M solution of trimethylamine, a weak base:
$\ce{(CH3)3N}(aq)+\ce{H2O}(l)⇌\ce{(CH3)3NH+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=6.3×10^{−5} \nonumber$
Solution This problem requires that we calculate an equilibrium concentration by determining concentration changes as the ionization of a base goes to equilibrium. The solution is approached in the same way as that for the ionization of formic acid in Example $6$. The reactants and products will be different and the numbers will be different, but the logic will be the same:
1. Determine x and equilibrium concentrations. The table shows the changes and concentrations:
2. Solve for $x$ and the equilibrium concentrations. At equilibrium:
$K_\ce{b}=\ce{\dfrac{[(CH3)3NH+][OH- ]}{[(CH3)3N]}}=\dfrac{(x)(x)}{0.25−x=}6.3×10^{−5} \nonumber$
If we assume that x is small relative to 0.25, then we can replace (0.25 − x) in the preceding equation with 0.25. Solving the simplified equation gives:
$x=4.0×10^{−3} \nonumber$
This change is less than 5% of the initial concentration (0.25), so the assumption is justified.
Recall that, for this computation, $x$ is equal to the equilibrium concentration of hydroxide ion in the solution (see earlier tabulation):
\begin{align*} (\ce{[OH- ]}=~0+x=x=4.0×10^{−3}\:M \[4pt] &=4.0×10^{−3}\:M \end{align*} \nonumber
Then calculate pOH as follows:
$\ce{pOH}=−\log(4.3×10^{−3})=2.40 \nonumber$
Using the relation introduced in the previous section of this chapter:
$\mathrm{pH + pOH=p\mathit{K}_w=14.00}\nonumber$
permits the computation of pH:
$\mathrm{pH=14.00−pOH=14.00−2.37=11.60} \nonumber$
Check the work. A check of our arithmetic shows that $K_b = 6.3 \times 10^{−5}$.
Exercise $7$
1. Show that the calculation in Step 2 of this example gives an x of 4.3 × 10−3 and the calculation in Step 3 shows Kb = 6.3 × 10−5.
2. Find the concentration of hydroxide ion in a 0.0325-M solution of ammonia, a weak base with a Kb of 1.76 × 10−5. Calculate the percent ionization of ammonia, the fraction ionized × 100, or $\ce{\dfrac{[NH4+]}{[NH3]}}×100 \%$
Answer a
$7.56 × 10^{−4}\, M$, 2.33%
Answer b
2.33%
Some weak acids and weak bases ionize to such an extent that the simplifying assumption that x is small relative to the initial concentration of the acid or base is inappropriate. As we solve for the equilibrium concentrations in such cases, we will see that we cannot neglect the change in the initial concentration of the acid or base, and we must solve the equilibrium equations by using the quadratic equation.
Example $8$: Equilibrium Concentrations in a Solution of a Weak Acid
Sodium bisulfate, NaHSO4, is used in some household cleansers because it contains the $\ce{HSO4-}$ ion, a weak acid. What is the pH of a 0.50-M solution of $\ce{HSO4-}$?
$\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \hspace{20px} K_\ce{a}=1.2×10^{−2} \nonumber$
Solution
We need to determine the equilibrium concentration of the hydronium ion that results from the ionization of $\ce{HSO4-}$ so that we can use $\ce{[H3O+]}$ to determine the pH. As in the previous examples, we can approach the solution by the following steps:
1. Determine $x$ and equilibrium concentrations. This table shows the changes and concentrations:
2. Solve for $x$ and the concentrations.
As we begin solving for $x$, we will find this is more complicated than in previous examples. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of $x$.
At equilibrium:
$K_\ce{a}=1.2×10^{−2}=\ce{\dfrac{[H3O+][SO4^2- ]}{[HSO4- ]}}=\dfrac{(x)(x)}{0.50−x} \nonumber$
If we assume that x is small and approximate (0.50 − x) as 0.50, we find:
$x=7.7×10^{−2} \nonumber$
When we check the assumption, we confirm:
$\dfrac{x}{\mathrm{[HSO_4^- ]_i}} \overset{?}{\le} 0.05 \nonumber$
which for this system is
$\dfrac{x}{0.50}=\dfrac{7.7×10^{−2}}{0.50}=0.15(15\%) \nonumber$
The value of $x$ is not less than 5% of 0.50, so the assumption is not valid. We need the quadratic formula to find $x$.
The equation:
$K_\ce{a}=1.2×10^{−2}=\dfrac{(x)(x)}{0.50−x}\nonumber$
gives
$6.0×10^{−3}−1.2×10^{−2}x=x^{2+} \nonumber$
or
$x^{2+}+1.2×10^{−2}x−6.0×10^{−3}=0 \nonumber$
This equation can be solved using the quadratic formula. For an equation of the form
$ax^{2+} + bx + c=0, \nonumber$
$x$ is given by the quadratic equation:
$x=\dfrac{−b±\sqrt{b^{2+}−4ac}}{2a} \nonumber$
In this problem, $a = 1$, $b = 1.2 × 10^{−3}$, and $c = −6.0 × 10^{−3}$.
Solving for x gives a negative root (which cannot be correct since concentration cannot be negative) and a positive root:
$x=7.2×10^{−2} \nonumber$
Now determine the hydronium ion concentration and the pH:
\begin{align*} \ce{[H3O+]} &=~0+x=0+7.2×10^{−2}\:M \[4pt] &=7.2×10^{−2}\:M \end{align*} \nonumber
The pH of this solution is:
$\mathrm{pH=−log[H_3O^+]=−log7.2×10^{−2}=1.14} \nonumber$
Exercise $8$
1. Show that the quadratic formula gives $x = 7.2 × 10^{−2}$.
2. Calculate the pH in a 0.010-M solution of caffeine, a weak base:
$\ce{C8H10N4O2}(aq)+\ce{H2O}(l)⇌\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=2.5×10^{−4} \nonumber$
Hint
It will be necessary to convert [OH] to $\ce{[H3O+]}$ or pOH to pH toward the end of the calculation.
Answer
pH 11.16
The Relative Strengths of Strong Acids and Bases
Strong acids, such as $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$, all exhibit the same strength in water. The water molecule is such a strong base compared to the conjugate bases Cl, Br, and I that ionization of these strong acids is essentially complete in aqueous solutions. In solvents less basic than water, we find $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$ differ markedly in their tendency to give up a proton to the solvent. For example, when dissolved in ethanol (a weaker base than water), the extent of ionization increases in the order $\ce{HCl < HBr < HI}$, and so $\ce{HI}$ is demonstrated to be the strongest of these acids. The inability to discern differences in strength among strong acids dissolved in water is known as the leveling effect of water.
Water also exerts a leveling effect on the strengths of strong bases. For example, the oxide ion, O2−, and the amide ion, $\ce{NH2-}$, are such strong bases that they react completely with water:
$\ce{O^2-}(aq)+\ce{H2O}(l)⟶\ce{OH-}(aq)+\ce{OH-}(aq) \nonumber$
$\ce{NH2-}(aq)+\ce{H2O}(l)⟶\ce{NH3}(aq)+\ce{OH-}(aq) \nonumber$
Thus, O2− and $\ce{NH2-}$ appear to have the same base strength in water; they both give a 100% yield of hydroxide ion.
In the absence of any leveling effect, the acid strength of binary compounds of hydrogen with nonmetals (A) increases as the H-A bond strength decreases down a group in the periodic table. For group 17, the order of increasing acidity is $\ce{HF < HCl < HBr < HI}$. Likewise, for group 16, the order of increasing acid strength is H2O < H2S < H2Se < H2Te. Across a row in the periodic table, the acid strength of binary hydrogen compounds increases with increasing electronegativity of the nonmetal atom because the polarity of the H-A bond increases. Thus, the order of increasing acidity (for removal of one proton) across the second row is $\ce{CH4 < NH3 < H2O < HF}$; across the third row, it is $\ce{SiH4 < PH3 < H2S < HCl}$ (see Figure $6$).
Compounds containing oxygen and one or more hydroxyl (OH) groups can be acidic, basic, or amphoteric, depending on the position in the periodic table of the central atom E, the atom bonded to the hydroxyl group. Such compounds have the general formula OnE(OH)m, and include sulfuric acid, $\ce{O2S(OH)2}$, sulfurous acid, $\ce{OS(OH)2}$, nitric acid, $\ce{O2NOH}$, perchloric acid, $\ce{O3ClOH}$, aluminum hydroxide, $\ce{Al(OH)3}$, calcium hydroxide, $\ce{Ca(OH)2}$, and potassium hydroxide, $\ce{KOH}$:
If the central atom, E, has a low electronegativity, its attraction for electrons is low. Little tendency exists for the central atom to form a strong covalent bond with the oxygen atom, and bond a between the element and oxygen is more readily broken than bond b between oxygen and hydrogen. Hence bond a is ionic, hydroxide ions are released to the solution, and the material behaves as a base—this is the case with Ca(OH)2 and KOH. Lower electronegativity is characteristic of the more metallic elements; hence, the metallic elements form ionic hydroxides that are by definition basic compounds.
If, on the other hand, the atom E has a relatively high electronegativity, it strongly attracts the electrons it shares with the oxygen atom, making bond a relatively strongly covalent. The oxygen-hydrogen bond, bond b, is thereby weakened because electrons are displaced toward E. Bond b is polar and readily releases hydrogen ions to the solution, so the material behaves as an acid. High electronegativities are characteristic of the more nonmetallic elements. Thus, nonmetallic elements form covalent compounds containing acidic −OH groups that are called oxyacids.
Increasing the oxidation number of the central atom E also increases the acidity of an oxyacid because this increases the attraction of E for the electrons it shares with oxygen and thereby weakens the O-H bond. Sulfuric acid, H2SO4, or O2S(OH)2 (with a sulfur oxidation number of +6), is more acidic than sulfurous acid, H2SO3, or OS(OH)2 (with a sulfur oxidation number of +4). Likewise nitric acid, HNO3, or O2NOH (N oxidation number = +5), is more acidic than nitrous acid, HNO2, or ONOH (N oxidation number = +3). In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure $7$).
Hydroxy compounds of elements with intermediate electronegativities and relatively high oxidation numbers (for example, elements near the diagonal line separating the metals from the nonmetals in the periodic table) are usually amphoteric. This means that the hydroxy compounds act as acids when they react with strong bases and as bases when they react with strong acids. The amphoterism of aluminum hydroxide, which commonly exists as the hydrate $\ce{Al(H2O)3(OH)3}$, is reflected in its solubility in both strong acids and strong bases. In strong bases, the relatively insoluble hydrated aluminum hydroxide, $\ce{Al(H2O)3(OH)3}$, is converted into the soluble ion, $\ce{[Al(H2O)2(OH)4]-}$, by reaction with hydroxide ion:
$[\ce{Al(H2O)3(OH)3}](aq)+\ce{OH-}(aq)⇌\ce{H2O}(l)+\ce{[Al(H2O)2(OH)4]-}(aq) \nonumber$
In this reaction, a proton is transferred from one of the aluminum-bound H2O molecules to a hydroxide ion in solution. The $\ce{Al(H2O)3(OH)3}$ compound thus acts as an acid under these conditions. On the other hand, when dissolved in strong acids, it is converted to the soluble ion $\ce{[Al(H2O)6]^3+}$ by reaction with hydronium ion:
$\ce{3H3O+}(aq)+\ce{Al(H2O)3(OH)3}(aq)⇌\ce{Al(H2O)6^3+}(aq)+\ce{3H2O}(l) \nonumber$
In this case, protons are transferred from hydronium ions in solution to $\ce{Al(H2O)3(OH)3}$, and the compound functions as a base.
Summary
The strengths of Brønsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons. Strong bases react with water to quantitatively form hydroxide ions. Weak bases give only small amounts of hydroxide ion. The strengths of the binary acids increase from left to right across a period of the periodic table (CH4 < NH3 < H2O < HF), and they increase down a group (HF < HCl < HBr < HI). The strengths of oxyacids that contain the same central element increase as the oxidation number of the element increases (H2SO3 < H2SO4). The strengths of oxyacids also increase as the electronegativity of the central element increases [H2SeO4 < H2SO4].
Key Equations
• $K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}$
• $K_\ce{b}=\ce{\dfrac{[HB+][OH- ]}{[B]}}$
• $K_a \times K_b = 1.0 \times 10^{−14} = K_w \,(\text{at room temperature})$
• $\textrm{Percent ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}×100$
Glossary
acid ionization constant (Ka)
equilibrium constant for the ionization of a weak acid
base ionization constant (Kb)
equilibrium constant for the ionization of a weak base
leveling effect of water
any acid stronger than $\ce{H3O+}$, or any base stronger than OH will react with water to form $\ce{H3O+}$, or OH, respectively; water acts as a base to make all strong acids appear equally strong, and it acts as an acid to make all strong bases appear equally strong
oxyacid
compound containing a nonmetal and one or more hydroxyl groups
percent ionization
ratio of the concentration of the ionized acid to the initial acid concentration, times 100 | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/16%3A_AcidBase_Equilibria/16.06%3A_Weak_Acids.txt |
The pH of a solution of a weak base can be calculated in a way which is very similar to that used for a weak acid. Instead of an acid constant Ka, a base constant Kb must be used. If a weak base B accepts protons from water according to the equation
$\text{B} + \text{ H}_{\text{2}}\text{O}\rightleftharpoons\text{BH}^{+} + \text{OH}^{-} \label{1}$
then the base constant is defined by the expression
$K_{b}=\dfrac{ \text{ BH}^{\text{+}} \text{ OH}^{-} }{ \text{ B } } \label{2}$
A list of Kb values for selected bases arranged in order of strength is given in the table below. This table is part of our larger collection of acid-base resources.
Table $1$: The Base Constants for Some Bases at 25°C. Taken from Hogfelt, E. Perrin, D. D. Stability Constants of Metal Ion Complexes, 1st ed. Oxford; New Pergamon, 1979-1982. International Union of Pure and Applied Chemistry, Commission on Equilibrium. ISBN: 0080209580
Base Formula and Ionization Equation Kb Molecular Shape
Ammonia $NH_3 + H_2O \rightleftharpoons NH^+_4 + OH^–$ 1.77 × 10–5
Aniline $C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5NH^+_3 + OH^–$ 3.9 × 10–10
Carbonate ion $CO_3^{2–} + H_2O \rightleftharpoons HCO^-_3 + OH^–$ 2.1 × 10–4
Hydrazine $N_2H_4 + H_2O \rightleftharpoons N_2H^+_5 + OH^–$
$N_2H^+_5 + H_2O \rightleftharpoons N_2H_6^{2+} + OH^–$
K1 = 1.2 × 10–6
K2 = 1.3 × 10–15
Hydride ion $H^– + H_2O \rightarrow H_2 + OH^–$ 1.0
Phosphate ion $PO_4^{3–} + H_2O \rightleftharpoons HPO^{2-}_4 + OH^–$ 5.9 × 10–3
Pyridine $C_5H_5N + H_2O \rightleftharpoons C_5H_5NH^+ + OH^–$ 1.6 × 10–9
To find the pH we follow the same general procedure as in the case of a weak acid. If the stoichiometric concentration of the base is indicated by cb, the result is entirely analogous to equation 4 in the section on the pH of weak acids; namely,
$K_{b}=\dfrac{ [\text{OH}^{-}]^2}{c_{b}- [\text{ OH}^{-}] } \label{3}$
Under most circumstances we can make the approximation
$c_b – [OH^–] \approx c_b \nonumber$
in which case Equation \ref{3} reduces to the approximation
$[OH^–] ≈ \sqrt{K_{b}c_{b}} \label{4}$
which is identical to the expression obtained in the acid case (approximation shown in equation 6 in the section on the pH of weak acids) except that OH replaces H3O+ and b replaces a. Once we have found the hydroxide-ion concentration from this approximation, we can then easily find the pOH, and from it the pH.
Example $1$: pH using Kb
Using the value for Kb listed in the table, find the pH of 0.100 M NH3.
Solution
It is not a bad idea to guess an approximate pH before embarking on the calculation. Since we have a dilute solution of a weak base, we expect the solution to be only mildly basic. A pH of 13 or 14 would be too basic, while a pH of 8 or 9 is too close to neutral. A pH of 10 or 11 seems reasonable. Using Equation \ref{4} we have
\begin{align*} [\text{ OH}^{-}] &=\sqrt{K_{b}c_{b}} \[4pt] & =\sqrt{\text{1.8 }\times \text{ 10}^{-\text{5}}\text{ mol L}^{-\text{1}} \times \text{ 0.100 mol L}^{-\text{1}}} \[4pt] &=\sqrt{\text{1.8 }\times \text{ 10}^{-\text{6}}\text{ mol}^{\text{2}}\text{ L}^{-2}} \[4pt] &=\text{1.34 }\times \text{ 10}^{-\text{3}}\text{ mol L}^{-\text{1}} \end{align*} \nonumber
Checking the accuracy of the approximation, we find
$\dfrac{ [\text{ OH}^{-} ]}{c_{\text{b}}}=\dfrac{\text{1.34 }\times \text{ 10}^{-\text{3}}}{\text{0.1}}\approx \text{1 percent}$
The approximation is valid, and we thus proceed to find the pOH.
$\text{pOH}=-\text{log}\dfrac{ [\text{ OH}^{-} ]}{\text{mol L}^{-\text{1}}}=-\text{log(1.34 }\times \text{ 10}^{-\text{3}}\text{)}=\text{2.87}$
From which
$pH = 14.00 – pOH = 14.00 – 2.87 = 11.13 \nonumber$
This calculated value checks well with our initial guess.
Occasionally we will find that the approximation
$c_b – [OH^{–}] ≈ c_b \nonumber$
is not valid, in which case we must use a series of successive approximations similar to that outlined above for acids. The appropriate formula can be derived from Equation \ref{3} and reads
$[OH^{-}] \approx \sqrt{K_{b} ( c_b - [OH^{-}] )} \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/16%3A_AcidBase_Equilibria/16.07%3A_Weak_Bases.txt |
Learning Objectives
• To know the relationship between acid or base strength and the magnitude of $K_a$, $K_b$, $pK_a$, and $pK_b$.
The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For example, the general equation for the ionization of a weak acid in water, where HA is the parent acid and A− is its conjugate base, is as follows:
$HA_{(aq)}+H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)}+A^−_{(aq)} \label{16.5.1}$
The equilibrium constant for this dissociation is as follows:
$K=\dfrac{[H_3O^+][A^−]}{[HA]} \label{16.5.2}$
The equilibrium constant for this reaction is the acid ionization constant $K_a$, also called the acid dissociation constant:
$K_a=\dfrac{[H_3O^+][A^−]}{[HA]} \label{16.5.3}$
Thus the numerical values of K and $K_a$ differ by the concentration of water (55.3 M). Again, for simplicity, $H_3O^+$ can be written as $H^+$ in Equation $\ref{16.5.3}$. Keep in mind, though, that free $H^+$ does not exist in aqueous solutions and that a proton is transferred to $H_2O$ in all acid ionization reactions to form $H^3O^+$. The larger the $K_a$, the stronger the acid and the higher the $H^+$ concentration at equilibrium. Like all equilibrium constants, acid–base ionization constants are actually measured in terms of the activities of $H^+$ or $OH^−$, thus making them unitless. The values of $K_a$ for a number of common acids are given in Table $1$.
Table $1$: Values of $K_a$, $pK_a$, $K_b$, and $pK_b$ for Selected Acids ($HA$ and Their Conjugate Bases ($A^−$)
Acid $HA$ $K_a$ $pK_a$ $A^−$ $K_b$ $pK_b$
*The number in parentheses indicates the ionization step referred to for a polyprotic acid.
hydroiodic acid $HI$ $2 \times 10^{9}$ −9.3 $I^−$ $5.5 \times 10^{−24}$ 23.26
sulfuric acid (1)* $H_2SO_4$ $1 \times 10^{2}$ −2.0 $HSO_4^−$ $1 \times 10^{−16}$ 16.0
nitric acid $HNO_3$ $2.3 \times 10^{1}$ −1.37 $NO_3^−$ $4.3 \times 10^{−16}$ 15.37
hydronium ion $H_3O^+$ $1.0$ 0.00 $H_2O$ $1.0 \times 10^{−14}$ 14.00
sulfuric acid (2)* $HSO_4^−$ $1.0 \times 10^{−2}$ 1.99 $SO_4^{2−}$ $9.8 \times 10^{−13}$ 12.01
hydrofluoric acid $HF$ $6.3 \times 10^{−4}$ 3.20 $F^−$ $1.6 \times 10^{−11}$ 10.80
nitrous acid $HNO_2$ $5.6 \times 10^{−4}$ 3.25 $NO2^−$ $1.8 \times 10^{−11}$ 10.75
formic acid $HCO_2H$ $1.78 \times 10^{−4}$ 3.750 $HCO_2−$ $5.6 \times 10^{−11}$ 10.25
benzoic acid $C_6H_5CO_2H$ $6.3 \times 10^{−5}$ 4.20 $C_6H_5CO_2^−$ $1.6 \times 10^{−10}$ 9.80
acetic acid $CH_3CO_2H$ $1.7 \times 10^{−5}$ 4.76 $CH_3CO_2^−$ $5.8 \times 10^{−10}$ 9.24
pyridinium ion $C_5H_5NH^+$ $5.9 \times 10^{−6}$ 5.23 $C_5H_5N$ $1.7 \times 10^{−9}$ 8.77
hypochlorous acid $HOCl$ $4.0 \times 10^{−8}$ 7.40 $OCl^−$ $2.5 \times 10^{−7}$ 6.60
hydrocyanic acid $HCN$ $6.2 \times 10^{−10}$ 9.21 $CN^−$ $1.6 \times 10^{−5}$ 4.79
ammonium ion $NH_4^+$ $5.6 \times 10^{−10}$ 9.25 $NH_3$ $1.8 \times 10^{−5}$ 4.75
water $H_2O$ $1.0 \times 10^{−14}$ 14.00 $OH^−$ $1.00$ 0.00
acetylene $C_2H_2$ $1 \times 10^{−26}$ 26.0 $HC_2^−$ $1 \times 10^{12}$ −12.0
ammonia $NH_3$ $1 \times 10^{−35}$ 35.0 $NH_2^−$ $1 \times 10^{21}$ −21.0
Weak bases react with water to produce the hydroxide ion, as shown in the following general equation, where B is the parent base and BH+ is its conjugate acid:
$B_{(aq)}+H_2O_{(l)} \rightleftharpoons BH^+_{(aq)}+OH^−_{(aq)} \label{16.5.4}$
The equilibrium constant for this reaction is the base ionization constant (Kb), also called the base dissociation constant:
$K_b=\dfrac{[BH^+][OH^−]}{[B]} \label{16.5.5}$
Once again, the concentration does not appear in the equilibrium constant expression.. The larger the $K_b$, the stronger the base and the higher the $OH^−$ concentration at equilibrium. The values of $K_b$ for a number of common weak bases are given in Table $2$.
Table $2$: Values of $K_b$, $pK_b$, $K_a$, and $pK_a$ for Selected Weak Bases (B) and Their Conjugate Acids (BH+)
Base $B$ $K_b$ $pK_b$ $BH^+$ $K_a$ $pK_a$
*As in Table $1$.
hydroxide ion $OH^−$ $1.0$ 0.00* $H_2O$ $1.0 \times 10^{−14}$ 14.00
phosphate ion $PO_4^{3−}$ $2.1 \times 10^{−2}$ 1.68 $HPO_4^{2−}$ $4.8 \times 10^{−13}$ 12.32
dimethylamine $(CH_3)_2NH$ $5.4 \times 10^{−4}$ 3.27 $(CH_3)_2NH_2^+$ $1.9 \times 10^{−11}$ 10.73
methylamine $CH_3NH_2$ $4.6 \times 10^{−4}$ 3.34 $CH_3NH_3^+$ $2.2 \times 10^{−11}$ 10.66
trimethylamine $(CH_3)_3N$ $6.3 \times 10^{−5}$ 4.20 $(CH_3)_3NH^+$ $1.6 \times 10^{−10}$ 9.80
ammonia $NH_3$ $1.8 \times 10^{−5}$ 4.75 $NH_4^+$ $5.6 \times 10^{−10}$ 9.25
pyridine $C_5H_5N$ $1.7 \times 10^{−9}$ 8.77 $C_5H_5NH^+$ $5.9 \times 10^{−6}$ 5.23
aniline $C_6H_5NH_2$ $7.4 \times 10^{−10}$ 9.13 $C_6H_5NH_3^+$ $1.3 \times 10^{−5}$ 4.87
water $H_2O$ $1.0 \times 10^{−14}$ 14.00 $H_3O^+$ $1.0^*$ 0.00
There is a simple relationship between the magnitude of $K_a$ for an acid and $K_b$ for its conjugate base. Consider, for example, the ionization of hydrocyanic acid ($HCN$) in water to produce an acidic solution, and the reaction of $CN^−$ with water to produce a basic solution:
$HCN_{(aq)} \rightleftharpoons H^+_{(aq)}+CN^−_{(aq)} \label{16.5.6}$
$CN^−_{(aq)}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+HCN_{(aq)} \label{16.5.7}$
The equilibrium constant expression for the ionization of HCN is as follows:
$K_a=\dfrac{[H^+][CN^−]}{[HCN]} \label{16.5.8}$
The corresponding expression for the reaction of cyanide with water is as follows:
$K_b=\dfrac{[OH^−][HCN]}{[CN^−]} \label{16.5.9}$
If we add Equations $\ref{16.5.6}$ and $\ref{16.5.7}$, we obtain the following (recall that the equilibrium constant for the sum of two reactions is the product of the equilibrium constants for the individual reactions):
$\cancel{HCN_{(aq)}} \rightleftharpoons H^+_{(aq)}+\cancel{CN^−_{(aq)}} \;\;\; K_a=[H^+]\cancel{[CN^−]}/\cancel{[HCN]}$
$\cancel{CN^−_{(aq)}}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+\cancel{HCN_{(aq)}} \;\;\; K_b=[OH^−]\cancel{[HCN]}/\cancel{[CN^−]}$
$H_2O_{(l)} \rightleftharpoons H^+_{(aq)}+OH^−_{(aq)} \;\;\; K=K_a \times K_b=[H^+][OH^−]$
In this case, the sum of the reactions described by $K_a$ and $K_b$ is the equation for the autoionization of water, and the product of the two equilibrium constants is $K_w$:
$K_aK_b = K_w \label{16.5.10}$
Thus if we know either $K_a$ for an acid or $K_b$ for its conjugate base, we can calculate the other equilibrium constant for any conjugate acid–base pair.
Just as with $pH$, $pOH$, and pKw, we can use negative logarithms to avoid exponential notation in writing acid and base ionization constants, by defining $pK_a$ as follows:
$pKa = −\log_{10}K_a \label{16.5.11}$
$K_a=10^{−pK_a} \label{16.5.12}$
and $pK_b$ as
$pK_b = −\log_{10}K_b \label{16.5.13}$
$K_b=10^{−pK_b} \label{16.5.14}$
Similarly, Equation 16.5.10, which expresses the relationship between $K_a$ and $K_b$, can be written in logarithmic form as follows:
$pK_a + pK_b = pK_w \label{16.5.15}$
At 25°C, this becomes
$pK_a + pK_b = 14.00 \label{16.5.16}$
The values of $pK_a$ and $pK_b$ are given for several common acids and bases in Table 16.5.1 and Table 16.5.2, respectively, and a more extensive set of data is provided in Tables E1 and E2. Because of the use of negative logarithms, smaller values of $pK_a$ correspond to larger acid ionization constants and hence stronger acids. For example, nitrous acid ($HNO_2$), with a $pK_a$ of 3.25, is about a 1000 times stronger acid than hydrocyanic acid (HCN), with a $pK_a$ of 9.21. Conversely, smaller values of $pK_b$ correspond to larger base ionization constants and hence stronger bases.
The relative strengths of some common acids and their conjugate bases are shown graphically in Figure 16.5. The conjugate acid–base pairs are listed in order (from top to bottom) of increasing acid strength, which corresponds to decreasing values of $pK_a$. This order corresponds to decreasing strength of the conjugate base or increasing values of $pK_b$. At the bottom left of Figure 16.5.2 are the common strong acids; at the top right are the most common strong bases. Notice the inverse relationship between the strength of the parent acid and the strength of the conjugate base. Thus the conjugate base of a strong acid is a very weak base, and the conjugate base of a very weak acid is a strong base.
The conjugate base of a strong acid is a weak base and vice versa.
We can use the relative strengths of acids and bases to predict the direction of an acid–base reaction by following a single rule: an acid–base equilibrium always favors the side with the weaker acid and base, as indicated by these arrows:
$\text{stronger acid + stronger base} \ce{ <=>>} \text{weaker acid + weaker base}$
In an acid–base reaction, the proton always reacts with the stronger base.
For example, hydrochloric acid is a strong acid that ionizes essentially completely in dilute aqueous solution to produce $H_3O^+$ and $Cl^−$; only negligible amounts of $HCl$ molecules remain undissociated. Hence the ionization equilibrium lies virtually all the way to the right, as represented by a single arrow:
$HCl_{(aq)} + H_2O_{(l)} \rightarrow \rightarrow H_3O^+_{(aq)}+Cl^−_{(aq)} \label{16.5.17}$
In contrast, acetic acid is a weak acid, and water is a weak base. Consequently, aqueous solutions of acetic acid contain mostly acetic acid molecules in equilibrium with a small concentration of $H_3O^+$ and acetate ions, and the ionization equilibrium lies far to the left, as represented by these arrows:
$\ce{ CH_3CO_2H_{(aq)} + H_2O_{(l)} <<=> H_3O^+_{(aq)} + CH_3CO_{2(aq)}^- }$
Similarly, in the reaction of ammonia with water, the hydroxide ion is a strong base, and ammonia is a weak base, whereas the ammonium ion is a stronger acid than water. Hence this equilibrium also lies to the left:
$H_2O_{(l)} + NH_{3(aq)} \ce{ <<=>} NH^+_{4(aq)} + OH^-_{(aq)}$
All acid–base equilibria favor the side with the weaker acid and base. Thus the proton is bound to the stronger base.
Example $1$: Butyrate and Dimethylammonium Ions
1. Calculate $K_b$ and $pK_b$ of the butyrate ion ($CH_3CH_2CH_2CO_2^−$). The $pK_a$ of butyric acid at 25°C is 4.83. Butyric acid is responsible for the foul smell of rancid butter.
2. Calculate $K_a$ and $pK_a$ of the dimethylammonium ion ($(CH_3)_2NH_2^+$). The base ionization constant $K_b$ of dimethylamine ($(CH_3)_2NH$) is $5.4 \times 10^{−4}$ at 25°C.
Given: $pK_a$ and $K_b$
Asked for: corresponding $K_b$ and $pK_b$, $K_a$ and $pK_a$
Strategy:
The constants $K_a$ and $K_b$ are related as shown in Equation 16.5.10. The $pK_a$ and $pK_b$ for an acid and its conjugate base are related as shown in Equation 16.5.15 and Equation 16.5.16. Use the relationships pK = −log K and K = 10−pK (Equation 16.5.11 and Equation 16.5.13) to convert between $K_a$ and $pK_a$ or $K_b$ and $pK_b$.
Solution:
We are given the $pK_a$ for butyric acid and asked to calculate the $K_b$ and the $pK_b$ for its conjugate base, the butyrate ion. Because the $pK_a$ value cited is for a temperature of 25°C, we can use Equation 16.5.16: $pK_a$ + $pK_b$ = pKw = 14.00. Substituting the $pK_a$ and solving for the $pK_b$,
$4.83+pK_b=14.00$
$pK_b=14.00−4.83=9.17$
Because $pK_b = −\log K_b$, $K_b$ is $10^{−9.17} = 6.8 \times 10^{−10}$.
In this case, we are given $K_b$ for a base (dimethylamine) and asked to calculate $K_a$ and $pK_a$ for its conjugate acid, the dimethylammonium ion. Because the initial quantity given is $K_b$ rather than $pK_b$, we can use Equation 16.5.10: $K_aK_b = K_w$. Substituting the values of $K_b$ and $K_w$ at 25°C and solving for $K_a$,
$K_a(5.4 \times 10^{−4})=1.01 \times 10^{−14}$
$K_a=1.9 \times 10^{−11}$
Because $pK_a$ = −log $K_a$, we have $pK_a = −\log(1.9 \times 10^{−11}) = 10.72$. We could also have converted $K_b$ to $pK_b$ to obtain the same answer:
$pK_b=−\log(5.4 \times 10^{−4})=3.27$
$pKa+pK_b=14.00$
$pK_a=10.73$
$K_a=10^{−pK_a}=10^{−10.73}=1.9 \times 10^{−11}$
If we are given any one of these four quantities for an acid or a base ($K_a$, $pK_a$, $K_b$, or $pK_b$), we can calculate the other three.
Exercise $1$: Lactic Acid
Lactic acid ($CH_3CH(OH)CO_2H$) is responsible for the pungent taste and smell of sour milk; it is also thought to produce soreness in fatigued muscles. Its $pK_a$ is 3.86 at 25°C. Calculate $K_a$ for lactic acid and $pK_b$ and $K_b$ for the lactate ion.
Answer
$K_a = 1.4 \times 10^{−4}$ for lactic acid;
$pK_b$ = 10.14 and $K_b = 7.2 \times 10^{−11}$ for the lactate ion
Summary
Two species that differ by only a proton constitute a conjugate acid–base pair. The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For an aqueous solution of a weak acid, the dissociation constant is called the acid ionization constant (Ka). Similarly, the equilibrium constant for the reaction of a weak base with water is the base ionization constant (Kb). For any conjugate acid–base pair, $K_aK_b = K_w$. Smaller values of $pK_a$ correspond to larger acid ionization constants and hence stronger acids. Conversely, smaller values of $pK_b$ correspond to larger base ionization constants and hence stronger bases. At 25°C, $pK_a + pK_b = 14.00$. Acid–base reactions always proceed in the direction that produces the weaker acid–base pair.
Key Takeaways
• The Ka and Kb values for a conjugated acid–base pairs are related through the Kw value: $K_aK_b = K_w$
• The conjugate base of a strong acid is a very weak base, and the conjugate base of a very weak acid is a strong base.
Key Equations
• Acid ionization constant: $K_a=\dfrac{[H_3O^+][A^−]}{[HA]}$
• Base ionization constant: $K_b=\dfrac{[BH^+][OH^−]}{[B]}$
• Relationship between $K_a$ and $K_b$ of a conjugate acid–base pair: $K_aK_b = K_w$
• Definition of $pK_a$: $pKa = −\log_{10}K_a \nonumber$ $K_a=10^{−pK_a}$
• Definition of $pK_b$: $pK_b = −\log_{10}K_b \nonumber$ $K_b=10^{−pK_b}$
• Relationship between $pK_a$ and $pK_b$ of a conjugate acid–base pair:
$pK_a + pK_b = pK_w$
$pK_a + pK_b = 14.00 \; \text{at 25°C}$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/16%3A_AcidBase_Equilibria/16.08%3A_Relationship_Between_Ka_and_Kb.txt |
Learning Objectives
• To recognize salts that will produce acidic, basic, or neutral solutions in water
• To understand the Lewis acidity of small, highly-charged metal ions in water
A neutralization reaction can be defined as the reaction of an acid and a base to produce a salt and water. That is, another cation, such as $Na^+$, replaces the proton on the acid. An example is the reaction of $CH_3CO_2H$, a weak acid, with $NaOH$, a strong base:
$\underset{acid}{CH_3CO_2H_{(l)}} +\underset{base}{NaOH_{(s)}} \overset{H_2O}{\longrightarrow} \underset{salt}{H_2OCH_3CO_2Na_{(aq)} }+\underset{water}{H_2O_{(l)}} \nonumber$
Depending on the acid–base properties of its component ions, however, a salt can dissolve in water to produce a neutral solution, a basic solution, or an acidic solution.
When a salt such as $NaCl$ dissolves in water, it produces $Na^+_{(aq)}$ and $Cl^−_{(aq)}$ ions. Using a Lewis approach, the $Na^+$ ion can be viewed as an acid because it is an electron pair acceptor, although its low charge and relatively large radius make it a very weak acid. The $Cl^−$ ion is the conjugate base of the strong acid $HCl$, so it has essentially no basic character. Consequently, dissolving $NaCl$ in water has no effect on the $pH$ of a solution, and the solution remains neutral.
Now let's compare this behavior to the behavior of aqueous solutions of potassium cyanide and sodium acetate. Again, the cations ($K^+$ and $Na^+$) have essentially no acidic character, but the anions ($CN^−$ and $CH_3CO_2^−$) are weak bases that can react with water because they are the conjugate bases of the weak acids $HCN$ and acetic acid, respectively.
$CN^-_{(aq)} + H_2O_{(l)} \ce{ <<=>} HCN_{(aq)} + OH^-_{(aq)} \nonumber$
$CH_3CO^2_{2(aq)} + H_2O_{(l)} \ce{<<=>} CH_3CO_2H_{(aq)} + OH^-_{(aq)} \nonumber$
Neither reaction proceeds very far to the right as written because the formation of the weaker acid–base pair is favored. Both $HCN$ and acetic acid are stronger acids than water, and hydroxide is a stronger base than either acetate or cyanide, so in both cases, the equilibrium lies to the left. Nonetheless, each of these reactions generates enough hydroxide ions to produce a basic solution. For example, the $pH$ of a 0.1 M solution of sodium acetate or potassium cyanide at 25°C is 8.8 or 11.1, respectively. From Table $1$ and Figure $1$, we can see that $CN^−$ is a stronger base ($pK_b = 4.79$) than acetate ($pK_b = 9.24$), which is consistent with $KCN$ producing a more basic solution than sodium acetate at the same concentration.
In contrast, the conjugate acid of a weak base should be a weak acid (Equation $\ref{16.2}$). For example, ammonium chloride and pyridinium chloride are salts produced by reacting ammonia and pyridine, respectively, with $HCl$. As you already know, the chloride ion is such a weak base that it does not react with water. In contrast, the cations of the two salts are weak acids that react with water as follows:
$NH^+_{4(aq)} + H_2O_{(l)} \ce{ <<=>} HH_{3(aq)} + H_3O^+_{(aq)} \label{16.2}$
$C_5H_5NH^+_{(aq)} + H_2O_{(l)} \ce{<<=>} C_5H_5NH_{(aq)} + H_3O^+_{(aq)} \label{16.3}$
Equation $\ref{16.2}$ indicates that $H_3O^+$ is a stronger acid than either $NH_4^+$ or $C_5H_5NH^+$, and conversely, ammonia and pyridine are both stronger bases than water. The equilibrium will therefore lie far to the left in both cases, favoring the weaker acid–base pair. The $H_3O^+$ concentration produced by the reactions is great enough, however, to decrease the $pH$ of the solution significantly: the $pH$ of a 0.10 M solution of ammonium chloride or pyridinium chloride at 25°C is 5.13 or 3.12, respectively. This is consistent with the information shown in Figure 16.2, indicating that the pyridinium ion is more acidic than the ammonium ion.
What happens with aqueous solutions of a salt such as ammonium acetate, where both the cation and the anion can react separately with water to produce an acid and a base, respectively? According to Figure 16.10, the ammonium ion will lower the $pH$, while according to Equation $\ref{16.3}$, the acetate ion will raise the $pH$. This particular case is unusual, in that the cation is as strong an acid as the anion is a base (pKa ≈ pKb). Consequently, the two effects cancel, and the solution remains neutral. With salts in which the cation is a stronger acid than the anion is a base, the final solution has a $pH$ < 7.00. Conversely, if the cation is a weaker acid than the anion is a base, the final solution has a $pH$ > 7.00.
Solutions of simple salts of metal ions can also be acidic, even though a metal ion cannot donate a proton directly to water to produce $H_3O^+$. Instead, a metal ion can act as a Lewis acid and interact with water, a Lewis base, by coordinating to a lone pair of electrons on the oxygen atom to form a hydrated metal ion (part (a) in Figure $1$). A water molecule coordinated to a metal ion is more acidic than a free water molecule for two reasons. First, repulsive electrostatic interactions between the positively charged metal ion and the partially positively charged hydrogen atoms of the coordinated water molecule make it easier for the coordinated water to lose a proton.
The magnitude of this effect depends on the following two factors (Figure $2$):
1. The charge on the metal ion. A divalent ion ($M^{2+}$) has approximately twice as strong an effect on the electron density in a coordinated water molecule as a monovalent ion ($M^+$) of the same radius.
2. The radius of the metal ion. For metal ions with the same charge, the smaller the ion, the shorter the internuclear distance to the oxygen atom of the water molecule and the greater the effect of the metal on the electron density distribution in the water molecule.
Thus aqueous solutions of small, highly charged metal ions, such as $Al^{3+}$ and $Fe^{3+}$, are acidic:
$[Al(H_2O)_6]^{3+}_{(aq)} \rightleftharpoons [Al(H_2O)_5(OH)]^{2+}_{(aq)}+H^+_{(aq)} \label{16.36}$
The $[Al(H_2O)_6]^{3+}$ ion has a $pK_a$ of 5.0, making it almost as strong an acid as acetic acid. Because of the two factors described previously, the most important parameter for predicting the effect of a metal ion on the acidity of coordinated water molecules is the charge-to-radius ratio of the metal ion. A number of pairs of metal ions that lie on a diagonal line in the periodic table, such as $Li^+$ and $Mg^{2+}$ or $Ca^{2+}$ and $Y^{3+}$, have different sizes and charges, but similar charge-to-radius ratios. As a result, these pairs of metal ions have similar effects on the acidity of coordinated water molecules, and they often exhibit other significant similarities in chemistry as well.
Solutions of small, highly charged metal ions in water are acidic.
Reactions such as those discussed in this section, in which a salt reacts with water to give an acidic or basic solution, are often called hydrolysis reactions. Using a separate name for this type of reaction is unfortunate because it suggests that they are somehow different. In fact, hydrolysis reactions are just acid–base reactions in which the acid is a cation or the base is an anion; they obey the same principles and rules as all other acid–base reactions.
A hydrolysis reaction is an acid–base reaction.
Example $1$
Predict whether aqueous solutions of these compounds are acidic, basic, or neutral.
1. $\ce{KNO_3}$
2. $\ce{CrBr_3} cdot \ce{H_2O}$
3. $\ce{Na_2SO_4}$
Given: compound
Asked for: acidity or basicity of aqueous solution
Strategy:
1. Assess the acid–base properties of the cation and the anion. If the cation is a weak Lewis acid, it will not affect the $pH$ of the solution. If the cation is the conjugate acid of a weak base or a relatively highly charged metal cation, however, it will react with water to produce an acidic solution.
2. f the anion is the conjugate base of a strong acid, it will not affect the $pH$ of the solution. If, however, the anion is the conjugate base of a weak acid, the solution will be basic.
Solution:
a
1. The $K^+$ cation has a small positive charge (+1) and a relatively large radius (because it is in the fourth row of the periodic table), so it is a very weak Lewis acid.
2. The $NO_3−$ anion is the conjugate base of a strong acid, so it has essentially no basic character (Table 16.1). Hence neither the cation nor the anion will react with water to produce $H^+$ or $OH^−$, and the solution will be neutral.
b.
1. The $Cr^{3+}$ ion is a relatively highly charged metal cation that should behave similarly to the $Al^{3+}$ ion and form the $[Cr(H2O)_6]^{3+}$ complex, which will behave as a weak acid: $Cr(H_2O)_6]^{3+}_{(aq)} \ce{ <=>>} Cr(H_2O)_5(OH)]^{2+}_{(aq)} + H^+_{(aq)}\nonumber$
2. The $Br^−$ anion is a very weak base (it is the conjugate base of the strong acid $HBr$), so it does not affect the $pH$ of the solution. Hence the solution will be acidic.
c.
1. The $Na^+$ ion, like the $K^+$, is a very weak acid, so it should not affect the acidity of the solution.
2. In contrast, $SO_4^{2−}$ is the conjugate base of $HSO_4^−$, which is a weak acid. Hence the $SO_4^{2−}$ ion will react with water as shown in Figure 16.6 to give a slightly basic solution.
Exercise $1$
Predict whether aqueous solutions of the following are acidic, basic, or neutral.
1. $KI$
2. $Mg(ClO_4)_2$
3. $NaHS$
Answer a
neutral
Answer b
acidic
Answer c
basic (due to the reaction of $\ce{HS^{-}}$ with water to form $\ce{H_2S}$ and $\ce{OH^{-}}$)
Summary
A salt can dissolve in water to produce a neutral, a basic, or an acidic solution, depending on whether it contains the conjugate base of a weak acid as the anion ($A^−$), the conjugate acid of a weak base as the cation ($BH^+$), or both. Salts that contain small, highly charged metal ions produce acidic solutions in water. The reaction of a salt with water to produce an acidic or a basic solution is called a hydrolysis reaction.
Key Takeaways
• Acid–base reactions always contain two conjugate acid–base pairs.
• Each acid and each base has an associated ionization constant that corresponds to its acid or base strength. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/16%3A_AcidBase_Equilibria/16.09%3A_Acid-Base_Properties_of_Salt_Solutions.txt |
Learning Objectives
• To understand how molecular structure affects the strength of an acid or base.
We have seen that the strengths of acids and bases vary over many orders of magnitude. In this section, we explore some of the structural and electronic factors that control the acidity or basicity of a molecule.
Bond Strengths
In general, the stronger the $\ce{A–H}$ or $\ce{B–H^+}$ bond, the less likely the bond is to break to form $H^+$ ions and thus the less acidic the substance. This effect can be illustrated using the hydrogen halides:
Relative Acid Strength HF HCl HBr HI
H–X Bond Energy (kJ/mol) 570 432 366 298
pKa 3.20 −6.1 −8.9 −9.3
The trend in bond energies is due to a steady decrease in overlap between the 1s orbital of hydrogen and the valence orbital of the halogen atom as the size of the halogen increases. The larger the atom to which H is bonded, the weaker the bond. Thus the bond between H and a large atom in a given family, such as I or Te, is weaker than the bond between H and a smaller atom in the same family, such as F or O. As a result, acid strengths of binary hydrides increase as we go down a column of the periodic table. For example, the order of acidity for the binary hydrides of Group 16 elements is as follows, with $pK_a$ values in parentheses:
$H_2O (14.00 = pK_w) < H_2S (7.05) < H_2Se (3.89) < H_2Te (2.6) \label{1}$
Stability of the Conjugate Base
Whether we write an acid–base reaction as $AH \rightleftharpoons A^−+H^+$ or as $BH^+ \rightleftharpoons B+H^+$, the conjugate base ($A^−$ or $B$) contains one more lone pair of electrons than the parent acid ($AH$ or $BH^+$). Any factor that stabilizes the lone pair on the conjugate base favors dissociation of $H^+$ and makes the parent acid a stronger acid. Let’s see how this explains the relative acidity of the binary hydrides of the elements in the second row of the periodic table. The observed order of increasing acidity is the following, with pKa values in parentheses:
$CH_4 (~50) \ll NH_3 (~36) < H_2O (14.00) < HF (3.20) \label{2}$
Consider, for example, the compounds at both ends of this series: methane and hydrogen fluoride. The conjugate base of $CH_4$ is $CH_3^−$, and the conjugate base of $HF$ is $F^−$. Because fluorine is much more electronegative than carbon, fluorine can better stabilize the negative charge in the $F^−$ ion than carbon can stabilize the negative charge in the CH3− ion. Consequently, $\ce{HF}$ has a greater tendency to dissociate to form $H^+$ and $F^−$ than does methane to form $H^+$ and $CH_3^−$, making HF a much stronger acid than $CH_4$.
The same trend is predicted by analyzing the properties of the conjugate acids. For a series of compounds of the general formula $HE$, as the electronegativity of E increases, the E–H bond becomes more polar, favoring dissociation to form $E^−$ and $H^+$. Due to both the increasing stability of the conjugate base and the increasing polarization of the E–H bond in the conjugate acid, acid strengths of binary hydrides increase as we go from left to right across a row of the periodic table.
Acid strengths of binary hydrides increase as we go down a column or from left to right across a row of the periodic table.
the strongest acid Known: The hydrohelium Cation
The stornger acid, the weaker the covalent bond to a hydrogen atom. So the strongest acid possible is the molecule with the weakest bond. That is the hydrohelium (1+) cation, $\ce{HeH^{+}}$, which is a positively charged ion formed by the reaction of a proton with a helium atom in the gas phase. It was first produced in the laboratory in 1925 and is isoelectronic with molecular hydrogen (\ce{H2}}). It is the strongest known acid, with a proton affinity of 177.8 kJ/mol.
Ball and stick model of the hydrohelium ion. (CC BY-SA 3.0; CCoil).
$\ce{HeH^{+}}$ cannot be prepared in a condensed phase, as it would protonate any anion, molecule or atom with which it were associated. However it is possible to estimate a hypothetical aqueous acidity using Hess's law:
HHe+(g) H+(g) + He(g) +178 kJ/mol
HHe+(aq) HHe+(g) +973 kJ/mol
H+(g) H+(aq) −1530 kJ/mol
He(g) He(aq) +19 kJ/mol
HHe+(aq) H+(aq) + He(aq) −360 kJ/mol
A free energy change of dissociation of −360 kJ/mol is equivalent to a pKa of −63.
It has been suggested that $\ce{HeH^{+}}$ should occur naturally in the interstellar medium, but it has not yet been detected.
Inductive Effects
Atoms or groups of atoms in a molecule other than those to which H is bonded can induce a change in the distribution of electrons within the molecule. This is called an inductive effect, and, much like the coordination of water to a metal ion, it can have a major effect on the acidity or basicity of the molecule. For example, the hypohalous acids (general formula HOX, with X representing a halogen) all have a hydrogen atom bonded to an oxygen atom. In aqueous solution, they all produce the following equilibrium:
$HOX_{(aq)} \rightleftharpoons H^+_{(aq)} + OX^−{(aq)} \label{3}$
The acidities of these acids vary by about three orders of magnitude, however, due to the difference in electronegativity of the halogen atoms:
HOX Electronegativity of X pKa
HOCl 3.0 7.40
HOBr 2.8 8.55
HOI 2.5 10.5
As the electronegativity of $X$ increases, the distribution of electron density within the molecule changes: the electrons are drawn more strongly toward the halogen atom and, in turn, away from the H in the O–H bond, thus weakening the O–H bond and allowing dissociation of hydrogen as $H^+$.
The acidity of oxoacids, with the general formula $HOXO_n$ (with $n$ = 0−3), depends strongly on the number of terminal oxygen atoms attached to the central atom $X$. As shown in Figure $1$, the $K_a$ values of the oxoacids of chlorine increase by a factor of about $10^4$ to $10^6$ with each oxygen as successive oxygen atoms are added. The increase in acid strength with increasing number of terminal oxygen atoms is due to both an inductive effect and increased stabilization of the conjugate base.
Any inductive effect that withdraws electron density from an O–H bond increases the acidity of the compound.
Because oxygen is the second most electronegative element, adding terminal oxygen atoms causes electrons to be drawn away from the O–H bond, making it weaker and thereby increasing the strength of the acid. The colors in Figure $1$ show how the electrostatic potential, a measure of the strength of the interaction of a point charge at any place on the surface of the molecule, changes as the number of terminal oxygen atoms increases. In Figure $1$ and Figure $2$, blue corresponds to low electron densities, while red corresponds to high electron densities. The oxygen atom in the O–H unit becomes steadily less red from $HClO$ to $HClO_4$ (also written as $HOClO_3$, while the H atom becomes steadily bluer, indicating that the electron density on the O–H unit decreases as the number of terminal oxygen atoms increases. The decrease in electron density in the O–H bond weakens it, making it easier to lose hydrogen as $H^+$ ions, thereby increasing the strength of the acid.
At least as important, however, is the effect of delocalization of the negative charge in the conjugate base. As shown in Figure $2$, the number of resonance structures that can be written for the oxoanions of chlorine increases as the number of terminal oxygen atoms increases, allowing the single negative charge to be delocalized over successively more oxygen atoms.
Electron delocalization in the conjugate base increases acid strength.
The electrostatic potential plots in Figure $2$ demonstrate that the electron density on the terminal oxygen atoms decreases steadily as their number increases. The oxygen atom in ClO− is red, indicating that it is electron rich, and the color of oxygen progressively changes to green in $ClO_4^+$, indicating that the oxygen atoms are becoming steadily less electron rich through the series. For example, in the perchlorate ion ($ClO_4^−$), the single negative charge is delocalized over all four oxygen atoms, whereas in the hypochlorite ion ($OCl^−$), the negative charge is largely localized on a single oxygen atom (Figure $2$). As a result, the perchlorate ion has no localized negative charge to which a proton can bind. Consequently, the perchlorate anion has a much lower affinity for a proton than does the hypochlorite ion, and perchloric acid is one of the strongest acids known.
As the number of terminal oxygen atoms increases, the number of resonance structures that can be written for the oxoanions of chlorine also increases, and the single negative charge is delocalized over more oxygen atoms. As these electrostatic potential plots demonstrate, the electron density on the terminal oxygen atoms decreases steadily as their number increases. As the electron density on the oxygen atoms decreases, so does their affinity for a proton, making the anion less basic. As a result, the parent oxoacid is more acidic.
Similar inductive effects are also responsible for the trend in the acidities of oxoacids that have the same number of oxygen atoms as we go across a row of the periodic table from left to right. For example, $H_3PO_4$ is a weak acid, $H_2SO_4$ is a strong acid, and $HClO_4$ is one of the strongest acids known. The number of terminal oxygen atoms increases steadily across the row, consistent with the observed increase in acidity. In addition, the electronegativity of the central atom increases steadily from P to S to $Cl$, which causes electrons to be drawn from oxygen to the central atom, weakening the $\ce{O–H}$ bond and increasing the strength of the oxoacid.
Careful inspection of the data in Table $1$ shows two apparent anomalies: carbonic acid and phosphorous acid. If carbonic acid $(H_2CO_3$) were a discrete molecule with the structure $\ce{(HO)_2C=O}$, it would have a single terminal oxygen atom and should be comparable in acid strength to phosphoric acid ($H_3PO_4$), for which pKa1 = 2.16. Instead, the tabulated value of $pK_{a1}$ for carbonic acid is 6.35, making it about 10,000 times weaker than expected. As we shall see, however, $H_2CO_3$ is only a minor component of the aqueous solutions of $CO_2$ that are referred to as carbonic acid. Similarly, if phosphorous acid ($H_3PO_3$) actually had the structure $(HO)_3P$, it would have no terminal oxygen atoms attached to phosphorous. It would therefore be expected to be about as strong an acid as $HOCl$ (pKa = 7.40). In fact, the $pK_{a1}$ for phosphorous acid is 1.30, and the structure of phosphorous acid is $\ce{(HO)_2P(=O)H}$ with one H atom directly bonded to P and one $\ce{P=O}$ bond. Thus the pKa1 for phosphorous acid is similar to that of other oxoacids with one terminal oxygen atom, such as $H_3PO_4$. Fortunately, phosphorous acid is the only common oxoacid in which a hydrogen atom is bonded to the central atom rather than oxygen.
Table $1$: Values of pKa for Selected Polyprotic Acids and Bases
*$H_2CO_3$ and $H_2SO_3$ are at best minor components of aqueous solutions of $CO_{2(g)}$ and $SO_{2(g)}$, respectively, but such solutions are commonly referred to as containing carbonic acid and sulfurous acid, respectively.
Polyprotic Acids Formula $pK_{a1}$ $pK_{a2}$ $pK_{a3}$
carbonic acid* “$H_2CO_3$” 6.35 10.33
citric acid $HO_2CCH-2C(OH)(CO_2H)CH_2CO_2H$ 3.13 4.76 6.40
malonic acid $HO-2CCH_2CO_2H$ 2.85 5.70
oxalic acid $HO_2CCO_2H$ 1.25 3.81
phosphoric acid $H_3PO_4$ 2.16 7.21 12.32
phosphorous acid $H_3PO_3$ 1.3 6.70
succinic acid $HO_2CCH_2CH_2CO_2H$ 4.21 5.64
sulfuric acid $H_2SO_4$ −2.0 1.99
sulfurous acid* “$H_2SO_3$” 1.85 7.21
Polyprotic Bases Formula $pK_{b1}$ $pK_{b2}$
ethylenediamine $H_2N(CH_2)_2NH_2$ 4.08 7.14
piperazine $HN(CH_2CH_2)_2NH$ 4.27 8.67
propylenediamine $H_2N(CH_2)_3NH_2$ 3.45 5.12
Inductive effects are also observed in organic molecules that contain electronegative substituents. The magnitude of the electron-withdrawing effect depends on both the nature and the number of halogen substituents, as shown by the pKa values for several acetic acid derivatives:
$pK_a CH_3CO_2H 4.76< CH_2ClCO_2H 2.87<CHCl_2CO_2H 1.35<CCl_3CO_2H 0.66<CF_3CO_2H 0.52 \nonumber$
As you might expect, fluorine, which is more electronegative than chlorine, causes a larger effect than chlorine, and the effect of three halogens is greater than the effect of two or one. Notice from these data that inductive effects can be quite large. For instance, replacing the $\ce{–CH_3}$ group of acetic acid by a $\ce{–CF_3}$ group results in about a 10,000-fold increase in acidity!
Example $1$
Arrange the compounds of each series in order of increasing acid or base strength.
1. sulfuric acid [$H_2SO_4$, or $(HO)_2SO_2$], fluorosulfonic acid ($FSO_3H$, or $FSO_2OH$), and sulfurous acid [$H_2SO_3$, or $(HO)_2SO$]
2. ammonia ($NH_3$), trifluoramine ($NF_3$), and hydroxylamine ($NH_2OH$)
The structures are shown here.
Given: series of compounds
Asked for: relative acid or base strengths
Strategy:
Use relative bond strengths, the stability of the conjugate base, and inductive effects to arrange the compounds in order of increasing tendency to ionize in aqueous solution.
Solution:
Although both sulfuric acid and sulfurous acid have two –OH groups, the sulfur atom in sulfuric acid is bonded to two terminal oxygen atoms versus one in sulfurous acid. Because oxygen is highly electronegative, sulfuric acid is the stronger acid because the negative charge on the anion is stabilized by the additional oxygen atom. In comparing sulfuric acid and fluorosulfonic acid, we note that fluorine is more electronegative than oxygen. Thus replacing an –OH by –F will remove more electron density from the central S atom, which will, in turn, remove electron density from the S–OH bond and the O–H bond. Because its O–H bond is weaker, $FSO_3H$ is a stronger acid than sulfuric acid. The predicted order of acid strengths given here is confirmed by the measured pKa values for these acids:
$pKa H_2SO_3 1.85<H_2SO_4^{−2} < FSO_3H−10 \nonumber$
The structures of both trifluoramine and hydroxylamine are similar to that of ammonia. In trifluoramine, all of the hydrogen atoms in NH3 are replaced by fluorine atoms, whereas in hydroxylamine, one hydrogen atom is replaced by OH. Replacing the three hydrogen atoms by fluorine will withdraw electron density from N, making the lone electron pair on N less available to bond to an $H^+$ ion. Thus $NF_3$ is predicted to be a much weaker base than $NH_3$. Similarly, because oxygen is more electronegative than hydrogen, replacing one hydrogen atom in $NH_3$ by $OH$ will make the amine less basic. Because oxygen is less electronegative than fluorine and only one hydrogen atom is replaced, however, the effect will be smaller. The predicted order of increasing base strength shown here is confirmed by the measured $pK_b$ values:
$pK_bNF_3—<<NH_2OH 8.06<NH_3 4.75 \nonumber$
Trifluoramine is such a weak base that it does not react with aqueous solutions of strong acids. Hence its base ionization constant has never been measured.
Exercise $1$
Arrange the compounds of each series in order of
1. decreasing acid strength: $H_3PO_4$, $CH_3PO_3H_2$, and $HClO_3$.
2. increasing base strength: $CH_3S^−$, $OH^−$, and $CF_3S^−$.
Answer a
$HClO-3 > CH_3PO_3H_2 > H_3PO_4$
Answer a
$CF_3S^− < CH_3S^− < OH^−$
Summary
Inductive effects and charge delocalization significantly influence the acidity or basicity of a compound. The acid–base strength of a molecule depends strongly on its structure. The weaker the A–H or B–H+ bond, the more likely it is to dissociate to form an $H^+$ ion. In addition, any factor that stabilizes the lone pair on the conjugate base favors the dissociation of $H^+$, making the conjugate acid a stronger acid. Atoms or groups of atoms elsewhere in a molecule can also be important in determining acid or base strength through an inductive effect, which can weaken an $\ce{O–H}$ bond and allow hydrogen to be more easily lost as $H^+$ ions.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/16%3A_AcidBase_Equilibria/16.10%3A_Acid-Base_Behavior_and_Chemical_Structure.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
• Write the equation for the proton transfer reaction involving a Brønsted-Lowry acid or base, and show how it can be interpreted as an electron-pair transfer reaction, clearly identifying the donor and acceptor.
• Give an example of a Lewis acid-base reaction that does not involve protons.
• Write equations illustrating the behavior of a given non-aqueous acid-base system.
The Brønsted-Lowry proton donor-acceptor concept has been one of the most successful theories of Chemistry. But as with any such theory, it is fair to ask if this is not just a special case of a more general theory that could encompass an even broader range of chemical science. In 1916, G.N. Lewis of the University of California proposed that the electron pair is the dominant actor in acid-base chemistry. The Lewis theory did not become very well known until about 1923 (the same year that Brønsted and Lowry published their work), but since then it has been recognized as a very powerful tool for describing chemical reactions of widely different kinds and is widely used in organic and inorganic chemistry. According to Lewis,
• An acid is a substance that accepts a pair of electrons, and in doing so, forms a covalent bond with the entity that supplies the electrons.
• A base is a substance that donates an unshared pair of electrons to a recipient species with which the electrons can be shared.
In modern chemistry, electron donors are often referred to as nucleophiles, while acceptors are electrophiles.
Proton-Transfer Reactions Involve Electron-Pair Transfer
Just as any Arrhenius acid is also a Brønsted acid, any Brønsted acid is also a Lewis acid, so the various acid-base concepts are all "upward compatible". Although we do not really need to think about electron-pair transfers when we deal with ordinary aqueous-solution acid-base reactions, it is important to understand that it is the opportunity for electron-pair sharing that enables proton transfer to take place.
This equation for a simple acid-base neutralization shows how the Brønsted and Lewis definitions are really just different views of the same process. Take special note of the following points:
• The arrow shows the movement of a proton from the hydronium ion to the hydroxide ion.
• Note carefully that the electron-pairs themselves do not move; they remain attached to their central atoms. The electron pair on the base is "donated" to the acceptor (the proton) only in the sense that it ends up being shared with the acceptor, rather than being the exclusive property of the oxygen atom in the hydroxide ion.
• Although the hydronium ion is the nominal Lewis acid here, it does not itself accept an electron pair, but acts merely as the source of the proton that coordinates with the Lewis base.
The point about the electron-pair remaining on the donor species is especially important to bear in mind. For one thing, it distinguishes a Lewis acid-base reaction from an oxidation-reduction reaction, in which a physical transfer of one or more electrons from donor to acceptor does occur. The product of a Lewis acid-base reaction is known formally as an "adduct" or "complex", although we do not ordinarily use these terms for simple proton-transfer reactions such as the one in the above example. Here, the proton combines with the hydroxide ion to form the "adduct" H2O. The following examples illustrate these points for some other proton-transfer reactions that you should already be familiar with.
Another example, showing the autoprotolysis of water. Note that the conjugate base is also the adduct.
Ammonia is both a Brønsted and a Lewis base, owing to the unshared electron pair on the nitrogen. The reverse of this reaction represents the hydrolysis of the ammonium ion.
Because $\ce{HF}$ is a weak acid, fluoride salts behave as bases in aqueous solution. As a Lewis base, F accepts a proton from water, which is transformed into a hydroxide ion.
The bisulfite ion is amphiprotic and can act as an electron donor or acceptor.
Acid-base Reactions without Transferring Protons
The major utility of the Lewis definition is that it extends the concept of acids and bases beyond the realm of proton transfer reactions. The classic example is the reaction of boron trifluoride with ammonia to form an adduct:
$\ce{BF_3 + NH_3 \rightarrow F_3B-NH_3}$
One of the most commonly-encountered kinds of Lewis acid-base reactions occurs when electron-donating ligands form coordination complexes with transition-metal ions.
Exercise $1$
Here are several more examples of Lewis acid-base reactions that cannot be accommodated within the Brønsted or Arrhenius models. Identify the Lewis acid and Lewis base in each reaction.
1. $\ce{Al(OH)_3 + OH^{–} \rightarrow Al(OH)_4^–}$
2. $\ce{SnS_2 + S^{2–} \rightarrow SnS_3^{2–}}$
3. $\ce{Cd(CN)_2 + 2 CN^– \rightarrow Cd(CN)_4^{2+}}$
4. $\ce{AgCl + 2 NH_3 \rightarrow Ag(NH_3)_2^+ + Cl^–}$
5. $\ce{Fe^{2+} + NO \rightarrow Fe(NO)^{2+}}$
6. $\ce{Ni^{2+} + 6 NH_3 \rightarrow Ni(NH_3)_5^{2+}}$
Applications to organic reaction mechanisms
Although organic chemistry is beyond the scope of these lessons, it is instructive to see how electron donors and acceptors play a role in chemical reactions. The following two diagrams show the mechanisms of two common types of reactions initiated by simple inorganic Lewis acids:
In each case, the species labeled "Complex" is an intermediate that decomposes into the products, which are conjugates of the original acid and base pairs. The electric charges indicated in the complexes are formal charges, but those in the products are "real".
In reaction 1, the incomplete octet of the aluminum atom in $\ce{AlCl3}$ serves as a better electron acceptor to the chlorine atom than does the isobutyl part of the base. In reaction 2, the pair of non-bonding electrons on the dimethyl ether coordinates with the electron-deficient boron atom, leading to a complex that breaks down by releasing a bromide ion.
Non-aqueous Protonic Acid-Base Systems
We ordinarily think of Brønsted-Lowry acid-base reactions as taking place in aqueous solutions, but this need not always be the case. A more general view encompasses a variety of acid-base solvent systems, of which the water system is only one (Table $1$). Each of these has as its basis an amphiprotic solvent (one capable of undergoing autoprotolysis), in parallel with the familiar case of water.
The ammonia system is one of the most common non-aqueous system in Chemistry. Liquid ammonia boils at –33° C, and can conveniently be maintained as a liquid by cooling with dry ice (–77° C). It is a good solvent for substances that also dissolve in water, such as ionic salts and organic compounds since it is capable of forming hydrogen bonds. However, many other familiar substances can also serve as the basis of protonic solvent systems as Table $1$ indicates:
Table $1$: Popular Solvent systems
solvent
autoprotolysis reaction
pKap
water 2 H2O → H3O+ + OH 14
ammonia 2 NH3 → NH4+ + NH2 33
acetic acid 2 CH3COOH → CH3COOH2+ + CH3COO 13
ethanol 2 C2H5OH → C2H5OH2+ + C2H5O 19
hydrogen peroxide 2 HO-OH → HO-OH2+ + HO-O 13
hydrofluoric acid 2 HF → H2F+ + F 10
sulfuric acid 2 H2SO4 → H3SO4+ + HSO4 3.5
One use of nonaqueous acid-base systems is to examine the relative strengths of the strong acids and bases, whose strengths are "leveled" by the fact that they are all totally converted into H3O+ or OH ions in water. By studying them in appropriate non-aqueous solvents which are poorer acceptors or donors of protons, their relative strengths can be determined. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/16%3A_AcidBase_Equilibria/16.11%3A_Lewis_Acids_and_Bases.txt |
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
16.2: Brønsted–Lowry Acids and Bases
Conceptual Problems
1. Identify the conjugate acid–base pairs in each equilibrium.
1. $HSO^−_{4}\,(aq)+H_2O\,(l) \rightleftharpoons SO^{2−}_{4}\,(aq)+H_3O^{+}\,(aq)$
2. $C_{3}H_{7}NO_{2}\,(aq)+H_{3}O^{+}\,(aq) \rightleftharpoons C_{3}H_{8}NO^{+}_{2}\,(aq)+H_{2}O\,(l)$
3. $CH_{3}O_{2}H\,(aq)+NH_{3}\,(aq) \rightleftharpoons CH_{3}CO^{−}_{2}\,(aq)+NH^{+}_{4}\,(aq)$
4. $SbF_{5}\,(aq)+2\,HF\,(aq) \rightleftharpoons H_{2}F^{+}\,(aq)+SbF^{−}_{6}\,(aq)$
2. Identify the conjugate acid–base pairs in each equilibrium.
1. $HF\,(aq)+H_{2}O\,(l) \rightleftharpoons H_3O^{+}\,(aq)+F^{−}\,(aq)$
2. $CH_3CH_2NH_{2}\,(aq)+H_{2}O\,(l) \rightleftharpoons CH_3CH_2NH^{+}_{3}\,(aq)+OH^{−}\,(aq)$
3. $C_3H_7NO_{2}\,(aq)+OH^{−}\,(aq) \rightleftharpoons C_3H_6NO^{−}_{2}\,(aq)+H_{2}O\,(l)$
4. $CH_3CO_2H\,(aq)+2\,HF\,(aq) \rightleftharpoons CH_3C(OH)_{2}^{+}\,(aq)+HF^{−}_{2}\,(aq)$
3. Salts such as NaH contain the hydride ion ($H^−$). When sodium hydride is added to water, it produces hydrogen gas in a highly vigorous reaction. Write a balanced chemical equation for this reaction and identify the conjugate acid–base pairs.
4. Write the expression for $K_a$ for each reaction.
1. $HCO^−_{3}\,(aq)+H_2O\,(l) \rightleftharpoons CO^{2−}_{3}\,(aq)+H_3O^{+}\,(aq)$
2. $formic\;acid\,(aq)+H_2O\,(l) \rightleftharpoons formate\,(aq)+H_3O^+\,(aq)$
3. $H_3PO_{4}\,(aq)+H_2O\,(l) \rightleftharpoons H_2PO^−_{4}\,(aq)+H_3O^+\,(aq)$
5. Write an expression for the ionization constant $K_b$ for each reaction.
1. $OCH^{−}_{3}\,(aq)+H_2O\,(l) \rightleftharpoons HOCH_{3}\,(aq)+OH^{-}\,(aq)$
2. $NH^−_{2}\,(aq)+H_2O\,(l) \rightleftharpoons NH_{3}\,(aq)+OH^{−}\,(aq)$
3. $S^{2−}\,(aq)+H_2O\,(l) \rightleftharpoons HS^−\,(aq)+OH^−\,(aq)$
6. Predict whether each equilibrium lies primarily to the left or to the right.
1. $HBr\,(aq)+H_2O\,(l) \rightleftharpoons H_3O^+\,(aq)+Br^−\,(aq)$
2. $NaH\,(s)+NH_{3}\,(aq) \rightleftharpoons H_{2}\,(g)+NaNH_{2}\,(s)$
3. $OCH^{−}_{3}\,(aq)+NH_{3}\,(aq) \rightleftharpoons CH_{3}OH\,(aq)+NH^−_{2}\,(aq)$
4. $NH_{3}\,(aq)+HCl\,(aq) \rightleftharpoons NH^{+}_{4}\,(aq)+Cl^−\,(aq)$
7. Species that are strong bases in water, such as $CH_3^−$, $NH_2^−$, and $S^{2−}$, are leveled to the strength of $OH^−$, the conjugate base of $H_2O$. Because their relative base strengths are indistinguishable in water, suggest a method for identifying which is the strongest base. How would you distinguish between the strength of the acids $HIO_3$, $H_2SO_4$, and $HClO_4$?
8. Is it accurate to say that a 2.0 M solution of $H_2SO_4$, which contains two acidic protons per molecule, is 4.0 M in $H^+$? Explain your answer.
9. The alkalinity of soil is defined by the following equation: alkalinity = $[HCO_3^−] + 2[CO_3^{2−}] + [OH^−] − [H^+]$. The source of both $HCO_3^−$ and $CO_3^{2−}$ is $H_2CO_3$. Explain why the basicity of soil is defined in this way.
10. Why are aqueous solutions of salts such as $CaCl_2$ neutral? Why is an aqueous solution of $NaNH_2$ basic?
11. Predict whether the aqueous solutions of the following are acidic, basic, or neutral.
1. $Li_3N$
2. $NaH$
3. $KBr$
4. $C_2H_5NH_3Cl$
12. When each compound is added to water, would you expect the $pH$ of the solution to increase, decrease, or remain the same?
1. $LiCH_3$
2. $MgCl_2$
3. $K_2O$
4. $(CH_3)_2NH_2^+Br^−$
13. Which complex ion would you expect to be more acidic: $Pb(H_2O)_4^{2+}$ or $Sn(H_2O)_4^{2+}$? Why?
14. Would you expect $Sn(H_2O)_4^{2+}$ or $Sn(H_2O)_6^{4+}$ to be more acidic in aqueous solutions? Why?
15. Is it possible to arrange the hydrides $LiH$, $RbH$, $KH$, $CsH$, and $NaH$ in order of increasing base strength in aqueous solution? Why or why not?
Conceptual Answer
1.
a. $\underset{\text{acid}}{HSO^−_{4}\,(aq)} + \underset{\text{base}}{H_2O\,(l)} \rightleftharpoons \underset{\text{conjugate base}}{SO^{2−}_{4}\,(aq)} + \underset{\text{conjugate acid}}{H_3O^+\,(aq)}$
b. $\underset{\text{base}}{C_3H_7NO_{2}\,(aq)} + \underset{\text{acid}}{H_3O^+\,(aq)} \rightleftharpoons \underset{\text{conjugate acid}}{C_3H_8NO^+_{2}\,(aq)} + \underset{\text{conjugate base}}{H_2O\,(l)}$
c. $\underset{\text{acid}}{HOAc\,(aq)} + \underset{\text{base}}{NH_{3}\,(aq)} \rightleftharpoons \underset{\text{conjugate base}}{CH_3CO^−_{2}\,(aq)} + \underset{\text{conjugate acid}}{NH^+_{4}\,(aq)}$
d. $\underset{\text{acid}}{SbF_{5}\,(aq)} + \underset{\text{base}}{2\,HF\,(aq)} \rightleftharpoons \underset{\text{conjugate acid}}{H_2F^+\,(aq)} + \underset{\text{conjugate base}}{SbF_6^−(aq)}$
2.
a. $\underset{\text{acid}}{HF\,(aq)} + \underset{\text{base}}{H_2O\,(l)} \rightleftharpoons \underset{\text{conjugate acid}}{H_{3}O^{+}\,(aq)} + \underset{\text{conjugate base}}{F^{-}\,(aq)}$
b. $\underset{\text{base}}{CH_{3}CH_{2}NH_{2}\,(aq)} + \underset{\text{acid}}{H_{2}O\,(l)} \rightleftharpoons \underset{\text{conjugate acid}}{CH_{3}CH_{2}NH_{3}^{+}\,(aq)} + \underset{\text{conjugate base}}{OH^{-}\,(aq)}$
c. $\underset{\text{acid}}{C_{3}H_{7}NO_{2}\,(aq)} + \underset{\text{base}}{OH^{-}\,(aq)} \rightleftharpoons \underset{\text{conjugate base}}{C_{3}H_{6}NO_{2}^{-}\,(aq)} + \underset{\text{conjugate acid}}{H_{2}O\,(l)}$
d. $\underset{\text{base}}{CH_{3}CO_{2}H\,(aq)} + \underset{\text{acid}}{2\,HF\,(aq)} \rightleftharpoons \underset{\text{conjugate acid}}{CH_{3}C(OH)_{2}^{+}\,(aq)} + \underset{\text{conjugate base}}{HF_{2}^{-}(aq)}$
3. $\underset{\text{base}}{NaH\,(s)} + \underset{\text{acid}}{H_{2}O\,(l)} \rightleftharpoons \underset{\text{conjugate acid}}{H_{2}\,(g)} + \underset{\text{conjugate base}}{NaOH\,(aq)}$
4.
a. $K_a=\frac{[CO_{3}^{2-}][H_{3}O^{+}]}{[HCO_{3}^{-}]}$
b. $K_a=\frac{[formate][H_{3}O^{+}]}{[formic\,acid]}$
c. $K_a=\frac{[H_{2}PO_{4}^{-}][H_{3}O^{+}]}{[H_{3}PO_{4}]}$
5.
a. $K_b=\frac{[CO_{3}^{2-}][H_{3}O^{+}]}{[HCO_{3}^{-}]}$
b. $K_b=\frac{[NH_{3}][OH^{-}]}{[NH_{2}^{-}]}$
c. $K_b=\frac{[HS^{-}][OH^{-}]}{[S^{2-}]}$
6. Strong acids have the smaller $pK_a$.
a. Equilibrium lies primarily to the right because $HBr$ ($pK_a=-8.7$) is a stronger acid than $H_{3}O^{+}$ ($pK_a=-1.7$) and $H_{2}O$ ($pK_a=14$) is a stronger base than $Br^-$ ( $pK_a=-8.7$).
b. Equilibrium lies primarily to the left because $H_{2}$ ($pK_a=36$) is a stronger acid than $NH_{3}$ ($pK_a=38$) and ($NaNH_2$) ($pK_a=38$) is a stronger base than $NaH$ ($pK_a=35$).
c. Equilibrium lies primarily to the left because $CH_{3}OH$ ($pK_a=17$) is a stronger acid than $NH_{3}$ ($pK_a=38$) and $NH_{2}^{-}$ ($pK_a=38$) is a stronger base than $OCH_{3}^{-}$ ($pK_a=25$).
d. Equilibrium lies to the right because $HCl$ ($pK_a=-7$) is a stronger acid than $NH_{4}^{+}$ ($pK_a=9.3$) and $NH_{3}$ is a stronger base than $Cl^{-}$ ($pK_a=-7$).
7. To identify the strongest base we can determine their weakest conjugate acid. The conjugate acids of $CH_{3}^{-}$, $NH_{2}^{-}$, and $S_{2}^{-}$ are $CH_{4}$, $NH_{3}$, and $HS^{-}$, respectively. Next, we consider that acidity increases with positive charge on the molecule, thus ruling out that $S_{2}^{-}$ is the weakest base. Finally, we consider that acidity increases with electronegativity, therefore $NH_{3}$ is the second most basic and $CH_{4}$ is the most basic. To distinguish between the strength of the acids $HIO_3$, $H_{2}SO_{4}$, and $HClO_4$ we can consider that the higher electronegativity and oxidation state of the central nonmetal is the more acidic, therefore the order of acidity is: $HIO_3$<$H_{2}SO_{4}$<$HClO_4$ because electronegativity and oxidation state increases as follows: $I(+5)<S(+6)<Cl(+7)$.
8. It is not accurate to say that a 2.0 M solution of $H_2SO_4$, which contains two acidic protons per molecule, is 4.0 M in $H^+$ because a 2.0 M solution of $H_2SO_4$ is equivalent to 4.0 N in $H^+$.
$\frac{2.0\,mol\,H_{2}SO_{4}}{1\,L} \times \frac{2\,eq\,H^{+}}{1\,mol\,H_{2}SO_{4}}=\frac{4\,eq\,H^{+}}{L}=4\,N\,H^{+}$
9. Alkalinity is a measure of acid neutralizing capability. The basicity of the soil is defined this way because bases such as $HCO_{3}^{-}$ and $CO_{3}^{2-}$ can neutralize acids in soil. Because most soil has a pH between 6 and 8, alkalinity can be estimated by its carbonate species alone. At a near neutral pH, most carbonate species are bicarbonate.
10. Aqueous solutions of salts such as $CaCl_{2}$ are neutral because it is created from hydrochloric acid (a strong acid) and calcium hydroxide (a strong base). An aqueous solution of $NaNH_2$ is basic because it can deprotonate alkynes, alcohols, and a host of other functional groups with acidic protons such as esters and ketones.
11.
a. $Li_3N$ is a base because the lone pair on the nitrogen can accept a proton.
b. $NaH$ is a base because the hydrogen has a negative charge.
c. $KBr$ is neutral because it is formed from $HBr$ (a strong acid) and $KOH$ (a strong base).
d. $C_2H_5NH_3Cl$ is acidic because it can donate a proton.
12.
a. The pH is expected to increase. $\underset{\text{acid}}{LiCH_{3}\,(aq)} + \underset{\text{base}}{H_2O\,(l)} \rightleftharpoons \underset{\text{conjugate base}}{LiOH\,(aq)} + \underset{\text{conjugate acid}}{CH_{4}\,(aq)}$
b. The pH is expected to increase. $\underset{\text{acid}}{MgCl_{2}\,(aq)} + \underset{\text{base}}{H_2O\,(l)} \rightleftharpoons \underset{\text{conjugate acid}}{2\,HCl\,(aq)} + \underset{\text{conjugate base}}{MgO\,(aq)}$
c. The pH is expected to remain the same. $K_{2}O\,(aq)+H_2O\,(l) \rightleftharpoons 2\,KOH\,(aq)$
d. The pH is expected to increase. $\underset{\text{acid}}{(CH_3)_2NH_2^+Br^−\,(aq)} + \underset{\text{base}}{H_2O\,(l)} \rightleftharpoons \underset{\text{conjugate acid}}{H_3O^{+}\,(aq)} + \underset{\text{conjugate base}}{(CH_{3})_{2}NH\,(aq)}$
13. $Sn(H_2O)_4^{2+}$ is expected to be more acidic than $Pb(H_2O)_4^{2+}$ because $Sn$ is more electronegative than $Pb$.
14. $Sn(H_2O)_6^{4+}$ is expected to be more acidic than $Sn(H_2O)_4^{2+}$ because the charge on $Sn$ is greater ($4^+>2^+$).
15. Yes, it is possible the order of increasing base strength is: $LiH<NaH<RbH<CsH$ because increasing base strength is dependent on decreasing electronegativity.
Numerical Problems
1. Arrange these acids in order of increasing strength.
• acid A: $pK_a = 1.52$
• acid B: $pK_a = 6.93$
• acid C: $pK_a = 3.86$
Given solutions with the same initial concentration of each acid, which would have the highest percent ionization?
1. Arrange these bases in order of increasing strength:
• base A: $pK_b = 13.10$
• base B: $pK_b = 8.74$
• base C: $pK_b = 11.87$
Given solutions with the same initial concentration of each base, which would have the highest percent ionization?
1. Calculate the $K_a$ and the $pK_a$ of the conjugate acid of a base with each $pK_b$ value.
1. 3.80
2. 7.90
3. 13.70
4. 1.40
5. −2.50
2. Benzoic acid is a food preservative with a $pK_a$ of 4.20. Determine the $K_b$ and the $pK_b$ for the benzoate ion.
3. Determine $K_a$ and $pK_a$ of boric acid $[B(OH)_3]$, solutions of which are occasionally used as an eyewash; the $pK_b$ of its conjugate base is 4.80.
Numerical Answers
1. Acids in order of increasing strength: $acid\,B<acid\,C<acid\,A$. Given the same initial concentration of each acid, the highest percent of ionization is acid A because it is the strongest acid.
2. Bases in order of increasing strength: $base\,A<base\,C<base\,B$. Given the solutions with the same initial concentration of each base, the higher percent of ionization is base A because it is the weakest base.
3.
a.
$pK_a+pK_b=14 \rightarrow pK_a=14-pK_b=14-3.80=10.2$
$K_a=10^{-pK_a}=10^{-10.2}=6.31 \times 10^{-11}$
b.
$pK_a+pK_b=14 \rightarrow pK_a=14-pK_b=14-7.90=6.10$
$K_a=10^{-pK_a}=10^{-6.10}=7.94 \times 10^{-7}$
c.
$pK_a+pK_b=14 \rightarrow pK_a=14-pK_b=14-7.90=3.000 \times 10^{-1}$
$K_a=10^{-pK_a}=10^{-3.000 \times 10^{-1}}=-5.012 \times 10^{-1}$
d.
$pK_a+pK_b=14 \rightarrow pK_a=14-pK_b=14-1.40=12.6$
$K_a=10^{-pK_a}=10^{-12.6}=2.51 \times 10^{-13}$
e. $pK_a+pK_b=14 \rightarrow pK_a=14-pK_b=14-7.90=16.5$
$K_a=10^{-pK_a}=10^{-16.5}=3.16 \times 10^{-17}$
4.
$pK_a+pK_b=14 \rightarrow pK_b=14-pK_a=14-4.20=9.80$
$K_b=10^{-pK_b}=10^{-9.80}=1.58 \times 10^{-10}$
5.
$pK_a+pK_b=14 \rightarrow pK_a=14-pK_b=14-4.80=9.20$
$K_a=10^{-pK_a}=10^{-9.20}=6.31 \times 10^{-10}$
16.3: The Autoionization of Water
Conceptual Problems
1. What is the relationship between the value of the equilibrium constant for the autoionization of liquid water and the tabulated value of the ion-product constant of liquid water ($K_w$)?
2. The density of liquid water decreases as the temperature increases from 25°C to 50°C. Will this effect cause $K_w$ to increase or decrease? Why?
3. Show that water is amphiprotic by writing balanced chemical equations for the reactions of water with $HNO_3$ and $NH_3$. In which reaction does water act as the acid? In which does it act as the base?
4. Write a chemical equation for each of the following.
1. Nitric acid is added to water.
2. Potassium hydroxide is added to water.
3. Calcium hydroxide is added to water.
4. Sulfuric acid is added to water.
5. Show that $K$ for the sum of the following reactions is equal to $K_w$.
$HMnO_{4}\,(aq) \rightleftharpoons H^+\,(aq) + MnO^−_{4}\,(aq)$
$MnO^−_{4}\,(aq)+H_2O\,(l) \rightarrow HMnO_{4}\,(aq) + OH^−\,(aq)$
Conceptual Answers
1.
$K_{auto} = \dfrac{[H_3O^+][OH^−]}{[H_2O]^2}$
$K_w = [H_3O^+][OH^−] = K_{auto}[H_2O]^2$
2. This will affect $K_w$ as it is dependent on temperature. As the temperature increases, an endothermic process occurs (energy must be absorbed to break the bonds). Consequently, according to Le Chatelier, an increase in temperature favors the forward reaction thus the position of equilibrium shifts toward the right-hand side and $K_w$ becomes larger.
3.
Water acts as the base: $H_2O\,(l) + HNO_{3}\,(g) \rightarrow H_3O^+\,(aq) + NO^−_{3}\,(aq)$
Water acts as the acid: $H_2O\,(l) + NH_{3}\,(g) \rightarrow OH^−\,(aq) + NH^−_{4}\,(aq)$
4.
a. $HNO_3\,(aq)+H_2O\,(l) \rightleftharpoons H_3O^{+}\,(aq)+ HNO_{2}^{-}\,(aq)$
b. $KOH\,(s)+H_2O\,(l) \rightleftharpoons K^{-}\,(aq)+OH^{-}\,(aq)$
c. $Ca(OH)_{2}\,(s)+H_2O\,(l) \rightleftharpoons Ca^{2+}\,(aq)+2\,OH^{-}\,(aq)$
d. $H_2SO_4\, (aq)+H_2O\,(l) \rightleftharpoons HSO_4^{-}\,(aq)+H^{+}\,(aq)$
5.
$H_{2}O\,(l) \rightleftharpoons H^{+}\,(aq)+OH^{-}\,(aq)$
$K_w=[H^{+}][OH^{-}]$
Numerical Problems
1. The autoionization of sulfuric acid can be described by the following chemical equation: $H_2SO_{4}\,(l)+H_2SO_{4}\,(aq) \rightleftharpoons H_3SO^+_{4}\,(aq)+HSO_{4}^{-}\,(aq)$ At 25°C, $K = 3 \times 10^{−4}$. Write an equilibrium constant expression for $K_{H_2SO_4}$ that is analogous to $K_w$. The density of $H_2SO_4$ is $1.8\frac{g}{cm^{3}}$ at 25°C. What is the concentration of $H_3SO_{4}^{+}$ ? What fraction of $H_2SO_4$ is ionized?
2. An aqueous solution of a substance is found to have $[H_3O]^+ = 2.48 \times 10^{−8}\; M$. Is the solution acidic, neutral, or basic?
3. The pH of a solution is 5.63. What is its pOH? What is the $[OH^{−}]$? Is the solution acidic or basic?
4. State whether each solution is acidic, neutral, or basic.
1. $[H_3O^+] = 8.6 \times 10^{−3}\; M$
2. $[H_3O^+] = 3.7 \times 10^{−9}\; M$
3. $[H_3O^+] = 2.1 \times 10^{−7}\; M$
4. $[H_3O^+] = 1.4 \times 10^{−6}\; M$
5. Calculate the pH and the pOH of each solution.
1. 0.15 $M\,HBr$
2. 0.03 $M\,KOH$
3. $2.3 \times 10^{−3}\; M\; HNO_3$
4. $9.78 \times 10^{−2} \;M\; NaOH$
5. 0.00017 $M\,HCl$
6. 5.78 $M\,HI$
6. Calculate the pH and the pOH of each solution.
1. 25.0 mL of $2.3 \times 10^{−2}\;M\;HCl$, diluted to 100 mL
2. 5.0 mL of $1.87\,M\,NaOH$, diluted to 125 mL
3. 5.0 mL of $5.98\,M\,HCl$ added to 100 mL of water
4. 25.0 mL of $3.7\,M\,HNO_3$ added to 250 mL of water
5. 35.0 mL of $0.046\,M\,HI$ added to 500 mL of water
6. 15.0 mL of $0.0087\,M\,KOH$ added to 250 mL of water.
7. The pH of stomach acid is approximately 1.5. What is the $[H^+]$?
8. Given the pH values in parentheses, what is the $[H^+]$ of each solution?
1. household bleach (11.4)
2. milk (6.5)
3. orange juice (3.5)
4. seawater (8.5)
5. tomato juice (4.2)
9. A reaction requires the addition of 250.0 mL of a solution with a pH of 3.50. What mass of HCl (in milligrams) must be dissolved in 250 mL of water to produce a solution with this pH?
10. If you require 333 mL of a pH 12.50 solution, how would you prepare it using a 0.500 M sodium hydroxide stock solution?
Numerical Answers
1.
$K_{H_2SO_4}=[H_3SO_4^+][HSO_4^−]=K[H_2SO_4]_2$
$[H_3SO_4^+] = 0.3\,M$
So the fraction ionized is 0.02.
2. The solution is basic because the $pH=-log([H_3O^{+}])=-log(2.48 \times 10^{−8})=7.61>7$.
3.
$pH+pOH=14 \rightarrow pOH=14-pH=14-5.63=8.37$
$[OH^{-}]=10^{-pOH}=-4.27 \times 10^{-9}$
The $pH=5.63<7$, therefore the solution is acidic.
4.
a. The solution is acidic. $pH=-log([H_3O^{+}])=-log(8.6 \times 10^{−3})=2.1<7$
b. The solution is basic. $pH=-log([H_3O^{+}])=-log(3.7 \times 10^{−9})=8.4>7$
c. The solution is acidic. $pH=-log([H_3O^{+}])=-log(2.1 \times 10^{−7})=6.7<7$
d. The solution is acidic. $pH=-log([H_3O^{+}])=-log(1.4 \times 10^{−6})=5.9<7$
5.
a.
$pH=-log([H_3O^{+}])=-log(0.15)=0.82$
$pH+pOH=14 \rightarrow pOH=14-pH=14-0.82=13$
b.
$pOH=-log([OH^{-}])=-log(0.03)=2$
$pH+pOH=14 \rightarrow pH=14-pOH=14-2=10$
c.
$pH=-log([H_3O^{+}])=-log(2.3 \times 10^{−3})=2.6$
$pH+pOH=14 \rightarrow pOH=14-pH=14-2.6=11$
d.
$pOH=-log([OH^{-}])=-log(9.78 \times 10^{−2})=1.01$
$pH+pOH=14 \rightarrow pH=14-pOH=14-1.01=13.0$
e.
$pH=-log([H_3O^{+}])=-log(0.00017)=3.8$
$pH+pOH=14 \rightarrow pOH=14-pH=14-3.8=10$
f.
$pH=-log([H_3O^{+}])=-log(5.78)=-0.762$
$pH+pOH=14 \rightarrow pOH=14-pH=14-(-0.762)=14.8$
6.
a. $25.0\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{2.3 \times 10^{-2}\,mol}{1\,L} \times \frac{1}{100\,mL \times \frac{1\,L}{1,000\,mL}}=0.060\,M\,HCl$
$pH=-log([H_3O^{+}])=-log(0.060)=1.22$
$pH+pOH=14 \rightarrow pOH=14-pH=14-1.22=12.78$
b. $5.0\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{1.87\,mol}{1\,L} \times \frac{1}{125\,mL \times \frac{1\,L}{1,000\,mL}}=7.5 \times 10^{-2}\,M\,NaOH$
$pOH=-log([OH^{-}])=-log(7.5 \times 10^{-2})=1.1$
$pH+pOH=14 \rightarrow pH=14-pOH=14-1.1=12.9$
c. $5.0\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{5.98\,mol}{1\,L} \times \frac{1}{100\,mL \times \frac{1\,L}{1,000\,mL}}=0.20\,M\,HCl$
$pH=-log([H_3O^{+}])=-log(0.20)=0.70$
$pH+pOH=14 \rightarrow pOH=14-pH=14-0.70=13.3$
d. $25.0\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{3.7\,mol}{1\,L} \times \frac{1}{250\,mL \times \frac{1\,L}{1,000\,mL}}=0.370\,M\,HNO_3$
$pH=-log([H_3O^{+}])=-log(0.370)=0.432$
$pH+pOH=14 \rightarrow pOH=14-pH=14-0.432=13.568$
e. $35.0\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{0.046\,mol}{1\,L} \times \frac{1}{500\,mL \times \frac{1\,L}{1,000\,mL}}=3 \times 10^{-3}\,M\,HI$
$pH=-log([H_3O^{+}])=-log(3 \times 10^{-3})=2.52$
$pH+pOH=14 \rightarrow pOH=14-pH=14-2.52=11.48$
f. $15.0\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{0.0087\,mol}{1\,L} \times \frac{1}{125\,mL \times \frac{1\,L}{1,000\,mL}}=5.20 \times 10^{-4}\,M\,KOH$
$pOH=-log([OH^{-}])=-log(5.20 \times 10^{-4})=3.28$
$pH+pOH=14 \rightarrow pH=14-pOH=14-3.28=10.72$
7. $[H^+]=10^{-pH}=10^{-1.5}=3.2 \times 10^{-2}\,M$
8.
a. $[H^{+}]=10^{-11.4}=3.98 \times 10^{-12}\,M$
b. $[H^{+}]=10^{-6.5}=3.2 \times 10^{-7}\,M$
c. $[H^{+}]=10^{-3.5}=3.2 \times 10^{-4}\,M$
d. $[H^{+}]=10^{-8.5}=3.2 \times 10^{-9}\,M$
e. $[H^{+}]=10^{-4.2}=6.3 \times 10^{-5}\,M$
9. 2.9 mg $HCl$
$[H^{+}]=10^{-pH}=10^{-3.50}=3.1622 \times 10^{-4}\,M$
$x\,mg\,HCl \times \frac{1\,g\,HCl}{1,000\,mg\,HCl} \times \frac{1\,mol\,HCl}{36.46\,g\,HCl} \times \frac{1}{250\,mL\,H_2O \times \frac{1\,L}{1,000\,mL\,H_2O}}=3.1622 \times 10^{-4}\,M\,HCl \rightarrow \frac{x\,mol\,HCl}{9115\,L\,HCl}=3.1622 \times 10^{-4}\,M\,HCl \rightarrow x\,mol\,HCl=3.1622\times 10^{-4}\,M \times 9115\,L\,HCl \rightarrow x\,mol\,HCl=2.9\,mol\,HCl \rightarrow x=2.9$
10. To prepare the stock solution, $2.11 \times 10^{-2}\,L$ of $0.500\,M\,NaOH$ solution is required.
$\frac{0.03162\,mol\,NaOH}{1\,L\,NaOH} \times \frac{1\,L\,NaOH}{0.5\,mol\,NaOH} \times 0.333\,L=2.11 \times 10^{-2}\,g\,NaOH$.
$[OH^{-}]=10^{-1.5}=0.03162\,M$
$pH+pOH=14 \rightarrow pOH=14-12.50=1.5$
16.10: Acid-Base Behavior and Chemical Structure
Conceptual Problems
1. Several factors affect the relative strengths of acids and bases. For each pair, identify the most important factor in determining which is the stronger acid or base in aqueous solution.
1. $CH_3CCl_2CH_2CO_2H$ versus $CH_3CH_2CH_2CO_2H$
2. $CH_3CO_2H$ versus $CH_3CH_2OH$
3. $HClO$ versus $HBrO$
4. $\ce{CH_3C(=O)NH_2}$ versus $CH_3CH_2NH_2$
5. $H_3AsO_4$ versus $H_3AsO_3$
2. The stability of the conjugate base is an important factor in determining the strength of an acid. Which would you expect to be the stronger acid in aqueous solution—$C_6H_5NH_3^+$ or $NH_4^+$? Justify your reasoning.
3. Explain why $H_2Se$ is a weaker acid than $HBr$.
4. Arrange the following in order of decreasing acid strength in aqueous solution: $H_3PO_4$, $CH_3PO_3H_2$, and $HClO_3$.
5. Arrange the following in order of increasing base strength in aqueous solution: $\ce{CH_3S−}$, $OH^−$, and $CF_3S^−$.
6. Arrange the following in order of increasing acid strength in aqueous solution: $HClO_2$, $HNO_2$, and $HNO_3$.
7. Do you expect $H_2SO_3$ or $H_2SeO_3$ to be the stronger acid? Why?
8. Give a plausible explanation for why $CF_3OH$ is a stronger acid than $CH_3OH$ in aqueous solution. Do you expect $CHCl_2CH_2OH$ to be a stronger or a weaker acid than $CH_3OH$? Why?
9. Do you expect $Cl_2NH$ or $NH_3$ to be the stronger base in aqueous solution? Why?
Conceptual Answers
1.
a. The most important factor in determining the stronger acid is considering the inductive effect. Chlorine is an electron-withdrawing group. It pulls electron density away from the compound by means of the inductive effect through the sigma bond. In considering the conjugate base of $CH_3CCl_2CH_2CO_2H$, Chlorine absorbs some of the electron density or excess negative charge on the oxygen atom. This causes the C bonded to the attached Chlorine atoms to be partially positive. The conjugate base of $CH_3CCl_2CH_2CO_2H$ is more stable, thus more acidic than the conjugate base of $CH_3CH_2CH_2CO_2H$.
b. The most important factor in determining the stronger acid is knowing the $pK_a$ values for functional groups. The $pK_a$ of alcohol is about 16 while the $pK_a$ of a carboxylic acid is about 5. Therefore, $CH_3CO_2H$ is more acidic than $CH_3CH_2OH$.
c. The most important factor in determining the stronger acid is electronegativity. The chlorine atom is more electronegative than the bromine atom, therefore $HClO$ is more acidic than $HBrO$.
d. The most important factor in determining the stronger acid is considering resonance. The $\ce{CH_3C(=O)NH_2}$ has a resonance which increases the stability of the conjugate base (therefore increasing acidity) because the negative charge can be delocalized. Thus, $\ce{CH_3C(=O)NH_2}$ is more acidic than $CH_3CH_2NH_2$.
e. The most important factor in determining the stronger acid is considering oxidation states on the central nonmetal. $H_3AsO_4$ has an oxidation state of +5 which is larger and thus more acidic than $H_3AsO_3$ which has an oxidation state of +3.
$CF_3S^− < CH_3S^− < OH^−$ (strongest base)
$NH_3$; $Cl$ atoms withdraw electron density from $N$ in $Cl_2NH$.
2. It is expected that the stronger acid is $C_6H_5NH_3^+$ because in considering the conjugate base the lone pair of electrons on nitrogen is involved in resonance, hence the molecule is stable.
3. $H_2Se$ is a weaker acid than HBr because $Br$ is more electronegative than $Se$ thus more stable.
4. $HClO_3>CH_3PO_3H_2>H_3PO_4$
This is because $H_3PO_4$ is a polyprotic acid which contains more than one ionizable proton, and the protons are lost in a stepwise manner. The fully protonated species is always the strongest acid because it is easier to remove a proton from a neutral molecule than from a negatively charged ion. Thus, acid strength decreases with the loss of subsequent protons, and, correspondingly, the $pK_a$ increases which indicate it is the most basic. The conjugate base of $ClO^{3-}$ has a much smaller charge to volume ratio, thus most stable and acidic.
5. $CF_{3}S^{-}<CH_{3}S^{-}<OH^{-}$
$CF_{3}S^{-}$ is the most acidic because of the three electronegative Fluorines. $OH^{-}$ is the strongest base in water. Thus, $CH_{3}S^{-}$ is in between these aqueous solutions.
6. $HNO_2<HClO_2<HNO_3$
$HNO_3$ has resonance stabilization, therefore it is the most acidic. Between $HClO_2$ and $HNO_2$, $Cl$ is the most electronegative therefore $HClO_2$ is more acidic than $HNO_2$.
7. I expect $H_2SO_3$ to be the stronger acid because it is more electronegative or has greater attraction which means it’ll be less inclined to share their electrons with a proton.
8. $CF_3OH$ is a stronger acid than $CH_3OH$ in aqueous solution because $F$ is more electronegative than $H$.
9. It would be expected that $NH_3$ be a stronger base than $Cl_2NH$ because it is electronegative due to the two $Cl$ atoms.
16.11: Lewis Acids and Bases
Problems
1. Identify the nature of each of the following as either a Lewis Acid or a Lewis Base:
1. $NH_3$
2. $Ag^+$
3. $Ni^{2+}$
4. $Pt^{4+}$
5. $H_2O$
6. $SO_2$
2. Explain why $SiF_4$ can act as a Lewis Acid.
3. Identify the nature of each of the following as either a Lewis Acid or a Lewis Base:
1. $[Fe(CN)_6]^{3-}$
2. $[Ni(NH_3)_6]^{2+}$
3. $CdBr_4^{2-}$
4. What is the product of the reaction of $CO_2 + OH^- \rightarrow$ ?
5. In the reactions below, which is the Lewis Acid and/or which is the Lewis Base?
1. $NH_3 + H^+ \rightarrow NH_4^+$
2. $H_2O + H^+ \rightarrow H_3O^+$
6. In the complex ion, $[PtCl_6]^{2-}$ which is the Lewis Acid and which is the Lewis base?
7. The reaction of $AgCl$ + $NH_3$ produces what complex ion?
Solutions
1.
a. $NH_3$ is a lewis base because nitrogen has a lone pair of electrons to "donate."
b. $Ag^{+}$ is a lewis acid because it has an unfilled octet and thus is able to accept a pair of electrons.
c. $Ni^{+}$ is a lewis acid because it has an unfilled octet and thus is able to accept a pair of electrons.
d. $Pt^{4+}$ is a lewis acid because it has an unfilled octet and thus is able to accept a pair of electrons.
e. $H_2O$ is a lewis base because oxygen has two lone pairs of electrons to "donate."
f. $SO_2$ is a lewis acid because sulfur has an unfilled octet and thus is able to accept a pair of electrons.
2. $SiF_4$ has a central Silicon Atom which can expand its octet to 12 (compared to the typical 8) so that it forms $[SiF_6]^{2-}$.
3.
1. Lewis Acid: $Fe^{3+}$, Lewis Base: $CN^-$
2. Lewis Acid: $Ni^{2+}$, Lewis Base: $NH_3$
3. Lewis Acid: $Cd^{2+}$, Lewis Base: $Br^-$
4. This reaction forms a bicarbonate ion. $CO_2 + OH^- \rightarrow O--COH=O$.
5.
a. Lewis Acid: $H^+$, Lewis Base: $NH_3$
b. Lewis Acid: $H^+$, Lewis Base: $H_2O$
6. $Pt^{4+}$ is the Lewis acid and $Cl^-$ is the Lewis base.
7. $AgCl + 2\,NH_3 \rightarrow [Ag(NH_3)_2]^+ + Cl^-$
Conceptual Problems
1. Construct a table comparing how OH, NH3, H2O, and BCl3 are classified according to the Arrhenius, the Brønsted–Lowry, and the Lewis definitions of acids and bases
2. Describe how the proton $(H^{+})$ can simultaneously behave as an Arrhenius acid, a Brønsted–Lowry acid, and a Lewis acid.
3. Would you expect aluminum to form compounds with covalent bonds or coordinate covalent bonds? Explain your answer.
4. Classify each compound as a Lewis acid or a Lewis base and justify your choice.
a. $AlCl_{3}$
b. $CH_{3}N$
c. $IO_{3}^{-}$
5. Explain how a carboxylate ion $(RCO_{2}^{-})$ can act as both a Brønsted–Lowry base and a Lewis base.
Conceptual Answers
1.
Arrhenius Acid
Arrhenius
Base
Brønsted–Lowry Acid Brønsted–Lowry Base Lewis Acid Lewis Base
$OH^{-}$ X X X
$NH_3$ X X
$H_2O$ X X X X X
$BCl_3$ X
An Arrhenius acid is a molecule that when dissolved in water it will donate an $H^{+}$ in solution.
An Arrhenius base is a molecule that when dissolved in water it will donate an $OH^{-}$ in solution.
A Brønsted–Lowry acid is a molecule that when dissolved in a solution it will donate an $H^{+}$ in solution.
A Brønsted–Lowry base is a molecule that when dissolved in a solution it will donate an atom or ion capable of accepting or bonding to a free proton in solution.
A Lewis acid is an atom or molecule that accepts an electron pair.
A Lewis base is an atom or molecule that donates an electron pair.
2.
The proton $(H^{+})$ can simultaneously behave as an Arrhenius acid because when it is dissolved in water it will donate itself. $H^{+}+H_{2}O \rightleftharpoons H_{3}O^{+}$
The proton $(H^{+})$ can simultaneously behave as a Brønsted–Lowry acid because when it is dissolved in solution it will donate itself.
$H^{+}+B^{-} \rightleftharpoons HB$
The proton $(H^{+})$ can simultaneously behave as a Lewis acid as it can accept an electron pair.
$H^{+}+B^{-} \rightleftharpoons HB$
3. It is expected that Aluminum forms a coordinate covalent bond as it can participate in a Lewis acid and a Lewis base interaction. For example $Al^{3+}+H_{2}O \rightleftharpoons [Al(OH_2)_{6}]^{3+}$
4.
a. $AlCl_{3}$ is a Lewis acid as $Al$ can accept an electron pair.
b. $CH_{3}N$ is a Lewis base as $N$ can donate an electron pair.
c. $IO_{3}^{-}$ is a Lewis base as $I$ can donate an electron pair.
5. The carboxylate ion $(RCO_{2}^{-})$ can act as Brønsted–Lowry base because when dissolved in a solution the electron rich $O$ is capable of accepting a proton. The carboxylate ion $(RCO_{2}^{-})$ can act as a Lewis base because the electron rich $O$ can donate an electron pair.
Numerical Problems
1. In each reaction, identify the Lewis acid and the Lewis base and complete the reaction by writing the products(s).
a. (CH3)2O + AlCl3
b. SnCl4 + 2 Cl
2. Use Lewis dot symbols to depict the reaction of BCl3 with dimethyl ether [(CH3)2O]. How is this reaction similar to that in which a proton is added to ammonia?
Answer
1.
a. The Lewis acid is $AlCl_{3}$ and the Lewis base is $(CH_{3})_{2}O$.
$(CH_{3})_{2}O+AlCl_{3} \rightleftharpoons AlCl_{3} \cdot O(CH_{3})_{2}$
b. The Lewis acid is $SnCl_{4}$ and the Lewis base is $Cl^{-}$.
$SnCl_{4}+2\,Cl^{-} \rightleftharpoons SnCl_{6}^{2-}$
2.
$BCl_{3}+(CH_{3})_{2}O \rightleftharpoons BCl_{3}\cdot O(CH_{3})_{2}$
This reaction is similar to that in which a proton is added to ammonia as it also involves a Lewis acid and a Lewis base interaction. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/16%3A_AcidBase_Equilibria/16.E%3A_AcidBase_Equilibria_%28Exercises%29.txt |
16.1: Acids and Bases: A Brief Review
• acids have sour taste and turn litmus paper red
• bases have a bitter taste and feel slippery
• Svante Arrhenius (1859-1927)
• Acids associated with H+ ions
• Bases associated with OH- ions
• Solution is acidic if [H+] > [OH-]
• Solution is basic if [OH-] > [H+]
16.2: Brønsted–Lowry Acids and Bases
• Arrhenius definition of acids and bases
• Acids when dissolved in water increase H+ concentration
• Bases when dissolved in water increase OH- concentration
16.2.1 Proton Transfer Reactions
• Brønsted-Lowry definition of acids an bases
• Acid is a proton donor
• Base is a proton acceptor
• Can be applied to non-aqueous solutions
• Brønsted-Lowry acid must be able to lose a H+ ion
• Brønsted-Lowry base must have at least one non-bonding pair (lone pair) of electrons to bind to H+ ion
• Amphoteric - substance that can act as an acid or base
16.2.2 Conjugate Acid-Base Pairs
• conjugate acid - product formed by adding a proton to base
• conjugate base - product formed by removal of a proton from acid
16.2.3 Related Strengths of Acids and Bases
• the stronger the acid, the weaker the conjugate base
• the stronger the base, the weaker the conjugate acid
• equilibrium favors transfer of proton from stronger acid to stronger base
16.3: The Autoionization of Water
• autoionization of water - dissociation of H2O molecules to H+ and OH- ions
• at room temperature only 1 out of 109 molecules are ionized
• exclude water from equilibrium expressions involving aqueous solutions
• ion-product constant
• kw = k[H2O] = [H+][OH-] = 1.0 x 10-14 (at 25° C)
• solution is neutral when [H+] = [OH-]
• solution is acidic when [H+] > [OH-]
• solution is basic when [H+] < [OH-]
16.3.1 The Proton in Water
• H+ ion is a proton with no valence electrons
• H+ ion react with H2O molecule to form H3O+, hydronium ion
• H3O+ ion can bond with other H2O molecules to form hydrated hydrogen ions
• H+ and H3O+ used interchangeably
16.4: The pH Scale
• concentration of [H+] expressed in terms of pH
• pH = -log [H+]
• acidic solutions [H+] > 1.0 x 10-7 [OH-] < 1.0 x 10-7 pH < 7.00
• neutral solutions [H+] = [OH-] = 1.0 x 10-7 pH = 7
• basic solutions [H+] < 1.0 x 10-7 [OH-] > 1.0 x 10-7 pH > 7
Other "p" Series
• pOH = -log [OH-]
• pH + pOH = -log Kw = 14.00
Measuring pH
• pH meter
• has a pair of electrodes connected to a meter that measures in millivolts
• voltage generated when electrodes placed in solution, and is measured by meter
• blue litmus paper turns red in acidic solution
• red litmus paper turns blue in basic solution
16.5: Strong Acids and Bases
strong acids and bases are strong electrolytes
16.5.1 Strong Acids
strongest monoprotic acids
• HCl, HBr, HI, HNO3, HclO3, HclO4, and diprotic H2SO4
• For strong monoprotic acid concentration of [H+] equals the original concentration of the acid
16.5.2 Strong Bases
• most common strong bases are ionic hydroxides of alkali metals and the heavier alkaline-earth metals
• complete dissociation
16.6: Weak Acids
• $HA_{(aq)} + H_2O_{(l)} \to H_3O^+ + A^-_{(aq)}$
• $HA_{(aq)} \to H^+_{(aq)} + A^-_{(aq)}$
• $K_a = \frac{[H^+][A^-]}{[HA]}$
• Ka = acid - dissociation constant
• The lager the Ka the stronger the acid
• Ka usually less than 10-3
16.6.1 Calculating pH for Solutions of Weak Acids
• 1) write ionization equilibrium
• 2) write equilibrium expression
• 3) I.C.E. Table
• 4) substitute equilibrium concentrations into equilibrium expression
• percent ionization = fraction of weak acid molecules that ionize * 100%
• in weak acids [H+] is small fraction of concentration of acid
• percent ionization depends on temperature, identity of acid and concentration
• as percent ionization decreases, concentration increases
16.6.2 Polyprotic Acids
• more than one ionizable H atom
• easier to remove first proton than second
• acid dissociation constants are Ka1, Ka2, etc…
• Ka values usually differ by 103
16.7: Weak Bases
• base-dissociation constant, Kb
• equilibrium at which base reacts with H2O to form a conjugate acid and OH-
• contain 1 or more lone pair of electrons
16.7.1 Types of Weak Bases
• weak bases have NH3 and anions of weak acids
16.8: Relationship Between Ka and Kb
• when two reactions are added together then equilibrium constant of third reaction is equal to the product of the equilibrium constants of the added reactions
• reaction 1 + reaction 2 = reaction 3
• K1 x K2 = K3
• Ka x Kb = [H+][OH-] = Kw
• Acid-dissociation constant times base-dissociation constant equals the ion-product constant for water
• Ka x Kb = Kw = 1.0 x 10-14
• pKa x pKb = pKw = 14; (pKa= -log Ka and pKb = -log Kb)
16.9: Acid-Base Properties of Salt Solutions
• hydrolysis - ions reacting with water to produce H+ and OH- ions
• anions from weak acids react with water to produce OH- ions which is basic
• anions of strong acids are not basic and do not influence pH
• anions that have ionizable protons are amphoteric
• behavior depends on Ka and Kb
• all cations except those of alkali metals and heavier alkaline earth (Ca2+, Sr2+ and Ba2+) are weak acids in water
• alkali metal and alkaline earth cations do not hydrolyze
• do not affect pH
• strengths of acids and bases from salts
• 1) salts derived from strong acid and base
• no hydrolysis and solution has pH of 7
• 2) salts derived from strong base and weak acid
• strong conjugate base
• anion hydrolyzes and produces OH- ions
• cation does not hydrolyze
• pH greater than 7
• 3) salts derived from weak base and strong acids
• cation is strong conjugate acid
• cation hydrolyzes to produce H+
• anion does not hydrolyze
• solution has pH below 7
• 4) salts derived from weak acid and base
• both cation and anion hydrolyze
• pH depends on extent on hydrolysis of each ion]
16.10: Acid-Base Behavior and Chemical Structure
16.10.1 Factors that Affect Acid Strength
• strength of acid depends on:
• 1) polarity of H-X bond
• 2) strength of H-X bond
• 3) stability of conjugate base, X-
• molecule will transfer proton if H-X bond is polarized
• in ionic hydrides H- acts as proton acceptor because of negative charge
• nonpolar bonds produce neither acidic nor basic solutions
• strong bonds less easily dissociated that weak bonds
• the greater the stability of conjugate base, the stronger the acid]
16.10.2 Binary Hydrides
• metal hydrides are basic or have no acid-base properties in water
• nonmetal hydrides can be between having no acid-base properties to being acidic
• in each group of nonmetallic elements, acidity increases with increasing atomic number
• bond strengths decrease as central atom gets larger and overlap of orbitals get smaller
16.10.3 Oxyacids
• Y-O-H bond
• Oxyacids - have OH bonded to central atom
• Base if bonded to a metal because pair of electrons shared between Y-O is completely transferred to O
• Ionic compound with OH- is formed
• When bonded to nonmetal the bond is covalent and compounds are acidic or neutral
• As electronegativity of Y increases , acidity also increases
• O-H bond becomes more polar
• Conjugate base usually an anion and stability increases as electronegativity of Y increases
• Relating acid strengths of oxyacids to electronegativity of Y and to number of groups attached to Y
• 1) same number of oxygen atoms, acid strength increases as electronegativity of central atom increases
• 2) same central atom Y, acid strength increases with increasing number of bonded oxygen atoms to central atom
• acidity increases as oxidation number of central atom increases
16.10.4 Carboxylic Acids
• carboxyl group - COOH
• acidic behavior of carboxylic acids
• addition oxygen atom in carboxyl group draws density from O-H bond which increases the polarity
• conjugate base ion have resonance forms
• acidity increases as number of electronegative atoms in acid increases
16.11: Lewis Acids and Bases
• Lewis acid - electron pair acceptor
• Lewis base - electron pair donor
• Any Bronsted-Lowry base is a Lewis base
• Lewis acids contain at least one atom with an incomplete octet
16.11.1 Hydrolysis of Metal Ions
• hydration - attraction of metal ions to water molecules
• metal ion acts as Lewis acid
• water molecule acts as Lewis base
• electron density drawn from oxygen atom to water molecule
• O-H bond becomes more polarized
• For hydrolysis reactions Ka increases with increasing charge and decreasing radius of ion | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/16%3A_AcidBase_Equilibria/16.S%3A_AcidBase_Equilibria_%28Summary%29.txt |
This chapter address equilibria with solutions containing more than one solute.
• 17.1: The Common-Ion Effect
The common-ion effect is used to describe the effect on an equilibrium involving a substance that adds an ion that is a part of the equilibrium.
• 17.2: Buffered Solutions
Buffers are solutions that resist a change in pH after adding an acid or a base. Buffers contain a weak acid (\(HA\)) and its conjugate weak base (\(A^−\)). Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium shifts the equilibrium in such a way as to reduce the concentration of the common ion. Buffers are characterized by their pH range and buffer capacity.
• 17.3: Acid-Base Titrations
The shape of a titration curve, a plot of pH versus the amount of acid or base added, provides important information about what is occurring in solution during a titration. The shapes of titration curves for weak acids and bases depend dramatically on the identity of the compound. The equivalence point of an acid–base titration is the point at which exactly enough acid or base has been added to react completely with the other component.
• 17.4: Solubility Equilibria
The solubility product (Ksp) is used to calculate equilibrium concentrations of the ions in solution, whereas the ion product (Q) describes concentrations that are not necessarily at equilibrium. The equilibrium constant for a dissolution reaction, called the solubility product (Ksp), is a measure of the solubility of a compound. Whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, Ksp is defined in terms of the molar concentrations of the component ions.
• 17.5: Factors that Affect Solubility
Ion-pair formation, the incomplete dissociation of molecular solutes, the formation of complex ions, and changes in pH all affect solubility. There are four explanations why the solubility of a compound can differ from the solubility indicated by the concentrations of ions: (1) ion pair formation, in which an anion and a cation are in intimate contact in solution and not separated by solvent, (2) the incomplete dissociation of molecular solutes, (3) the formation of complex ions, and (4) changes
• 17.6: Precipitation and Separation of Ions
A mixture of metal ions in a solution can be separated by precipitation with anions such as \(\ce{Cl-}\), \(\ce{Br-}\), \(\ce{SO4^2-}\), \(\ce{CO3^2-}\), \(\ce{S^2-}\), \(\ce{Cr2O4^2-}\), \(\ce{PO4^2-}\), \(\ce{OH-}\) etc. When a metal ion or a group of metal ions form insoluble salts with a particular anion, they can be separated from others by precipitation. We can also separate the anions by precipitating them with appropriate metal ions.
• 17.7: Qualitative Analysis for Metallic Elements
In qualitative analysis, the identity, not the amount, of metal ions present in a mixture is determined. The technique consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all the metal ions are precipitated. Other additional steps are needed to separate metal ions that precipitate together.
• 17.E: Additional Aspects of Aqueous Equilibria (Exercises)
These are homework exercises for Chapter 17 "Additional Aspects of Aqueous Equilibria" of the Brown et al. Textmap, which addresses more complex equilibria with solutions containing more than one solute.
• 17.S: Additional Aspects of Aqueous Equilibria (Summary)
This is the summary for Chapter 17 "Additional Aspects of Aqueous Equilibria" of Brown et al. Textmap, which addresses more complex equilibria with solutions containing more than one solute.
17: Additional Aspects of Aqueous Equilibria
Learning Objectives
• Recognize common ions from various salts, acids, and bases.
• Calculate concentrations involving common ions.
• Calculate ion concentrations involving chemical equilibrium.
The common-ion effect is used to describe the effect on an equilibrium involving a substance that adds an ion that is a part of the equilibrium.
Introduction
The solubility products Ksp's are equilibrium constants in heterogeneous equilibria (i.e., between two different phases). If several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion. For example, a solution containing sodium chloride and potassium chloride will have the following relationship:
$\mathrm{[Na^+] + [K^+] = [Cl^-]} \label{1}$
Consideration of charge balance or mass balance or both leads to the same conclusion.
Common Ions
When $\ce{NaCl}$ and $\ce{KCl}$ are dissolved in the same solution, the $\mathrm{ {\color{Green} Cl^-}}$ ions are common to both salts. In a system containing $\ce{NaCl}$ and $\ce{KCl}$, the $\mathrm{ {\color{Green} Cl^-}}$ ions are common ions.
$\mathrm{NaCl \rightleftharpoons Na^+ + {\color{Green} Cl^-}} \nonumber$
$\mathrm{KCl \rightleftharpoons K^+ + {\color{Green} Cl^-}} \nonumber$
$\mathrm{CaCl_2 \rightleftharpoons Ca^{2+} + {\color{Green} 2 Cl^-}} \nonumber$
$\mathrm{AlCl_3 \rightleftharpoons Al^{3+} + {\color{Green} 3 Cl^-}} \nonumber$
$\mathrm{AgCl \rightleftharpoons Ag^+ + {\color{Green} Cl^-}} \nonumber$
For example, when $\ce{AgCl}$ is dissolved into a solution already containing $\ce{NaCl}$ (actually $\ce{Na+}$ and $\ce{Cl-}$ ions), the $\ce{Cl-}$ ions come from the ionization of both $\ce{AgCl}$ and $\ce{NaCl}$. Thus, $\ce{[Cl- ]}$ differs from $\ce{[Ag+]}$. The following examples show how the concentration of the common ion is calculated.
Example $1$
What are $\ce{[Na+]}$, $\ce{[Cl- ]}$, $\ce{[Ca^2+]}$, and $\ce{[H+]}$ in a solution containing 0.10 M each of $\ce{NaCl}$, $\ce{CaCl2}$, and $\ce{HCl}$?
Solution
Due to the conservation of ions, we have
$\mathrm{[Na^+] = [Ca^{2+}] = [H^+] = 0.10\: \ce M} \nonumber$
but
\begin{alignat}{3} \nonumber &\ce{[Cl- ]} &&= && && \:\textrm{0.10 (due to NaCl)}\ \nonumber & && && + &&\mathrm{\:0.20\: (due\: to\: CaCl_2)}\ \nonumber & && && + &&\mathrm{\:0.10\: (due\: to\: HCl)}\ \nonumber & &&= && &&\mathrm{\:0.40\: M} \nonumber \end{alignat}
Exercise $1$
John poured 10.0 mL of 0.10 M $\ce{NaCl}$, 10.0 mL of 0.10 M $\ce{KOH}$, and 5.0 mL of 0.20 M $\ce{HCl}$ solutions together and then he made the total volume to be 100.0 mL. What is $\ce{[Cl- ]}$ in the final solution?
Solution
$\mathrm{[Cl^-] = \dfrac{0.1\: M\times 10\: mL+0.2\: M\times 5.0\: mL}{100.0\: mL} = 0.020\: M} \nonumber$
Le Châtelier's Principle states that if an equilibrium becomes unbalanced, the reaction will shift to restore the balance. If a common ion is added to a weak acid or weak base equilibrium, then the equilibrium will shift towards the reactants, in this case the weak acid or base.
Example $2$: Solubility of Lead Chloride
Consider the lead(II) ion concentration in this saturated solution of PbCl2. The balanced reaction is
$PbCl_{2 (s)} \rightleftharpoons Pb^{2+} _{(aq)} + 2Cl^-_{(aq)} \nonumber$
Defining $s$ as the concentration of dissolved lead(II) chloride, then:
$[Pb^{2+}] = s \nonumber$
$[Cl^- ] = 2s \nonumber$
These values can be substituted into the solubility product expression, which can be solved for $s$:
$\begin{eqnarray} K_{sp} &=& [Pb^{2+}] [Cl^-]^2 \ &=& s \times (2s)^2 \ 1.7 \times 10^{-5} &=& 4s^3 \ s^3 &=& \frac{1.7 \times 10^{-5}}{4} \ &=& 4.25 \times 10^{-6} \ s &=& \sqrt[3]{4.25 \times 10^{-6}} \ &=& 1.62 \times 10^{-2}\ mol\ dm^{-3} \end{eqnarray} \nonumber$The concentration of lead(II) ions in the solution is 1.62 x 10-2 M. Consider what happens if sodium chloride is added to this saturated solution. Sodium chloride shares an ion with lead(II) chloride. The chloride ion is common to both of them; this is the origin of the term "common ion effect".
Look at the original equilibrium expression again:
$PbCl_2 \; (s) \rightleftharpoons Pb^{2+} \; (aq) + 2Cl^- \; (aq) \nonumber$
What happens to that equilibrium if extra chloride ions are added? According to Le Châtelier, the position of equilibrium will shift to counter the change, in this case, by removing the chloride ions by making extra solid lead(II) chloride.
Of course, the concentration of lead(II) ions in the solution is so small that only a tiny proportion of the extra chloride ions can be converted into solid lead(II) chloride. The lead(II) chloride becomes even less soluble, and the concentration of lead(II) ions in the solution decreases. This type of response occurs with any sparingly soluble substance: it is less soluble in a solution which contains any ion which it has in common. This is the common ion effect.
Example $3$
If an attempt is made to dissolve some lead(II) chloride in some 0.100 M sodium chloride solution instead of in water, what is the equilibrium concentration of the lead(II) ions this time? As before, define s to be the concentration of the lead(II) ions.
$[Pb^{2+}] = s \label{2}$
The calculations are different from before. This time the concentration of the chloride ions is governed by the concentration of the sodium chloride solution. The number of ions coming from the lead(II) chloride is going to be tiny compared with the 0.100 M coming from the sodium chloride solution.
In calculations like this, it can be assumed that the concentration of the common ion is entirely due to the other solution. This simplifies the calculation.
So we assume:
$[Cl^- ] = 0.100\; M \label{3}$
The rest of the mathematics looks like this:
$\begin{split} K_{sp}& = [Pb^{2+}][Cl^-]^2 \ & = s \times (0.100)^2 \ 1.7 \times 10^{-5} & = s \times 0.00100 \end{split} \nonumber$
therefore:
$\begin{split} s & = \dfrac{1.7 \times 10^{-5}}{0.0100} \ & = 1.7 \times 10^{-3} \, \text{M} \end{split} \label{4}$
Finally, compare that value with the simple saturated solution:
Original solution:
$[Pb^{2+}] = 0.0162 \, M \label{5}$
Solution in 0.100 M NaCl solution:
$[Pb^{2+}] = 0.0017 \, M \label{6}$
The concentration of the lead(II) ions has decreased by a factor of about 10. If more concentrated solutions of sodium chloride are used, the solubility decreases further.
A Video Discussing Finding the Solubility of a Salt: Finding the Solubility of a Salt(opens in new window) [youtu.be]
Common Ion Effect with Weak Acids and Bases
Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.
Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant.
The common ion effect of H3O+ on the ionization of acetic acid
The common ion effect suppresses the ionization of a weak base by adding more of an ion that is a product of this equilibrium. Now consider the common ion effect of $\ce{OH^{-}}$ on the ionization of ammonia
Adding the common ion of hydroxide shifts the reaction towards the left to decrease the stress (in accordance with Le Chatelier's Principle), forming more reactants. This decreases the reaction quotient, because the reaction is being pushed towards the left to reach equilibrium. The equilibrium constant, $K_b=1.8 \times 10^{-5}$, does not change. The reaction is put out of balance, or equilibrium.
$Q_a = \frac{\ce{[NH_4^{+}][OH^{-}]}}{\ce{[NH3]}} \nonumber$
At first, when more hydroxide is added, the quotient is greater than the equilibrium constant. The reaction then shifts right, causing the denominator to increase, decreasing the reaction quotient and pulling towards equilibrium and causing Q to decrease towards K.
Common Ion Effect on Solubility
When a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions. For the dissolution of calcium phosphate, one of the two main components of kidney stones, the equilibrium can be written as follows, with the solid salt on the left:
$\ce{Ca3(PO4)2(s) <=> 3Ca^{2+}(aq) + 2PO^{3−}4(aq)} \label{17.4.1}$
As you will discover in more advanced chemistry courses, basic anions, such as S2−, PO43, and CO32, react with water to produce OH and the corresponding protonated anion. Consequently, their calculated molarities, assuming no protonation in aqueous solution, are only approximate. The equilibrium constant for the dissolution of a sparingly soluble salt is the solubility product (Ksp) of the salt. Because the concentration of a pure solid such as Ca3(PO4)2 is a constant, it does not appear explicitly in the equilibrium constant expression. The equilibrium constant expression for the dissolution of calcium phosphate is therefore
$K=\dfrac{[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2}{[\mathrm{Ca_3(PO_4)_2}]} \label{17.4.2a}$
$[\mathrm{Ca_3(PO_4)_2}]K=K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2 \label{17.4.2b}$
At 25°C and pH 7.00, $K_{sp}$ for calcium phosphate is 2.07 × 10−33, indicating that the concentrations of Ca2+ and PO43 ions in solution that are in equilibrium with solid calcium phosphate are very low. The values of Ksp for some common salts vary dramatically for different compounds (Table E3). Although $K_{sp}$ is not a function of pH in Equation $\ref{17.4.2a}$, changes in pH can affect the solubility of a compound.
The solubility product expression tells us that the equilibrium concentrations of the cation and the anion are inversely related. That is, as the concentration of the anion increases, the maximum concentration of the cation needed for precipitation to occur decreases—and vice versa—so that $K_{sp}$ is constant. Consequently, the solubility of an ionic compound depends on the concentrations of other salts that contain the same ions. This dependency is another example of the common ion effect where adding a common cation or anion shifts a solubility equilibrium in the direction predicted by Le Chatelier’s principle. As a result, the solubility of any sparingly soluble salt is almost always decreased by the presence of a soluble salt that contains a common ion.
Consider, for example, the effect of adding a soluble salt, such as CaCl2, to a saturated solution of calcium phosphate [Ca3(PO4)2]. We have seen that the solubility of Ca3(PO4)2 in water at 25°C is 1.14 × 10−7 M (Ksp = 2.07 × 10−33). Thus a saturated solution of Ca3(PO4)2 in water contains
• $3 \times (1.14 \times 10^{−7}\; M) = 3.42 \times 10^{−7} M \; of \; Ca^{2+}$
• $2 \times (1.14 \times 10^{−7} M) = 2.28 \times 10^{−7} M \; of \; PO_4^{3−}$
according to the stoichiometry shown in Equation $\ref{17.4.2a}$ (neglecting hydrolysis to form HPO42). If CaCl2 is added to a saturated solution of Ca3(PO4)2, the Ca2+ ion concentration will increase such that [Ca2+] > 3.42 × 10−7 M, making Q > Ksp. The only way the system can return to equilibrium is for the reaction in Equation $\ref{17.4.2a}$ to proceed to the left, resulting in precipitation of Ca3(PO4)2. This will decrease the concentration of both Ca2+ and PO43 until Q = Ksp.
Note
Adding a common ion decreases solubility, as the reaction shifts toward the left to relieve the stress of the excess product. Adding a common ion to a dissociation reaction causes the equilibrium to shift left, toward the reactants, causing precipitation.
Example $5$
Consider the reaction:
$PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq) \nonumber$
What happens to the solubility of PbCl2(s) when 0.1 M NaCl is added?
Solution
$K_{sp}=1.7 \times 10^{-5} \nonumber$
$Q_{sp}= 1.8 \times 10^{-5}\nonumber$
Identify the common ion: Cl-
Notice: Qsp > Ksp The addition of NaCl has caused the reaction to shift out of equilibrium because there are more dissociated ions. Typically, solving for the molarities requires the assumption that the solubility of PbCl2 is equivalent to the concentration of Pb2+ produced because they are in a 1:1 ratio.
Because Ksp for the reaction is 1.7×10-5, the overall reaction would be (s)(2s)2= 1.7×10-5. Solving the equation for s gives s= 1.62×10-2 M. The coefficient on Cl- is 2, so it is assumed that twice as much Cl- is produced as Pb2+, hence the '2s.' The solubility equilibrium constant can be used to solve for the molarities of the ions at equilibrium.
The molarity of Cl- added would be 0.1 M because Na+ and Cl- are in a 1:1 ration in the ionic salt, NaCl. Therefore, the overall molarity of Cl- would be 2s + 0.1, with 2s referring to the contribution of the chloride ion from the dissociation of lead chloride.
$\begin{eqnarray} Q_{sp} &=& [Pb^{2+}][Cl^-]^2 \ 1.8 \times 10^{-5} &=& (s)(2s + 0.1)^2 \ s &=& [Pb^{2+}] \ &=& 1.8 \times 10^{-3} M \ 2s &=& [Cl^-] \ &\approx & 0.1 M \end{eqnarray} \nonumber$
Notice that the molarity of Pb2+ is lower when NaCl is added. The equilibrium constant remains the same because of the increased concentration of the chloride ion. To simplify the reaction, it can be assumed that [Cl-] is approximately 0.1M since the formation of the chloride ion from the dissociation of lead chloride is so small. The reaction quotient for PbCl2 is greater than the equilibrium constant because of the added Cl-. This therefore shift the reaction left towards equilibrium, causing precipitation and lowering the current solubility of the reaction. Overall, the solubility of the reaction decreases with the added sodium chloride.
Exercise $5$
Calculate the solubility of silver carbonate in a 0.25 M solution of sodium carbonate. The solubility of silver carbonate in pure water is 8.45 × 10−12 at 25°C.
Answer
2.9 × 10−6 M (versus 1.3 × 10−4 M in pure water)
A Video Discussing the Common Ion Effect in Solubility Products: The Common Ion Effect in Solubility Products (opens in new window) [youtu.be]
Contributors and Attributions
• Emmellin Tung, Mahtab Danai (UCD)
• Jim Clark (ChemGuide)
• Chung (Peter) Chieh (Professor Emeritus, Chemistry @ University of Waterloo) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/17%3A_Additional_Aspects_of_Aqueous_Equilibria/17.01%3A_The_Common-Ion_Effect.txt |
Learning Objectives
• To understand how adding a common ion affects the position of an acid–base equilibrium.
• To know how to use the Henderson-Hasselbalch approximation to calculate the pH of a buffer.
Buffers are solutions that maintain a relatively constant pH when an acid or a base is added. They therefore protect, or “buffer,” other molecules in solution from the effects of the added acid or base. Buffers contain either a weak acid ($HA$) and its conjugate base $(A^−$) or a weak base ($B$) and its conjugate acid ($BH^+$), and they are critically important for the proper functioning of biological systems. In fact, every biological fluid is buffered to maintain its physiological pH.
The Common Ion Effect: Weak Acids Combined with Conjugate Bases
To understand how buffers work, let’s look first at how the ionization equilibrium of a weak acid is affected by adding either the conjugate base of the acid or a strong acid (a source of $\ce{H^{+}}$). Le Chatelier’s principle can be used to predict the effect on the equilibrium position of the solution. A typical buffer used in biochemistry laboratories contains acetic acid and a salt such as sodium acetate. The dissociation reaction of acetic acid is as follows:
$\ce{CH3COOH (aq) <=> CH3COO^{−} (aq) + H^{+} (aq)} \label{Eq1}$
and the equilibrium constant expression is as follows:
$K_a=\dfrac{[\ce{H^{+}}][\ce{CH3COO^{-}}]}{[\ce{CH3CO2H}]} \label{Eq2}$
Sodium acetate ($\ce{CH_3CO_2Na}$) is a strong electrolyte that ionizes completely in aqueous solution to produce $\ce{Na^{+}}$ and $\ce{CH3CO2^{−}}$ ions. If sodium acetate is added to a solution of acetic acid, Le Chatelier’s principle predicts that the equilibrium in Equation \ref{Eq1} will shift to the left, consuming some of the added $\ce{CH_3COO^{−}}$ and some of the $\ce{H^{+}}$ ions originally present in solution.
Because $\ce{Na^{+}}$ is a spectator ion, it has no effect on the position of the equilibrium and can be ignored. The addition of sodium acetate produces a new equilibrium composition, in which $[\ce{H^{+}}]$ is less than the initial value. Because $[\ce{H^{+}}]$ has decreased, the pH will be higher. Thus adding a salt of the conjugate base to a solution of a weak acid increases the pH. This makes sense because sodium acetate is a base, and adding any base to a solution of a weak acid should increase the pH.
If we instead add a strong acid such as $\ce{HCl}$ to the system, $[\ce{H^{+}}]$ increases. Once again the equilibrium is temporarily disturbed, but the excess $\ce{H^{+}}$ ions react with the conjugate base ($\ce{CH_3CO_2^{−}}$), whether from the parent acid or sodium acetate, to drive the equilibrium to the left. The net result is a new equilibrium composition that has a lower [$\ce{CH_3CO_2^{−}}$] than before. In both cases, only the equilibrium composition has changed; the ionization constant $K_a$ for acetic acid remains the same. Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium, in this case $\ce{CH3CO2^{−}}$, will therefore shift the equilibrium in the direction that reduces the concentration of the common ion. The shift in equilibrium is via the common ion effect.
Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant.
Example $1$
A 0.150 M solution of formic acid at 25°C (pKa = 3.75) has a pH of 2.28 and is 3.5% ionized.
1. Is there a change to the pH of the solution if enough solid sodium formate is added to make the final formate concentration 0.100 M (assume that the formic acid concentration does not change)?
2. What percentage of the formic acid is ionized if 0.200 M HCl is added to the system?
Given: solution concentration and pH, $pK_a$, and percent ionization of acid; final concentration of conjugate base or strong acid added
Asked for: pH and percent ionization of formic acid
Strategy:
1. Write a balanced equilibrium equation for the ionization equilibrium of formic acid. Tabulate the initial concentrations, the changes, and the final concentrations.
2. Substitute the expressions for the final concentrations into the expression for Ka. Calculate $[\ce{H^{+}}]$ and the pH of the solution.
3. Construct a table of concentrations for the dissociation of formic acid. To determine the percent ionization, determine the anion concentration, divide it by the initial concentration of formic acid, and multiply the result by 100.
Solution:
A Because sodium formate is a strong electrolyte, it ionizes completely in solution to give formate and sodium ions. The $\ce{Na^{+}}$ ions are spectator ions, so they can be ignored in the equilibrium equation. Because water is both a much weaker acid than formic acid and a much weaker base than formate, the acid–base properties of the solution are determined solely by the formic acid ionization equilibrium:
$\ce{HCO2H (aq) <=> HCO^{−}2 (aq) + H^{+} (aq)} \nonumber$
The initial concentrations, the changes in concentration that occur as equilibrium is reached, and the final concentrations can be tabulated.
Final Concentration
ICE $[HCO_2H (aq) ]$ $[H^+ (aq) ]$ $[HCO^−_2 (aq) ]$
Initial 0.150 $1.00 \times 10^{−7}$ 0.100
Change −x +x +x
Equilibrium (0.150 − x) x (0.100 + x)
B We substitute the expressions for the final concentrations into the equilibrium constant expression and make our usual simplifying assumptions, so
\begin{align*} K_a=\dfrac{[H^+][HCO_2^−]}{[HCO_2H]} &=\dfrac{(x)(0.100+x)}{0.150−x} \[4pt] &\approx \dfrac{x(0.100)}{0.150} \[4pt] &\approx 10^{−3.75} \[4pt] &\approx 1.8 \times 10^{−4} \end{align*} \nonumber
Rearranging and solving for $x$,
\begin{align*} x &=(1.8 \times 10^{−4}) \times \dfrac{0.150 \;M}{ 0.100 \;M} \[4pt] &=2.7 \times 10^{−4}\[4pt] &=[H^+] \end{align*} \nonumber
The value of $x$ is small compared with 0.150 or 0.100 M, so our assumption about the extent of ionization is justified. Moreover,
$K_aC_{HA} = (1.8 \times 10^{−4})(0.150) = 2.7 \times 10^{−5} \nonumber$
which is greater than $1.0 \times 10^{−6}$, so again, our assumption is justified. The final pH is:
$pH= −\log(2.7 \times 10^{−4}) = 3.57 \nonumber$
compared with the initial value of 2.29. Thus adding a salt containing the conjugate base of the acid has increased the pH of the solution, as we expect based on Le Chatelier’s principle; the stress on the system has been relieved by the consumption of $\ce{H^{+}}$ ions, driving the equilibrium to the left.
C Because $HCl$ is a strong acid, it ionizes completely, and chloride is a spectator ion that can be neglected. Thus the only relevant acid–base equilibrium is again the dissociation of formic acid, and initially the concentration of formate is zero. We can construct a table of initial concentrations, changes in concentration, and final concentrations.
$HCO_2H (aq) \leftrightharpoons H^+ (aq) +HCO^−_2 (aq) \nonumber$
initial concentrations, changes in concentration, and final concentrations
$[HCO_2H (aq) ]$ $[H^+ (aq) ]$ $[HCO^−_2 (aq) ]$
initial 0.150 0.200 0
change −x +x +x
final (0.150 − x) (0.200 + x) x
To calculate the percentage of formic acid that is ionized under these conditions, we have to determine the final $[\ce{HCO2^{-}}]$. We substitute final concentrations into the equilibrium constant expression and make the usual simplifying assumptions, so
$K_a=\dfrac{[H^+][HCO_2^−]}{[HCO_2H]}=\dfrac{(0.200+x)(x)}{0.150−x} \approx \dfrac{x(0.200)}{0.150}=1.80 \times 10^{−4} \nonumber$
Rearranging and solving for $x$,
\begin{align*} x &=(1.80 \times 10^{−4}) \times \dfrac{ 0.150\; M}{ 0.200\; M} \[4pt] &=1.35 \times 10^{−4}=[HCO_2^−] \end{align*} \nonumber
Once again, our simplifying assumptions are justified. The percent ionization of formic acid is as follows:
$\text{percent ionization}=\dfrac{1.35 \times 10^{−4} \;M} {0.150\; M} \times 100\%=0.0900\% \nonumber$
Adding the strong acid to the solution, as shown in the table, decreased the percent ionization of formic acid by a factor of approximately 38 (3.45%/0.0900%). Again, this is consistent with Le Chatelier’s principle: adding $\ce{H^{+}}$ ions drives the dissociation equilibrium to the left.
Exercise $1$
A 0.225 M solution of ethylamine ($\ce{CH3CH2NH2}$ with $pK_b = 3.19$) has a pH of 12.08 and a percent ionization of 5.4% at 20°C. Calculate the following:
1. the pH of the solution if enough solid ethylamine hydrochloride ($\ce{EtNH3Cl}$) is added to make the solution 0.100 M in $\ce{EtNH3^{+}}$
2. the percentage of ethylamine that is ionized if enough solid $\ce{NaOH}$ is added to the original solution to give a final concentration of 0.050 M $\ce{NaOH}$
Answer a
11.16
Answer b
1.3%
A Video Discussing the Common Ion Effect: The Common Ion Effecr(opens in new window) [youtu.be]
The Common Ion Effect: Weak Bases Combined with Conjugate Acids
Now let’s suppose we have a buffer solution that contains equimolar concentrations of a weak base ($B$) and its conjugate acid ($BH^+$). The general equation for the ionization of a weak base is as follows:
$B (aq) +H_2O (l) \leftrightharpoons BH^+ (aq) +OH^− (aq) \label{Eq3}$
If the equilibrium constant for the reaction as written in Equation $\ref{Eq3}$ is small, for example $K_b = 10^{−5}$, then the equilibrium constant for the reverse reaction is very large: $K = \dfrac{1}{K_b} = 10^5$. Adding a strong base such as $OH^-$ to the solution therefore causes the equilibrium in Equation $\ref{Eq3}$ to shift to the left, consuming the added $OH^-$. As a result, the $OH^-$ ion concentration in solution remains relatively constant, and the pH of the solution changes very little. Le Chatelier’s principle predicts the same outcome: when the system is stressed by an increase in the $OH^-$ ion concentration, the reaction will proceed to the left to counteract the stress.
If the $pK_b$ of the base is 5.0, the $pK_a$ of its conjugate acid is
$pK_a = pK_w − pK_b = 14.0 – 5.0 = 9.0. \nonumber$
Thus the equilibrium constant for ionization of the conjugate acid is even smaller than that for ionization of the base. The ionization reaction for the conjugate acid of a weak base is written as follows:
$BH^+ (aq) +H_2O (l) \leftrightharpoons B (aq) +H_3O^+ (aq) \label{Eq4}$
Again, the equilibrium constant for the reverse of this reaction is very large: K = 1/Ka = 109. If a strong acid is added, it is neutralized by reaction with the base as the reaction in Equation $\ref{Eq4}$ shifts to the left. As a result, the $H^+$ ion concentration does not increase very much, and the pH changes only slightly. In effect, a buffer solution behaves somewhat like a sponge that can absorb $H^+$ and $OH^-$ ions, thereby preventing large changes in pH when appreciable amounts of strong acid or base are added to a solution.
Buffers are characterized by the pH range over which they can maintain a more or less constant pH and by their buffer capacity, the amount of strong acid or base that can be absorbed before the pH changes significantly. Although the useful pH range of a buffer depends strongly on the chemical properties of the weak acid and weak base used to prepare the buffer (i.e., on $K$), its buffer capacity depends solely on the concentrations of the species in the buffered solution. The more concentrated the buffer solution, the greater its buffer capacity. As illustrated in Figure $1$, when $NaOH$ is added to solutions that contain different concentrations of an acetic acid/sodium acetate buffer, the observed change in the pH of the buffer is inversely proportional to the concentration of the buffer. If the buffer capacity is 10 times larger, then the buffer solution can absorb 10 times more strong acid or base before undergoing a significant change in pH.
A buffer maintains a relatively constant pH when acid or base is added to a solution. The addition of even tiny volumes of 0.10 M $NaOH$ to 100.0 mL of distilled water results in a very large change in pH. As the concentration of a 50:50 mixture of sodium acetate/acetic acid buffer in the solution is increased from 0.010 M to 1.00 M, the change in the pH produced by the addition of the same volume of $NaOH$ solution decreases steadily. For buffer concentrations of at least 0.500 M, the addition of even 25 mL of the $NaOH$ solution results in only a relatively small change in pH.
Calculating the pH of a Buffer
The pH of a buffer can be calculated from the concentrations of the weak acid and the weak base used to prepare it, the concentration of the conjugate base and conjugate acid, and the $pK_a$ or $pK_b$ of the weak acid or weak base. The procedure is analogous to that used in Example $1$ to calculate the pH of a solution containing known concentrations of formic acid and formate.
An alternative method frequently used to calculate the pH of a buffer solution is based on a rearrangement of the equilibrium equation for the dissociation of a weak acid. The simplified ionization reaction is $HA \leftrightharpoons H^+ + A^−$, for which the equilibrium constant expression is as follows:
$K_a=\dfrac{[H^+][A^-]}{[HA]} \label{Eq5}$
This equation can be rearranged as follows:
$[H^+]=K_a\dfrac{[HA]}{[A^−]} \label{Eq6}$
Taking the logarithm of both sides and multiplying both sides by −1,
\begin{align} −\log[H^+] &=−\log K_a−\log\left(\dfrac{[HA]}{[A^−]}\right) \[4pt] &=−\log{K_a}+\log\left(\dfrac{[A^−]}{[HA]}\right) \label{Eq7} \end{align}
Replacing the negative logarithms in Equation $\ref{Eq7}$,
$pH=pK_a+\log \left( \dfrac{[A^−]}{[HA]} \right) \label{Eq8}$
or, more generally,
$pH=pK_a+\log\left(\dfrac{[base]}{[acid]}\right) \label{Eq9}$
Equation $\ref{Eq8}$ and Equation $\ref{Eq9}$ are both forms of the Henderson-Hasselbalch approximation, named after the two early 20th-century chemists who first noticed that this rearranged version of the equilibrium constant expression provides an easy way to calculate the pH of a buffer solution. In general, the validity of the Henderson-Hasselbalch approximation may be limited to solutions whose concentrations are at least 100 times greater than their $K_a$ values.
There are three special cases where the Henderson-Hasselbalch approximation is easily interpreted without the need for calculations:
• $[base] = [acid]$: Under these conditions, $\dfrac{[base]}{[acid]} = 1 \nonumber$ in Equation \ref{Eq9}. Because $\log 1 = 0$, $pH = pK_a \nonumber$ regardless of the actual concentrations of the acid and base. Recall that this corresponds to the midpoint in the titration of a weak acid or a weak base.
• $[base]/[acid] = 10$: In Equation $\ref{Eq9}$, because $\log 10 = 1$, $pH = pK_a + 1. \nonumber$
• $[base]/[acid] = 100$: In Equation $\ref{Eq9}$, because $\log 100 = 2$, $pH = pK_a + 2. \nonumber$
Each time we increase the [base]/[acid] ratio by 10, the pH of the solution increases by 1 pH unit. Conversely, if the [base]/[acid] ratio is 0.1, then pH = $pK_a$ − 1. Each additional factor-of-10 decrease in the [base]/[acid] ratio causes the pH to decrease by 1 pH unit.
If [base] = [acid] for a buffer, then pH = $pK_a$. Changing this ratio by a factor of 10 either way changes the pH by ±1 unit.
Example $2$
What is the pH of a solution that contains
1. 0.135 M $\ce{HCO2H}$ and 0.215 M $\ce{HCO2Na}$? (The $pK_a$ of formic acid is 3.75.)
2. 0.0135 M $\ce{HCO2H}$ and 0.0215 M $\ce{HCO2Na}$?
3. 0.119 M pyridine and 0.234 M pyridine hydrochloride? (The $pK_b$ of pyridine is 8.77.)
Given: concentration of acid, conjugate base, and $pK_a$; concentration of base, conjugate acid, and $pK_b$
Asked for: pH
Strategy:
Substitute values into either form of the Henderson-Hasselbalch approximation (Equations \ref{Eq8} or \ref{Eq9}) to calculate the pH.
Solution:
According to the Henderson-Hasselbalch approximation (Equation \ref{Eq8}), the pH of a solution that contains both a weak acid and its conjugate base is
$pH = pK_a + \log([A−]/[HA]). \nonumber$
A
Inserting the given values into the equation,
\begin{align*} pH &=3.75+\log\left(\dfrac{0.215}{0.135}\right) \[4pt] &=3.75+\log 1.593 \[4pt] &=3.95 \end{align*} \nonumber
This result makes sense because the $[A^−]/[HA]$ ratio is between 1 and 10, so the pH of the buffer must be between the $pK_a$ (3.75) and $pK_a + 1$, or 4.75.
B
This is identical to part (a), except for the concentrations of the acid and the conjugate base, which are 10 times lower. Inserting the concentrations into the Henderson-Hasselbalch approximation,
\begin{align*} pH &=3.75+\log\left(\dfrac{0.0215}{0.0135}\right) \[4pt] &=3.75+\log 1.593 \[4pt] &=3.95 \end{align*} \nonumber
This result is identical to the result in part (a), which emphasizes the point that the pH of a buffer depends only on the ratio of the concentrations of the conjugate base and the acid, not on the magnitude of the concentrations. Because the [A]/[HA] ratio is the same as in part (a), the pH of the buffer must also be the same (3.95).
C
In this case, we have a weak base, pyridine (Py), and its conjugate acid, the pyridinium ion ($HPy^+$). We will therefore use Equation $\ref{Eq9}$, the more general form of the Henderson-Hasselbalch approximation, in which “base” and “acid” refer to the appropriate species of the conjugate acid–base pair. We are given [base] = [Py] = 0.119 M and $[acid] = [HPy^{+}] = 0.234\, M$. We also are given $pK_b = 8.77$ for pyridine, but we need $pK_a$ for the pyridinium ion. Recall from Equation 16.23 that the $pK_b$ of a weak base and the $pK_a$ of its conjugate acid are related:
$pK_a + pK_b = pK_w. \nonumber$
Thus $pK_a$ for the pyridinium ion is $pK_w − pK_b = 14.00 − 8.77 = 5.23$. Substituting this $pK_a$ value into the Henderson-Hasselbalch approximation,
\begin{align*} pH=pK_a+\log \left(\dfrac{[base]}{[acid]}\right) \[4pt] &=5.23+\log\left(\dfrac{0.119}{0.234}\right) \[4pt] & =5.23 −0.294 \[4pt] &=4.94 \end{align*} \nonumber
Once again, this result makes sense: the $[B]/[BH^+]$ ratio is about 1/2, which is between 1 and 0.1, so the final pH must be between the $pK_a$ (5.23) and $pK_a − 1$, or 4.23.
Exercise $2$
What is the pH of a solution that contains
1. 0.333 M benzoic acid and 0.252 M sodium benzoate?
2. 0.050 M trimethylamine and 0.066 M trimethylamine hydrochloride?
The $pK_a$ of benzoic acid is 4.20, and the $pK_b$ of trimethylamine is also 4.20.
Answer a
4.08
Answer b
9.68
A Video Discussing Using the Henderson Hasselbalch Equation: Using the Henderson Hasselbalch Equation(opens in new window) [youtu.be] (opens in new window)
The Henderson-Hasselbalch approximation ((Equation $\ref{Eq8}$) can also be used to calculate the pH of a buffer solution after adding a given amount of strong acid or strong base, as demonstrated in Example $3$.
Example $3$
The buffer solution in Example $2$ contained 0.135 M $\ce{HCO2H}$ and 0.215 M $\ce{HCO2Na}$ and had a pH of 3.95.
1. What is the final pH if 5.00 mL of 1.00 M $HCl$ are added to 100 mL of this solution?
2. What is the final pH if 5.00 mL of 1.00 M $NaOH$ are added?
Given: composition and pH of buffer; concentration and volume of added acid or base
Asked for: final pH
Strategy:
1. Calculate the amounts of formic acid and formate present in the buffer solution using the procedure from Example $1$. Then calculate the amount of acid or base added.
2. Construct a table showing the amounts of all species after the neutralization reaction. Use the final volume of the solution to calculate the concentrations of all species. Finally, substitute the appropriate values into the Henderson-Hasselbalch approximation (Equation \ref{Eq9}) to obtain the pH.
Solution:
The added $\ce{HCl}$ (a strong acid) or $\ce{NaOH}$ (a strong base) will react completely with formate (a weak base) or formic acid (a weak acid), respectively, to give formic acid or formate and water. We must therefore calculate the amounts of formic acid and formate present after the neutralization reaction.
A We begin by calculating the millimoles of formic acid and formate present in 100 mL of the initial pH 3.95 buffer:
$100 \, \cancel{mL} \left( \dfrac{0.135 \, mmol\; \ce{HCO2H}}{\cancel{mL}} \right) = 13.5\, mmol\, \ce{HCO2H} \nonumber$
$100\, \cancel{mL } \left( \dfrac{0.215 \, mmol\; \ce{HCO2^{-}}}{\cancel{mL}} \right) = 21.5\, mmol\, \ce{HCO2^{-}} \nonumber$
The millimoles of $\ce{H^{+}}$ in 5.00 mL of 1.00 M $\ce{HCl}$ is as follows:
$5.00 \, \cancel{mL } \left( \dfrac{1.00 \,mmol\; \ce{H^{+}}}{\cancel{mL}} \right) = 5\, mmol\, \ce{H^{+}} \nonumber$
B Next, we construct a table of initial amounts, changes in amounts, and final amounts:
$\ce{HCO^{2−}(aq) + H^{+} (aq) <=> HCO2H (aq)} \nonumber$
initial amounts, changes in amounts, and final amounts:
$HCO^{2−} (aq)$ $H^+ (aq)$ $HCO_2H (aq)$
Initial 21.5 mmol 5.00 mmol 13.5 mmol
Change −5.00 mmol −5.00 mmol +5.00 mmol
Final 16.5 mmol ∼0 mmol 18.5 mmol
The final amount of $H^+$ in solution is given as “∼0 mmol.” For the purposes of the stoichiometry calculation, this is essentially true, but remember that the point of the problem is to calculate the final $[H^+]$ and thus the pH. We now have all the information we need to calculate the pH. We can use either the lengthy procedure of Example $1$ or the Henderson–Hasselbach approximation. Because we have performed many equilibrium calculations in this chapter, we’ll take the latter approach. The Henderson-Hasselbalch approximation requires the concentrations of $HCO_2^−$ and $HCO_2H$, which can be calculated using the number of millimoles ($n$) of each and the total volume ($VT$). Substituting these values into the Henderson-Hasselbalch approximation (Equation $\ref{Eq9}$):
\begin{align*} pH &=pK_a+\log \left( \dfrac{[HCO_2^−]}{[HCO_2H]} \right) \[4pt] &=pK_a+\log\left(\dfrac{n_{HCO_2^−}/V_f}{n_{HCO_2H}/V_f}\right) \[4pt] &=pK_a+\log \left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \end{align*} \nonumber
Because the total volume appears in both the numerator and denominator, it cancels. We therefore need to use only the ratio of the number of millimoles of the conjugate base to the number of millimoles of the weak acid. So
\begin{align*} pH &=pK_a+\log\left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \[4pt] &=3.75+\log\left(\dfrac{16.5\; mmol}{18.5\; mmol}\right) \[4pt] &=3.75 −0.050=3.70 \end{align*} \nonumber
Once again, this result makes sense on two levels. First, the addition of $HCl$has decreased the pH from 3.95, as expected. Second, the ratio of $HCO_2^−$ to $HCO_2H$ is slightly less than 1, so the pH should be between the $pK_a$ and $pK_a$ − 1.
A The procedure for solving this part of the problem is exactly the same as that used in part (a). We have already calculated the numbers of millimoles of formic acid and formate in 100 mL of the initial pH 3.95 buffer: 13.5 mmol of $HCO_2H$ and 21.5 mmol of $HCO_2^−$. The number of millimoles of $OH^-$ in 5.00 mL of 1.00 M $NaOH$ is as follows:
B With this information, we can construct a table of initial amounts, changes in amounts, and final amounts.
$\ce{HCO2H (aq) + OH^{−} (aq) <=> HCO^{−}2 (aq) + H2O (l)} \nonumber$
initial amounts, changes in amounts, and final amounts
$HCO_2H (aq)$ $OH^−$ $HCO^−_2 (aq)$
Initial 13.5 mmol 5.00 mmol 21.5 mmol
Change −5.00 mmol −5.00 mmol +5.00 mmol
Final 8.5 mmol ∼0 mmol 26.5 mmol
The final amount of $OH^-$ in solution is not actually zero; this is only approximately true based on the stoichiometric calculation. We can calculate the final pH by inserting the numbers of millimoles of both $HCO_2^−$ and $HCO_2H$ into the simplified Henderson-Hasselbalch expression used in part (a) because the volume cancels:
\begin{align*} pH &=pK_a+\log \left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \[4pt] &=3.75+\log \left(\dfrac{26.5\; mmol}{8.5\; mmol} \right) \[4pt] &=3.75+0.494 =4.24 \end{align*} \nonumber
Once again, this result makes chemical sense: the pH has increased, as would be expected after adding a strong base, and the final pH is between the $pK_a$ and $pK_a$ + 1, as expected for a solution with a $HCO_2^−/HCO_2H$ ratio between 1 and 10.
Exercise $3$
The buffer solution from Example $2$ contained 0.119 M pyridine and 0.234 M pyridine hydrochloride and had a pH of 4.94.
1. What is the final pH if 12.0 mL of 1.5 M $\ce{NaOH}$ are added to 250 mL of this solution?
2. What is the final pH if 12.0 mL of 1.5 M $\ce{HCl}$ are added?
Answer a
5.30
Answer b
4.42
Only the amounts (in moles or millimoles) of the acidic and basic components of the buffer are needed to use the Henderson-Hasselbalch approximation, not their concentrations.
A Video Discussing the Change in pH with the Addition of a Strong Acid to a Buffer: The Change in pH with the Addition of a Strong Acid to a Buffer(opens in new window) [youtu.be]
The Change in pH with the Addition of a Strong Base to a Buffer:
The results obtained in Example $3$ and its corresponding exercise demonstrate how little the pH of a well-chosen buffer solution changes despite the addition of a significant quantity of strong acid or strong base. Suppose we had added the same amount of $HCl$ or $NaOH$ solution to 100 mL of an unbuffered solution at pH 3.95 (corresponding to $1.1 \times 10^{−4}$ M HCl). In this case, adding 5.00 mL of 1.00 M $HCl$ would lower the final pH to 1.32 instead of 3.70, whereas adding 5.00 mL of 1.00 M $NaOH$ would raise the final pH to 12.68 rather than 4.24. (Try verifying these values by doing the calculations yourself.) Thus the presence of a buffer significantly increases the ability of a solution to maintain an almost constant pH.
The most effective buffers contain equal concentrations of an acid and its conjugate base.
A buffer that contains approximately equal amounts of a weak acid and its conjugate base in solution is equally effective at neutralizing either added base or added acid. This is shown in Figure $2$ for an acetic acid/sodium acetate buffer. Adding a given amount of strong acid shifts the system along the horizontal axis to the left, whereas adding the same amount of strong base shifts the system the same distance to the right. In either case, the change in the ratio of $CH_3CO_2^−$ to $CH_3CO_2H$ from 1:1 reduces the buffer capacity of the solution.
A Video Discussing The Buffer Region: The Buffer Region (opens in new window) [youtu.be]
The Relationship between Titrations and Buffers
There is a strong correlation between the effectiveness of a buffer solution and the titration curves discussed in Section 16.5. Consider the schematic titration curve of a weak acid with a strong base shown in Figure $3$. As indicated by the labels, the region around $pK_a$ corresponds to the midpoint of the titration, when approximately half the weak acid has been neutralized. This portion of the titration curve corresponds to a buffer: it exhibits the smallest change in pH per increment of added strong base, as shown by the nearly horizontal nature of the curve in this region. The nearly flat portion of the curve extends only from approximately a pH value of 1 unit less than the $pK_a$ to approximately a pH value of 1 unit greater than the $pK_a$, which is why buffer solutions usually have a pH that is within ±1 pH units of the $pK_a$ of the acid component of the buffer.
This schematic plot of pH for the titration of a weak acid with a strong base shows the nearly flat region of the titration curve around the midpoint, which corresponds to the formation of a buffer. At the lower left, the pH of the solution is determined by the equilibrium for dissociation of the weak acid; at the upper right, the pH is determined by the equilibrium for reaction of the conjugate base with water.
In the region of the titration curve at the lower left, before the midpoint, the acid–base properties of the solution are dominated by the equilibrium for dissociation of the weak acid, corresponding to $K_a$. In the region of the titration curve at the upper right, after the midpoint, the acid–base properties of the solution are dominated by the equilibrium for reaction of the conjugate base of the weak acid with water, corresponding to $K_b$. However, we can calculate either $K_a$ or $K_b$ from the other because they are related by $K_w$.
Blood: A Most Important Buffer
Metabolic processes produce large amounts of acids and bases, yet organisms are able to maintain an almost constant internal pH because their fluids contain buffers. This is not to say that the pH is uniform throughout all cells and tissues of a mammal. The internal pH of a red blood cell is about 7.2, but the pH of most other kinds of cells is lower, around 7.0. Even within a single cell, different compartments can have very different pH values. For example, one intracellular compartment in white blood cells has a pH of around 5.0.
Because no single buffer system can effectively maintain a constant pH value over the entire physiological range of approximately pH 5.0 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the $\ce{CO2}/\ce{HCO3^{−}}$ system, which dominates the buffering action of blood plasma.
The acid–base equilibrium in the $\ce{CO2}/\ce{HCO3^{−}}$ buffer system is usually written as follows:
$\ce{H2CO3 (aq) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{Eq10}$
with $K_a = 4.5 \times 10^{−7}$ and $pK_a = 6.35$ at 25°C. In fact, Equation $\ref{Eq10}$ is a grossly oversimplified version of the $\ce{CO2}/\ce{HCO3^{-}}$ system because a solution of $\ce{CO2}$ in water contains only rather small amounts of $H_2CO_3$. Thus Equation $\ref{Eq10}$ does not allow us to understand how blood is actually buffered, particularly at a physiological temperature of 37°C.
As shown in Equation $\ref{Eq11}$, $\ce{CO2}$ is in equilibrium with $\ce{H2CO3}$, but the equilibrium lies far to the left, with an $\ce{H2CO3}/\ce{CO2}$ ratio less than 0.01 under most conditions:
$\ce{CO2 (aq) + H2O (l) <=> H2CO3 (aq)} \label{Eq11}$
with $K′ = 4.0 \times 10^{−3}$ at 37°C. The true $pK_a$ of carbonic acid at 37°C is therefore 3.70, not 6.35, corresponding to a $K_a$ of $2.0 \times 10^{−4}$, which makes it a much stronger acid than Equation \ref{Eq10} suggests. Adding Equation \ref{Eq10} and Equation \ref{Eq11} and canceling $\ce{H2CO3}$ from both sides give the following overall equation for the reaction of $\ce{CO2}$ with water to give a proton and the bicarbonate ion:
$\ce{CO2 (aq) + H2O (l) <=> H2CO3 (aq)} \label{16.65a}$
with $K'=4.0 \times 10^{−3} (37°C)$
$\ce{H2CO3 (aq) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{16.65b}$
with $K_a=2.0 \times 10^{−4} (37°C)$
$\ce{CO2 (aq) + H2O (l) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{16.65c}$
with $K=8.0 \times 10^{−7} (37°C)$
The $K$ value for the reaction in Equation \ref{16.65c} is the product of the true ionization constant for carbonic acid ($K_a$) and the equilibrium constant (K) for the reaction of $\ce{CO2 (aq)}$ with water to give carbonic acid. The equilibrium equation for the reaction of $\ce{CO2}$ with water to give bicarbonate and a proton is therefore
$K=\dfrac{[\ce{H^{+}}][\ce{HCO3^{-}}]}{[\ce{CO2}]}=8.0 \times 10^{−7} \label{eq13}$
The presence of a gas in the equilibrium constant expression for a buffer is unusual. According to Henry’s law,
$[\ce{CO2}]=k P_{\ce{CO2}} \nonumber$
where $k$ is the Henry’s law constant for $\ce{CO2}$, which is $3.0 \times 10^{−5} \;M/mmHg$ at 37°C. Substituting this expression for $[\ce{CO2}]$ in Equation \ref{eq13},
$K=\dfrac{[\ce{H^{+}}][\ce{HCO3^{-}}]}{(3.0 \times 10^{−5}\; M/mmHg)(P_{\ce{CO2}})} \nonumber$
where $P_{\ce{CO2}}$ is in mmHg. Taking the negative logarithm of both sides and rearranging,
$pH=6.10+\log \left( \dfrac{ [\ce{HCO3^{−}}]}{(3.0 \times 10^{−5} M/mm \;Hg)\; (P_{\ce{CO2}}) } \right) \label{Eq15}$
Thus the pH of the solution depends on both the $\ce{CO2}$ pressure over the solution and $[\ce{HCO3^{−}}]$. Figure $4$ plots the relationship between pH and $[\ce{HCO3^{−}}]$ under physiological conditions for several different values of $P_{\ce{CO2}}$, with normal pH and $[\ce{HCO3^{−}}]$ values indicated by the dashed lines.
According to Equation \ref{Eq15}, adding a strong acid to the $\ce{CO2}/\ce{HCO3^{−}}$ system causes $[\ce{HCO3^{−}}]$ to decrease as $\ce{HCO3^{−}}$ is converted to $\ce{CO2}$. Excess $\ce{CO2}$ is released in the lungs and exhaled into the atmosphere, however, so there is essentially no change in $P_{\ce{CO2}}$. Because the change in $[\ce{HCO3^{−}}]/P_{CO_2}$ is small, Equation \ref{Eq15} predicts that the change in pH will also be rather small. Conversely, if a strong base is added, the $\ce{OH^{-}}$ reacts with $\ce{CO2}$ to form $\ce{HCO3^{−}}$, but $\ce{CO2}$ is replenished by the body, again limiting the change in both $[\ce{HCO3^{−}}]/P_{\ce{CO2}}$ and pH. The $\ce{CO2}/\ce{HCO3^{−}}$ buffer system is an example of an open system, in which the total concentration of the components of the buffer change to keep the pH at a nearly constant value.
If a passenger steps out of an airplane in Denver, Colorado, for example, the lower $P_{\ce{CO2}}$ at higher elevations (typically 31 mmHg at an elevation of 2000 m versus 40 mmHg at sea level) causes a shift to a new pH and $[\ce{HCO3^{-}}]$. The increase in pH and decrease in $[\ce{HCO3^{−}}]$ in response to the decrease in $P_{\ce{CO2}}$ are responsible for the general malaise that many people experience at high altitudes. If their blood pH does not adjust rapidly, the condition can develop into the life-threatening phenomenon known as altitude sickness.
A Video Summary of the pH Curve for a Strong Acid/Strong Base Titration:
Summary
Buffers are solutions that resist a change in pH after adding an acid or a base. Buffers contain a weak acid ($HA$) and its conjugate weak base ($A^−$). Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium shifts the equilibrium in such a way as to reduce the concentration of the common ion. The shift in equilibrium is called the common ion effect. Buffers are characterized by their pH range and buffer capacity. The useful pH range of a buffer depends strongly on the chemical properties of the conjugate weak acid–base pair used to prepare the buffer (the $K_a$ or $K_b$), whereas its buffer capacity depends solely on the concentrations of the species in the solution. The pH of a buffer can be calculated using the Henderson-Hasselbalch approximation, which is valid for solutions whose concentrations are at least 100 times greater than their $K_a$ values. Because no single buffer system can effectively maintain a constant pH value over the physiological range of approximately 5 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the $CO_2/HCO_3^−$ system, which dominates the buffering action of blood plasma. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/17%3A_Additional_Aspects_of_Aqueous_Equilibria/17.02%3A_Buffered_Solutions.txt |
Learning Objectives
• To calculate the pH at any point in an acid–base titration.
In an acid–base titration, a buret is used to deliver measured volumes of an acid or a base solution of known concentration (the titrant) to a flask that contains a solution of a base or an acid, respectively, of unknown concentration (the unknown). If the concentration of the titrant is known, then the concentration of the unknown can be determined. The following discussion focuses on the pH changes that occur during an acid–base titration. Plotting the pH of the solution in the flask against the amount of acid or base added produces a titration curve. The shape of the curve provides important information about what is occurring in solution during the titration.
Titrations of Strong Acids and Bases
Figure $\PageIndex{1a}$ shows a plot of the pH as 0.20 M $\ce{HCl}$ is gradually added to 50.00 mL of pure water. The pH of the sample in the flask is initially 7.00 (as expected for pure water), but it drops very rapidly as $\ce{HCl}$ is added. Eventually the pH becomes constant at 0.70—a point well beyond its value of 1.00 with the addition of 50.0 mL of $\ce{HCl}$ (0.70 is the pH of 0.20 M HCl). In contrast, when 0.20 M $\ce{NaOH}$ is added to 50.00 mL of distilled water, the pH (initially 7.00) climbs very rapidly at first but then more gradually, eventually approaching a limit of 13.30 (the pH of 0.20 M NaOH), again well beyond its value of 13.00 with the addition of 50.0 mL of $\ce{NaOH}$ as shown in Figure $\PageIndex{1b}$. As you can see from these plots, the titration curve for adding a base is the mirror image of the curve for adding an acid.
Suppose that we now add 0.20 M $\ce{NaOH}$ to 50.0 mL of a 0.10 M solution of $\ce{HCl}$. Because $\ce{HCl}$ is a strong acid that is completely ionized in water, the initial $[H^+]$ is 0.10 M, and the initial pH is 1.00. Adding $\ce{NaOH}$ decreases the concentration of H+ because of the neutralization reaction (Figure $\PageIndex{2a}$):
$\ce{OH^{−} + H^{+} <=> H_2O}. \nonumber$
Thus the pH of the solution increases gradually. Near the equivalence point, however, the point at which the number of moles of base (or acid) added equals the number of moles of acid (or base) originally present in the solution, the pH increases much more rapidly because most of the $\ce{H^{+}}$ ions originally present have been consumed. For the titration of a monoprotic strong acid ($\ce{HCl}$) with a monobasic strong base ($\ce{NaOH}$), we can calculate the volume of base needed to reach the equivalence point from the following relationship:
$moles\;of \;base=(volume)_b(molarity)_bV_bM_b= moles \;of \;acid=(volume)_a(molarity)_a=V_aM_a \label{Eq1}$
If 0.20 M $\ce{NaOH}$ is added to 50.0 mL of a 0.10 M solution of $\ce{HCl}$, we solve for $V_b$:
$V_b(0.20 Me)=0.025 L=25 mL \nonumber$
At the equivalence point (when 25.0 mL of $\ce{NaOH}$ solution has been added), the neutralization is complete: only a salt remains in solution (NaCl), and the pH of the solution is 7.00. Adding more $\ce{NaOH}$ produces a rapid increase in pH, but eventually the pH levels off at a value of about 13.30, the pH of 0.20 M $NaOH$.
As shown in Figure $\PageIndex{2b}$, the titration of 50.0 mL of a 0.10 M solution of $\ce{NaOH}$ with 0.20 M $\ce{HCl}$ produces a titration curve that is nearly the mirror image of the titration curve in Figure $\PageIndex{2a}$. The pH is initially 13.00, and it slowly decreases as $\ce{HCl}$ is added. As the equivalence point is approached, the pH drops rapidly before leveling off at a value of about 0.70, the pH of 0.20 M $\ce{HCl}$.
The titration of either a strong acid with a strong base or a strong base with a strong acid produces an S-shaped curve. The curve is somewhat asymmetrical because the steady increase in the volume of the solution during the titration causes the solution to become more dilute. Due to the leveling effect, the shape of the curve for a titration involving a strong acid and a strong base depends on only the concentrations of the acid and base, not their identities.
The shape of the titration curve involving a strong acid and a strong base depends only on their concentrations, not their identities.
Example $1$: Hydrochloric Acid
Calculate the pH of the solution after 24.90 mL of 0.200 M $\ce{NaOH}$ has been added to 50.00 mL of 0.100 M $\ce{HCl}$.
Given: volumes and concentrations of strong base and acid
Asked for: pH
Strategy:
1. Calculate the number of millimoles of $\ce{H^{+}}$ and $\ce{OH^{-}}$ to determine which, if either, is in excess after the neutralization reaction has occurred. If one species is in excess, calculate the amount that remains after the neutralization reaction.
2. Determine the final volume of the solution. Calculate the concentration of the species in excess and convert this value to pH.
Solution
A Because 0.100 mol/L is equivalent to 0.100 mmol/mL, the number of millimoles of $\ce{H^{+}}$ in 50.00 mL of 0.100 M $\ce{HCl}$ can be calculated as follows:
$50.00 \cancel{mL} \left ( \dfrac{0.100 \;mmol \;HCl}{\cancel{mL}} \right )= 5.00 \;mmol \;HCl=5.00 \;mmol \;H^{+} \nonumber$
The number of millimoles of $\ce{NaOH}$ added is as follows:
$24.90 \cancel{mL} \left ( \dfrac{0.200 \;mmol \;NaOH}{\cancel{mL}} \right )= 4.98 \;mmol \;NaOH=4.98 \;mmol \;OH^{-} \nonumber$
Thus $\ce{H^{+}}$ is in excess. To completely neutralize the acid requires the addition of 5.00 mmol of $\ce{OH^{-}}$ to the $\ce{HCl}$ solution. Because only 4.98 mmol of $OH^-$ has been added, the amount of excess $\ce{H^{+}}$ is 5.00 mmol − 4.98 mmol = 0.02 mmol of $H^+$.
B The final volume of the solution is 50.00 mL + 24.90 mL = 74.90 mL, so the final concentration of $\ce{H^{+}}$ is as follows:
$\left [ H^{+} \right ]= \dfrac{0.02 \;mmol \;H^{+}}{74.90 \; mL}=3 \times 10^{-4} \; M \nonumber$
Hence,
$pH \approx −\log[\ce{H^{+}}] = −\log(3 \times 10^{-4}) = 3.5 \nonumber$
This is significantly less than the pH of 7.00 for a neutral solution.
Exercise $1$
Calculate the pH of a solution prepared by adding $40.00\; mL$ of $0.237\; M$ $HCl$ to $75.00\; mL$ of a $0.133 M$ solution of $NaOH$.
Answer
11.6
pH after the addition of 10 ml of Strong Base to a Strong Acid:
https://youtu.be/_cM1_-kdJ20 (opens in new window)
pH at the Equivalence Point in a Strong Acid/Strong Base Titration:
https://youtu.be/7POGDA5Ql2M
Titrations of Weak Acids and Bases
In contrast to strong acids and bases, the shape of the titration curve for a weak acid or a weak base depends dramatically on the identity of the acid or the base and the corresponding $K_a$ or $K_b$. As we shall see, the pH also changes much more gradually around the equivalence point in the titration of a weak acid or a weak base. As you learned previously, $[\ce{H^{+}}]$ of a solution of a weak acid (HA) is not equal to the concentration of the acid but depends on both its $pK_a$ and its concentration. Because only a fraction of a weak acid dissociates, $[\(\ce{H^{+}}]$ is less than $[\ce{HA}]$. Thus the pH of a solution of a weak acid is greater than the pH of a solution of a strong acid of the same concentration.
Figure $\PageIndex{3a}$ shows the titration curve for 50.0 mL of a 0.100 M solution of acetic acid with 0.200 M $\ce{NaOH}$ superimposed on the curve for the titration of 0.100 M $\ce{HCl}$ shown in part (a) in Figure $2$. Below the equivalence point, the two curves are very different. Before any base is added, the pH of the acetic acid solution is greater than the pH of the $\ce{HCl}$ solution, and the pH changes more rapidly during the first part of the titration. Note also that the pH of the acetic acid solution at the equivalence point is greater than 7.00. That is, at the equivalence point, the solution is basic. In addition, the change in pH around the equivalence point is only about half as large as for the $\ce{HCl}$ titration; the magnitude of the pH change at the equivalence point depends on the $pK_a$ of the acid being titrated. Above the equivalence point, however, the two curves are identical. Once the acid has been neutralized, the pH of the solution is controlled only by the amount of excess $\ce{NaOH}$ present, regardless of whether the acid is weak or strong.
The shape of the titration curve of a weak acid or weak base depends heavily on their identities and the $K_a$ or $K_b$.
The titration curve in Figure $\PageIndex{3a}$ was created by calculating the starting pH of the acetic acid solution before any $\ce{NaOH}$ is added and then calculating the pH of the solution after adding increasing volumes of $NaOH$. The procedure is illustrated in the following subsection and Example $2$ for three points on the titration curve, using the $pK_a$ of acetic acid (4.76 at 25°C; $K_a = 1.7 \times 10^{-5}$.
Calculating the pH of a Solution of a Weak Acid or a Weak Base
As explained discussed, if we know $K_a$ or $K_b$ and the initial concentration of a weak acid or a weak base, we can calculate the pH of a solution of a weak acid or a weak base by setting up a ICE table (i.e, initial concentrations, changes in concentrations, and final concentrations). In this situation, the initial concentration of acetic acid is 0.100 M. If we define $x$ as $[\ce{H^{+}}]$ due to the dissociation of the acid, then the table of concentrations for the ionization of 0.100 M acetic acid is as follows:
$\ce{CH3CO2H(aq) <=> H^{+}(aq) + CH3CO2^{−}} \nonumber$
table of concentrations for the ionization of 0.100 M acetic acid
ICE $[CH_3CO_2H]$ $[H^+]$ $[CH_3CO_2^−]$
initial 0.100 $1.00 \times 10^{−7}$ 0
change −x +x +x
final 0.100 − x x x
In this and all subsequent examples, we will ignore $[H^+]$ and $[OH^-]$ due to the autoionization of water when calculating the final concentration. However, you should use Equation 16.45 and Equation 16.46 to check that this assumption is justified.
Inserting the expressions for the final concentrations into the equilibrium equation (and using approximations),
\begin{align*} K_a &=\dfrac{[H^+][CH_3CO_2^-]}{[CH_3CO_2H]} \[4pt] &=\dfrac{(x)(x)}{0.100 - x} \[4pt] &\approx \dfrac{x^2}{0.100} \[4pt] &\approx 1.74 \times 10^{-5} \end{align*} \nonumber
Solving this equation gives $x = [H^+] = 1.32 \times 10^{-3}\; M$. Thus the pH of a 0.100 M solution of acetic acid is as follows:
$pH = −\log(1.32 \times 10^{-3}) = 2.879 \nonumber$
pH at the Start of a Weak Acid/Strong Base Titration: https://youtu.be/AtdBKfrfJNg
Calculating the pH during the Titration of a Weak Acid or a Weak Base
Now consider what happens when we add 5.00 mL of 0.200 M $\ce{NaOH}$ to 50.00 mL of 0.100 M $CH_3CO_2H$ (part (a) in Figure $3$). Because the neutralization reaction proceeds to completion, all of the $OH^-$ ions added will react with the acetic acid to generate acetate ion and water:
$CH_3CO_2H_{(aq)} + OH^-_{(aq)} \rightarrow CH_3CO^-_{2\;(aq)} + H_2O_{(l)} \label{Eq2}$
All problems of this type must be solved in two steps: a stoichiometric calculation followed by an equilibrium calculation. In the first step, we use the stoichiometry of the neutralization reaction to calculate the amounts of acid and conjugate base present in solution after the neutralization reaction has occurred. In the second step, we use the equilibrium equation to determine $[\ce{H^{+}}]$ of the resulting solution.
Step 1
To determine the amount of acid and conjugate base in solution after the neutralization reaction, we calculate the amount of $\ce{CH_3CO_2H}$ in the original solution and the amount of $\ce{OH^{-}}$ in the $\ce{NaOH}$ solution that was added. The acetic acid solution contained
$50.00 \; \cancel{mL} (0.100 \;mmol (\ce{CH_3CO_2H})/\cancel{mL} )=5.00\; mmol (\ce{CH_3CO_2H}) \nonumber$
The $\ce{NaOH}$ solution contained
5.00 mL=1.00 mmol $NaOH$
Comparing the amounts shows that $CH_3CO_2H$ is in excess. Because $OH^-$ reacts with $CH_3CO_2H$ in a 1:1 stoichiometry, the amount of excess $CH_3CO_2H$ is as follows:
5.00 mmol $CH_3CO_2H$ − 1.00 mmol $OH^-$ = 4.00 mmol $CH_3CO_2H$
Each 1 mmol of $OH^-$ reacts to produce 1 mmol of acetate ion, so the final amount of $CH_3CO_2^−$ is 1.00 mmol.
The stoichiometry of the reaction is summarized in the following ICE table, which shows the numbers of moles of the various species, not their concentrations.
$\ce{CH3CO2H(aq) + OH^{−} (aq) <=> CH3CO2^{-}(aq) + H2O(l)} \nonumber$
ICE table
ICE $[\ce{CH_3CO_2H}]$ $[\ce{OH^{−}}]$ $[\ce{CH_3CO_2^{−}}]$
initial 5.00 mmol 1.00 mmol 0 mmol
change −1.00 mmol −1.00 mmol +1.00 mmol
final 4.00 mmol 0 mmol 1.00 mmol
This ICE table gives the initial amount of acetate and the final amount of $OH^-$ ions as 0. Because an aqueous solution of acetic acid always contains at least a small amount of acetate ion in equilibrium with acetic acid, however, the initial acetate concentration is not actually 0. The value can be ignored in this calculation because the amount of $CH_3CO_2^−$ in equilibrium is insignificant compared to the amount of $OH^-$ added. Moreover, due to the autoionization of water, no aqueous solution can contain 0 mmol of $OH^-$, but the amount of $OH^-$ due to the autoionization of water is insignificant compared to the amount of $OH^-$ added. We use the initial amounts of the reactants to determine the stoichiometry of the reaction and defer a consideration of the equilibrium until the second half of the problem.
Step 2
To calculate $[\ce{H^{+}}]$ at equilibrium following the addition of $NaOH$, we must first calculate [$\ce{CH_3CO_2H}$] and $[\ce{CH3CO2^{−}}]$ using the number of millimoles of each and the total volume of the solution at this point in the titration:
$final \;volume=50.00 \;mL+5.00 \;mL=55.00 \;mL \nonumber$ $\left [ CH_{3}CO_{2}H \right ] = \dfrac{4.00 \; mmol \; CH_{3}CO_{2}H }{55.00 \; mL} =7.27 \times 10^{-2} \;M \nonumber$ $\left [ CH_{3}CO_{2}^{-} \right ] = \dfrac{1.00 \; mmol \; CH_{3}CO_{2}^{-} }{55.00 \; mL} =1.82 \times 10^{-2} \;M \nonumber$
Knowing the concentrations of acetic acid and acetate ion at equilibrium and $K_a$ for acetic acid ($1.74 \times 10^{-5}$), we can calculate $[H^+]$ at equilibrium:
$K_{a}=\dfrac{\left [ CH_{3}CO_{2}^{-} \right ]\left [ H^{+} \right ]}{\left [ CH_{3}CO_{2}H \right ]} \nonumber$
$\left [ H^{+} \right ]=\dfrac{K_{a}\left [ CH_{3}CO_{2}H \right ]}{\left [ CH_{3}CO_{2}^{-} \right ]} = \dfrac{\left ( 1.72 \times 10^{-5} \right )\left ( 7.27 \times 10^{-2} \;M\right )}{\left ( 1.82 \times 10^{-2} \right )}= 6.95 \times 10^{-5} \;M \nonumber$
Calculating $−\log[\ce{H^{+}}]$ gives
$pH = −\log(6.95 \times 10^{−5}) = 4.158. \nonumber$
Comparing the titration curves for $\ce{HCl}$ and acetic acid in Figure $\PageIndex{3a}$, we see that adding the same amount (5.00 mL) of 0.200 M $\ce{NaOH}$ to 50 mL of a 0.100 M solution of both acids causes a much smaller pH change for $\ce{HCl}$ (from 1.00 to 1.14) than for acetic acid (2.88 to 4.16). This is consistent with the qualitative description of the shapes of the titration curves at the beginning of this section. In Example $2$, we calculate another point for constructing the titration curve of acetic acid.
pH Before the Equivalence Point of a Weak Acid/Strong Base Titration:
https://youtu.be/znpwGCsefXc
Example $2$
What is the pH of the solution after 25.00 mL of 0.200 M $\ce{NaOH}$ is added to 50.00 mL of 0.100 M acetic acid?
Given: volume and molarity of base and acid
Asked for: pH
Strategy:
1. Write the balanced chemical equation for the reaction. Then calculate the initial numbers of millimoles of $OH^-$ and $CH_3CO_2H$. Determine which species, if either, is present in excess.
2. Tabulate the results showing initial numbers, changes, and final numbers of millimoles.
3. If excess acetate is present after the reaction with $\ce{OH^{-}}$, write the equation for the reaction of acetate with water. Use a tabular format to obtain the concentrations of all the species present.
4. Calculate $K_b$ using the relationship $K_w = K_aK_b$. Calculate [OH−] and use this to calculate the pH of the solution.
Solution
A Ignoring the spectator ion ($Na^+$), the equation for this reaction is as follows:
$CH_3CO_2H_{ (aq)} + OH^-(aq) \rightarrow CH_3CO_2^-(aq) + H_2O(l) \nonumber$
The initial numbers of millimoles of $OH^-$ and $CH_3CO_2H$ are as follows:
25.00 mL(0.200 mmol OH−mL=5.00 mmol $OH-$
$50.00\; mL (0.100 CH_3CO_2 HL=5.00 mmol \; CH_3CO_2H \nonumber$
The number of millimoles of $OH^-$ equals the number of millimoles of $CH_3CO_2H$, so neither species is present in excess.
B Because the number of millimoles of $OH^-$ added corresponds to the number of millimoles of acetic acid in solution, this is the equivalence point. The results of the neutralization reaction can be summarized in tabular form.
$CH_3CO_2H_{(aq)}+OH^-_{(aq)} \rightleftharpoons CH_3CO_2^{-}(aq)+H_2O(l) \nonumber$
results of the neutralization reaction
ICE $[\ce{CH3CO2H}]$ $[\ce{OH^{−}}]$ $[\ce{CH3CO2^{−}}]$
initial 5.00 mmol 5.00 mmol 0 mmol
change −5.00 mmol −5.00 mmol +5.00 mmol
final 0 mmol 0 mmol 5.00 mmol
C Because the product of the neutralization reaction is a weak base, we must consider the reaction of the weak base with water to calculate [H+] at equilibrium and thus the final pH of the solution. The initial concentration of acetate is obtained from the neutralization reaction:
$[\ce{CH_3CO_2}]=\dfrac{5.00 \;mmol \; CH_3CO_2^{-}}{(50.00+25.00) \; mL}=6.67\times 10^{-2} \; M \nonumber$
The equilibrium reaction of acetate with water is as follows:
$\ce{CH_3CO^{-}2(aq) + H2O(l) <=> CH3CO2H(aq) + OH^{-} (aq)} \nonumber$
The equilibrium constant for this reaction is
$K_b = \dfrac{K_w}{K_a} \label{16.18}$
where $K_a$ is the acid ionization constant of acetic acid. We therefore define x as $[\ce{OH^{−}}]$ produced by the reaction of acetate with water. Here is the completed table of concentrations:
$H_2O_{(l)}+CH_3CO^−_{2(aq)} \rightleftharpoons CH_3CO_2H_{(aq)} +OH^−_{(aq)} \nonumber$
completed table of concentrations
$[\ce{CH3CO2^{−}}]$ $[\ce{CH3CO2H}]$ $[\ce{OH^{−}}]$
initial 0.0667 0 1.00 × 10−7
change −x +x +x
final (0.0667 − x) x x
D We can obtain $K_b$ by substituting the known values into Equation \ref{16.18}:
$K_{b}= \dfrac{K_w}{K_a} =\dfrac{1.01 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.80 \times 10^{-10} \label{16.23}$
Substituting the expressions for the final values from the ICE table into Equation \ref{16.23} and solving for $x$:
\begin{align*} \dfrac{x^{2}}{0.0667} &= 5.80 \times 10^{-10} \[4pt] x &= \sqrt{(5.80 \times 10^{-10})(0.0667)} \[4pt] &= 6.22 \times 10^{-6}\end{align*} \nonumber
Thus $[OH^{−}] = 6.22 \times 10^{−6}\, M$ and the pH of the final solution is 8.794 (Figure $\PageIndex{3a}$). As expected for the titration of a weak acid, the pH at the equivalence point is greater than 7.00 because the product of the titration is a base, the acetate ion, which then reacts with water to produce $\ce{OH^{-}}$.
Exercise $2$
Calculate the pH of a solution prepared by adding 45.0 mL of a 0.213 M $\ce{HCl}$ solution to 125.0 mL of a 0.150 M solution of ammonia. The $pK_b$ of ammonia is 4.75 at 25°C.
Answer
9.23
As shown in part (b) in Figure $3$, the titration curve for NH3, a weak base, is the reverse of the titration curve for acetic acid. In particular, the pH at the equivalence point in the titration of a weak base is less than 7.00 because the titration produces an acid.
The identity of the weak acid or weak base being titrated strongly affects the shape of the titration curve. Figure $4$ illustrates the shape of titration curves as a function of the $pK_a$ or the $pK_b$. As the acid or the base being titrated becomes weaker (its $pK_a$ or $pK_b$ becomes larger), the pH change around the equivalence point decreases significantly. With very dilute solutions, the curve becomes so shallow that it can no longer be used to determine the equivalence point.
One point in the titration of a weak acid or a weak base is particularly important: the midpoint of a titration is defined as the point at which exactly enough acid (or base) has been added to neutralize one-half of the acid (or the base) originally present and occurs halfway to the equivalence point. The midpoint is indicated in Figures $\PageIndex{4a}$ and $\PageIndex{4b}$ for the two shallowest curves. By definition, at the midpoint of the titration of an acid, [HA] = [A−]. Recall that the ionization constant for a weak acid is as follows:
$K_a=\dfrac{[H_3O^+][A^−]}{[HA]} \nonumber$
If $[HA] = [A^−]$, this reduces to $K_a = [H_3O^+]$. Taking the negative logarithm of both sides,
$−\log K_a = −\log[H_3O+] \nonumber$
From the definitions of $pK_a$ and pH, we see that this is identical to
$pK_a = pH \label{16.52}$
Thus the pH at the midpoint of the titration of a weak acid is equal to the $pK_a$ of the weak acid, as indicated in part (a) in Figure $4$ for the weakest acid where we see that the midpoint for $pK_a$ = 10 occurs at pH = 10. Titration methods can therefore be used to determine both the concentration and the $pK_a$ (or the $pK_b$) of a weak acid (or a weak base).
The pH at the midpoint of the titration of a weak acid is equal to the $pK_a$ of the weak acid.
Titrations of Polyprotic Acids or Bases
When a strong base is added to a solution of a polyprotic acid, the neutralization reaction occurs in stages. The most acidic group is titrated first, followed by the next most acidic, and so forth. If the $pK_a$ values are separated by at least three $pK_a$ units, then the overall titration curve shows well-resolved “steps” corresponding to the titration of each proton. A titration of the triprotic acid $H_3PO_4$ with $\ce{NaOH}$ is illustrated in Figure $5$ and shows two well-defined steps: the first midpoint corresponds to $pK_a$1, and the second midpoint corresponds to $pK_a$2. Because HPO42 is such a weak acid, $pK_a$3 has such a high value that the third step cannot be resolved using 0.100 M $\ce{NaOH}$ as the titrant.
The titration curve for the reaction of a polyprotic base with a strong acid is the mirror image of the curve shown in Figure $5$. The initial pH is high, but as acid is added, the pH decreases in steps if the successive $pK_b$ values are well separated. Table E1 lists the ionization constants and $pK_a$ values for some common polyprotic acids and bases.
Example $3$
Calculate the pH of a solution prepared by adding 55.0 mL of a 0.120 M $\ce{NaOH}$ solution to 100.0 mL of a 0.0510 M solution of oxalic acid ($\ce{HO_2CCO_2H}$), a diprotic acid (abbreviated as $\ce{H2ox}$). Oxalic acid, the simplest dicarboxylic acid, is found in rhubarb and many other plants. Rhubarb leaves are toxic because they contain the calcium salt of the fully deprotonated form of oxalic acid, the oxalate ion ($\ce{O2CCO2^{2−}}$, abbreviated $\ce{ox^{2-}}$).Oxalate salts are toxic for two reasons. First, oxalate salts of divalent cations such as $\ce{Ca^{2+}}$ are insoluble at neutral pH but soluble at low pH. As a result, calcium oxalate dissolves in the dilute acid of the stomach, allowing oxalate to be absorbed and transported into cells, where it can react with calcium to form tiny calcium oxalate crystals that damage tissues. Second, oxalate forms stable complexes with metal ions, which can alter the distribution of metal ions in biological fluids.
Given: volume and concentration of acid and base
Asked for: pH
Strategy:
1. Calculate the initial millimoles of the acid and the base. Use a tabular format to determine the amounts of all the species in solution.
2. Calculate the concentrations of all the species in the final solution. Determine $\ce{[H{+}]}$ and convert this value to pH.
Solution:
A Table E5 gives the $pK_a$ values of oxalic acid as 1.25 and 3.81. Again we proceed by determining the millimoles of acid and base initially present:
$100.00 \cancel{mL} \left ( \dfrac{0.510 \;mmol \;H_{2}ox}{\cancel{mL}} \right )= 5.10 \;mmol \;H_{2}ox \nonumber$
$55.00 \cancel{mL} \left ( \dfrac{0.120 \;mmol \;NaOH}{\cancel{mL}} \right )= 6.60 \;mmol \;NaOH \nonumber$
The strongest acid ($H_2ox$) reacts with the base first. This leaves (6.60 − 5.10) = 1.50 mmol of $OH^-$ to react with Hox−, forming ox2 and H2O. The reactions can be written as follows:
$\underset{5.10\;mmol}{H_{2}ox}+\underset{6.60\;mmol}{OH^{-}} \rightarrow \underset{5.10\;mmol}{Hox^{-}}+ \underset{5.10\;mmol}{H_{2}O} \nonumber$
$\underset{5.10\;mmol}{Hox^{-}}+\underset{1.50\;mmol}{OH^{-}} \rightarrow \underset{1.50\;mmol}{ox^{2-}}+ \underset{1.50\;mmol}{H_{2}O} \nonumber$
In tabular form,
Solutions to Example 17.3.3
$\ce{H2ox}$ $\ce{OH^{-}}$ $\ce{Hox^{−}}$ $\ce{ox^{2−}}$
initial 5.10 mmol 6.60 mmol 0 mmol 0 mmol
change (step 1) −5.10 mmol −5.10 mmol +5.10 mmol 0 mmol
final (step 1) 0 mmol 1.50 mmol 5.10 mmol 0 mmol
change (step 2) −1.50 mmol −1.50 mmol +1.50 mmol
final 0 mmol 0 mmol 3.60 mmol 1.50 mmol
B The equilibrium between the weak acid ($\ce{Hox^{-}}$) and its conjugate base ($\ce{ox^{2-}}$) in the final solution is determined by the magnitude of the second ionization constant, $K_{a2} = 10^{−3.81} = 1.6 \times 10^{−4}$. To calculate the pH of the solution, we need to know $\ce{[H^{+}]}$, which is determined using exactly the same method as in the acetic acid titration in Example $2$:
$\text{final volume of solution} = 100.0\, mL + 55.0\, mL = 155.0 \,mL \nonumber$
Thus the concentrations of $\ce{Hox^{-}}$ and $\ce{ox^{2-}}$ are as follows:
$\left [ Hox^{-} \right ] = \dfrac{3.60 \; mmol \; Hox^{-}}{155.0 \; mL} = 2.32 \times 10^{-2} \;M \nonumber$
$\left [ ox^{2-} \right ] = \dfrac{1.50 \; mmol \; ox^{2-}}{155.0 \; mL} = 9.68 \times 10^{-3} \;M \nonumber$
We can now calculate [H+] at equilibrium using the following equation:
$K_{a2} =\dfrac{\left [ ox^{2-} \right ]\left [ H^{+} \right ] }{\left [ Hox^{-} \right ]} \nonumber$
Rearranging this equation and substituting the values for the concentrations of $\ce{Hox^{−}}$ and $\ce{ox^{2−}}$,
$\left [ H^{+} \right ] =\dfrac{K_{a2}\left [ Hox^{-} \right ]}{\left [ ox^{2-} \right ]} = \dfrac{\left ( 1.6\times 10^{-4} \right ) \left ( 2.32\times 10^{-2} \right )}{\left ( 9.68\times 10^{-3} \right )}=3.7\times 10^{-4} \; M \nonumber$
So
$pH = -\log\left [ H^{+} \right ]= -\log\left ( 3.7 \times 10^{-4} \right )= 3.43 \nonumber$
This answer makes chemical sense because the pH is between the first and second $pK_a$ values of oxalic acid, as it must be. We added enough hydroxide ion to completely titrate the first, more acidic proton (which should give us a pH greater than $pK_{a1}$), but we added only enough to titrate less than half of the second, less acidic proton, with $pK_{a2}$. If we had added exactly enough hydroxide to completely titrate the first proton plus half of the second, we would be at the midpoint of the second step in the titration, and the pH would be 3.81, equal to $pK_{a2}$.
Exercise $3$: Piperazine
Piperazine is a diprotic base used to control intestinal parasites (“worms”) in pets and humans. A dog is given 500 mg (5.80 mmol) of piperazine ($pK_{b1}$ = 4.27, $pK_{b2}$ = 8.67). If the dog’s stomach initially contains 100 mL of 0.10 M $\ce{HCl}$ (pH = 1.00), calculate the pH of the stomach contents after ingestion of the piperazine.
Answer
pH=4.9
Indicators
In practice, most acid–base titrations are not monitored by recording the pH as a function of the amount of the strong acid or base solution used as the titrant. Instead, an acid–base indicator is often used that, if carefully selected, undergoes a dramatic color change at the pH corresponding to the equivalence point of the titration. Indicators are weak acids or bases that exhibit intense colors that vary with pH. The conjugate acid and conjugate base of a good indicator have very different colors so that they can be distinguished easily. Some indicators are colorless in the conjugate acid form but intensely colored when deprotonated (phenolphthalein, for example), which makes them particularly useful.
We can describe the chemistry of indicators by the following general equation:
$\ce{ HIn (aq) <=> H^{+}(aq) + In^{-}(aq)} \nonumber$
where the protonated form is designated by $\ce{HIn}$ and the conjugate base by $\ce{In^{−}}$. The ionization constant for the deprotonation of indicator $\ce{HIn}$ is as follows:
$K_{In} =\dfrac{ [\ce{H^{+}} ][ \ce{In^{-}}]}{[\ce{HIn}]} \label{Eq3}$
The $pK_{in}$ (its $pK_a$) determines the pH at which the indicator changes color.
Many different substances can be used as indicators, depending on the particular reaction to be monitored. For example, red cabbage juice contains a mixture of colored substances that change from deep red at low pH to light blue at intermediate pH to yellow at high pH. Similarly, Hydrangea macrophylla flowers can be blue, red, pink, light purple, or dark purple depending on the soil pH (Figure $6$). Acidic soils will produce blue flowers, whereas alkaline soils will produce pinkish flowers.
Irrespective of the origins, a good indicator must have the following properties:
• The color change must be easily detected.
• The color change must be rapid.
• The indicator molecule must not react with the substance being titrated.
• To minimize errors, the indicator should have a $pK_{in}$ that is within one pH unit of the expected pH at the equivalence point of the titration.
Synthetic indicators have been developed that meet these criteria and cover virtually the entire pH range. Figure $7$ shows the approximate pH range over which some common indicators change color and their change in color. In addition, some indicators (such as thymol blue) are polyprotic acids or bases, which change color twice at widely separated pH values.
It is important to be aware that an indicator does not change color abruptly at a particular pH value; instead, it actually undergoes a pH titration just like any other acid or base. As the concentration of HIn decreases and the concentration of In− increases, the color of the solution slowly changes from the characteristic color of HIn to that of In−. As we will see later, the [In−]/[HIn] ratio changes from 0.1 at a pH one unit below pKin to 10 at a pH one unit above pKin. Thus most indicators change color over a pH range of about two pH units.
We have stated that a good indicator should have a pKin value that is close to the expected pH at the equivalence point. For a strong acid–strong base titration, the choice of the indicator is not especially critical due to the very large change in pH that occurs around the equivalence point. In contrast, using the wrong indicator for a titration of a weak acid or a weak base can result in relatively large errors, as illustrated in Figure $8$. This figure shows plots of pH versus volume of base added for the titration of 50.0 mL of a 0.100 M solution of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M $NaOH$. The pH ranges over which two common indicators (methyl red, $pK_{in} = 5.0$, and phenolphthalein, $pK_{in} = 9.5$) change color are also shown. The horizontal bars indicate the pH ranges over which both indicators change color cross the $\ce{HCl}$ titration curve, where it is almost vertical. Hence both indicators change color when essentially the same volume of $\ce{NaOH}$ has been added (about 50 mL), which corresponds to the equivalence point. In contrast, the titration of acetic acid will give very different results depending on whether methyl red or phenolphthalein is used as the indicator. Although the pH range over which phenolphthalein changes color is slightly greater than the pH at the equivalence point of the strong acid titration, the error will be negligible due to the slope of this portion of the titration curve. Just as with the $\ce{HCl}$ titration, the phenolphthalein indicator will turn pink when about 50 mL of $\ce{NaOH}$ has been added to the acetic acid solution. In contrast, methyl red begins to change from red to yellow around pH 5, which is near the midpoint of the acetic acid titration, not the equivalence point. Adding only about 25–30 mL of $\ce{NaOH}$ will therefore cause the methyl red indicator to change color, resulting in a huge error.
The graph shows the results obtained using two indicators (methyl red and phenolphthalein) for the titration of 0.100 M solutions of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M $NaOH$. Due to the steepness of the titration curve of a strong acid around the equivalence point, either indicator will rapidly change color at the equivalence point for the titration of the strong acid. In contrast, the pKin for methyl red (5.0) is very close to the $pK_a$ of acetic acid (4.76); the midpoint of the color change for methyl red occurs near the midpoint of the titration, rather than at the equivalence point.
In general, for titrations of strong acids with strong bases (and vice versa), any indicator with a pKin between about 4.0 and 10.0 will do. For the titration of a weak acid, however, the pH at the equivalence point is greater than 7.0, so an indicator such as phenolphthalein or thymol blue, with pKin > 7.0, should be used. Conversely, for the titration of a weak base, where the pH at the equivalence point is less than 7.0, an indicator such as methyl red or bromocresol blue, with pKin < 7.0, should be used.
The existence of many different indicators with different colors and pKin values also provides a convenient way to estimate the pH of a solution without using an expensive electronic pH meter and a fragile pH electrode. Paper or plastic strips impregnated with combinations of indicators are used as “pH paper,” which allows you to estimate the pH of a solution by simply dipping a piece of pH paper into it and comparing the resulting color with the standards printed on the container (Figure $9$).
pH Indicators: pH Indicators(opens in new window) [youtu.be]
Summary and Takeaway
Plots of acid–base titrations generate titration curves that can be used to calculate the pH, the pOH, the $pK_a$, and the $pK_b$ of the system. The shape of a titration curve, a plot of pH versus the amount of acid or base added, provides important information about what is occurring in solution during a titration. The shapes of titration curves for weak acids and bases depend dramatically on the identity of the compound. The equivalence point of an acid–base titration is the point at which exactly enough acid or base has been added to react completely with the other component. The equivalence point in the titration of a strong acid or a strong base occurs at pH 7.0. In titrations of weak acids or weak bases, however, the pH at the equivalence point is greater or less than 7.0, respectively. The pH tends to change more slowly before the equivalence point is reached in titrations of weak acids and weak bases than in titrations of strong acids and strong bases. The pH at the midpoint, the point halfway on the titration curve to the equivalence point, is equal to the $pK_a$ of the weak acid or the $pK_b$ of the weak base. Thus titration methods can be used to determine both the concentration and the $pK_a$ (or the $pK_b$) of a weak acid (or a weak base). Acid–base indicators are compounds that change color at a particular pH. They are typically weak acids or bases whose changes in color correspond to deprotonation or protonation of the indicator itself. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/17%3A_Additional_Aspects_of_Aqueous_Equilibria/17.03%3A_Acid-Base_Titrations.txt |
Learning Objectives
• To calculate the solubility of an ionic compound from its Ksp
We begin our discussion of solubility and complexation equilibria—those associated with the formation of complex ions—by developing quantitative methods for describing dissolution and precipitation reactions of ionic compounds in aqueous solution. Just as with acid–base equilibria, we can describe the concentrations of ions in equilibrium with an ionic solid using an equilibrium constant expression.
The Solubility Product
When a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions. For the dissolution of calcium phosphate, one of the two main components of kidney stones, the equilibrium can be written as follows, with the solid salt on the left:
$Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}_{(aq)} + 2PO^{3−}_{4(aq)} \label{Eq1}$
As you will discover in Section 17.4 and in more advanced chemistry courses, basic anions, such as S2−, PO43, and CO32, react with water to produce OH and the corresponding protonated anion. Consequently, their calculated molarities, assuming no protonation in aqueous solution, are only approximate.
The equilibrium constant for the dissolution of a sparingly soluble salt is the solubility product (Ksp) of the salt. Because the concentration of a pure solid such as Ca3(PO4)2 is a constant, it does not appear explicitly in the equilibrium constant expression. The equilibrium constant expression for the dissolution of calcium phosphate is therefore
$K=\dfrac{[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2}{[\mathrm{Ca_3(PO_4)_2}]} \label{Eq2a}$
$[\mathrm{Ca_3(PO_4)_2}]K=K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2 \label{Eq2b}$
At 25°C and pH 7.00, Ksp for calcium phosphate is 2.07 × 10−33, indicating that the concentrations of Ca2+ and PO43 ions in solution that are in equilibrium with solid calcium phosphate are very low. The values of Ksp for some common salts are listed in Table $1$, which shows that the magnitude of Ksp varies dramatically for different compounds. Although Ksp is not a function of pH in Equations $\ref{Eq2a}$ and $\ref{Eq2b}$, changes in pH can affect the solubility of a compound as discussed later.
As with any K, the concentration of a pure solid does not appear explicitly in Ksp.
Table $1$: Solubility Products for Selected Ionic Substances at 25°C
Solid Color $K_{sp}$ Solid Color $K_{sp}$
*These contain the Hg22+ ion.
Acetates Iodides
Ca(O2CCH3)2·3H2O white 4 × 10−3 Hg2I2* yellow 5.2 × 10−29
Bromides PbI2 yellow 9.8 × 10−9
AgBr off-white 5.35 × 10−13 Oxalates
Hg2Br2* yellow 6.40 × 10−23 Ag2C2O4 white 5.40 × 10−12
Carbonates MgC2O4·2H2O white 4.83 × 10−6
CaCO3 white 3.36 × 10−9 PbC2O4 white 4.8 × 10−10
PbCO3 white 7.40 × 10−14 Phosphates
Chlorides Ag3PO4 white 8.89 × 10−17
AgCl white 1.77 × 10−10 Sr3(PO4)2 white 4.0 × 10−28
Hg2Cl2* white 1.43 × 10−18 FePO4·2H2O pink 9.91 × 10−16
PbCl2 white 1.70 × 10−5 Sulfates
Chromates Ag2SO4 white 1.20 × 10−5
CaCrO4 yellow 7.1 × 10−4 BaSO4 white 1.08 × 10−10
PbCrO4 yellow 2.8 × 10−13 PbSO4 white 2.53 × 10−8
Fluorides Sulfides
BaF2 white 1.84 × 10−7 Ag2S black 6.3 × 10−50
PbF2 white 3.3 × 10−8 CdS yellow 8.0 × 10−27
Hydroxides PbS black 8.0 × 10−28
Ca(OH)2 white 5.02 × 10−6 ZnS white 1.6 × 10−24
Cu(OH)2 pale blue 1 × 10−14
Mn(OH)2 light pink 1.9 × 10−13
Cr(OH)3 gray-green 6.3 × 10−31
Fe(OH)3 rust red 2.79 × 10−39
Definition of a Solubility Product: Definition of a Solubility Product(opens in new window) [youtu.be]
Solubility products are determined experimentally by directly measuring either the concentration of one of the component ions or the solubility of the compound in a given amount of water. However, whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, $K_{sp}$, like $K$, is defined in terms of the molar concentrations of the component ions.
Example $1$
Calcium oxalate monohydrate [Ca(O2CCO2)·H2O, also written as CaC2O4·H2O] is a sparingly soluble salt that is the other major component of kidney stones [along with Ca3(PO4)2]. Its solubility in water at 25°C is 7.36 × 10−4 g/100 mL. Calculate its Ksp.
Given: solubility in g/100 mL
Asked for: Ksp
Strategy:
1. Write the balanced dissolution equilibrium and the corresponding solubility product expression.
2. Convert the solubility of the salt to moles per liter. From the balanced dissolution equilibrium, determine the equilibrium concentrations of the dissolved solute ions. Substitute these values into the solubility product expression to calculate Ksp.
Solution
A We need to write the solubility product expression in terms of the concentrations of the component ions. For calcium oxalate monohydrate, the balanced dissolution equilibrium and the solubility product expression (abbreviating oxalate as ox2) are as follows:
$\mathrm{Ca(O_2CCO_2)}\cdot\mathrm{H_2O(s)}\rightleftharpoons \mathrm{Ca^{2+}(aq)}+\mathrm{^-O_2CCO_2^-(aq)}+\mathrm{H_2O(l)}\hspace{5mm}K_{\textrm{sp}}=[\mathrm{Ca^{2+}}][\mathrm{ox^{2-}}]$
Neither solid calcium oxalate monohydrate nor water appears in the solubility product expression because their concentrations are essentially constant.
B Next we need to determine [Ca2+] and [ox2] at equilibrium. We can use the mass of calcium oxalate monohydrate that dissolves in 100 mL of water to calculate the number of moles that dissolve in 100 mL of water. From this we can determine the number of moles that dissolve in 1.00 L of water. For dilute solutions, the density of the solution is nearly the same as that of water, so dissolving the salt in 1.00 L of water gives essentially 1.00 L of solution. Because each 1 mol of dissolved calcium oxalate monohydrate dissociates to produce 1 mol of calcium ions and 1 mol of oxalate ions, we can obtain the equilibrium concentrations that must be inserted into the solubility product expression. The number of moles of calcium oxalate monohydrate that dissolve in 100 mL of water is as follows:
$\dfrac{7.36\times10^{-4}\textrm{ g}}{146.1\textrm{ g/mol}}=5.04\times10^{-6}\textrm{ mol }\mathrm{Ca(O_2CCO_2)\cdot H_2O}$
The number of moles of calcium oxalate monohydrate that dissolve in 1.00 L of the saturated solution is as follows:
$\left(\dfrac{5.04\times10^{-6}\textrm{ mol }\mathrm{Ca(O_2CCO_2\cdot)H_2O}}{\textrm{100 mL}}\right)\left(\dfrac{\textrm{1000 mL}}{\textrm{1.00 L}}\right)=5.04\times10^{-5}\textrm{ mol/L}=5.04\times10^{-5}\textrm{ M}$
Because of the stoichiometry of the reaction, the concentration of Ca2+ and ox2 ions are both 5.04 × 10−5 M. Inserting these values into the solubility product expression,
$K_{sp} = [Ca^{2+}][ox^{2−}] = (5.04 \times 10^{−5})(5.04 \times10^{−5}) = 2.54 \times 10^{−9} \nonumber$
In our calculation, we have ignored the reaction of the weakly basic anion with water, which tends to make the actual solubility of many salts greater than the calculated value.
Exercise $1$: Calcite
One crystalline form of calcium carbonate (CaCO3) is "calcite", found as both a mineral and a structural material in many organisms. Calcite is found in the teeth of sea urchins. The urchins create depressions in limestone that they can settle in by grinding the rock with their teeth. Limestone, however, also consists of calcite, so how can the urchins grind the rock without also grinding their teeth? Researchers have discovered that the teeth are shaped like needles and plates and contain magnesium. The concentration of magnesium increases toward the tip, which contributes to the hardness. Moreover, each tooth is composed of two blocks of the polycrystalline calcite matrix that are interleaved near the tip. This creates a corrugated surface that presumably increases grinding efficiency. Toolmakers are particularly interested in this approach to grinding.
The solubility of calcite in water is 0.67 mg/100 mL. Calculate its Ksp.
Answer
4.5 × 10−9
The reaction of weakly basic anions with H2O tends to make the actual solubility of many salts higher than predicted.
Finding Ksp from Ion Concentrations: Finding Ksp from Ion Concentrations(opens in new window) [youtu.be]
Tabulated values of Ksp can also be used to estimate the solubility of a salt with a procedure that is essentially the reverse of the one used in Example $1$. In this case, we treat the problem as a typical equilibrium problem and set up a table of initial concentrations, changes in concentration, and final concentrations (ICE Tables), remembering that the concentration of the pure solid is essentially constant.
Example $2$
We saw that the Ksp for Ca3(PO4)2 is 2.07 × 10−33 at 25°C. Calculate the aqueous solubility of Ca3(PO4)2 in terms of the following:
1. the molarity of ions produced in solution
2. the mass of salt that dissolves in 100 mL of water at 25°C
Given: Ksp
Asked for: molar concentration and mass of salt that dissolves in 100 mL of water
Strategy:
1. Write the balanced equilibrium equation for the dissolution reaction and construct a table showing the concentrations of the species produced in solution. Insert the appropriate values into the solubility product expression and calculate the molar solubility at 25°C.
2. Calculate the mass of solute in 100 mL of solution from the molar solubility of the salt. Assume that the volume of the solution is the same as the volume of the solvent.
Solution:
1. A The dissolution equilibrium for Ca3(PO4)2 (Equation $\ref{Eq2a}$) is shown in the following ICE table. Because we are starting with distilled water, the initial concentration of both calcium and phosphate ions is zero. For every 1 mol of Ca3(PO4)2 that dissolves, 3 mol of Ca2+ and 2 mol of PO43 ions are produced in solution. If we let x equal the solubility of Ca3(PO4)2 in moles per liter, then the change in [Ca2+] will be +3x, and the change in [PO43] will be +2x. We can insert these values into the table.
Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO43(aq)
Solutions to Example 17.4.2
Ca3(PO4)2 [Ca2+] [PO43]
initial pure solid 0 0
change +3x +2x
final pure solid 3x 2x
Although the amount of solid Ca3(PO4)2 changes as some of it dissolves, its molar concentration does not change. We now insert the expressions for the equilibrium concentrations of the ions into the solubility product expression (Equation 17.2):
\begin{align}K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2&=(3x)^3(2x)^2 \2.07\times10^{-33}&=108x^5 \1.92\times10^{-35}&=x^5 \1.14\times10^{-7}\textrm{ M}&=x\end{align}
This is the molar solubility of calcium phosphate at 25°C. However, the molarity of the ions is 2x and 3x, which means that [PO43] = 2.28 × 10−7 and [Ca2+] = 3.42 × 10−7.
1. B To find the mass of solute in 100 mL of solution, we assume that the density of this dilute solution is the same as the density of water because of the low solubility of the salt, so that 100 mL of water gives 100 mL of solution. We can then determine the amount of salt that dissolves in 100 mL of water:
$\left(\dfrac{1.14\times10^{-7}\textrm{ mol}}{\textrm{1 L}}\right)\textrm{100 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}} \right )\left(\dfrac{310.18 \textrm{ g }\mathrm{Ca_3(PO_4)_2}}{\textrm{1 mol}}\right)=3.54\times10^{-6}\textrm{ g }\mathrm{Ca_3(PO_4)_2}$
Exercise $2$
The solubility product of silver carbonate (Ag2CO3) is 8.46 × 10−12 at 25°C. Calculate the following:
1. the molarity of a saturated solution
2. the mass of silver carbonate that will dissolve in 100 mL of water at this temperature
Answer
1. 1.28 × 10−4 M
2. 3.54 mg
Finding the Solubility of a Salt: Finding the Solubility of a Salt (opens in new window) [youtu.be]
The Ion Product
The ion product (Q) of a salt is the product of the concentrations of the ions in solution raised to the same powers as in the solubility product expression. It is analogous to the reaction quotient (Q) discussed for gaseous equilibria. Whereas Ksp describes equilibrium concentrations, the ion product describes concentrations that are not necessarily equilibrium concentrations.
The ion product Q is analogous to the reaction quotient Q for gaseous equilibria.
As summarized in Figure $1$, there are three possible conditions for an aqueous solution of an ionic solid:
• Q < Ksp. The solution is unsaturated, and more of the ionic solid, if available, will dissolve.
• Q = Ksp. The solution is saturated and at equilibrium.
• Q > Ksp. The solution is supersaturated, and ionic solid will precipitate.
The process of calculating the value of the ion product and comparing it with the magnitude of the solubility product is a straightforward way to determine whether a solution is unsaturated, saturated, or supersaturated. More important, the ion product tells chemists whether a precipitate will form when solutions of two soluble salts are mixed.
Example $3$
We mentioned that barium sulfate is used in medical imaging of the gastrointestinal tract. Its solubility product is 1.08 × 10−10 at 25°C, so it is ideally suited for this purpose because of its low solubility when a “barium milkshake” is consumed by a patient. The pathway of the sparingly soluble salt can be easily monitored by x-rays. Will barium sulfate precipitate if 10.0 mL of 0.0020 M Na2SO4 is added to 100 mL of 3.2 × 10−4 M BaCl2? Recall that NaCl is highly soluble in water.
Given: Ksp and volumes and concentrations of reactants
Asked for: whether precipitate will form
Strategy:
1. Write the balanced equilibrium equation for the precipitation reaction and the expression for Ksp.
2. Determine the concentrations of all ions in solution when the solutions are mixed and use them to calculate the ion product (Q).
3. Compare the values of Q and Ksp to decide whether a precipitate will form.
Solution
A The only slightly soluble salt that can be formed when these two solutions are mixed is BaSO4 because NaCl is highly soluble. The equation for the precipitation of BaSO4 is as follows:
$BaSO_{4(s)} \rightleftharpoons Ba^{2+}_{(aq)} + SO^{2−}_{4(aq)} \nonumber$
The solubility product expression is as follows:
Ksp = [Ba2+][SO42] = 1.08×10−10
B To solve this problem, we must first calculate the ion product—Q = [Ba2+][SO42]—using the concentrations of the ions that are present after the solutions are mixed and before any reaction occurs. The concentration of Ba2+ when the solutions are mixed is the total number of moles of Ba2+ in the original 100 mL of BaCl2 solution divided by the final volume (100 mL + 10.0 mL = 110 mL):
$\textrm{moles Ba}^{2+}=\textrm{100 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}}\right)\left(\dfrac{3.2\times10^{-4}\textrm{ mol}}{\textrm{1 L}} \right )=3.2\times10^{-5}\textrm{ mol Ba}^{2+}$
$[\mathrm{Ba^{2+}}]=\left(\dfrac{3.2\times10^{-5}\textrm{ mol Ba}^{2+}}{\textrm{110 mL}}\right)\left(\dfrac{\textrm{1000 mL}}{\textrm{1 L}}\right)=2.9\times10^{-4}\textrm{ M Ba}^{2+}$
Similarly, the concentration of SO42 after mixing is the total number of moles of SO42 in the original 10.0 mL of Na2SO4 solution divided by the final volume (110 mL):
$\textrm{moles SO}_4^{2-}=\textrm{10.0 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}}\right)\left(\dfrac{\textrm{0.0020 mol}}{\textrm{1 L}}\right)=2.0\times10^{-5}\textrm{ mol SO}_4^{2-}$
$[\mathrm{SO_4^{2-}}]=\left(\dfrac{2.0\times10^{-5}\textrm{ mol SO}_4^{2-}}{\textrm{110 mL}} \right )\left(\dfrac{\textrm{1000 mL}}{\textrm{1 L}}\right)=1.8\times10^{-4}\textrm{ M SO}_4^{2-}$
We can now calculate Q:
Q = [Ba2+][SO42] = (2.9×10−4)(1.8×10−4) = 5.2×10−8
C We now compare Q with the Ksp. If Q > Ksp, then BaSO4 will precipitate, but if Q < Ksp, it will not. Because Q > Ksp, we predict that BaSO4 will precipitate when the two solutions are mixed. In fact, BaSO4 will continue to precipitate until the system reaches equilibrium, which occurs when [Ba2+][SO42] = Ksp = 1.08 × 10−10.
Exercise $3$
The solubility product of calcium fluoride (CaF2) is 3.45 × 10−11. If 2.0 mL of a 0.10 M solution of NaF is added to 128 mL of a 2.0 × 10−5M solution of Ca(NO3)2, will CaF2 precipitate?
Answer
yes (Q = 4.7 × 10−11 > Ksp)
Determining if a Precipitate forms (The Ion Product): Determining if a Precipitate forms (The Ion Product)(opens in new window) [youtu.be]
The Common Ion Effect and Solubility
The solubility product expression tells us that the equilibrium concentrations of the cation and the anion are inversely related. That is, as the concentration of the anion increases, the maximum concentration of the cation needed for precipitation to occur decreases—and vice versa—so that Ksp is constant. Consequently, the solubility of an ionic compound depends on the concentrations of other salts that contain the same ions. Adding a common cation or anion shifts a solubility equilibrium in the direction predicted by Le Chatelier’s principle. As a result, the solubility of any sparingly soluble salt is almost always decreased by the presence of a soluble salt that contains a common ion. The exceptions generally involve the formation of complex ions, which is discussed later.
Consider, for example, the effect of adding a soluble salt, such as CaCl2, to a saturated solution of calcium phosphate [Ca3(PO4)2]. We have seen that the solubility of Ca3(PO4)2 in water at 25°C is 1.14 × 10−7 M (Ksp = 2.07 × 10−33). Thus a saturated solution of Ca3(PO4)2 in water contains 3 × (1.14 × 10−7 M) = 3.42 × 10−7 M Ca2+ and 2 × (1.14 × 10−7 M) = 2.28 × 10−7 M PO43, according to the stoichiometry shown in Equation $\ref{Eq1}$ (neglecting hydrolysis to form HPO42 as described in Chapter 16). If CaCl2 is added to a saturated solution of Ca3(PO4)2, the Ca2+ ion concentration will increase such that [Ca2+] > 3.42 × 10−7 M, making Q > Ksp. The only way the system can return to equilibrium is for the reaction in Equation $\ref{Eq1}$ to proceed to the left, resulting in precipitation of Ca3(PO4)2. This will decrease the concentration of both Ca2+ and PO43 until Q = Ksp.
The common ion effect usually decreases the solubility of a sparingly soluble salt.
Example $4$
Calculate the solubility of calcium phosphate [Ca3(PO4)2] in 0.20 M CaCl2.
Given: concentration of CaCl2 solution
Asked for: solubility of Ca3(PO4)2 in CaCl2 solution
Strategy:
1. Write the balanced equilibrium equation for the dissolution of Ca3(PO4)2. Tabulate the concentrations of all species produced in solution.
2. Substitute the appropriate values into the expression for the solubility product and calculate the solubility of Ca3(PO4)2.
Solution
A The balanced equilibrium equation is given in the following table. If we let x equal the solubility of Ca3(PO4)2 in moles per liter, then the change in [Ca2+] is once again +3x, and the change in [PO43] is +2x. We can insert these values into the ICE table.
$Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}_{(aq)} + 2PO^{3−}_{4(aq)} \nonumber$
Solutions to Example 17.4.4
Ca3(PO4)2 [Ca2+] [PO43]
initial pure solid 0.20 0
change +3x +2x
final pure solid 0.20 + 3x 2x
B The Ksp expression is as follows:
Ksp = [Ca2+]3[PO43]2 = (0.20 + 3x)3(2x)2 = 2.07×10−33
Because Ca3(PO4)2 is a sparingly soluble salt, we can reasonably expect that x << 0.20. Thus (0.20 + 3x) M is approximately 0.20 M, which simplifies the Ksp expression as follows:
\begin{align*}K_{\textrm{sp}}=(0.20)^3(2x)^2&=2.07\times10^{-33} \x^2&=6.5\times10^{-32} \x&=2.5\times10^{-16}\textrm{ M}\end{align*} \nonumber
This value is the solubility of Ca3(PO4)2 in 0.20 M CaCl2 at 25°C. It is approximately nine orders of magnitude less than its solubility in pure water, as we would expect based on Le Chatelier’s principle. With one exception, this example is identical to Example $2$—here the initial [Ca2+] was 0.20 M rather than 0.
Exercise $4$
Calculate the solubility of silver carbonate in a 0.25 M solution of sodium carbonate. The solubility of silver carbonate in pure water is 8.45 × 10−12 at 25°C.
Answer
2.9 × 10−6 M (versus 1.3 × 10−4 M in pure water)
The Common Ion Effect in Solubility Products: The Common Ion Effect in Solubility Products(opens in new window) [youtu.be]
Summary
The solubility product (Ksp) is used to calculate equilibrium concentrations of the ions in solution, whereas the ion product (Q) describes concentrations that are not necessarily at equilibrium. The equilibrium constant for a dissolution reaction, called the solubility product (Ksp), is a measure of the solubility of a compound. Whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, Ksp is defined in terms of the molar concentrations of the component ions. In contrast, the ion product (Q) describes concentrations that are not necessarily equilibrium concentrations. Comparing Q and Ksp enables us to determine whether a precipitate will form when solutions of two soluble salts are mixed. Adding a common cation or common anion to a solution of a sparingly soluble salt shifts the solubility equilibrium in the direction predicted by Le Chatelier’s principle. The solubility of the salt is almost always decreased by the presence of a common ion. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/17%3A_Additional_Aspects_of_Aqueous_Equilibria/17.04%3A_Solubility_Equilibria.txt |
Learning Objectives
• To understand the factors that determine the solubility of ionic compounds.
The solubility product of an ionic compound describes the concentrations of ions in equilibrium with a solid, but what happens if some of the cations become associated with anions rather than being completely surrounded by solvent? Then predictions of the total solubility of the compound based on the assumption that the solute exists solely as discrete ions would differ substantially from the actual solubility, as would predictions of ionic concentrations. In general, four situations explain why the solubility of a compound may be other than expected: ion pair formation, the incomplete dissociation of molecular solutes, the formation of complex ions, and changes in pH.
Ion-Pair Formation
An ion pair consists of a cation and an anion that are in intimate contact in solution, rather than separated by solvent (Figure $1$). The ions in an ion pair are held together by the same attractive electrostatic force in ionic solids. As a result, the ions in an ion pair migrate as a single unit, whose net charge is the sum of the charges on the ions. In many ways, we can view an ion pair as a species intermediate between the ionic solid (in which each ion participates in many cation–anion interactions that hold the ions in a rigid array) and the completely dissociated ions in solution (where each is fully surrounded by water molecules and free to migrate independently).
As illustrated for calcium sulfate in the following equation, a second equilibrium must be included to describe the solubility of salts that form ion pairs:
$\mathrm{CaSO_4(s)}\rightleftharpoons \underbrace{\mathrm{Ca^{2+}}\cdot \mathrm{SO_4^{2-}(aq)}}_ {\textrm{ion pair}} \rightleftharpoons \mathrm{Ca^{2+}(aq)}+\mathrm{SO_4^{2-}(aq)} \label{17.5.3.1}$
The ion pair is represented by the symbols of the individual ions separated by a dot, which indicates that they are associated in solution. The formation of an ion pair is a dynamic process, just like any other equilibrium, so a particular ion pair may exist only briefly before dissociating into the free ions, each of which may later associate briefly with other ions.
Ion-pair formation can have a major effect on the measured solubility of a salt. For example, the measured Ksp for calcium sulfate is 4.93 × 10−5 at 25°C. The solubility of CaSO4 should be 7.02 × 10−3 M if the only equilibrium involved were as follows:
$\ce{ CaSO4(s) <=> Ca^{2+}(aq) + SO^{2−}4(aq)} \label{17.5.2}$
In fact, the experimentally measured solubility of calcium sulfate at 25°C is 1.6 × 10−2 M, almost twice the value predicted from its Ksp. The reason for the discrepancy is that the concentration of ion pairs in a saturated $\ce{CaSO4}$ solution is almost as high as the concentration of the hydrated ions. Recall that the magnitude of attractive electrostatic interactions is greatest for small, highly charged ions. Hence ion pair formation is most important for salts that contain M2+ and M3+ ions, such as Ca2+ and La3+, and is relatively unimportant for salts that contain monopositive cations, except for the smallest, Li+. We therefore expect a saturated solution of $\ce{CaSO4}$ to contain a high concentration of ion pairs and its solubility to be greater than predicted from its Ksp.
The formation of ion pairs increases the solubility of a salt.
Incomplete Dissociation
A molecular solute may also be more soluble than predicted by the measured concentrations of ions in solution due to incomplete dissociation. This is particularly common with weak organic acids. Although strong acids (HA) dissociate completely into their constituent ions (H+ and A) in water, weak acids such as carboxylic acids do not (Ka = 1.5 × 10−5). However, the molecular (undissociated) form of a weak acid (HA) is often quite soluble in water; for example, acetic acid (CH3CO2H) is completely miscible with water. Many carboxylic acids, however, have only limited solubility in water, such as benzoic acid (C6H5CO2H), with Ka = 6.25 × 10−5. Just as with calcium sulfate, we need to include an additional equilibrium to describe the solubility of benzoic acid:
$\ce{ C6H5CO2H(s) \rightleftharpoons C6H5CO2H(aq) \rightleftharpoons C6H5CO^{-}2(aq) + H^{+}(aq)} \nonumber$
In a case like this, measuring only the concentration of the ions grossly underestimates the total concentration of the organic acid in solution. In the case of benzoic acid, for example, the pH of a saturated solution at 25°C is 2.85, corresponding to [H+] = [C6H5CO2] = 1.4 × 10−3 M. The total concentration of benzoic acid in the solution, however, is 2.8 × 10−2 M. Thus approximately 95% of the benzoic acid in solution is in the form of hydrated neutral molecules—$C_6H_5CO_2H_{(aq)}$—and only about 5% is present as the dissociated ions (Figure $2$).
Although ion pairs, such as Ca2+·SO42, and undissociated electrolytes, such as C6H5CO2H, are both electrically neutral, there is a major difference in the forces responsible for their formation. Simple electrostatic attractive forces between the cation and the anion hold the ion pair together, whereas a polar covalent O−H bond holds together the undissociated electrolyte.
Incomplete dissociation of a molecular solute that is miscible with water can increase the solubility of the solute.
Complex Ion Formation
Previously, you learned that metal ions in aqueous solution are hydrated—that is, surrounded by a shell of usually four or six water molecules. A hydrated ion is one kind of a complex ion (or, simply, complex), a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons, such as the [Al(H2O)6]3+ ion.
A complex ion forms from a metal ion and a ligand because of a Lewis acid–base interaction. The positively charged metal ion acts as a Lewis acid, and the ligand, with one or more lone pairs of electrons, acts as a Lewis base. Small, highly charged metal ions, such as Cu2+ or Ru3+, have the greatest tendency to act as Lewis acids, and consequently, they have the greatest tendency to form complex ions.
As an example of the formation of complex ions, consider the addition of ammonia to an aqueous solution of the hydrated Cu2+ ion {[Cu(H2O)6]2+}. Because it is a stronger base than H2O, ammonia replaces the water molecules in the hydrated ion to form the [Cu(NH3)4(H2O)2]2+ ion. Formation of the [Cu(NH3)4(H2O)2]2+ complex is accompanied by a dramatic color change, as shown in Figure $1$. The solution changes from the light blue of [Cu(H2O)6]2+ to the blue-violet characteristic of the [Cu(NH3)4(H2O)2]2+ ion.
The Formation Constant
The replacement of water molecules from $\ce{[Cu(H2O)6]^{2+}}$ by ammonia occurs in sequential steps. Omitting the water molecules bound to $\ce{Cu^{2+}}$ for simplicity, we can write the equilibrium reactions as follows:
\begin{align*}\mathrm{Cu^{2+}(aq)}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)]^{2+}_{(aq)}}\quad \quad K_1 \[4pt] \mathrm{[Cu(NH_3)]^{2+}_{(aq)}}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)_2]^{2+}_{(aq)}} \quad \quad K_2 \[4pt] \mathrm{[Cu(NH_3)_2]^{2+}_{(aq)}}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)_3]^{2+}_{(aq)}} \quad \quad K_3 \[4pt] \mathrm{[Cu(NH_3)_3]^{2+}_{(aq)}}+\mathrm{NH_{3(aq)}}&\rightleftharpoons \mathrm{[Cu(NH_3)_4]^{2+}_{(aq)}} \quad \quad K_4 \end{align*} \nonumber
The sum of the stepwise reactions is the overall equation for the formation of the complex ion: The hydrated $\ce{Cu^{2+}}$ ion contains six H2O ligands, but the complex ion that is produced contains only four $NH_3$ ligands, not six.
$\ce{Cu^{2+}(aq) + 4NH3(aq) <=> [Cu(NH3)4]^{2+}(aq)} \label{17.3.2}$
The equilibrium constant for the formation of the complex ion from the hydrated ion is called the formation constant ($K_f$). The equilibrium constant expression for $K_f$ has the same general form as any other equilibrium constant expression. In this case, the expression is as follows:
$K_\textrm f=\dfrac{\left[[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4}=2.1\times10^{13}=K_1K_2K_3K_4\label{17.3.3}$
The formation constant (Kf) has the same general form as any other equilibrium constant expression.
Water, a pure liquid, does not appear explicitly in the equilibrium constant expression, and the hydrated Cu2+(aq) ion is represented as Cu2+ for simplicity. As for any equilibrium, the larger the value of the equilibrium constant (in this case, Kf), the more stable the product. With Kf = 2.1 × 1013, the [Cu(NH3)4(H2O)2]2+ complex ion is very stable. The formation constants for some common complex ions are listed in Table $1$.
Table $1$: Formation Constants for Selected Complex Ions in Aqueous Solution*
Complex Ion Equilibrium Equation Kf
*Reported values are overall formation constants. Source: Data from Lange’s Handbook of Chemistry, 15th ed. (1999).
Ammonia Complexes [Ag(NH3)2]+ Ag+ + 2NH3 ⇌ [Ag(NH3)2]+ 1.1 × 107
[Cu(NH3)4]2+ Cu2+ + 4NH3 ⇌ [Cu(NH3)4]2+ 2.1 × 1013
[Ni(NH3)6]2+ Ni2+ + 6NH3 ⇌ [Ni(NH3)6]2+ 5.5 × 108
Cyanide Complexes [Ag(CN)2] Ag+ + 2CN ⇌ [Ag(CN)2] 1.1 × 1018
[Ni(CN)4]2− Ni2+ + 4CN ⇌ [Ni(CN)4]2− 2.2 × 1031
[Fe(CN)6]3− Fe3+ + 6CN ⇌ [Fe(CN)6]3− 1 × 1042
Hydroxide Complexes [Zn(OH)4]2− Zn2+ + 4OH ⇌ [Zn(OH)4]2− 4.6 × 1017
[Cr(OH)4] Cr3+ + 4OH ⇌ [Cr(OH)4] 8.0 × 1029
Halide Complexes [HgCl4]2− Hg2+ + 4Cl ⇌ [HgCl4]2− 1.2 × 1015
[CdI4]2− Cd2+ + 4I ⇌ [CdI4]2− 2.6 × 105
[AlF6]3− Al3+ + 6F ⇌ [AlF6]3− 6.9 × 1019
Other Complexes [Ag(S2O3)2]3− Ag+ + 2S2O32 ⇌ [Ag(S2O3)2]3− 2.9 × 1013
[Fe(C2O4)3]3− Fe3+ + 3C2O42 ⇌ [Fe(C2O4)3]3− 2.0 × 1020
Example $1$
If 12.5 g of Cu(NO3)2•6H2O is added to 500 mL of 1.00 M aqueous ammonia, what is the equilibrium concentration of Cu2+(aq)?
Given: mass of Cu2+ salt and volume and concentration of ammonia solution
Asked for: equilibrium concentration of Cu2+(aq)
Strategy:
1. Calculate the initial concentration of Cu2+ due to the addition of copper(II) nitrate hexahydrate. Use the stoichiometry of the reaction shown in Equation $\ref{17.3.2}$ to construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations of all species in solution.
2. Substitute the final concentrations into the expression for the formation constant (Equation $\ref{17.3.3}$) to calculate the equilibrium concentration of Cu2+(aq).
Solution
Adding an ionic compound that contains Cu2+ to an aqueous ammonia solution will result in the formation of [Cu(NH3)4]2+(aq), as shown in Equation $\ref{17.3.2}$. We assume that the volume change caused by adding solid copper(II) nitrate to aqueous ammonia is negligible.
A The initial concentration of Cu2+ from the amount of added copper nitrate prior to any reaction is as follows:
$12.5\mathrm{\;\cancel{g}\;Cu(NO_3)_2}\cdot\mathrm{6H_2O}\left(\dfrac{\textrm{1 mol}}{\textrm{295.65} \cancel{g}} \right )\left(\dfrac{1}{\textrm{500}\; \cancel{mL}} \right )\left(\dfrac{\textrm{1000}\; \cancel{mL}}{\textrm{1 L}} \right )=\textrm{0.0846 M} \nonumber$
Because the stoichiometry of the reaction is four NH3 to one Cu2+, the amount of NH3 required to react completely with the Cu2+ is 4(0.0846) = 0.338 M. The concentration of ammonia after complete reaction is 1.00 M − 0.338 M = 0.66 M. These results are summarized in the first two lines of the following table. Because the equilibrium constant for the reaction is large (2.1 × 1013), the equilibrium will lie far to the right. Thus we will assume that the formation of [Cu(NH3)4]2+ in the first step is complete and allow some of it to dissociate into Cu2+ and NH3 until equilibrium has been reached. If we define x as the amount of Cu2+ produced by the dissociation reaction, then the stoichiometry of the reaction tells us that the change in the concentration of [Cu(NH3)4]2+ is −x, and the change in the concentration of ammonia is +4x, as indicated in the table. The final concentrations of all species (in the bottom row of the table) are the sums of the concentrations after complete reaction and the changes in concentrations.
$\ce{Cu^{2+} + 4NH3 <=> [Cu(NH3)4]^{2+}} \nonumber$
Solutions to Example 17.5.1
[Cu2+] [NH3] [[Cu(NH3)4]2+]
initial 0.0846 1.00 0
after complete reaction 0 0.66 0.0846
change +x +4x x
final x 0.66 + 4x 0.0846 − x
B Substituting the final concentrations into the expression for the formation constant (Equation $\ref{17.3.3}$) and assuming that $x \ll 0.0846$, which allows us to remove x from the sum and difference,
\begin{align*}K_\textrm f&=\dfrac{\left[[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4}=\dfrac{0.0846-x}{x(0.66+4x)^4}\approx\dfrac{0.0846}{x(0.66)^4}=2.1\times10^{13} \x&=2.1\times 10^{-14}\end{align*} \nonumber
The value of $x$ indicates that our assumption was justified. The equilibrium concentration of $\ce{Cu^{2+}(aq)}$ in a 1.00 M ammonia solution is therefore $2.1 \times 10^{−14} M$.
Exercise $1$
The ferrocyanide ion $\ce{[Fe(CN)6]^{4−}}$ is very stable, with a Kf = 1 × 1035. Calculate the concentration of cyanide ion in equilibrium with a 0.65 M solution of $\ce{K4[Fe(CN)6]}$.
Answer
2 × 10−6 M
The Effect of the Formation of Complex Ions on Solubility
What happens to the solubility of a sparingly soluble salt if a ligand that forms a stable complex ion is added to the solution? One such example occurs in conventional black-and-white photography. Recall that black-and-white photographic film contains light-sensitive microcrystals of AgBr, or mixtures of AgBr and other silver halides. AgBr is a sparingly soluble salt, with a Ksp of 5.35 × 10−13 at 25°C. When the shutter of the camera opens, the light from the object being photographed strikes some of the crystals on the film and initiates a photochemical reaction that converts AgBr to black Ag metal. Well-formed, stable negative images appear in tones of gray, corresponding to the number of grains of AgBr converted, with the areas exposed to the most light being darkest. To fix the image and prevent more AgBr crystals from being converted to Ag metal during processing of the film, the unreacted $\ce{AgBr}$ on the film is removed using a complexation reaction to dissolve the sparingly soluble salt.
The reaction for the dissolution of silver bromide is as follows:
$\ce{AgBr(s) <=> Ag^{+}(aq) + Br^{-}(aq)} \nonumber$
with $K_{sp} = 5.35 \times 10^{−13}$ at 25°C.
The equilibrium lies far to the left, and the equilibrium concentrations of Ag+ and Br ions are very low (7.31 × 10−7 M). As a result, removing unreacted AgBr from even a single roll of film using pure water would require tens of thousands of liters of water and a great deal of time. Le Chatelier’s principle tells us, however, that we can drive the reaction to the right by removing one of the products, which will cause more AgBr to dissolve. Bromide ion is difficult to remove chemically, but silver ion forms a variety of stable two-coordinate complexes with neutral ligands, such as ammonia, or with anionic ligands, such as cyanide or thiosulfate (S2O32). In photographic processing, excess AgBr is dissolved using a concentrated solution of sodium thiosulfate.
The reaction of Ag+ with thiosulfate is as follows:
$\ce{Ag^{+}(aq) + 2S2O^{2-}3(aq) <=> [Ag(S2O3)2]^{3-}(aq)} \nonumber$
with $K_f = 2.9 \times 10^{13}$.
The magnitude of the equilibrium constant indicates that almost all $\ce{Ag^{+}}$ ions in solution will be immediately complexed by thiosulfate to form [Ag(S2O3)2]3−. We can see the effect of thiosulfate on the solubility of AgBr by writing the appropriate reactions and adding them together:
\begin{align}\mathrm{AgBr(s)}\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Br^-(aq)}\hspace{3mm}K_{\textrm{sp}}&=5.35\times10^{-13} \ \mathrm{Ag^+(aq)}+\mathrm{2S_2O_3^{2-}(aq)}\rightleftharpoons\mathrm{[Ag(S_2O_3)_2]^{3-}(aq)}\hspace{3mm}K_\textrm f&=2.9\times10^{13} \ \mathrm{AgBr(s)}+\mathrm{2S_2O_3^{2-}(aq)}\rightleftharpoons\mathrm{[Ag(S_2O_3)_2]^{3-}(aq)}+\mathrm{Br^-(aq)}\hspace{3mm}K&=K_{\textrm{sp}}K_{\textrm f}=15\end{align} \label{17.3.6}
Comparing K with Ksp shows that the formation of the complex ion increases the solubility of AgBr by approximately 3 × 1013. The dramatic increase in solubility combined with the low cost and the low toxicity explains why sodium thiosulfate is almost universally used for developing black-and-white film. If desired, the silver can be recovered from the thiosulfate solution using any of several methods and recycled.
If a complex ion has a large Kf, the formation of a complex ion can dramatically increase the solubility of sparingly soluble salts.
Example $2$
Due to the common ion effect, we might expect a salt such as AgCl to be much less soluble in a concentrated solution of KCl than in water. Such an assumption would be incorrect, however, because it ignores the fact that silver ion tends to form a two-coordinate complex with chloride ions (AgCl2). Calculate the solubility of AgCl in each situation:
1. in pure water
2. in 1.0 M KCl solution, ignoring the formation of any complex ions
3. the same solution as in part (b) except taking the formation of complex ions into account, assuming that AgCl2 is the only Ag+ complex that forms in significant concentrations
At 25°C, Ksp = 1.77 × 10−10 for AgCl and Kf = 1.1 × 105 for AgCl2.
Given: Ksp of AgCl, Kf of AgCl2, and KCl concentration
Asked for: solubility of AgCl in water and in KCl solution with and without the formation of complex ions
Strategy:
1. Write the solubility product expression for AgCl and calculate the concentration of Ag+ and Cl in water.
2. Calculate the concentration of Ag+ in the KCl solution.
3. Write balanced chemical equations for the dissolution of AgCl and for the formation of the AgCl2 complex. Add the two equations and calculate the equilibrium constant for the overall equilibrium.
4. Write the equilibrium constant expression for the overall reaction. Solve for the concentration of the complex ion.
Solution
1. A If we let x equal the solubility of AgCl, then at equilibrium [Ag+] = [Cl] = x M. Substituting this value into the solubility product expression,
Ksp = [Ag+][Cl] = (x)(x) = x2 =1.77×10−10
x = 1.33×10−5
Thus the solubility of AgCl in pure water at 25°C is 1.33 × 10−5 M.
1. B If x equals the solubility of AgCl in the KCl solution, then at equilibrium [Ag+] = x M and [Cl] = (1.0 + x) M. Substituting these values into the solubility product expression and assuming that x << 1.0,
Ksp = [Ag+][Cl] = (x)(1.0 + x) ≈ x(1.0) = 1.77×10−10 = x
If the common ion effect were the only important factor, we would predict that AgCl is approximately five orders of magnitude less soluble in a 1.0 M KCl solution than in water.
1. C To account for the effects of the formation of complex ions, we must first write the equilibrium equations for both the dissolution and the formation of complex ions. Adding the equations corresponding to Ksp and Kf gives us an equation that describes the dissolution of AgCl in a KCl solution. The equilibrium constant for the reaction is therefore the product of Ksp and Kf:
\begin{align}\mathrm{AgCl(s)}\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Cl^-(aq)}\hspace{3mm}K_{\textrm{sp}}&=1.77\times10^{-10} \ \mathrm{Ag^+(aq)}+\mathrm{2Cl^{-}}\rightleftharpoons\mathrm{[AgCl_2]^{-}}\hspace{3mm}K_\textrm f&=1.1\times10^{5} \ \mathrm{AgCl(s)}+\mathrm{Cl^{-}}\rightleftharpoons\mathrm{[AgCl_2]^{-}}\hspace{3mm}K&=K_{\textrm{sp}}K_{\textrm f}=1.9\times10^{-5}\end{align}
D If we let x equal the solubility of AgCl in the KCl solution, then at equilibrium [AgCl2] = x and [Cl] = 1.0 − x. Substituting these quantities into the equilibrium constant expression for the net reaction and assuming that x << 1.0,
$K=\dfrac{[\mathrm{AgCl_2^-}]}{[\mathrm{Cl^-}]}=\dfrac{x}{1.0-x}\approx1.9\times10^{-5}=x$
That is, AgCl dissolves in 1.0 M KCl to produce a 1.9 × 10−5 M solution of the AgCl2 complex ion. Thus we predict that AgCl has approximately the same solubility in a 1.0 M KCl solution as it does in pure water, which is 105 times greater than that predicted based on the common ion effect. (In fact, the measured solubility of AgCl in 1.0 M KCl is almost a factor of 10 greater than that in pure water, largely due to the formation of other chloride-containing complexes.)
Exercise $2$
Calculate the solubility of mercury(II) iodide ($\ce{HgI2}$) in each situation:
1. pure water
2. a 3.0 M solution of NaI, assuming [HgI4]2− is the only Hg-containing species present in significant amounts
Ksp = 2.9 × 10−29 for HgI2 and Kf = 6.8 × 1029 for [HgI4]2−.
Answer
1. 1.9 × 10−10 M
2. 1.4 M
Solubility of Complex Ions: Solubility of Complex Ions(opens in new window) [youtu.be]
Complexing agents, molecules or ions that increase the solubility of metal salts by forming soluble metal complexes, are common components of laundry detergents. Long-chain carboxylic acids, the major components of soaps, form insoluble salts with Ca2+ and Mg2+, which are present in high concentrations in “hard” water. The precipitation of these salts produces a bathtub ring and gives a gray tinge to clothing. Adding a complexing agent such as pyrophosphate (O3POPO34, or P2O74) or triphosphate (P3O105) to detergents prevents the magnesium and calcium salts from precipitating because the equilibrium constant for complex-ion formation is large:
$\ce{Ca^{2+}(aq) + O_3POPO^{4−}4(aq) <=> [Ca(O_3POPO_3)]^{2−}(aq)} \label{17.3.7a}$
with $K_f = 4\times 10^4$.
However, phosphates can cause environmental damage by promoting eutrophication, the growth of excessive amounts of algae in a body of water, which can eventually lead to large decreases in levels of dissolved oxygen that kill fish and other aquatic organisms. Consequently, many states in the United States have banned the use of phosphate-containing detergents, and France has banned their use beginning in 2007. “Phosphate-free” detergents contain different kinds of complexing agents, such as derivatives of acetic acid or other carboxylic acids. The development of phosphate substitutes is an area of intense research.
Commercial water softeners also use a complexing agent to treat hard water by passing the water over ion-exchange resins, which are complex sodium salts. When water flows over the resin, sodium ion is dissolved, and insoluble salts precipitate onto the resin surface. Water treated in this way has a saltier taste due to the presence of Na+, but it contains fewer dissolved minerals.
Another application of complexing agents is found in medicine. Unlike x-rays, magnetic resonance imaging (MRI) can give relatively good images of soft tissues such as internal organs. MRI is based on the magnetic properties of the 1H nucleus of hydrogen atoms in water, which is a major component of soft tissues. Because the properties of water do not depend very much on whether it is inside a cell or in the blood, it is hard to get detailed images of these tissues that have good contrast. To solve this problem, scientists have developed a class of metal complexes known as “MRI contrast agents.” Injecting an MRI contrast agent into a patient selectively affects the magnetic properties of water in cells of normal tissues, in tumors, or in blood vessels and allows doctors to “see” each of these separately (Figure $4$). One of the most important metal ions for this application is Gd3+, which with seven unpaired electrons is highly paramagnetic. Because Gd3+(aq) is quite toxic, it must be administered as a very stable complex that does not dissociate in the body and can be excreted intact by the kidneys. The complexing agents used for gadolinium are ligands such as DTPA5 (diethylene triamine pentaacetic acid), whose fully protonated form is shown here.
Summary
Ion-pair formation, the incomplete dissociation of molecular solutes, the formation of complex ions, and changes in pH all affect solubility. There are four explanations why the solubility of a compound can differ from the solubility indicated by the concentrations of ions: (1) ion pair formation, in which an anion and a cation are in intimate contact in solution and not separated by solvent, (2) the incomplete dissociation of molecular solutes, (3) the formation of complex ions, and (4) changes in pH. An ion pair is held together by electrostatic attractive forces between the cation and the anion, whereas incomplete dissociation results from intramolecular forces, such as polar covalent O–H bonds.
The formation of complex ions can substantially increase the solubility of sparingly soluble salts if the complex ion has a large Kf. A complex ion is a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons. Small, highly charged metal ions have the greatest tendency to act as Lewis acids and form complex ions. The equilibrium constant for the formation of the complex ion is the formation constant (Kf). The formation of a complex ion by adding a complexing agent increases the solubility of a compound. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/17%3A_Additional_Aspects_of_Aqueous_Equilibria/17.05%3A_Factors_that_Affect_Solubility.txt |
Learning Objectives
• Calculate ion concentrations to maintain a heterogeneous equilibrium.
• Calculate pH required to precipitate a metal hydroxide.
• Design experiments to separate metal ions in a solution of mixtures of metals.
A mixture of metal ions in a solution can be separated by precipitation with anions such as $\ce{Cl-}$, $\ce{Br-}$, $\ce{SO4^2-}$, $\ce{CO3^2-}$, $\ce{S^2-}$, $\ce{Cr2O4^2-}$, $\ce{PO4^2-}$, $\ce{OH-}$ etc. When a metal ion or a group of metal ions form insoluble salts with a particular anion, they can be separated from others by precipitation. We can also separate the anions by precipitating them with appropriate metal ions. There are no definite dividing lines between insoluble salts, sparingly soluble, and soluble salts, but concentrations of their saturated solutions are small, medium, and large. Solubility products are usually listed for insoluble and sparingly soluble salts, but they are not given for soluble salts. Solubility products for soluble salts are very large.
What type of salts are usually soluble, sparingly soluble and insoluble? The following are some general guidelines, but these are not precise laws.
• All nitrates are soluble. The singly charged large $\ce{NO3-}$ ions form salts with high solubilities. So do $\ce{ClO4-}$, $\ce{ClO3-}$, $\ce{NO2-}$, $\ce{HCOO-}$, and $\ce{CH3COO-}$.
• All chlorides, bromides, and iodides are soluble except those of $\ce{Ag+}$, $\ce{Hg2^2+}$, and $\ce{Pb^2+}$. $\ce{CaF2}$, $\ce{BaF2}$, and $\ce{PbF2}$ are also insoluble.
• All sulfates are soluble, except those of $\ce{Ba^2+}$, $\ce{Sr^2+}$, and $\ce{Pb^2+}$. The doubly charged sulfates are usually less soluble than halides and nitrates.
• Most singly charge cations $\ce{K+}$, $\ce{Na+}$, $\ce{NH4+}$ form soluble salts. However, $\ce{K3Co(NO2)6}$ and $\ce{(NH4)3Co(NO2)6}$ are insoluble.
These are handy rules for us to have if we deal with salts often. On the other hand, solubility is an important physical property of a substance, and these properties are listed in handbooks.
Chemical Separation of Metal Ions
Formation of crystals from a saturated solution is a heterogeneous equilibrium phenomenon, and it can be applied to separate various chemicals or ions in a solution. When solubilities of two metal salts are very different, they can be separated by precipitation. The Ksp values for various salts are valuable information, and some data are given in Table E3. In the first two examples, we show how barium and strontium can be separated as chromate.
Example $1$
The Ksp for strontium chromate is $3.6 \times 10^{-5}$ and the Ksp for barium chromate is $1.2 \times10^{-10}$. What concentration of potassium chromate will precipitate the maximum amount of either the barium or the strontium chromate from an equimolar 0.30 M solution of barium and strontium ions without precipitating the other?
Solution
Since the Ksp for barium chromate is smaller, we know that $\ce{BaCrO4}$ will form a precipitate first as $\ce{[CrO4^2- ]}$ increases so that Qsp for $\ce{BaCrO4}$ also increases from zero to Ksp of $\ce{BaCrO4}$, at which point, $\ce{BaCrO4}$ precipitates. As $\ce{[CrO4^2- ]}$ increases, $\ce{[Ba^2+]}$ decreases. Further increase of $\ce{[CrO4^2- ]}$ till Qsp for $\ce{SrCrO4}$ increases to Ksp of $\ce{SrCrO4}$; it then precipitates.
Let us write the equilibrium equations and data down to help us think. Let $x$ be the concentration of chromate to precipitate $\ce{Sr^2+}$, and $y$ be that to precipitate $\ce{Ba^2+}$:
$\ce{SrCrO4(s) \rightarrow Sr^{2+}(aq) + CrO4^{2-}(aq)} \nonumber$
According to the definition of Ksp we have we have $K_{\ce{sp}} = (0.30)(x) = 3.6 \times 10^{-5}$. Solving for $x$ gives
$x = \dfrac{3.6 \times 10^{-5}}{0.30} = 1.2 \times 10^{-4} M \nonumber$
Further, let $y$ be the concentration of chromate to precipitate precipitate $\ce{Ba^2+}$:
$\ce{BaCrO4(s) \rightarrow Ba^{2+}(aq) + CrO4^{2-}(aq)} \nonumber$
with $K_{\ce{sp}} = (0.30)(y) = 1.2 \times 10^{-10}$. Solving for $y$ gives
$y = \dfrac{1.2 \times 10^{-10}}{0.30} = 4.0 \times 10^{-10} \;M \nonumber$
The Ksp's for the two salts indicate $\ce{BaCrO4}$ to be much less soluble, and it will precipitate before any $\ce{SrCrO4}$ precipitates. If chromate concentration is maintained less than $1.2 \times 10^{-4} M$, then all $\ce{Sr^2+}$ ions will remain in the solution.
Discussion
In reality, controling the increase of $\ce{[CrO4^2- ]}$ is very difficult.
Example $2$
The Ksp for strontium chromate is $3.6\times 10^{-5}$ and the Ksp for barium chromate is $1.2\times 10^{-10}$. Potassium chromate is added a small amount at a time to first precipitate $\ce{BaCrO4}$. Calculate $\ce{[Ba^2+]}$ when the first trace of $\ce{SrCrO4}$ precipitate starts to form in a solution that contains 0.30 M each of $\ce{Ba^2+}$ and $\ce{Sr^2+}$ ions.
Solution
From the solution given in Example $1$, $\ce{[CrO4^2- ]} = 3.6\times 10^{-4}\; M$ when $\ce{SrCrO4}$ starts to form. At this concentration, the $\ce{[Ba^2+]}$ is estimated at $3.6 \times 10^{-4} = 1.2\times 10^{-10}$.
The Ksp of $\ce{BaCrO4}$.
Thus,
$\ce{[Ba^2+]} = 3.33 \times 10^{-7}\, M \nonumber$
Very small indeed, compared to 0.30. In the fresh precipitate of $\ce{SrCrO4}$, the molar ratio of $\ce{SrCrO4}$ to $\ce{BaCrO4}$ is
$\dfrac{0.30}{3.33 \times 10^{-7}} = 9.0 \times 10^{5}. \nonumber$
Hence, the amount of $\ce{Ba^2+}$ ion in the solid is only $1 \times 10^{-6}$ (i.e., 1 ppm) of all metal ions, providing that all the solid was removed when
$\ce{[CrO4^{2-}]} = 3.6 \times 10^{-4} M. \nonumber$
Discussion
The calculation shown here indicates that the separation of $\ce{Sr}$ and $\ce{Ba}$ is pretty good. In practice, an impurity level of 1 ppm is a very small value.
Example $3$
What reagent should you use to separate silver and lead ions that are present in a solution? What data or information will be required for this task?
Solution
The Ksp's for salts of silver and lead are required. We list the Ksp's for chlorides and sulfates in a table here. These value are found in the Handbook Menu of our website as Salts Ksp.
Solutions to Example 17.6.3
Salt Ksp Salt Ksp
$\ce{AgCl}$ $1.8 \times 10^{-10}$ $\ce{Ag2SO4}$ $1.4\times 10^{-5}$
$\ce{Hg2Cl2}$ $1.3\times 10^{-18}$ $\ce{BaSO4}$ $1.1\times 10^{-10}$
$\ce{PbCl2}$ $1.7 \times 10^{-5}$ $\ce{CaSO4}$ $2.4\times 10^{-5}$
$\ce{PbSO4}$ $6.3\times 10^{-7}$
$\ce{SrSO4}$ $3.2\times 10^{-7}$
Because the Ksp's $\ce{AgCl}$ and $\ce{PbCl2}$ are very different, chloride, $\ce{Cl-}$, apppears a good choice of negative ions for their separation.
The literature also indicates that $\ce{PbCl2}$ is rather soluble in warm water, and by heating the solution to 350 K (80oC), you can keep $\ce{Pb^2+}$ ions in solution and precipitate $\ce{AgCl}$ as a solid. The solubility of $\ce{AgCl}$ is very small even at high temperatures.
Discussion
Find more detailed information about the solubility of lead chloride as a function of temperature.
Can sulfate be used to separate silver and lead ions? Which one will form a precipitate first as the sulfate ion concentration increases? What is the $\ce{[Pb^2+]}$ when $\ce{Ag2SO4}$ begins to precipitate in a solution that contains 0.10 M $\ce{Ag+}$?
The Separation of Two Ions by a Difference in Solubility: The Separation of Two Ions by a Difference in Solubility(opens in new window) [youtu.be] | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/17%3A_Additional_Aspects_of_Aqueous_Equilibria/17.06%3A_Precipitation_and_Separation_of_Ions.txt |
Learning Objectives
• To know how to separate metal ions by selective precipitation.
• To understand how several common metal cations can be identified in a solution using selective precipitation.
The composition of relatively complex mixtures of metal ions can be determined using qualitative analysis, a procedure for discovering the identity of metal ions present in the mixture (rather than quantitative information about their amounts). The procedure used to separate and identify more than 20 common metal cations from a single solution consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all of the metal ions are precipitated, as illustrated in Figure $1$.
Group 1: Insoluble Chlorides
Most metal chloride salts are soluble in water; only $\ce{Ag^{+}}$, $\ce{Pb^{2+}}$, and $\ce{Hg2^{2+}}$ form chlorides that precipitate from water. Thus the first step in a qualitative analysis is to add about 6 M $\ce{HCl}$, thereby causing $\ce{AgCl}$, $\ce{PbCl2}$, and/or $\ce{Hg2Cl2}$ to precipitate. If no precipitate forms, then these cations are not present in significant amounts. The precipitate can be collected by filtration or centrifugation.
Group 2: Acid-Insoluble Sulfides
Next, the acidic solution is saturated with $\ce{H2S}$ gas. Only those metal ions that form very insoluble sulfides, such as $\ce{As^{3+}}$, $\ce{Bi^{3+}}$, $\ce{Cd^{2+}}$, $\ce{Cu^{2+}}$, $\ce{Hg^{2+}}$, $\ce{Sb^{3+}}$, and $\ce{Sn^{2+}}$, precipitate as their sulfide salts under these acidic conditions. All others, such as $\ce{Fe^{2+}}$ and $\ce{Zn^{2+}}$, remain in solution. Once again, the precipitates are collected by filtration or centrifugation.
Group 3: Base-Insoluble Sulfides (and Hydroxides)
Ammonia or $\ce{NaOH}$ is now added to the solution until it is basic, and then $\ce{(NH4)2S}$ is added. This treatment removes any remaining cations that form insoluble hydroxides or sulfides. The divalent metal ions $\ce{Co^{2+}}$, $\ce{Fe^{2+}}$, $\ce{Mn^{2+}}$, $\ce{Ni^{2+}}$, and $\ce{Zn^{2+}}$ precipitate as their sulfides, and the trivalent metal ions $\ce{Al^{3+}}$ and $\ce{Cr^{3+}}$ precipitate as their hydroxides: $\ce{Al(OH)3}$ and $\ce{Cr(OH)3}$. If the mixture contains $\ce{Fe^{3+}}$, sulfide reduces the cation to $\ce{Fe^{2+}}$, which precipitates as $\ce{FeS}$.
Group 4: Insoluble Carbonates or Phosphates
The next metal ions to be removed from solution are those that form insoluble carbonates and phosphates. When $\ce{Na2CO3}$ is added to the basic solution that remains after the precipitated metal ions are removed, insoluble carbonates precipitate and are collected. Alternatively, adding $\ce{(NH4)2HPO4}$ causes the same metal ions to precipitate as insoluble phosphates.
Group 5: Alkali Metals
At this point, we have removed all the metal ions that form water-insoluble chlorides, sulfides, carbonates, or phosphates. The only common ions that might remain are any alkali metals ($\ce{Li^{+}}$, $\ce{Na^{+}}$, $\ce{K^{+}}$, $\ce{Rb^{+}}$, and $\ce{Cs^{+}}$) and ammonium ($\ce{NH4^{+}}$). We now take a second sample from the original solution and add a small amount of $\ce{NaOH}$ to neutralize the ammonium ion and produce $\ce{NH3}$. (We cannot use the same sample we used for the first four groups because we added ammonium to that sample in earlier steps.) Any ammonia produced can be detected by either its odor or a litmus paper test. A flame test on another original sample is used to detect sodium, which produces a characteristic bright yellow color. The other alkali metal ions also give characteristic colors in flame tests, which allows them to be identified if only one is present.
Metal ions that precipitate together are separated by various additional techniques, such as forming complex ions, changing the pH of the solution, or increasing the temperature to redissolve some of the solids. For example, the precipitated metal chlorides of group 1 cations, containing $\ce{Ag^{+}}$, $\ce{Pb^{2+}}$, and $\ce{Hg2^{2+}}$, are all quite insoluble in water. Because $\ce{PbCl2}$ is much more soluble in hot water than are the other two chloride salts, however, adding water to the precipitate and heating the resulting slurry will dissolve any $\ce{PbCl2}$ present. Isolating the solution and adding a small amount of $\ce{Na2CrO4}$ solution to it will produce a bright yellow precipitate of $\ce{PbCrO4}$ if $\ce{Pb^{2+}}$ were in the original sample (Figure $2$).
As another example, treating the precipitates from group 1 cations with aqueous ammonia will dissolve any $\ce{AgCl}$ because $\ce{Ag^{+}}$ forms a stable complex with ammonia: $\ce{[Ag(NH3)2]^{+}}$. In addition, $\ce{Hg2Cl2}$ disproportionates in ammonia.
$\ce{2Hg2^{2+} \rightarrow Hg + Hg^{2+}} \nonumber$
to form a black solid that is a mixture of finely divided metallic mercury and an insoluble mercury(II) compound, which is separated from solution:
$\ce{Hg2Cl2(s) + 2NH3(aq) \rightarrow Hg(l) + Hg(NH_2)Cl(s) + NH^{+}4(aq) + Cl^{−}(aq)} \nonumber$
Any silver ion in the solution is then detected by adding $\ce{HCl}$, which reverses the reaction and gives a precipitate of white $\ce{AgCl}$ that slowly darkens when exposed to light:
$\ce{[Ag(NH3)2]^{+} (aq) + 2H^{+}(aq) + Cl^{−}(aq) \rightarrow AgCl(s) + 2NH^{+}4(aq)} \nonumber$
Similar but slightly more complex reactions are also used to separate and identify the individual components of the other groups.
Summary
In qualitative analysis, the identity, not the amount, of metal ions present in a mixture is determined. The technique consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all the metal ions are precipitated. Other additional steps are needed to separate metal ions that precipitate together. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/17%3A_Additional_Aspects_of_Aqueous_Equilibria/17.07%3A_Qualitative_Analysis_for_Metallic_Elements.txt |
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
17.2: Buffered Solutions
Conceptual Problems
1. Explain why buffers are crucial for the proper functioning of biological systems.
2. What is the role of a buffer in chemistry and biology? Is it correct to say that buffers prevent a change in $[H_3O^+]$? Explain your reasoning.
3. Explain why the most effective buffers are those that contain approximately equal amounts of the weak acid and its conjugate base.
4. Which region of the titration curve of a weak acid or a weak base corresponds to the region of the smallest change in pH per amount of added strong acid or strong base?
5. If you were given a solution of sodium acetate, describe two ways you could convert the solution to a buffer.
6. Why are buffers usually used only within approximately one pH unit of the $pK_a$ or $pK_b$ of the parent weak acid or base?
7. The titration curve for a monoprotic acid can be divided into four regions: the starting point, the region around the midpoint of the titration, the equivalence point, and the region after the equivalence point. For which region would you use each approach to describe the behavior of the solution?
1. a buffer
2. a solution of a salt of a weak base
3. a solution of a weak acid
4. diluting a strong base
8. Which of the following will produce a buffer solution? Explain your reasoning in each case.
1. mixing 100 mL of 0.1 M $HCl$ and 100 mL of 0.1 M sodium fluoride
2. mixing 50 mL of 0.1 M $HCl$ and 100 mL of 0.1 M sodium fluoride
3. mixing 100 mL of 0.1 M hydrofluoric acid and 100 mL of 0.1 M $HCl$
4. mixing 100 mL of 0.1 M hydrofluoric acid and 50 mL of 0.1 M $NaOH$
5. mixing 100 mL of 0.1 M sodium fluoride and 50 mL of 0.1 M $NaOH$.
9. Which of the following will produce a buffer solution? Explain your reasoning in each case.
1. mixing 100 mL of 0.1 M $HCl$ and 100 mL of 0.1 M sodium acetate
2. mixing 50 mL of 0.1 M $HCl$ and 100 mL of 0.1 M sodium acetate
3. mixing 100 mL of 0.1 M acetic acid and 100 mL of 0.1 M $NaOH$
4. mixing 100 mL of 0.1 M acetic acid and 50 mL of 0.1 M $NaOH$
5. mixing 100 mL of 0.1 M sodium acetate and 50 mL of 0.1 M acetic acid
10. Use the definition of Kb for a weak base to derive the following expression, which is analogous to the Henderson-Hasselbalch approximation but for a weak base (B) rather than a weak acid (HA):
$pOH=pK_b−\log\left(\dfrac{[base]}{[acid]}\right)$
11. Why do biological systems use overlapping buffer systems to maintain a constant pH?
12. The $CO_2/HCO_3^−$ buffer system of blood has an effective $pK_a$ of approximately 6.1, yet the normal pH of blood is 7.4. Why is $CO_2/HCO_3^−$ an effective buffer when the $pK_a$ is more than 1 unit below the pH of blood? What happens to the pH of blood when the $CO_2$ pressure increases? when the $O_2$ pressure increases?
13. Carbon dioxide produced during respiration is converted to carbonic acid ($H_2CO_3$). The $pK_a1$ of carbonic acid is 6.35, and its $pK_a2$ is 10.33. Write the equations corresponding to each pK value and predict the equilibrium position for each reaction.
Answer
1. Buffers are crucial for the proper function of biological systems because they help maintain a relatively constant $pH$ that maintains homeostasis and this is required for bodily function. For example, buffers can prevent stomach acid from being too acidic in which tissues and other parts of the body would be harmed.
2. The role of a buffer in chemistry and biology is to resist pH changes upon the addition of an acid or base. It is correct to say that buffers prevent a change in hydronium ion concentration because the pH is a measure of the acidity or alkalinity of a solution.
3. The most effective buffers are those that contain approximately equal amounts of the weak acid and its conjugate base because when placed in the Henderson-Hasselbalch equation the $pH$ approximately equals the $pK_{a}$. This is the goal because the more the ratio needs to differ to achieve the desired pH, the less effective the buffer and it should not differ by more than tenfold.
4. The region after the equivalence point on a titration curve of a weak acid or a weak base corresponds to the smallest change in pH per amount of added strong acid or strong base. This is because the strong acid or strong base completely neutralized the weak acid or weak base in which the pH remains constant relative to the strong acid or strong base.
5. Sodium acetate is a salt, therefore it can’t be regarded as an acid or a base. One way to convert the solution to a buffer is to combine sodium acetate with a weak acid such as acetic acid. The sodium acetate would react with the hydronium ions and the acetic acid would react with hydroxide ions to prevent drastic changes in pH. Another way to convert the solution to a buffer is by titrating a weak acid and strong base or strong base and weak acid. For example, combining acetic acid to sodium hydroxide to produce sodium acetate (a neutralization reaction).
6.
Buffers are usually used only within approximately one $pH$ unit of the $pK_a$ or $pK_b$ of the parent weak acid or base because the ability of a buffer solution to maintain a nearly constant pH due to a small amount of acid or base is greatest at the $pK_a$ and decreases as the pH of the solution goes above or below the $pK_a$.
$pH=pK_a+log(\frac{[B]}{[A]})$
When $\frac{[B]}{[A]}=10$: $pH=pK_a+1$
When $\frac{[B]}{[A]}=\frac{1}{10}$: $pH=pK_a-1$
7.
a. I would use the region around the midpoint of the titration to describe the behavior of a buffer because the midpoint of the titration is defined as the point at which exactly enough acid or base has been added to neutralize one-half of the acid or the base originally present while maintaining a relatively constant pH.
b. I would use the equivalence point region to describe the behavior of a solution of a salt of a weak base because at this point the amount of titrant added is enough to completely neutralize the solution.
c. I would use the starting point region of the titration curve to describe the behavior of the solution of a weak acid because it is expected to be at a higher pH than a strong acid but lower than $pH$ 7.
d. I would use the region after the equivalence point to describe the dilution of a strong base because in this region one can predict the pH by simply taking into account the amount of excess base.
8.
a. This will not produce a buffer solution because the hydrochloric acid completely neutralizes the sodium fluoride to give sodium chloride.
b. This will produce a buffer solution because the hydrochloric acid neutralizes only half of the acetic acid to give a solution containing equal amounts of hydrofluoric acid and sodium chloride.
c. This will not produce a buffer solution because hydrofluoric acid is a weak acid and hydrochloric acid is a strong acid.
d. This will produce a buffer solution because the solution will contain a 2:1 ratio of hydrofluoric acid and sodium hydroxide.
e. This will not produce a buffer solution solution because sodium fluoride is a weak base and sodium hydroxide is a strong base.
9.
a. This will not produce a buffer because the hydrochloric acid completely neutralizes the sodium acetate to give acetic acid and sodium chloride.
b. This will produce a buffer because the HCl neutralizes only half of the sodium acetate to give a solution containing equal amounts of acetic acid and sodium acetate.
c. This will not produce a buffer because the sodium hydroxide completely neutralizes the acetic acid to give sodium acetate.
d. This will produce a buffer because the sodium hydroxide neutralizes only half of the acetic acid to give a solution containing equal amounts of acetic acid and sodium acetate.
e. This will produce a buffer because the solution will contain a 2:1 ratio of sodium acetate and acetic acid.
10.
For $pH$:
$HA \overset{K_a}{\rightleftharpoons} A^{-}+H^{+}$
$K_a= \frac{[H^{+}][A^{-}]}{[HA]}$
$log(K_a)=log([H^{+}])+log(\frac{[A^{-}]}{[HA]})$
$-pK_a=-pH+log(\frac{[A^{-}]}{[HA]})$
$pH=pK_a+log(\frac{[A^{-}]}{[HA]})$
For pOH:
$A^{-} \overset{K_b}{\rightleftharpoons} HA+OH^{-}$
$K_b= \frac{[HA][OH^{-}]}{[A^{-}]}$
$log(K_b)=log([OH^{-}])+log(\frac{[HA]}{[A^{-}]})$
$-pK_b=-pOH-log(\frac{[A^{-}]}{[HA]})$
$pOH=pK_b-log(\frac{[A^{-}]}{[HA]})$
Helpful Equations:
$pK_a=-log(K_a) \rightarrow -pK_a=log(K_a)$
$-log([H^{+}])=pH \rightarrow log([H^{+}])=-pH$
$pK_b=-log(K_b) \rightarrow -pK_a=log(K_b)$
$-log([OH^{-}])=pOH \rightarrow log([OH^{-}])=-pOH$
11. Biological systems use overlapping buffer systems to maintain a constant pH because no single buffer system can effectively maintain a constant pH value over the entire physiological range of approximately $pH$ 5.0 to 7.4.
12. $CO_2/HCO_3^−$ is an effective buffer when the $pK_a$ is more than 1 unit below the $pH$ of blood because the kidneys help control acid-base balance by excreting hydrogen ions and generating bicarbonate that helps maintain blood plasma $pH$ within the normal range. The bicarbonate ion can combine with a proton to form carbonic acid and the more protons are absorbed from the solution the more the $pH$ increases, resulting in hemoglobin absorbing oxygen. Carbonic acid, which can be formed from carbon dioxide and water, can dissociate into a proton and bicarbonate ion which increases the hydrogen ion concentration and lowers the blood $pH$, resulting in hemoglobin proteins releasing oxygen. The carbonic acid, which can be formed from bicarbonate, is converted to carbon dioxide and water via a very fast enzymatic reaction. Carbon dioxide, being volatile, can be rapidly expelled from the body at varying rates by respiration.
13.
$H_{2}CO_{3} \rightleftharpoons HCO_{3}^{-}+H^{+}$ $pK_{a1}=6.35$ left
$HCO_{3}^{-} \rightleftharpoons CO_{3}^{2-}+H^{+}$ $pK_{a2}=10.33$ right
Numerical Problems
1.Benzenesulfonic acid ($pKa = 0.70$) is synthesized by treating benzene with concentrated sulfuric acid. Calculate the following:
1. the $pH$ of a 0.286 M solution of benzenesulfonic acid
2. the pH after adding enough sodium benzenesulfonate to give a final benzenesulfonate concentration of 0.100 M
2. Phenol has a $pK_a$ of 9.99. Calculate the following:
1. the $pH$ of a 0.195 M solution
2. the percent increase in the concentration of phenol after adding enough solid sodium phenoxide (the sodium salt of the conjugate base) to give a total phenoxide concentration of 0.100 M
3. Salicylic acid is used in the synthesis of acetylsalicylic acid, or aspirin. One gram dissolves in 460 mL of water to create a saturated solution with a $pH$ of 2.40.
1. What is the $K_a$ of salicylic acid?
2. What is the final pH of a saturated solution that is also 0.238 M in sodium salicylate?
3. What is the final pH if 10.00 mL of 0.100 M HCl are added to 150.0 mL of the buffered solution?
4. What is the final pH if 10.00 mL of 0.100 M NaOH are added to 150.0 mL of the buffered solution?
4. An intermediate used in the synthesis of perfumes is valeric acid, also called pentanoic acid. The $pK_a$ of pentanoic acid is 4.84 at 25°C.
1. What is the $pH$ of a 0.259 M solution of pentanoic acid?
2. Sodium pentanoate is added to make a buffered solution. What is the pH of the solution if it is 0.210 M in sodium pentanoate?
3. What is the final $pH$ if 8.00 mL of 0.100 M HCl are added to 75.0 mL of the buffered solution?
4. What is the final $pH$ if 8.00 mL of 0.100 M NaOH are added to 75.0 mL of the buffered solution?
Answer
1.
a. $pH=-log([H^{+}])=-log(0.2388)=0.62$
$K_a=\frac{[H^{+}][A^{-}]}{[HA]} \rightarrow K_a=\frac{[H^{+}]^{2}}{[HA]} \rightarrow [H^{+}]= \sqrt{K_a\times[HA]}=\sqrt{10^{-0.70} \times 0.286\,M}=0.2388$
b. $pH=-log([H^{+}])=-log(0.14125)=0.850$
$K_a=\frac{[H^{+}][A^{-}]}{[HA]} \rightarrow K_a=\frac{[H^{+}]^{2}}{[HA]} \rightarrow [H^{+}]= \sqrt{K_a\times[HA]}=\sqrt{10^{-0.70} \times 0.100\,M}=0.14125$
2.
a. $pH=-log(4.46 \times 10^{-6})=5.35$
$K_a=\frac{[C_6H_5O^{-}][H_3O^{+}]}{[C_6H_5OH]}=\frac{x^2}{0.195} \rightarrow 10^{-9.99}=\frac{x^2}{0.195} \rightarrow x=4.46 \times 10^{-6}$
$Assumption\,Valid: \frac{4.46 \times 10^{-6}}{0.195\,M} \times 100 \% <5 \%$
$C_6H_5OH$
$H_2O$
$C_6H_5O^{-}$
$H_3O^{+}$
I
0.195
-
0
0
C
-x
-
+x
+x
E
0.195-x
-
x
x
b. $Percent\,increase=\frac{3.20 \times 10^{-6}}{4.46 \times 10^{-6}}x100 \% =71.6 \%$
$K_a=\frac{[C_6H_5O^{-}][H_3O^{+}]}{C_6H_5OH}=\frac{x^2}{0.100} \rightarrow 10^{-9.99}=\frac{x^2}{0.100} \rightarrow x=3.20 \times 10^{-6}$
3.
1. $K_a=\frac{[C_9H_8O_4][H_3O^{+}]}{[C_7H_6O_3]}=\frac{x^{2}}{0.015739} \rightarrow \frac{(3.98 \times 10^{-3})^{2}}{0.015739}= 1.0 \times 10^{-3}$
$C_{7}H_{6}O_{3}:1\,g \times \frac{1\,mol}{138.121\,g\,mol} \times \frac{1}{460\,mL \times \frac {1\,L}{1000\,mL}}=0.015739\,M$
$pH=-log([H^{+}]) \rightarrow [H^{+}]=10^{-pH}=10^{-2.40}=3.98 \times 10^{-3}$
2. $pH=pK_a+log(\frac{[C_{7}H_{5}NaO_{3}]}{[C_{7}H_{6}O_{3}]})=-log(1.01 \times 10^{-3})+log(\frac{0.238}{0.015739})=4.17$
3. $pH=pK_a+log(\frac{mol\,C_{7}H_{5}NaO_{3}-mol\,HCl}{mol\,C_9H_8O_4+mol\,HCl})=-log(1.01 \times 10^{-3})+log(\frac{0.0357-0.001}{0.00236085+0.001})=4.01$
$HCl:.01\,L \times 0.100\,M=0.001\,mol$
$C_{7}H_{5}NaO_{3}: 0.15\,L \times 0.238\,M=0.0357\,mol$
$C_{7}H_{6}O_{3}:0.15\,L \times 0.015739\,M=0.00236085\,mol$
4. $pH=pK_a+log(\frac{mol\,C_{7}H_{5}NaO_{3}+mol\,NaOH}{mol\,C_9H_8O_4-mol\,NaOH})=-log(1.01 \times 10^{-3})+log(\frac{0.0357+0.001}{0.00236085-0.001})=4.43$
$NaOH:.01\,L \times 0.100\,M=0.001\,mol$
$C_{7}H_{5}NaO_{3}: 0.15\,L \times 0.238\,M=0.0357\,mol$
$C_{7}H_{6}O_{3}:0.15\,L \times 0.015739\,M=0.00236085\,mol$
4.
a. $pH=-log([H^{+}])=-log(3.7 \times 10^{-6})=5.43$
$K_a=\frac{[H^{+}][A^{-}]}{[HA]} \rightarrow K_a=\frac{[H^{+}]^{2}}{[HA]} \rightarrow [H^{+}]=\sqrt{K_a\times[HA]}=\sqrt{10^{-4.84} \times 0.259\,M}=3.7 \times 10^{-6}$
b. $pH=pK_a+log(\frac{B}{A})=4.84+log(\frac{0.210}{0.259})=4.75$
c. $pH=pK_a+log(\frac{B}{A})=4.84+log(\frac{mol\,C_{5}H_{9}NaO_{2}-mol\,HCl}{mol\,C_{5}H_{10}O_{2}+mol\,HCl})=4.84+log(\frac{0.01575-8 \times 10^{-4}}{0.019425+8\times 10^{-4}})=4.71$
$HCl:8.00\,mL \times \frac{1\,L}{1,000\,mL} \times 0.100\,M=8 \times 10^{-4}\,mol$
$C_{5}H_{9}NaO_{2}:75.0\,mL \times \frac{1\,L}{1,000\,mL} \times 0.210\,M=0.01575\,mol$
$C_{5}H_{10}O_{2}: 75.0 \,mL \times \frac{1\,L}{1,000\,mL} \times 0.259\,M=0.019425\,mol$
d. $pH=pK_a+log(\frac{B}{A})=4.84+log(\frac{mol\,C_{5}H_{9}NaO_{2}+mol\,NaOH}{mol\,C_{5}H_{10}O_{2}-mol\,NaOH})=4.84+log(\frac{0.01575+8 \times 10^{-4}}{0.019425-8\times 10^{-4}})=6.07$
$NaOH:8.00\,mL \times \frac{1\,L}{1,000\,mL} \times 0.100\,M=8 \times 10^{-4}\,mol$
$C_{5}H_{9}NaO_{2}:75.0\,mL \times \frac{1\,L}{1,000\,mL} \times 0.210\,M=0.01575\,mol$
$C_{5}H_{10}O_{2}: 75.0 \,mL \times \frac{1\,L}{1,000\,mL} \times 0.259\,M=0.019425\,mol$
17.3: Acid-Base Titrations
Conceptual Problems
1. Why is the portion of the titration curve that lies below the equivalence point of a solution of a weak acid displaced upward relative to the titration curve of a strong acid? How are the slopes of the curves different at the equivalence point? Why?
2. Predict whether each solution will be neutral, basic, or acidic at the equivalence point of each titration.
1. An aqueous solution of $NaOH$ is titrated with 0.100 M $HCl$.
2. An aqueous solution of ethylamine ($CH_3CH_2NH_2$) is titrated with 0.150 M $HNO_{3}$
3. An aqueous solution of aniline hydrochloride ($C_6H_5NH_3^+Cl^−$) is titrated with 0.050 M $KOH$.
3. The pKa values of phenol red, bromophenol blue, and phenolphthalein are 7.4, 4.1, and 9.5, respectively. Which indicator is best suited for each acid–base titration?
1. titrating a solution of $Ba(OH)_2$ with 0.100 M $HCl$
2. titrating a solution of trimethylamine ($Me_3N$) with 0.150 M $HNO_3$
3. titrating a solution of aniline hydrochloride ($C_6H_5NH_3^+Cl^−$) with 0.050 M $KOH$
4. For the titration of any strong acid with any strong base, the $pH$ at the equivalence point is 7.0. Why is this not usually the case in titrations of weak acids or weak bases?
5. Why are the titration curves for a strong acid with a strong base and a weak acid with a strong base identical in shape above the equivalence points but not below?
6. Describe what is occurring on a molecular level during the titration of a weak acid, such as acetic acid, with a strong base, such as $NaOH$, at the following points along the titration curve. Which of these points corresponds to $pH=pK_{a}$?
1. at the beginning of the titration
2. at the midpoint of the titration
3. at the equivalence point
4. when excess titrant has been added
7. On a molecular level, describe what is happening during the titration of a weak base, such as ammonia, with a strong acid, such as $HCl$, at the following points along the titration curve. Which of these points corresponds to $pOH=pK_{b}$?
1. at the beginning of the titration
2. at the midpoint of the titration
3. at the equivalence point
4. when excess titrant has been added
8. For the titration of a weak acid with a strong base, use the $K_{a}$ expression to show that $pH=pK_{a}$ at the midpoint of the titration.
9. Chemical indicators can be used to monitor $pH$ rapidly and inexpensively. Nevertheless, electronic methods are generally preferred. Why?
10. Why does adding ammonium chloride to a solution of ammonia in water decrease the pH of the solution?
11. Given the equilibrium system $CH_3CO_2H\,(aq) \rightleftharpoons CH_3CO_2^{-}\,(aq) + H^{+}\,(aq)$, explain what happens to the position of the equilibrium and the $pH$ in each case.
1. Dilute $HCl$ is added.
2. Dilute $NaOH$ is added.
3. Solid sodium acetate is added.
12. Given the equilibrium system $CH_3NH_2\,(aq) + H_2O\,(l) \rightleftharpoons CH_{3}NH_3^{+}\,(aq) + OH^{-}\,(aq)$, explain what happens to the position of the equilibrium and the $pH$ in each case.
1. Dilute $HCl$ is added.
2. Dilute $NaOH$ is added.
3. Solid $CH_3NH_3^{+}Cl^{−}$ is added.
Answers
1. The portion of the titration curve that lies below the equivalence point of a solution of a weak acid is displaced upward relative to the titration curve of a strong acid because the starting point of a weak acid is at a higher $pH$ compared to that of a strong acid. The slope for a strong acid is greater compared to that of a weak acid because when a weak acid is neutralized, the solution that remains is basic because the acid’s conjugate base remains in solution.
2.
a. The solution will be neutral at the equivalence point because sodium hydroxide is a strong base and hydrochloric acid is a strong acid.
b. The solution will be acidic at the equivalence point because ethylamine is a weak base and nitric acid is a strong acid.
c. The solution will be basic at the equivalence point because aniline hydrochloride is a weak acid and potassium hydroxide is a strong base.
3.
a. Barium hydroxide is a strong base and hydrochloric acid is a strong acid thus phenol red with the $pK_a=7.4$ is best suited for the reaction.
b. Trimethylamine is a weak base and nitric acid is a strong acid thus bromophenol blue with the $pK_a=4.1$ is best suited for the reaction.
c. Aniline hydrochloride is a weak acid and potassium hydroxide is a strong base thus phenolphthalein with $pK_a=9.5$ is best suited for the reaction.
4. This is not usually the case in the titration of weak acids and weak bases because weak acids and weak bases would only ionize partially, thus complete neutralization does not occur.
5. The titration curve for a strong acid with a strong base is identical in shape above the equivalence point of the titration curve with a weak acid and strong base because both cases involve the addition of strong base. Not only does the strong base completely neutralize the acid but there is an excess of a strong base that makes the solution basic.
6.
a. The titration begins with a $pH$ is higher than the titration of a strong acid. At the beginning of the titration, there is a sharp increase in $pH$ because the anion of the weak acid becomes a common ion that reduces the ionization of the acid.
b. There is a sharp increase at the beginning of the titration that changes gradually due to the solution becoming a buffer. This continues until the strong base has overcome the buffer capacity. At the midpoint of the titration, the concentration of the weak acid is equal to the concentration of its conjugate base. This point is also known as half-neutralization because half the acid has been neutralized by the strong base.
c. At the equivalence point the pH is greater than 7 because the acid ($HA$) has been converted to its conjugate base ($A^{-}$) by sodium hydroxide and equilibrium moves backward toward the acid and produces hydroxide:
$A^{-}+H_2O \rightleftharpoons AH+OH^{-}$
d. When the excess titrant has been added the solution becomes basic because sodium hydroxide completely neutralized the weak acid.
7.
a. The titration begins with a $pH$ that is lower than the titration of a strong base but higher than the $pH=7$. At the beginning of the titration, there is a sharp decrease in $pH$ because the cation of the weak base becomes a common ion that reduces the ionization of the base.
b. The sharp decrease at the beginning of the titration changes gradually due to the solution becoming a buffer. This continues until the strong acid has overcome the buffer capacity. At the midpoint of the titration, the concentration of the weak base is equal to the concentration of its conjugate acid $pH=pK_a$. This point is also known as half-neutralization because half the base has been neutralized by the strong acid.
c. At the equivalence point the $pH$ is less than 7 because the base $NH_{3}$ has been converted to its conjugate acid $NH_{4}^{+}$ by hydrochloric acid and equilibrium moves forwards toward the base and produces $NH_{4}Cl^{-}$:
$NH_{3}+HCl \rightleftharpoons NH_{4}Cl$
d. When excess titrant has been added the solution becomes acidic because hydrochloric acid completely neutralized the weak base.
8.
$HA \overset{K_a}{\rightleftharpoons} A^{-}+H^{+}$
$K_a= \frac{[H^{+}][A^{-}]}{[HA]}$
$pH=pK_a+log(\frac{[A^{-}]}{[HA]}$
Assume $[A^{-}]=[HA]=1$
$pH=pK_a+log(1)$
$pH=pK_a+0$
$pH=pK_a$
9. Electronic methods are preferred over chemical indicators because it is a more accurate method at determining the pH. Also, chemical indicators must be selected to observe a narrow pH range while electronic methods observe a wider range.
10.
Ammonia is a weak base and dissociates in water as:
$NH_3\,(aq)+H_2O\,(l) \rightleftharpoons NH_4^{+}\,(aq)+OH^{-}\,(aq)$
When $NH_4Cl$ is added, it 100% dissociates into $NH_4^{+}$ and $Cl^{-}$ in water:
$NH_{4}Cl\,(aq) \rightleftharpoons NH_{4}^{+}\,(aq)+Cl^{-}\,(aq)$
Due to common ion, $NH_4^{+}\,(aq)$ the dissociation of $NH_{3}$ will decrease. Thus, the concentration of $OH^{-}\,(aq)$ decreases and the concentration of $H_{3}O^{+}$ increases.
Since $pH=-log(H_{3}O^{+})$ the $pH$ and $H_{3}O^{+}$ are inversely related, the concentration of $H_{3}O^{+}$ concentration increases and $pH$ decreases.
11.
a. The position of equilibrium shift to the left and the pH decreases.
b. The position of equilibrium shifts to the right and the pH increases.
c. The position of equilibrium shifts to the left and the pH increases.
12.
a. The position of equilibrium shifts to the left and the pH decreases.
b. The position of equilibrium shifts to the left and the pH increases.
c. The position of equilibrium shifts to the left and the pH increases.
Numerical Problems
1. Calculate the pH of each solution.
1. A volume of 25.0 mL of 6.09 M HCl is added to 100.0 mL of distilled water
2. A volume of 5.0 mL of 2.55 M $NaOH$ is added to 75.0 mL of distilled water.
2. What is the pH of a solution prepared by mixing 50.0 mL of 0.225 M HCl with 100.0 mL of a 0.184 M solution of $NaOH$?
3. What volume of 0.50 M HCl is needed to completely neutralize 25.00 mL of 0.86 M $NaOH$?
4.Calculate the final pH when each pair of solutions is mixed.
1. 100 mL of 0.105 M HCl and 100 mL of 0.115 M sodium acetate
2. 50 mL of 0.10 M HCl and 100 mL of 0.15 M sodium acetate
3. 100 mL of 0.109 M acetic acid and 100 mL of 0.118 M $NaOH$
4. 100 mL of 0.998 M acetic acid and 50.0 mL of 0.110 M $NaOH$
5. Calculate the final pH when each pair of solutions is mixed.
1. 100 mL of 0.983 M HCl and 100 mL of 0.102 M sodium fluoride
2. 50.0 mL of 0.115 M HCl and 100 mL of 0.109 M sodium fluoride
3. 100 mL of 0.106 M hydrofluoric acid and 50.0 mL of 0.996 M $NaOH$
4. 100 mL of 0.107 M sodium acetate and 50.0 mL of 0.987 M acetic acid
6. Calcium carbonate is a major contributor to the “hardness” of water. The amount of CaCO3 in a water sample can be determined by titrating the sample with an acid, such as HCl, which produces water and CO2. Write a balanced chemical equation for this reaction. Generate a plot of solution pH versus volume of 0.100 M HCl added for the titration of a solution of 250 mg of CaCO3 in 200.0 mL of water with 0.100 M HCl; assume that the HCl solution is added in 5.00 mL increments. What volume of HCl corresponds to the equivalence point?
7. For a titration of 50.0 mL of 0.288 M $NaOH$, you would like to prepare a 0.200 M HCl solution. The only HCl solution available to you, however, is 12.0 M.
1. How would you prepare 500 mL of a 0.200 M HCl solution?
2. Approximately what volume of your 0.200 M HCl solution is needed to neutralize the $NaOH$ solution?
3. After completing the titration, you find that your “0.200 M” HCl solution is actually 0.187 M. What was the exact volume of titrant used in the neutralization?
8. While titrating 50.0 mL of a 0.582 M solution of HCl with a solution labeled “0.500 M KOH,” you overshoot the endpoint. To correct the problem, you add 10.00 mL of the HCl solution to your flask and then carefully continue the titration. The total volume of titrant needed for neutralization is 71.9 mL.
1. What is the actual molarity of your KOH solution?
2. What volume of titrant was needed to neutralize 50.0 mL of the acid?
9. Complete the following table and generate a titration curve showing the pH versus volume of added base for the titration of 50.0 mL of 0.288 M HCl with 0.321 M $NaOH$. Clearly indicate the equivalence point.
Base Added (mL) 10.0 30.0 40.0 45.0 50.0 55.0 65.0 75.0
pH
10. The following data were obtained while titrating 25.0 mL of 0.156 M $NaOH$ with a solution labeled “0.202 M HCl.” Plot the pH versus volume of titrant added. Then determine the equivalence point from your graph and calculate the exact concentration of your HCl solution.
Volume of HCl (mL) 5 10 15 20 25 30 35
pH 11.46 11.29 10.98 4.40 2.99 2.70 2.52
11. Fill in the data for the titration of 50.0 mL of 0.241 M formic acid with 0.0982 M KOH. The pKa of formic acid is 3.75. What is the pH of the solution at the equivalence point?
Volume of Base Added (mL) 0 5 10 15 20 25
pH
12. Glycine hydrochloride, which contains the fully protonated form of the amino acid glycine, has the following structure:
It is a strong electrolyte that completely dissociates in water. Titration with base gives two equivalence points: the first corresponds to the deprotonation of the carboxylic acid group and the second to loss of the proton from the ammonium group. The corresponding equilibrium equations are as follows:
$^{+}NH_{3}-CH_{2}-CO_{2}H\left ( aq \right ) \rightleftharpoons \;\;\;\;\; pK_{a1}=2.3$
$^{+}NH_{3}-CH_{2}-CO_{2}\left ( aq \right )+ H^{+}$
$^{+}NH_{3}-CH_{2}-CO_{2}\left ( aq \right ) \rightleftharpoons \;\;\;\;\; pK_{a2}=9.6$
$NH_{2}-CH_{2}-COO\left ( aq \right )+ H^{+}$
Given 50.0 mL of solution that is 0.430 M glycine hydrochloride, how many milliliters of 0.150 M KOH are needed to fully deprotonate the carboxylic acid group?
1. How many additional milliliters of KOH are needed to deprotonate the ammonium group?
2. What is the pH of the solution at each equivalence point?
3. How many milliliters of titrant are needed to obtain a solution in which glycine has no net electrical charge? The pH at which a molecule such as glycine has no net charge is its isoelectric point. What is the isoelectric point of glycine?
13. What is the pH of a solution prepared by adding 38.2 mL of 0.197 M HCl to 150.0 mL of 0.242 M pyridine? The pKb of pyridine is 8.77.
14. What is the pH of a solution prepared by adding 40.3 mL of 0.289 M $NaOH$ to 150.0 mL of 0.564 M succinic acid ($HO_2CCH_2CH_2CO_2H$)? (For succinic acid, pKa1 = 4.21 and pKa2 = 5.64).
14. Calculate the pH of a 0.15 M solution of malonic acid ($HO_2CCH_2CO_2H$), whose pKa values are as follows: pKa1 = 2.85 and pKa2 = 5.70.
Answers
1.
a.$pH=-log([H_3O^{+}])=-log(1.218)=-0.086$
$[HCl]=[H^{+}]$
$M_{1}V_{1}=M_{2}V_{2} \rightarrow (6.09\,M)(25\,mL)=(x)(125\,mL) \rightarrow x=1.218$
b. $pH+pOH=14 \rightarrow pH=14-pOH=14-0.80=13$
$pOH=-log([OH^{-}])=-log(0.159)=0.80$
$[NaOH]=[OH^{-}]$
$M_{1}V_{1}=M_{2}V_{2} \rightarrow (2.55\,M)(5.0\,mL)=(x)(80\,mL) \rightarrow x=0.159$
2.
$pH+pOH=14 \rightarrow pH=14-pOH=14-1.321=12.7$
$HCl:50.0\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{0.225\,mol}{1\,L}=0.01125\,mol$
$NaOH:100.0\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{0.184\,mol}{1\,L}=0.0184\,mol$
Note the number of moles for $NaOH$ is greater than $HCl$, therefore we calculate $pOH$.
$0.0184-0.01125=0.00715\,mol$
$\frac{0.00715\,mol}{.15\,L}=0.0477$
$pOH=-log([OH^{-}])=-log(0.0477)=1.321$
3. $M_{1}V_{1}=M_{2}V_{2} \rightarrow (0.50\,M)(x\,L)=(0.86\,M)(0.25\,L) \rightarrow x\,L=4.3 \times 10^{-2}\,L$
4. Video Solution
a. $pH=pK_a+log(\frac{C_{2}H_{3}NaO_{2}}{C_{2}H_{4}O_{2}})=4.76+log(\frac{0.0935}{0.0105})=5.71$
$HCl: 100\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{0.0105\,mol}{1\,L}=0.0105\,mol$
$C_2H_3NaO_2: 100\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{0.115\,mol}{1\,L}=0.0115\,mol$
$C_{2}H_{3}NaO_{2}\,(s)+HCl\,(aq) \rightarrow C_{2}H_{4}O_{2}\,(aq)+NaCl\,(s)$
$C_{2}H_{3}NaO_{2}$
$HCl$
$C_{2}H_{4}O_{2}$
$NaCl$
I
0.0115
0.0105
0
-
C
-0.0105
-0.0105
+0.0105
-
E
0.0935
0
0.0105
-
b. $pH=pK_a+log(\frac{C_{2}H_{3}NaO_{2}}{C_{2}H_{4}O_{2}})=4.76+log(\frac{0.01}{0.005})=5.05$
$HCl: 100\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{0.0105\,mol}{1\,L}=0.005\,mol$
$C_2H_3NaO_2: 100\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{0.115\,mol}{1\,L}=0.015\,mol$
$C_{2}H_{3}NaO_{2}\,(s)+HCl\,(aq) \rightarrow C_{2}H_{4}O_{2}\,(aq)+NaCl\,(s)$
$C_{2}H_{3}NaO_{2}$
$HCl$
$C_{2}H_{4}O_{2}$
$NaCl$
I
0.015
0.005
0
0
C
-0.005
-0.005
+0.005
+0.005
E
0.01
0
0.005
0.005
c. $pH=14-pOH=14-(-log(\frac{0.0118-0.0109}{0.2\,L}))=11.7$
$CH_{3}COOH: 0.1\,L \times 0.109\,M=0.0109\,mol$
$NaOH: 0.1\,L \times 0.118\,M=0.0118\,mol$
$CH_{3}COOH\,(aq)+NaOH\,(aq) \rightarrow H_{2}O\,(l)+CH_{3}COO^{-}Na\,(aq)$
$CH_{3}COOH$
$NaOH$
$H_{2}O$
$CH_{3}COONa$
I
0.0109
0.0118
-
0
C
-0.0109
-0.0109
-
+0.0109
E
0
$9 \times 10^{-4}$
-
0.0109
d. $pH=4.76+log(\frac{0.0055}{0.0943})=3.53$
$CH_{3}COOH:0.1\,L \times 0.998\,M=0.0998\,mol$
$NaOH: 0.05\,L \times 0.110\,M=0.0055\,mol$
$CH_{3}COOH\,(aq)+NaOH\,(aq) \rightarrow H_{2}O\,(l)+CH_{3}COONa\,(aq)$
$CH_{3}COOH$
$NaOH$
$H_{2}O$
$CH_{3}COONa$
I
0.0998
0.0055
-
0
C
-0.0055
-0.0055
-
+0.0055
E
0.0943
0
-
0.0055
5. Video Solution
a. $pH=-log(\frac{0.9728}{0.2\,L})=4.86$
$HCl:100\,mL \times \frac{1\,L}{1,000\,L} \times0.983\,M=0.983\,mol$
$NaF:100\,mL \times \frac{1\,L}{1,000\,L} \times 0.102\,M=0.0102\,mol$
$HCl\,(aq)+NaF\,(s) \rightarrow NaCl\,(aq)+HF\,(aq)$
$HCl$
$NaF$
$NaCl$
$HF$
I
0.983
0.0102
-
0
C
-0.0102
-0.0102
-
-0.0102
E
0.9728
0
-
-0.0102
b. $pH=pK_a+log(\frac{[NaF]}{[HF]})=3.14+log(\frac{0.00515}{0.00575})=3.09$
$HCl:50\,mL \times \frac{1\,L}{1,000\,L} \times 0.115\,M=0.00575\,mol$
$NaF:100\,mL \times \frac{1\,L}{1,000\,L} \times 0.109\,M=0.0109\,mol$
$HCl\,(aq)+NaF\,(s) \rightarrow NaCl\,(aq)+HF\,(aq)$
$HCl$
$NaF$
$NaCl$
$HF$
I
0.00575
0.0109
-
0
C
-0.00575
-0.00575
-
+0.00575
E
0
0.00515
-
0.00575
c. $pH=14-pOH=14-(-log(0.0498-0.0106))=12.6$
$HF:100\,mL \times \frac{1\,L}{1,000\,L} \times 0.106\,M=0.0106\,mol$
$NaOH: 50\,mL \times \frac{1\,L}{1,000\,L} \times 0.996\,M=0.0498\,mol$
$HF\,(aq)+NaOH\,(aq) \rightarrow NaF\,(aq)+H_{2}O\,(aq)$
$HCl(aq)+NaOH(aq) \rightarrow NaCl\,(aq)+H_2O$
d.$pH=pK_a+log(\frac{[C_2H_3NaO_2]}{[CH_{3}COOH]})=4.76+log(\frac{0.0107}{0.04935})=4.10$
$C_2H_3NaO_2: 0.1\,L \times 0.107\,M=0.0107$
$CH_3COOH: 0.05\,mL \times 0.987\,M=0.04935$
$CH_{3}COOH\,(aq)+H_{2}O \rightarrow C_2H_3O^{-}+H_3O^{+}$
6.
$CaCO_{3}\,(aq)+2\,HCl\,(aq) \rightarrow CaCl_{2}\,(aq)+H_2O\,(l)+CO_{2}\,(g)$
$CO_{3}^{2-}\,(aq)+H^{+}\,(aq) \rightleftharpoons HCO_{3}^{-}\,(aq)$
$HCO_{3}^{2-}\,(aq)+H^{+}\,(aq) \rightleftharpoons H_{2}CO_{3}\,(aq)$
$K_{a1}=4.3 \times 10^{-7}$
$K_{a2}=5.0 \times 10^{-11}$
At 0 mL addition of $HCl$: $pH=14-pOH=14-(-log(\sqrt{0.0125 \times \frac{10^{-14}}{5.0 \times 10^{-11}}})=11.20$
$CaCO_{3}:250\,mg \times \frac{1\,g}{1,000\,mg} \times \frac{1\,mol}{100.0869\,g} \times \frac{1}{200\,mL \times \frac{1\,L}{1,000\,mL}}=0.0125\,M$
At 5 mL addition of $HCl$: $pH=-log(5 \times 10^{-11})+log(\frac{0.002}{0.0005})=10.9$
$CaCO_{3}:0.2\,L \times 0.0125\,M=0.0025\,mol$
$HCl: 0.005\,L \times 0.1\,M=0.0005\,mol$
$CO_{3}^{2-}$
$H^{+}$
$HCO_{3}^{-}$
I
0.0025
0.0005
0
C
-0.0005
-0.0005
0.0005
E
0.002
0
0.0005
At 10 mL addition of $HCl$: $pH=-log(5 \times 10^{-11})+log(\frac{0.001}{0.0015})=10.5$
$CaCO_{3}:0.2\,L \times 0.0125\,M=0.0025\,mol$
$HCl: 0.01\,L \times 0.1\,M=0.001\,mol$
$CO_{3}^{2-}$
$H^{+}$
$HCO_{3}^{-}$
I
0.0025
0.001
0
C
-0.001
-0.001
0.001
E
0.0015
0
0.001
At 15 mL addition of $HCl$:$pH=-log(5 \times 10^{-11})+log(\frac{0.001}{0.0015})=10.1$
$CaCO_{3}:0.2\,L \times 0.0125\,M=0.0025\,mol$
$HCl: 0.015\,L \times 0.1\,M=0.0015\,mol$
$CO_{3}^{2-}$
$H^{+}$
$HCO_{3}^{-}$
I
0.0025
0.0015
0
C
-0.0015
-0.0015
0.0015
E
0.001
0
0.0015
At 20 mL addition of $HCl$:$pH=-log(5 \times 10^{-11})+log(\frac{5 \times 10^{-4}}{0.002})=9.7$
$CaCO_{3}:0.2\,L \times 0.0125\,M=0.0025\,mol$
$HCl: 0.02\,L \times 0.1\,M=0.002\,mol$
$CO_{3}^{2-}$
$H^{+}$
$HCO_{3}^{-}$
I
0.0025
0.002
0
C
-0.002
-0.002
0.002
E
$5 \times 10^{-4}$
0
0.002
At 25 mL addition of $HCl$: first equivalence point: $pH=\frac{-log(K_{a1})-log(K_{a2})}{2}=\frac{-log(4.3 \times 10^{-7})-log(5.0 \times 10^{-11})}{2}=8.33$
$HCO_{3}^{-}:0.2\,L \times 0.0125\,M=0.0025\,mol$
$HCl: 0.025\,L \times 0.1\,M=0.0025\,mol$
$HCO_{3}^{-}$
$H^{+}$
$H_{2}CO_{3}$
I
0.0025
0.0025
0
C
-0.0025
-0.0025
0.0025
E
0
0
0.0025
At 30 mL addition of $HCl$: $pH=-log(4.3 \times 10^{-7})+log(\frac{0.0025}{0.0005})=7.07$
$HCO_{3}^{-}:0.2\,L \times 0.0125\,M=0.0025\,mol$
$HCl: 0.03\,L \times 0.1\,M=0.003$
$HCO_{3}^{-}$
$H^{+}$
$H_{2}CO_{3}$
I
0.0025
0.003
0
C
-0.0025
-0.0025
0.0025
E
0
$5 \times 10^{-4}$
0.0025
At 35 mL addition of $HCl$: $pH=-log(4.3 \times 10^{-7})+log(\frac{0.0025}{0.001})=6.76$
$HCO_{3}^{-}:0.2\,L \times 0.0125\,M=0.0025\,mol$
$HCl: 0.035\,L \times 0.1\,M=0.0035$
$HCO_{3}^{-}$
$H^{+}$
$H_{2}CO_{3}$
I
0.0025
0.0035
0
C
-0.0025
-0.0025
0.0025
E
0
0.001
0.0025
At 40 mL addition of $HCl$: $pH=-log(4.3 \times 10^{-7})+log(\frac{0.0025}{0.003})=6.59$
$HCO_{3}^{-}:0.2\,L \times 0.0125\,M=0.0025\,mol$
$HCl: 0.040\,L \times 0.1\,M=0.004$
$HCO_{3}^{-}$
$H^{+}$
$H_{2}CO_{3}$
I
0.0025
0.004
0
C
-0.0025
-0.0025
0.0025
E
0
0.0015
0.0025
At 45 mL addition of $HCl$: $pH=-log(4.3 \times 10^{-7})+log(\frac{0.0025}{0.002})=6.46$
$HCO_{3}^{-}:0.2\,L \times 0.0125\,M=0.0025\,mol$
$HCl: 0.045\,L \times 0.1\,M=0.0045$
$HCO_{3}^{-}$
$H^{+}$
$H_{2}CO_{3}$
I
0.0025
0.0045
0
C
-0.0025
-0.0025
0.0025
E
0
0.002
0.0025
At 50 mL addition of $HCl$: second equivalence point: $pH=3.86$
$HCO_{3}^{-}:0.2\,L \times 0.0125\,M=0.0025\,mol$
$HCl: 0.05\,L \times 0.1\,M=0.005$
$HCO_{3}^{-}$
$H^{+}$
$H_{2}CO_{3}$
I
0.0025
0.005
0
C
-0.0025
-0.0025
0.0025
E
0
0.0025
0.0025
At 55 mL addition of $HCl$: $pH=-log(\frac{0.003}{0.055})=1.26$
$HCO_{3}^{-}:0.2\,L \times 0.0125\,M=0.0025\,mol$
$HCl: 0.055\,L \times 0.1\,M=0.0055$
$HCO_{3}^{-}$
$H^{+}$
$H_{2}CO_{3}$
I
0.0025
0.0055
0
C
-0.0025
-0.0025
0.0025
E
0
0.003
0.0025
At 60 mL addition of $HCl$: $pH=-log(\frac{0.0035}{0.06})=1.23$
$HCO_{3}^{-}:0.2\,L \times 0.0125\,M=0.0025\,mol$
$HCl: 0.06\,L \times 0.1\,M=0.006$
$HCO_{3}^{-}$
$H^{+}$
$H_{2}CO_{3}$
I
0.0025
0.006
0
C
-0.0025
-0.0025
0.0025
E
0
0.0035
0.0025
At 65 mL addition of $HCl$: $pH=-log(\frac{0.0040}{0.065})=1.21$
$HCO_{3}^{-}:0.2\,L \times 0.0125\,M=0.0025\,mol$
$HCl: 0.06\,L \times 0.1\,M=0.006$
$HCO_{3}^{-}$
$H^{+}$
$H_{2}CO_{3}$
I
0.0025
0.0065
0
C
-0.0025
-0.0025
0.0025
E
0
0.0040
0.0025
7. Video Solution
1. $M_1V_1=M_2V_2 \rightarrow V_2=\frac{M_1V_1}{V_2}=\frac{(0.2\,M)(0.5\,L)}{12.0\,M}=8.33 \times10^{-3}\,L$ Therefore, dilute 8.33 mL of 12.0 M HCl to 500.0 mL.
2. $M_1V_1=M_2V_2 \rightarrow V_2=\frac{M_1V_1}{V_2}=\frac{(0.288\,M)(0.05\,L)}{0.2\,M}=7.20 \times 10^{-3}\,L$
3. $M_1V_1=M_2V_2 \rightarrow V_2=\frac{M_1V_1}{V_2}=\frac{(0.288\,M)(0.05\,L)}{0.187\,M}=7.70 \times 10^{-2}$
8.
a. $M_{1}V_{1}=M_{2}V_{2} \rightarrow M_2= \frac{M_1V_1}{V_{2}}=\frac{(0.582\,M)(0.060\,L)}{0.0719\,L} =0.49\,M$
b. $M_{1}V_{1}=M_{2}V_{2} \rightarrow V_{2}=\frac{M_{1}V_{1}}{M_{2}} =\frac{(0.582\,M)(0.050\,L)}{0.49\,M}=59.4 \times 10^{-2}\,L$
9.
Volume of Base Added (mL) 10.0 30.0 40.0 45.0 50.0 55.0 65.0 75.0
pH 0.73 1.22 1.76 7 12.2 12.5 13.0 13.1
$NaOH\,(aq)+HCl\,(aq) \rightarrow NaCl\,(s)+H_{2}O\,(l)$
At 0 mL of base added: $pH=-log(0.288)=0.54$
At 10 mL of base added: $pH=-log(\frac{0.0144-0.00321}{0.06})=0.73$
$HCl:0.05\,L \times 0.288\,M=0.0144$
$NaOH:0.01\,L \times 0.321\,M=0.00321$
At 30 mL of base added: $pH=-log(\frac{0.0144-0.00963}{0.08})=1.22$
$HCl:0.05\,L \times 0.288\,M=0.0144$
$NaOH:0.03\,L \times 0.321\,M=0.00963$
At 40.0 mL of base added: $pH=-log(\frac{0.0144-0.01284}{0.09})=1.76$
$HCl:0.05\,L \times 0.288\,M=0.0144$
$NaOH:0.04\,L \times 0.321\,M=0.01284$
At 45.0 mL of base added: $pH=7$
$HCl:0.05\,L \times 0.288\,M=0.0144$
$NaOH:0.045\,L \times 0.321\,M=0.0144$
All $HCl$ will neutralize at the equivalence point $[H^{+}]=[OH^{-}]$ and the pH of the solution is 7.
At 50.0 mL base added: $pH=14-pOH=14-log(\frac{0.01605-0.0144}{0.1})=12.2$
$HCl:0.05\,L \times 0.288\,M=0.0144$
$NaOH:0.050\,L \times 0.321\,M=0.01605$
At 55 mL base added: $pH=14-pOH=14-log(\frac{0.01605-0.0144}{0.1})=12.5$
$HCl:0.05\,L \times 0.288\,M=0.0144$
$NaOH:0.055\,L \times 0.321\,M=0.017655$
At 65 mL base added: $pH=14-pOH=14-log(\frac{0.020865-0.0144}{0.07})=13.0$
$HCl:0.05\,L \times 0.288\,M=0.0144$
$NaOH:0.065\,L \times 0.321\,M=0.020865$
At 75 mL base added: $pH=14-pOH=14-log(\frac{0.024075-0.0144}{0.08})=13.1$
$HCl:0.05\,L \times 0.288\,M=0.0144$
$NaOH:0.075\,L \times 0.321\,M=0.024075$
10.
The equivalence point is at $pH$ 7 and this occurs at 0.0193 L.
$V_{2}=\frac{(0.156\,M)(0.025\,L)}{0.202}=0.0193\,L$
11.
pH at equivalence point: $14-(-log\sqrt{(\frac{10^{-14}}{10^{-3.75}} \times \frac{0.01205}{0.05+0.1227}}))=8.30$
$0.05\,L \times 0.241\,M=0.01205\,mol$
$V_2=\frac{(0.05\,L)(0.241\,M)}{0.0982}=0.1227\,L$
Volume of Base Added (mL) 0 5 10 15 20 25
pH 2.18 2.38 2.70 2.89 3.04 3.16
At 0 mL of base added: $pH=-log([H^{+}])=-log(2.52 \times 10^{-6})=2.18$
$[H^{+}]= \sqrt{10^{-3.75} \times 0.241\,M}=2.52 \times 10^{-6}$
At 5 mL of base added: $pH=3.75+log(\frac{4.91 \times 10^{-4}}{0.01205-4.91 \times 10^{-4}})=2.38$
$KOH:0.05\,L \times 0.0982\,M=4.91 \times 10^{-4}\,mol$
$CH_{2}O_{2}: 0.05\,L \times 0.241\,M=0.01205\,mol$
At 10 mL of base added: $pH=3.75+log(\frac{9.82 \times 10^{-4}}{0.01205-9.82 \times 10^{-4}})=2.70$
$KOH:0.01\,L \times 0.0982\,M=9.82 \times 10^{-4}\,mol$
$CH_{2}O_{2}: 0.05\,L \times 0.241\,M=0.01205\,mol$
At 15 mL of base added:$pH=3.75+log(\frac{0.001473}{0.01205-0.001473}=2.89$
$KOH: 0.015\,L \times 0.0982\,M=0.001473$
$CH_{2}O_{2}: 0.05\,L \times 0.241\,M=0.01205$
At 20 mL of base added: $pH=3.75+log(\frac{0.001964}{0.01205-0.001964})=3.04$
$KOH: 0.02\,L \times 0.0982\,M=0.001964$
$CH_{2}O_{2}: 0.05\,L \times 0.241\,M=0.01205$
At 25 mL base added: $pH=3.75+log(\frac{0.002455}{0.01205-0.002455})=3.16$
$KOH:0.025\,L \times 0.0982\,M=0.002455\,mol$
$CH_{2}O_{2}: 0.05\,L \times 0.241\,M=0.01205$
12. $M_1V_1=M_2V_2 \rightarrow V_2=\frac{M_1V1}{M_2}=\frac{(0.430\,M)(0.05\,L)}{0.150\,M}=1.42 \times 10^{-1}\,L$
143 mL are needed to fully deprotonate the carboxylic acid group.
a. 143 additional milliliters of KOH are needed to deprotonate the ammonium group.
b. At the first equivalence point: $pH=\frac{(pKa_{1}+pKa_{2})}{2}=5.95$
c. 143 mL of titrant are needed to obtain a solution in which glycine has no electrical charge. The isoelectric point of glycine 5.95.
13.
$pH=pK_a+log(\frac{[C_5H_5N]}{[C_5H_5NH^{+}]})=(14-8.77)+log(\frac{0.02996}{0.0063434})=5.90$
$C_5H_5N\,(aq)+HCl\,(aq) \rightarrow C_5H_5NH^{+}\,(aq)+Cl^{-}\,(aq)$
$HCl: 32.2\,mL \times \frac{1\,L}{1,000\,mL} \times 0.197\,M=0.0063434\,mol$
$C_5H_5N: 150\,mL \times \frac{1\,L}{1,000\,mL} \times 0.242\,M=0.0363\,mol$
$C_5H_5N$
$HCl$
$C_5H_5NH^{+}$
$Cl^{-}$
I
0.0363
0.0063434
0
-
C
-0.0063434
-0.0063434
+0.0063434
-
E
0.02996
0
0.0063434
-
14. $pH=\frac{(4.21+5.64)}{2}=4.93$
15. $pH=\frac{(2.85+5.70)}{2}=4.25$
17.4: Solubility Equilibria
Conceptual Problems
1. Write an expression for Ksp for each salt.
1. $AgI$
2. $CaF_{2}$
3. $PbCl_2$
4. $Ag_{2}CrO_{4}$
1. Some species are not represented in a solubility product expression. Why?
1. Describe the differences between $Q$ and $K_{sp}$.
1. How can an ion product be used to determine whether a solution is saturated?
1. When using Ksp to directly compare the solubilities of compounds, why is it important to compare only the Ksp values of salts that have the same stoichiometry?
1. Describe the effect of a common ion on the solubility of a salt. Is this effect similar to the common ion effect found in buffers? Explain your answer.
1. Explain why the presence of MgCl2 decreases the molar solubility of the sparingly soluble salt MgCO3.
Conceptual Answers
1.
a. $K_{sp}=[Ag^{+}][I^{-}]$
b. $K_{sp}=[Ca^{2+}][F^{-}]^{2}$
c. $K_{sp}=[Pb^{2+}][Cl^{-}]^{2}$
d. $K_{sp}=[Ag^{+}]^{2}[CrO_{4}^{2-}]$
2. The solubility constant expression, $K_{sp}$ is an equilibrium constant for a solid substance dissolving in a an aqueous solution. Thus, it is a measure of solubility and species that do not dissolve are not represented in the solubility product expression.
3. The main difference between $Q$ and $K_sp$ is that $Q$ describes a reaction that is not at equilibrium unlike $K_{sp}$.
4. An ion product can be used to determine whether a solution is saturated as it is compared to $K_{sp}$ where there are three possible conditions for an aqueous solution of an ionic solid:
1. $Q<K_{sp}$. The solution is unsaturated, and more of the ionic solid, if available, will dissolve.
2. $Q=K_{sp}$. The solution is saturated and at equilibrium.
3. $Q>K_{sp}$. The solution is supersaturated, and ionic solid will precipitate.
1. For a 1:1 salt, the molar solubility is simply $\sqrt{K_{\textrm{sp}}}$; for a 2:1 salt, the molar solubility is $\sqrt[3]{K_{\textrm{sp}}/4}$. Consequently, the magnitudes of Ksp can be correlated with molar solubility only if the salts have the same stoichiometry.
2. By the presence of a common ion, the solubility of any sparingly soluble salt is always decreased. This is similar to the common ion effect found in buffers as adding a common cation or anion shifts the solubility equilibrium in the direction predicted by Le Chatelier’s principle.
1. Because of the common ion effect. Adding a soluble Mg2+ salt increases [Mg2+] in solution, and Le Chatelier’s principle predicts that this will shift the solubility equilibrium of MgCO3 to the left, decreasing its solubility.
Numerical Problems
1. Predict the molar solubility of each compound using the Ksp values given in Chapter 26.
1. $Cd(IO_{3})_{2}$
2. $AgCN$
3. $HgI_{2}$
1. Predict the molar solubility of each compound using the Ksp values given.
1. Li3PO4: 2.37 × 10−11
2. Ca(IO3)2: 6.47 × 10−6
3. Y(IO3)3: 1.12 × 10−10
1. A student prepared 750 mL of a saturated solution of silver sulfate (Ag2SO4). How many grams of Ag2SO4 does the solution contain? Ksp = 1.20 × 10−5.
1. Given the Ksp values in Table 17.1 "Solubility Products for Selected Ionic Substances at 25°C" and Appendix B, predict the molar concentration of each species in a saturated aqueous solution.
1. silver bromide
2. lead oxalate
3. iron(II) carbonate
4. silver phosphate
5. copper(I) cyanide
1. Given the Ksp values in Table 17.1 "Solubility Products for Selected Ionic Substances at 25°C" and Appendix B predict the molar concentration of each species in a saturated aqueous solution.
1. copper(I) chloride
2. lanthanum(III) iodate
3. magnesium phosphate
4. silver chromate
5. strontium sulfate
1. Silicon dioxide, the most common binary compound of silicon and oxygen, constitutes approximately 60% of Earth’s crust. Under certain conditions, this compound can react with water to form silicic acid, which can be written as either H4SiO4 or Si(OH)4. Write a balanced chemical equation for the dissolution of SiO2 in basic solution. Write an equilibrium constant expression for the reaction.
1. The Kspof Mg(OH)2 is 5.61 × 10−12. If you tried to dissolve 24.0 mg of Mg(OH)2 in 250 mL of water and then filtered the solution and dried the remaining solid, what would you predict to be the mass of the undissolved solid? You discover that only 1.0 mg remains undissolved. Explain the difference between your expected value and the actual value.
1. The Ksp of lithium carbonate is 8.15 × 10−4. If 2.34 g of the salt is stirred with 500 mL of water and any undissolved solid is filtered from the solution and dried, what do you predict to be the mass of the solid? You discover that all of your sample dissolves. Explain the difference between your predicted value and the actual value.
1. You have calculated that 24.6 mg of BaSO4 will dissolve in 1.0 L of water at 25°C. After adding your calculated amount to 1.0 L of water and stirring for several hours, you notice that the solution contains undissolved solid. After carefully filtering the solution and drying the solid, you find that 22.1 mg did not dissolve. According to your measurements, what is the Ksp of barium sulfate?
1. In a saturated silver chromate solution, the molar solubility of chromate is 6.54 × 10−5. What is the Ksp?
1. A saturated lead(II) chloride solution has a chloride concentration of 1.62 × 10−2 mol/L. What is the Ksp?
1. From the solubility data given, calculate Ksp for each compound.
1. AgI: 2.89 × 10−7 g/100 mL
2. SrF2: 1.22 × 10−2 g/100 mL
3. Pb(OH)2: 78 mg/500 mL
4. BiAsO4: 14.4 mg/2.0 L
1. From the solubility data given, calculate Ksp for each compound.
1. BaCO3: 10.0 mg/500 mL
2. CaF2: 3.50 mg/200 mL
3. Mn(OH)2: 6.30 × 10−4 g/300 mL
4. Ag2S: 1.60 × 10−13 mg/100 mL
1. Given the following solubilities, calculate Ksp for each compound.
1. BaCO3: 7.00 × 10−5 mol/L
2. CaF2: 1.70 mg/100 mL
3. Pb(IO3)2: 2.30 mg/100 mL
4. SrC2O4: 1.58 × 107mol/L
1. Given the following solubilities, calculate Ksp for each compound.
1. Ag2SO4: 4.2 × 10−1 g/100 mL
2. SrSO4: 1.5 × 10−3 g/100 mL
3. CdC2O4: 6.0 × 10−3 g/100 mL
4. Ba(IO3)2: 3.96 × 10−2 g/100 mL
1. The Ksp of the phosphate fertilizer CaHPO4·2H2O is 2.7 × 10−7 at 25°C. What is the molar concentration of a saturated solution? What mass of this compound will dissolve in 3.0 L of water at this temperature?
1. The Ksp of zinc carbonate monohydrate is 5.5 × 10−11 at 25°C. What is the molar concentration of a saturated solution? What mass of this compound will dissolve in 2.0 L of water at this temperature?
1. Silver nitrate eye drops were formerly administered to newborn infants to guard against eye infections contracted during birth. Although silver nitrate is highly water soluble, silver sulfate has a Ksp of 1.20 × 10−5 at 25°C. If you add 25.0 mL of 0.015 M AgNO3to 150 mL of 2.8 × 10−3 M Na2SO4, will you get a precipitate? If so, what will its mass be?
1. Use the data in Appendix B to predict whether precipitation will occur when each pair of solutions is mixed.
1. 150 mL of 0.142 M Ba(NO3)2 with 200 mL of 0.089 M NaF
2. 250 mL of 0.079 M K2CrO4 with 175 mL of 0.087 M CaCl2
3. 300 mL of 0.109 M MgCl2 with 230 mL of 0.073 M Na2(C2O4)
1. What is the maximum volume of 0.048 M Pb(NO3)2 that can be added to 250 mL of 0.10 M NaSCN before precipitation occurs?Ksp = 2.0 × 10−5 for Pb(SCN)2.
1. Given 300 mL of a solution that is 0.056 M in lithium nitrate, what mass of solid sodium carbonate can be added before precipitation occurs (assuming that the volume of solution does not change after adding the solid)? Ksp = 8.15 × 10−4 for Li2CO3.
1. Given the information in the following table, calculate the molar solubility of each sparingly soluble salt in 0.95 M MgCl2.
Saturated Solution Ksp
MgCO3·3H2O 2.4 × 10−6
Mg(OH)2 5.6 × 10−12
Mg3(PO4)2 1.04 × 10−24
Numerical Answers
1.
a. $K_{sp}=[Cd^{2+}][IO_{3}]^2 \rightarrow 2.5 \times 10^{-8}=x(2x)^{2} \rightarrow 2.5 \times 10^{-8}=4x^{3} \rightarrow x=1.84 \times 10^{-3}$
b. $K_{sp}=[Ag^{+}][CN^{-}] \rightarrow 1.6 \times 10^{-14}=(x)(x) \rightarrow x=7.73 \times 10^{-9}$
c. $K_{sp}=[Hg^{2+}][I^{-}]^2 \rightarrow 2.9 \times 10^{-29}=(x)(2x)^{2} \rightarrow 1.94\times 10^{-10}$
2.
a. $K_{sp}=[Li^{+}]^{3}[PO_{4}^{3-}] \rightarrow 2.37 \times 10^{-11}=(3x)^{3}(x) \rightarrow x=9.68 \times 10^{-4}$
b. $K_{sp}=[Ca^{2+}][IO_{3}^{-}]^{2} \rightarrow 6.47 \times 10^{-6}=(x)(2x)^{2} \rightarrow 1.17 \times 10^{-2}$
c. $K_{sp}=[Y^{3+}][IO_{3}]^{-}]^{3} \rightarrow 1.12 \times 10^{-10}=(x)(3x)^{3} \rightarrow x=1.42 \times 10^{-3}$
3. $0.750\,L \times \frac{0.0144\,mol}{1\,L} \times \frac{311.199\,g}{1\,mol}=3.37\,g$
$K_{sp}=[Ag^{+}]^{2}[SO_{4}^{2-}] \rightarrow 1.20 \times 10^{-5}=(2x)^{2}(x) \rightarrow x=0.0144$
4.
a. $K_{sp}=[Ag^{+}][Br^{-}] \rightarrow 5.35 \times 10^{-13}=(x)(x) \rightarrow x=7.31 \times 10^{-7}$
b. $K_{sp}=[PB^{2+}][C_{2}O_{4}] \rightarrow 8.5 \times 10^{-9}=(x)(x) \rightarrow x=9.2 \times 10^{-5}$
c. $K_{sp}=[Fe^{2+}][CO_{3}^{2-}] \rightarrow 3.13 \times 10^{-11}=(x)(x) \rightarrow x=5.60 \times 10^{-6}$
d. $K_{sp}=[Ag^{+}]^{3}[PO_{4}^{3-}] \rightarrow 8.89 \times 10^{-17}=(3x)^{3}(x) \rightarrow x=4.26 \times 10^{-5}$
e. $K_{sp}=[Cu][CN] \rightarrow 3.47 \times 10^{-20}=(x)(x) \rightarrow x=1.86 \times 10^{-10}$
5.
a. $K_{sp}=[Cu][CN] \rightarrow 1.72 \times 10^{-20}=(x)(x) \rightarrow x=4.15 \times 10^{-4}$
b. $K_{sp}=[La^{3+}][IO_{3}^{-}]^{3}] \rightarrow 7.50 \times 10^{-12}=(x)(3x)^{3} \rightarrow x=7.26 \times 10^{-4}$
c. $K_{sp}=[Mg^{2+}]^{3}[PO_{4}^{3-}]^{2} \rightarrow 1.04 \times 10^{-24}=(3x)^{3}(2x)^{2} \rightarrow x=6.26 \times 10^{-6}$
d. $K_{sp}=[Ag^{+}]^{2}[CrO_{4}^{2-}] \rightarrow 1.12 \times 10^{-12}=(2x)^{2}(x) \rightarrow 6.54 \times 10^{-5}$
e. $K_{sp}=[Sr^{2+}][SO_{4}^{2-}] \rightarrow 3.44 \times 10^{-7}=(x)(x) \rightarrow x=5.87 \times 10^{-4}$
6.
$SiO_{2}\,(g)+H_{2}O\,(l) \rightarrow Si(OH)_{4}\,(aq)$
$K_{sp}=[Si^{4+}][OH^{-}]^{4}$
1. 22.4 mg; a secondary reaction occurs, where OH from the dissociation of the salt reacts with H+ from the dissociation of water. This reaction causes further dissociation of the salt (Le Chatelier’s principle).
$K_{sp}=[Mg^{2+}][OH^{-}]^{2} \rightarrow 5.61 \times 10^{-12}=(x)(2x)^{2} \rightarrow x=1.12 \times 10^{-4}$
8.
$Predicted\,mass: \frac{5.88 \times 10^{-2}\,mol}{1\,L} \times \frac{73.891\,g}{1\,mol} \times 0.5\,L=2.17\,g$
$K_{sp}=[Li^{+}]^2[CO_{3}^{2-}] \rightarrow 8.15 \times 10^{-4}=(2x)^{2}(x) \rightarrow x=5.88 \times 10^{-2}$
The difference between the predicted value and the actual value occurs because the carbonate from the dissociation of the salt reacts with the $H^{+}$ from the dissociation of water. This reaction causes further dissociation of the salt (Le Chatelier’s principle).
9.
$K_{sp}=[Ba^{2+}][SO_{4}^{2-}]=(x)(x)=x^{2}=(1.07 \times 10^{-5}\,M)^{2}=1.15 \times 10^{-10}$
$BaSO_{4}:(24.6-22.1\,mg) \times \frac{1\,g}{1,000\,mg} \times \frac{1\,mol}{233.38\,g} \times \frac{1}{1.0\,L}=1.07 \times 10^{-5}\,M$
10. $K_{sp}=[Ag^{+}]^{2}[CrO_{4}^{2-}]=(2x)^{2}(x)=4x^{3}=4(6.54 \times 10^{-5})=1.12 \times 106{-12}$
1. $K_{sp}=[Pb^{2+}][Cl^{-}]^{2}=(x)(2x)^{2}=4x^{3}=4(3.24 \times 10^{-2})^{3}=1.70 \times 10^{-5}$
2.
a. $K_{sp}=[Ag^{+}][I^{-}]=x^{2}=(1.23 \times 10^{-8})^{2})=1.52 \times 10^{-16}$
$AgI: (2.89 \times 10^{-7}\,g \times \frac{1\,mol}{234.77\,g} \times \frac{1}{0.1\,L}=1.23 \times 10^{-8}$
b. $K_{sp}=[Sr^{2+}][F^{-}]^{2}]=(x)(2x)^{2}=4x^{3}=4(9.72 \times 10^{-4})^{3}=3.66 \times 10^{-9}$
$SrF_{2}=1.22 \times 10^{-2}\,g \times \frac{1\,mol}{125.62\,g} \times \frac{1}{0.1\,L}=9.72 \times 10^{-4}$
c. $K_{sp}=[Pb^{2+}][OH^{-}]^{2}=4x^{3}=4(6.47 \times 10^{-4})^{3}=1.08 \times 10^{-9}$
$Pb(OH)_{2}=0.078\,g \times \frac{1\,mol}{241.21\,g} \times \frac{1}{0.1\,L}=6.47 \times 10^{-4}$
d. $K_{sp}=[Bi^{3+}][AsO_{4}^{3-}]=x^{2}=(2.05 \times 10^{-5})^{2} =4.21 \times 10^{-10}$
$BiAsO_{4}=0.0144\,g \times \frac{1\,mol}{350.924\,g} \times \frac{1}{2\,L}=2.05 \times 10^{-5}$
13.
a. $K_{sp}=[Ba^{2+}][CO_{3}^{2-}]=x^{2}=(1.01 \times 10^{-4})^{2}=1.03 \times 10^{-8}$
$BaCO_{3}:0.01\,g \times \frac{1\,mol}{197.34\,g} \times \frac{1}{0.5\,L}=1.01 \times 10^{-4}$
b. $K_{sp}=[Ca^{2+}][F^{-}]^{2}=4x^{3}=4(2.24 \times 10^{-4})^{3}=4.51 \times 10^{-11}$
$CaF_{2}: 0.0035\,g \times \frac{1\,mol}{78.07\,g} \times \frac{1}{0.2\,L}=2.24 \times 10^{-4}$
c. $K_{sp}=[Mn^{2+}][OH^{-}]^2=4x^{3}=4(2.36 \times 10^{-5})^{3}=5.26 \times 10^{-14}$
$Mn(OH)_{2}: 6.30 \times 10^{-4}\,g \times \frac{1\,mol}{88.95\,g} \times \frac{1}{0.3\,L}=2.36 \times 10^{-5}$
d. $K_{sp}=[Ag^{+}]^{2}[S^{2-}]=4x^{3}=4(6.46 \times 10^{-18})^{3}=1.08 \times 10^{-51}$
$Ag_{2}S: 1.60 \times 10^{-16}\,g \times \frac{1\,mol}{247.8\,g} \times \frac{1}{0.1\,L}=6.46 \times 10^{-18}$
14.
a. $K_{sp}=[Ba^{2+}][CO_{3}^{2-}]=x^{2}=(7.0 \times 10^{-5})^{2}=4.90 \times 10^{-9}$
b. $K_{sp}=[Ca^{2+}][F^{-}]^{2}=4x^{3}=4(2.18 \times 10^{-4})^{3}=4.13 \times 10^{-11}$
$CaF_{2}: 0.0017\,g \times \frac{1\,mol}{78.07\,g} \times \frac{1}{0.1\,L}=2.18 \times 10^{-4}$
c. $K_{sp}=[Pb^{2+}][IO_{3}^{-}]^{2}=4x^{3}=4(4.13 \times 10^{-5})^{3}=2.82 \times 10^{-13}$
$Pb(IO_{3})_{2}: 0.0023\,g \times \frac{1\,mol}{557.0053\,g} \times \frac{1}{0.1\,L}=4.13 \times 10^{-5}$
d. $K_{sp}=[Sr^{2+}][C_2O_4^{2-}]=x^{2}=(1.58 \times 10^{-7})^{2}=2.50 \times 10^{-14}$
1.
a. $K_{sp}=[Ag^{+}]^{2}[SO_{4}]^{2-}=4x^{3}=4(1.35 \times 10^{-2})^{3}=9.8 \times 10^{-6}$
$Ag_{2}SO_{4}: 4.2 \times 10^{-1}\,g \times \frac{1\,mol}{311.799\,g} \times \frac{1}{0.1\,L}=1.35 \times 10^{-2}$
b. $K_{sp}=[Sr^{2+}][SO_{4}^{2-}]=x^{2}=(8.17 \times 10^{-5})^{2}=6.7 \times 10^{-9}$
$SrSO_{4}: 1.5 \times 10^{-3}\,g \times \frac{1\,mol}{183.68\,g} \times \frac{1}{0.1\,L}=8.17 \times 10^{-5}$
c. $K_{sp}=[Cd^{2+}][C_2O_4^{2-}]=x^{2}=(2.99 \times 10^{-4})^{2}=8.9 \times 10^{-8}$
$CdC_{2}O_{4}: 6.0 \times 10^{-3}\,g \times \frac{1\,mol}{200.43\,g} \times \frac{1}{0.1\,L}=2.99 \times 10^{-4}$
d. $K_{sp}=[Ba^{2+}][IO_{3}^{-}]^{2}=4x^{3}=4(8.13 \times 10^{-4})^{3}=2.15 \times 10^{-9}$
$Ba(IO_{3})_{2}: 3.96 \times 10^{-2}\,g \times \frac{1\,mol}{487.15\,g} \times \frac{1}{0.1\,L}=8.13 \times 10^{-4}$
16.
$K_{sp}=[Ca^{2+}][HPO_{4}^{2-}] \rightarrow 2.7 \times 10^{-7}=x^{2} \rightarrow x=5.2 \times 10^{-4}$
$CaHPO4·2H2O: 3.0\,L \times \frac{2.7 \times 10^{-7}\,mol}{1\,L} \times \frac{290.1299\,g}{1\,mol}=2.35 \times 10^{-4}\,g$
17.
$K_{sp}=[Zn^{2+}][CO_{3}^{2+}] \rightarrow 5.5 \times 10^{-11}=x^{2} \rightarrow x=7.4 \times 10^{-6}$
2.1 mg
18. $Q=[Ag^{+}]^{2}[SO_{4}^{2-}]=(2.14 \times 10^{-3})^{2} \times 2.4 \times 10^{-3}=1.1 \times 10^{-8}<1.20 \times 10^{-5}$ Therefore, no precipitate will form.
$Ag^{+}: \frac{(0.025\,L)(0.015\,M)}{0.175\,L}=2.14 \times 10^{-3}$
$SO_{4}^{2-}: \frac{(0.15\,L)(2.8 \times 10^{-3}}{0.175\,L}=2.4 \times 10^{-3}$
19.
a. $Q=[Ba^{2+}][F^{-}]^{2}=(0.060857)(0.05643)^{2}=1.94 \times 10^{-4}>1.84 \times 10^{-7}$ Therefore, precipitate will form.
$Ba^{2+}: \frac{(0.150)(0.142)}{0.35}=0.060857$
$F^{-}: \frac{(0.25)(0.079)}{0.35}=0.05643$
b. $Q=[K^{+}][Cl^{-}]=(0.0465)(0.0358)=1.66 \times 10^{-3}<21.7. Therefore, precipitate will not form. \(K^{+}: \frac{(0.25)(0.079)}{0.425}=0.0465$
$Cl^{-}: \frac{(0.175)(0.087)}{0.425}=0.0358$
c. $Q=[Mg^{2+}][C_{2}O_{4}^{2-}]=(0.0617)(0.0317)=1.95 \times 10^{-3}>8.5 \times 10^{-5}$ Therefore, precipitate will form.
$Mg^{2+}:\frac{(0.3)(0.109)}{0.53}=0.0617$
$C_{2}O_{4}: \frac{(0.23)(0.073)}{0.53}=0.0317$
20. 520 mL
1. 8.27 g
2.
a. $K_{sp}=\sqrt{2.4 \times 10^{-6}}=1.55 \times 10^{-3}$
b. $K_{sp}=(\frac{5.6 \times 10^{-12}}{4})^{\frac{1}{3}}=1.12 \times 10^{-4}$
c. $K_{sp}=(\frac{1.04 \times 10^{-24}}{108})^{\frac{1}{5}}=6.26 \times 10^{-6}$
17.5: Factors that Affect Solubility
Conceptual Problems
1. Do you expect the actual molar solubility of LaPO4 to be greater than, the same as, or less than the value calculated from its Ksp? Explain your reasoning.
2. Do you expect the difference between the calculated molar solubility and the actual molar solubility of Ca3(PO4)2 to be greater than or less than the difference in the solubilities of Mg3(PO4)2? Why?
3. Write chemical equations to describe the interactions in a solution that contains Mg(OH)2, which forms ion pairs, and in one that contains propanoic acid (CH3CH2CO2H), which forms a hydrated neutral molecule.
4. Draw representations of Ca(IO3)2 in solution
1. as an ionic solid.
2. in the form of ion pairs.
3. as discrete ions.
Conceptual Answers
1. It is expected that the molar solubility of \LaPO_{4}\) be less than the value from its $K_{sp}$ because the $K_{sp}=[La^{3+}][PO_{4}^{3-}]=x^{2}$ in which the molar solubility would be the square root of of $K_{sp}$.
2. It would be expected that there would be no difference in the calculated molar solubility and actual molar solubility of $Ca_{3}(PO_{4})_{2}$ and $Mg_{3}(PO_{4})_{2}$ follow the similar equation: $K_{sp}=108x^{5}$ . It would ultimately depend on the K_{sp} values as $x=(\frac{K_{sp}}{108})^{\frac{1}{5}} in which the larger \(K_{sp} gives the larger molar solubility. 3. \(Mg(OH)_{2}\,(s) \rightleftharpoons Mg^{2+}\,(aq)+2\,OH^{-}\,(aq)$
$CH_{3}CH_{2}CO_{2}H\,(aq) \rightleftharpoons CH_{3}CH_{2}CO_{2}^{-}\,(aq)+H^{+}\,(aq)$
Numerical Problem
1. Ferric phosphate has a molar solubility of 5.44 × 10−16 in 1.82 M Na3PO4. Predict its Ksp. The actual Ksp is 1.3 × 10−22. Explain this discrepancy.
Numerical Answer
1. 9.90 × 10−16; the solubility is much higher than predicted by Ksp due to the formation of ion pairs (and/or phosphate complexes) in the sodium phosphate solution.
17.6: Precipitation and Separation of Ions
Questions
1. Iron(II) hydroxide is only sparingly soluble in water at 25oC; its Ksp is $7.9 \times 10^{-16}$. Calculate the solubility of iron(II) hydroxide in a buffer solution with pH = 7.00.
2. A solution contains 0.60 M $\ce{Ba^2+}$ and 0.30 M $\ce{Ca^2+}$; Ksp values for $\ce{BaCrO4}$ and $\ce{CaCrO4}$ are $1.2 \times 10^{-10}$ and $7.1 \times 10^{-4}$ respectively. What value of $\ce{[CrO4^2- ]}$ will result in a maximum separation of these two ions?
3. A solution contains 0.60 M $\ce{Ba^2+}$ and 0.30 M $\ce{Ca^2+}$; Ksp values for $\ce{BaCrO4}$ and $\ce{CaCrO4}$ are $1.2 \times 10^{-10}$ and $7.1 \times 10^{-4}$ respectively. Calculate the $\ce{[Ca^2+]/[Ba^2+]}$ ratio in the solution when $\ce{[CrO4^2- ]}$ is maintained at $1.2 \times 10^{-3}\; M$.
Solutions
1. Answer $\ce{[Fe^2+]} = \textrm{0.079 M}$
Consider...
$\ce{[OH- ]} = 10^{(-14+7)} = \textrm{1.00e-7 (buffer)}$.
$\ce{[Fe^2+]} (\textrm{1.00e-7})^2 = K_{\ce{sp}}$; $\ce{[Fe^2+]} =\: ?$
This $\ce{Fe^2+}$ concentration is low; it is not very soluble in a neutral solution (pH = 7).
What is $\ce{[Fe^2+]}$ in a solution whose pH = 6.00?
2. Answer $\ce{[CrO4^2- ]} = \textrm{2.37e-3 M}$
Consider...
Solid $\ce{BaCrO4}$ will form first as $\ce{[CrO4^2- ]}$ increases. The maximum $\ce{[CrO4^2- ]}$ to precipitate $\ce{CaCrO4}$ is estimated as follows.
$\ce{[CrO4^2- ]} = \dfrac{\textrm{7.1e-4}}{0.30} = \textrm{2.37e-3 M}$
Estimate $\ce{[Ba^2+]}$ when $\ce{[CrO4^2- ]} = \textrm{2.3e-3 M}$, slightly below the maximum concentration.
3. Answer $\ce{\dfrac{[Ca^2+]}{[Ba^2+]}} = \textrm{3e6}$
Consider...
$\ce{[Ba^2+]} = \dfrac{\textrm{1.2e-10}}{\textrm{1.2e-3}} = \textrm{1e-7}$;
$\ce{\dfrac{[Ca^2+]}{[Ba^2+]}} = \dfrac{0.3}{\textrm{1e-7}} =\: ?$ The ratio of three million is large!
17.7: Qualitative Analysis for Metallic Elements
Conceptual Problem
1. Given a solution that contains a mixture of $NaCl$, $CuCl_{2}$, and $ZnCl_{2}$, propose a method for separating the metal ions.
Conceptual Solution
1. Given a solution that contains a mixture of $NaCl$, $CuCl_{2}$, and $ZnCl_{2}$, we can first use $H_{2}S$ to separate the $Zn^{2+}$ through filtration, then use $ZnS$ to separate the $Cu^{2+}$ through filtration, and be left with $Na^{+}$. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/17%3A_Additional_Aspects_of_Aqueous_Equilibria/17.E%3A_Additional_Aspects_of_Aqueous_Equilibria_%28Exercises%29.txt |
17.1: The Common-Ion Effect
The common-ion effect argues that the dissociation of a weak electrolyte is decreased by adding a strong electrolyte to the solution that has a common ion with the weak electrolyte.
17.2: Buffered Solutions
Buffers are solutions that resist a change in pH
17.2.1 Composition and Action of Buffered Solutions
• buffers have both acidic and basic species to neutralize H+ and OH- ions
• acid dissociation equilibrium in buffered solution $HX(aq) \rightleftharpoons H^+ (aq) + X^-(aq) \nonumber$ with $K_a = \dfrac{[H^+][X^-]}{[HX]} \nonumber$ or $[H^+]= K_a \dfrac{[HX]}{[X^-]} \nonumber$
• pH determined by: value of Ka and the ratio of [HX]/[X-]
• if OH- added:
• $OH^-(aq) + HX(aq) \rightleftharpoons H_2O(l) + X^-(aq) \nonumber$
• Therefore [HX] decreases and [X-] increases
• if amounts of HX and X- present are very much larger than the amount of OH- added, then the ratio of [HX]/[X-] will not change much, and so the increase in pH due to the added hydroxide ion is rather small
• when [HX] and [X-] are about the same, buffers are most effective: i.e., when $[H^+] = K_a$
17.2.2 Buffer Capacity and pH
• buffer capacity – amount of acid or base buffer can neutralize before the pH changes considerably
• capacity depends on amount of acid or base in buffer
• pH depends on Ka for acid and relative concentrations of the acid and base
• Henderson-Hasselbalch Approximation: $pH = pK_a + \log_{10} \dfrac{[base]}{[acid]} \nonumber$
• [base] and [acid] = concentrations of conjugate acid-base pair
• when [base]=[acid], pH = pKa
• can use initial concentrations of acid and base components of buffer directly into equation
17.2.3 Addition of Strong Acids or Bases to Buffers
reactions between strong acids and bases go to completion
Either strong base or acid assumed to be completely consumed by reaction with buffer if buffering capacity is not exceeded
17.3: Acid-Base Titrations
• solution containing a known [base] added to an acid or acid solution added to base
• acid-base indicators used to signal equivalence point
• titration curve – pH vs Volume
17.3.1 Strong Acid – Strong base Titrations
• pH starts out low ends high
• pH before equivalence point is pH of acid not neutralized by base
• pH at equivalence point is pH of solution
• pH equals 7.00
• for strong base titrations, the pH starts high ends low
17.3.2 The Addition of a Strong Base to a Weak Acid
Reactions between weak acid and strong base goes to completion
• calculating pH before equivalence point
• stoichiometric calculations: allow strong base to react to completion producing a solution containing a weak acid and its conjugate base
• equilibrium calculation: use Ka and equilibrium expression to find equilibrium concentrations of the weak acid and its conjugate base, and H+
17.3.3 Titration Curves for Weak Acids or Weak Bases
Differences between strong acid-strong base titrations
1. solution of weak acid as higher initial pH than solution of a strong acid with same concentration
2. solution of weak acid rises more rapidly in early part of titration and more slowly as it reached the equivalence point
3. pH is not 7.00 at equivalence point
• before equivalence point solution has mixture of weak acid and its salt
• also called the buffer region of curve
• at equivalence point solution contains only salt
• weakly basic due to hydrolysis of anion
• after equivalence point solution has mixture of salt and excess strong base
• pH determined by [base]
17.3.4 Titrations of Polyprotic Acids
• reaction occurs in series of steps
• titration curve shows multiple equivalence points
17.4: Solubility Equilibria
17.4.1 The Solubility-Product Constant, Ksp
• saturated solution – dissolved and undissolved solute are at equilibrium
• expressed by g/L
• molar solubility – moles of solute dissolved to form a liter of saturated solution (mol/L)
• Ksp equilibrium constant for the equilibrium between an ionic solid and its saturated solution
• Solubility of compound (g/L) à molar solubility of compound (mol/L) à [molar] of ions à Ksp of ions
17.5: Factors that Affect Solubility
solubility affected by temperature and presence of other solutes. The solubility of ionic compound affected by:
• the presence of common ions
• pH of solution
• presence of complexing agent
17.5.1 Common-Ion Effect
• solubility of slightly soluble salt decreases when a second solute has a common ion
17.5.2 Solubility and pH
• solubility of any ionic compound affected if solution is acidic or basic
• change only noticeable if both ions are moderately acidic or basic
• solubility of slightly soluble salts containing basic anions increase as [H+] increases (as pH is lowered)
• the more basic an anion is, the greater the solubility will be affected by pH
17.5.3 Formation of Complex Ions
• metal ions act as Lewis acids in water
• complex ion – metal ion and Lewis base bonded together
• Kf – formation constant, equilibrium expression for formation of a complex ion
• Solubility of metal salts increases in acceptable Lewis bases if metal forms a complex base
• Lewis bases: $NH_3$, $CN^-$, $OH^-$
17.5.4 Amphoterism
• amphoteric substances include hydroxides and oxides of: Al3+, Cr2+, Zn2+, and Sn2+
• dissolve in strongly basic solutions
• formation of complex anions containing, typically four, hydroxides bound to metal ion
• amphoterism also associated with behavior of water molecules that surround and bond to metal ions by Lewis acid-base interactions
17.6: Precipitation and Separation of Ions
Q = ion product
• If Q > Ksp, precipitation occurs until Q = Ksp
• If Q = Ksp, equilibrium exists, have a saturated solution
• If Q < Ksp, solid dissolves until Q = Ksp
17.6.1 Selective Precipitation of Ions
• separation of ions in aqueous solution using a reagent that precipitates only with selected ions
17.7: Qualitative Analysis for Metallic Elements
Qualitative analysis: determines presence or absence of a particular metal ion
• ions separated into broad groups on basis of solubility
• ions separated by dissolving selected members in group
• ions identified by specific tests | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/17%3A_Additional_Aspects_of_Aqueous_Equilibria/17.S%3A_Additional_Aspects_of_Aqueous_Equilibria_%28Summary%29.txt |
Environmental chemistry is the scientific study of the chemical and biochemical phenomena that occur in natural places. It should not be confused with green chemistry, which seeks to reduce potential pollution at its source.
• 18.1: Earth's Atmosphere
• 18.2: Outer Regions of the Atmosphere
• 18.3: Ozone in the Upper Atmostphere
Most of the ozone in the atmosphere is in the stratosphere of the atmosphere, with about 8% in the lower troposphere. As mentioned there, the ozone is formed due to photo reaction. The ozone level is measured in Dobson Unit (DU), named after G.M.B. Dobson, who investigated the ozone between 1920 and 1960. One Dobson Unit (DU) is defined to be 0.01 mm thickness of ozone at STP when all the ozone in the air column above an area is collected and spread over the entire area.
• 18.4: Chemistry of the Troposphere
• 18.5: The World Ocean
Water is the most important resource. Without water life is not possible. From a chemical point of view, water is a pure compound, but in reality, you seldom drink, see, touch or use pure water. Water from various sources contains dissolved gases, minerals, organic and inorganic substances.
• 18.6: Fresh Water
In addition to water, many inorganic substances or minerals are essential to life. These substances ionize in water to form ions and their solutions conduct electricity. Therefore, they are called electrolytes. Because most of these substances are already dissoved in natural water, we list ions instead of the mineral they come from.
• 18.7: Green Chemistry
Green chemistry, also called sustainable chemistry, is an area of chemistry and chemical engineering focused on the design of products and processes that minimize the use and generation of hazardous substances. Whereas environmental chemistry focuses on the effects of polluting chemicals on nature, green chemistry focuses on technological approaches to preventing pollution and reducing consumption of nonrenewable resources.
• 18.E: Chemistry of the Environment (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al.
Thumbnail: Acid mine drainage in the Rio Tinto River. (Public Domain; Carol Stoker @ NASA via Wikipedia(opens in new window)).
18: Chemistry of the Environment
Discussion Questions
• How do atmospheric scientists view the atmosphere?
• What gases are pollutants in the atmosphere?
The atmospheric chemistry studies the chemical composition of the natural atmosphere, the way gases, liquids, and solids in the atmosphere interact with each other and with the earth's surface and associated biota, and how human activities may be changing the chemical and physical characteristics of the atmosphere. It is interesting to note that the 1995 Nobel Prize in Chemistry 1995 was awarded to the atmospheric scientists P. Crutzen, M. Molina and F. S. Rowland. For convenience of study, atmospheric scientists divide the atmosphere as if it consists of four layers. The division is mainly due to temperature variations as the altitude increases. The four layers according to the variation of temperature are.
• Ionosphere (Aurora) or Thermosphere
• Mesosphere
• Stratosphere
• Troposphere
Above 100 km is the thermosphere and ionosphere where the temperature increases from 200 K at 100 km to 500 K at 300 km. The temperature goes even higher as the altitude increases. activity as the altitude decrease. In the outer space, most particles consist of single atoms, H, He, and O etc. At lower altitude (200 - 100 km), diatomic molecules N2, O2, NO etc are present. The ionosphere is full of electrically charged ions. The UV rays ionizes these gases. The major reactions are
In the ionosphere:
$O + h v \rightarrow O^+ + e^- \label{18.1.1}$
$N + h v \rightarrow N^+ + e^- \label{18.1.2}$
In the neutral thermosphere:
$N + O_2 \rightarrow NO + O \label{18.1.3}$
$N + NO \rightarrow N_2 + O \label{18.1.4}$
$O + O \rightarrow O_2 \label{18.1.5}$
Beyond the neutral thermosphere is the ionosphere and exosphere. These layers are of course interesting for space explorations and environmental concerns and space sciences. The atmosphere in the outer space is more like a plasma than a gas. Below the thermosphere is the mesosphere (100 - 50 km) in which the temperature decreases as the altitude increase. In this region, OH, H, NO, HO2, O2, and O3 are common, and the most prominent chemical reactions are:
$H_2O + h\nu \rightarrow OH + H \label{18.1.6}$
$H_2O_2 + O \rightarrow OH + OH. \label{18.1.7}$
Below the mesosphere is the stratosphere, in which the temperature increases as the altitude increase from 10 km to 50 km. In this region, the following reactions are common:
$NO_2 \rightarrow NO + O \label{18.1.8}$
$N_2O \rightarrow N_2 + O \label{18.1.9}$
$H_2 + O \rightarrow OH + H \label{18.1.10}$
$CH_4 + O \rightarrow OH + CH_3 \label{18.1.11}$
Air flow is horizontal in the stratosphere. A thin ozone layer in the upper stratosphere has a high concentration of ozone. This layer is primarily responsible for absorbing the ultraviolet radiation from the sun. The ozone is generated by these reactions:
$O_2 + h\nu \rightarrow O + O \label{18.1.12}$
$O_2 + O \rightarrow O_3 \label{18.1.13}$
The troposphere is where all weather takes place; it is the region of rising and falling packets of air. The air pressure at the top of the troposphere is only 10% of that at sea level (0.1 atmospheres). There is a thin buffer zone between the troposphere and the next layer called the tropopause.
The major components in the region close to the surface of the Earth are N2 (78%), O2 (21%), Ar (1%) with variable amounts of H2O, CO2, CH4, NO2, NO2, CO, N2O, and O3. The ozone concentration in this layer is low, about 8% of the total ozone in the atmosphere is in the troposphere.
What gases are pollutants in the atmosphere?
From the atmospheric science viewpoint, interactions of all gasses among themselves and their interaction with the environmental elements are of interest. However, for identification purposes, we need to identify the gases produced by man-made process (industry).
Some of the gases due to human activities are:
• Carbon dioxide result from the excess burning of carbon-containing fuel.
• Carbon monoxide produced by automobiles. This orderless and colorless gas is very toxic.
• Ozone produced in the exhaust of internal combustion engine, and the variation of ozone concentration in the stratosphere.
• Nitrogen oxides such as NO, NO2, N2O4; due to the production of NO in the internal combustion engine.
• Methane gas produced due to treatments of large amount of waste.
• Sulfur oxides produced in mining operation and in the combustion of sulfur containing fuel. Sulfur oxide causes the so called acid rain problem.
• Chlorofluorocarbons (CFC) are gases used as refrigerant. When disposed into the atmosphere, they cause the ozone concentration to decrease.
Water vapor is also considered a greenhouse gas, but it is also generated by nature continuously due to radiation from the Sun. Of course, when water vapor condense into a liquid, much energy is released in the exothermal process. Condensation of water vapor causes storms and many of the weather phenomena.
Questions
1. According to what is the atmosphere divided into 4 layers?
Skill -
Describe the structure of the atmosphere.
2. Which layer contains the most ozone?
Skill - Describe all the details of ozone?
3. How thick is the troposphere?
Discussion - At the top of the world highest mountain, ~10 km in altitude, the atmosphere is only 0.1 of that at sea level. This is the top of the troposphere.
4. What type of gas is present in the thermosphere?
Skill - Explain the chemistry taken place in the thermosphere?
5. How is the ionosphere different from other layers?
Discussion - The aurora is related to the ions in the atmosphere.
6. What causes the gas molecules in the ionosphere to ionize and become charged particles?
Skill - Describe the chemistry in the ionosphere.
7. If ozone is a beneficial gas in the atmosphere, why is ozone also a gaseous pollutant?
Discussion - Decomposition of ozone releases O, OOH, OH radicals and they are harmful to many living organisms.
8. Why are chlorofluorocarbons a gases pollutant?
Discussion - Ozone in the stratosphere absorbs harmful UV C and UV B, which are harmful to humans and plants.
9. Why is warming up of the Earth a bad thing?
Skill - Give your opinion please on an issue.
Solutions
1. The division is made according to patterns of temperature variation.
2. The stratosphere, between 15 and 50 km.
3. The troposphere ranges from 8 to 15 km.
4. A very dilute concentration of monoatomic gas.
5. The ionosphere contains high concentration of charged particles.
6. Radiation or high energy photons from the sun caused ionization of atoms.
7. Ozone is very reactive, causing harm to living organisms.
8. Because they catalyze the decomposition of ozone in the stratosphere.
9. Because we do not know what is the consequence in the future.
Learning Guide
• How is the atmosphere divided into layers? What are the names of the layers?
• Where is the ozone layer located in the atmosphere? What is the molecular structure of ozone? How is the ozone formed? | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/18%3A_Chemistry_of_the_Environment/18.01%3A_Earth%27s_Atmosphere.txt |
The mesosphere starts just above the stratosphere and extends to 85 kilometers (53 miles) high. Temperatures in the mesosphere decrease with altitude. Because there are few gas molecules in the mesosphere to absorb the Sun’s radiation, the heat source is the stratosphere below. The mesosphere is extremely cold, especially at its top, about -90 degrees C (-130 degrees F).
The air in the mesosphere has extremely low density: 99.9 percent of the mass of the atmosphere is below the mesosphere. As a result, air pressure is very low. A person traveling through the mesosphere would experience severe burns from ultraviolet light since the ozone layer which provides UV protection is in the stratosphere below. There would be almost no oxygen for breathing. Stranger yet, an unprotected traveler’s blood would boil at normal body temperature because the pressure is so low. Meteors burn up in this layer.
The thermosphere starts just above the mesosphere and extends to 600 kilometers (372 miles) high. Thermosphere is taken from the Greek word thermos which means heat. The density of molecules is so low in the thermosphere that one gas molecule can go about 1 km before it collides with another molecule. Since so little energy is transferred, the air feels very cold.
Within the thermosphere is the ionosphere. The ionosphere gets its name from the solar radiation that ionizes gas molecules to create a positively charged ion and one or more negatively charged electrons. The freed electrons travel within the ionosphere as electric currents. This dynamic region grows and shrinks based on solar conditions and divides further into the sub-regions: D, E and F; based on what wavelength of solar radiation is absorbed. The ionosphere is a critical link in the chain of Sun-Earth interactions.
Because of the free ions, the ionosphere has many interesting characteristics. At night, radio waves bounce off the ionosphere and back to Earth. This is why you can often pick up an AM radio station far from its source at night.The Van Allen radiation belts are two doughnut-shaped zones of highly charged particles that are located beyond the atmosphere in the magnetosphere. The particles originate in solar flares and fly to Earth on the solar wind. Once trapped by Earth’s magnetic field, they follow along the field’s magnetic lines of force. These lines extend from above the equator to the North Pole and also to the South Pole then return to the equator.
When massive solar storms cause the Van Allen belts to become overloaded with particles, the result is the most spectacular feature of the ionosphere — the aurora. The particles spiral along magnetic field lines toward the poles. The charged particles energize oxygen and nitrogen gas molecules, causing them to light up. Each gas emits a particular color of light.
There is no real outer limit to the exosphere, the outermost layer of the atmosphere; the gas molecules finally become so scarce that at some point there are no more. Beyond the atmosphere is the solar wind. The solar wind is made of high-speed particles, mostly protons and electrons, traveling rapidly outward from the Sun.
Figure 18.2.1 The Earth's atmosphere Credit: NASA/Goddard
Contributions from Open Geography. Located at: Open Geography(opens in new window) [www.opengeography.org]. License: CC BY: Attribution(opens in new window) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/18%3A_Chemistry_of_the_Environment/18.02%3A_Outer_Regions_of_the_Atmosphere.txt |
Discussion Questions
• What is UV?
• How is ozone produced in the atmosphere?
• How much ozone is in the atmosphere, and where is the ozone layer?
• What is the interaction of ozone and UV?
• What is ozone depletion?
• What is ozone hole and how does it vary over time?
• What are CFCs?
• How do CFCs help depleting ozone?
• How is ozone depletion in the polar region different from other regions?
• What has been done and what can be done to reduce ozone depletion?
Most of the ozone in the atmosphere is in the stratosphere of the atmosphere, with about 8% in the lower troposphere. As mentioned there, the ozone is formed due to photo reaction. The ozone level is measured in Dobson Unit (DU), named after G.M.B. Dobson, who investigated the ozone between 1920 and 1960. One Dobson Unit (DU) is defined to be 0.01 mm thickness of ozone at STP when all the ozone in the air column above an area is collected and spread over the entire area. Thus, 100 DU is 1 mm thick.
What is UV?
In the electromagnetic radiation spectrum, the region beyond the violet (wavelength ~ 400 nanometer nm) invisible to eye detection is called ultraviolet (UV) rays. Its wavelength is shorter than 400 nm.
UV is divided into three regions:
• UV A, wavelength = 400 - 320 nm
• UV B, wavelength = 320 - 280 nm
• UV C, wavelength = < 280 nm
Obviously, photons of UV C are the most energetic. UV-A radiation is needed by humans for the synthesis of vitamin-D; however, too much UV-A causes photoaging (toughening of the skin), suppression of the immune system and, to a lesser degree, reddening of the skin, and cataract formation. Ozone strongly absorbs UV B and C, but the absorption decreases as the wavelength increases to 320 nm. Very little UV C reaches the Earth surface due to ozone absorption.
How is Ozone produced in the Atmosphere?
When an oxygen molecule receive a photon (h\nu), it dissociates into monoatomic (reactive) atoms. These atoms attack an oxygen molecule to form ozone, O3.
$\ce{O2 + h\nu \rightarrow O + O}\label{1}$
$\ce{O2 + O \rightarrow O3} \label{2}$
The last reaction requires a third molecule to take away the energy associated with the free radical $O^{\cdot}$ and $O_2$, and the reaction can be represented by
$\ce{O2 + O + M \rightarrow O3 + M*} \label{3}$
The over all reaction between oxygen and ozone formation is:
$\ce{3 O2 \rightleftharpoons 2 O3} \label{4}$
The absorption of UV B and C leads to the destruction of ozone
$\ce{O3 + h\nu \rightarrow O + O2} \label{5}$
$\ce{O3 + O \rightarrow 2 O2} \label{6}$
A dynamic equilibrium is established in these reactions. The ozone concentration varies due to the amount of radiation received from the sun.
$1$
The enthalpy of formation of ozone is 142.7 kJ / mol. The bond energy of O2 is 498 kJ / mol. What is the average O=O bond energy of the bent ozone molecule O=O=O?
Solution
The overall reaction is
$\ce{3 O2 \rightarrow 2 O3} \;\;\; \Delta H = 286 kJ$
Note that 3 O=O bonds of oxygen are broken, and 4 O-O bonds of ozone are formed. If the bond energy of ozone is E, then
\begin{align*} E &= (3*498 + 286) kJ / 4 mol \[4pt] &= 445 kJ / mol \end{align*}
DISCUSSION
The ozone bonds are slightly weaker than the oxygen bonds. The average bond energy is not the bond energy for the removal of one oxygen from ozone.
$\ce{O3 + h\nu \rightarrow O + O2}$
Can the energy to remove one oxygen be estimated from the data given here?
The techniques used in this calculation is based on the principle of conservation of energy.
Example $2$
The bond energy of O2 is 498 kJ / mol. What is the maximum wavelength of the photon that has enough energy to break the O=O bond of oxygen?
Solution
The energy per O=O bond is:
$(498000 J/mol) / (6.022x1023 bonds/mol) = 8.27x10-19 J/bonds$
The wavelength $\lambda$ of the photons can be evaluated using
$E = \dfrac{h c}{\lambda}$
\begin{align*} \lambda &= \dfrac{(6.626 \times 10^{-34}\, J \cdot s)*(3 \times 10^8\, m/s)}{8.27 \times 10^{-19} J} \[4pt] &= 2.403 \times 10^{-7} \,m = 240 nm \end{align*}
DISCUSSION
The visible region range from 300 nm to 700 nm, and radiation with a wavelength of 240 nm is in the ultraviolet region (Figure $1$). Visible light cannot break the O=O bond, and UV light has enough energy to break the O=O bond.
Chlorofluorohydrocarbons (CFCs)
Chemist Roy J. Plunkett discovered tetrafluoroethylene resin while researching refrigerants at DuPont. Known by its trade name, Teflon, Plunkett's discovery was found to be extremely heat-tolerant and stick-resistant. After ten years of research, Teflon was introduced in 1949. His continued research led to the usage of chlorofluorohydrocarbons known as CFCs or freon as refrigerants.
CFCs are made up of carbon, hydrogen, fluorine, and chlorine. DuPont used a number system to distinguish their product based on three digits. The digits are related to the molecular formulas.
• The first digit is the number of carbon atoms minus 1.
• The second digit is the number of hydrogen atoms plus 1.
• The third digit is the number of fluorine atoms minus 1.
For example, CFC (or freon) 123 should have a formula C2HF3Cl2. The number of chlorine atoms can be deduced from the structural formula of saturated carbon chains. CFC's containing only one carbon atom per molecule has only two digits. Freon 12 used for fridge and automobil air conditioners has a formula of CF2Cl2. The nontoxic and nonflammable CFCs have been widely used as refrigerants, in aerosol spray, and dry cleaning liquid, foam blowing agents, cleansers for electronic components in the 70s, 80s and early 90s.
In 1973, James Lovelock demonstrated that all the CFCs produced up to that time have not been destroyed, but spread globally throughout the troposphere. (Lovelock's report was later published: J. E. Lovelock, R.J.Maggs, and R.J. Wade, (1974); Nature, 241, 194) In the article, concentrations of CFCs at some parts per 1011 by volume was measured, and they deducted that with such a concentration, CFCs are not destroyed over the years. In 1974, Mario J. Molina published an article in Nature describing the ozone depletion by CFCs. (see M.J. Malina and F.S. Rowland, (1974); Nature, 249, 810) NASA later confirmed that HF was present in the stratosphere, and this compound had no natural source but from the decomposition of CFCs. Molina and Rowland suggested that the chlorine radicals in CFCs catalyze the decomposition of ozone as discussed below.
How do CFCs help depleting ozone?
A relatively recent concern is the depletion of ozone, O3 due to the presence of chlorine in the troposphere, and eventually their migration to the stratosphere. A major source of chlorine is Freons: CFCl3 (Freon 11), CF2Cl2 (Freon 12), C2F3Cl3 (Freon 113), C2F4Cl2 (Freon 114). Freons decompose in the troposphere. For example,
$\ce{CFCl3 \rightarrow CFCl2 + Cl}$
$\ce{CF2Cl3 \rightarrow CF2Cl + Cl^.}$
The chlorine atoms catalyze the decomposition of ozone,
$\ce{Cl + O3 \rightarrow ClO + O2}$
and ClO molecules further react with O generated due to photochemical decomposition of ozone:
$\ce{O3 + h\nu \rightarrow O + O2}$
$\ce{ClO + O \rightarrow Cl + O2}$
$\ce{O + O3 \rightarrow O2 + O2.}$
The net result or reaction is
$\ce{2 O3 \rightarrow 3 O2}$
Thus, the use of CFCs is now a world wide concern. In 1987, one hundred and forty nine (149) nations signed the Montreal Protocol. They agreed to reduce the manufacturing of CFCs by half in 1998; they also agree to phase out CFCs.
Ozone depletion in the polar region is different from other regions. The debate of ozone depletion often involves the North and South Poles. In these regions when temperatures drop to 190 K, ice cloud is formed. The ice crystals act as heterogeneous catalyst converting HCl and ClONO2 into $HNO_3$ and $Cl_2$,
$\ce{Cl + ClONO2 \rightarrow HNO3 + Cl2}$
$\ce{H2O + ClONO2 \rightarrow HNO3 + HOCl.}$
Both Cl2, and HOCl are easily photolyzed to Cl atoms, which catalyze the depletion of ozone. This has just been discussed in the previous section.
What has been done and what can be done to reduce ozone depletion?
The U.S. and Canadian governments have banned the use of Freons in aerosol sprays, but their use in air conditioner and cooling machines continue. In order to eliminate Freon in the atmosphere, international concerted effort and determination is required. However, sound and reliable scientific information is required. The banning of CFCs opens a research opportunity for another invention to find its substitute. Who knows what other problems will the new product bring?
Questions
1. What is the unit used for measuring ozone layers?
Skill - Define a unit you use.
2. What is the wavelength range of the UV radiation?
Skill - Describe UV radiation.
3. How is ozone different from oxygen?
Skill - Describe the formation of ozone.
4. When CFCs are exposed to UV or sun light, what species are produced?
Skill - Explain a photodecomposition reaction.
5. What is the role of chlorine radical in the ozone formation or reactions.
Skill - Explain the mechanism of the catalytic reaction.
6. What in the polar zone makes the depletion of ozone more serious?
Solutions
1. Dobson unit.
2. UV radiation is electromagnetic radiation with wavelength between 100 and 400 nm.
3. The ozone molecules consist of 3 atoms whereas the usual oxygen molecules 2.
4. Reactive radicals are produced, including monoatomic chlorine radical.
5. Chlorine radical catalyze the decomposition of ozone.
6. The ice clouds act as heterogeneous catalysts for the formation of chlorine gas.
Discussion - The chlorine gas is photodissociated into Cl radicals that catalyze the decomposition of ozone.
Learning Guide
• Arrange the regions of the electromagnetic spectrum in increasing energies of their photons: X-rays, visible, gamma rays, ultraviolet, infrared, microwave, etc.
• Examples on this page can be testing questions.
• What are CFCs?
What are freon 12, 123, and 114?
Explain how CFC help destroy the atmosphere ozone layer, particularly the polar region? | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/18%3A_Chemistry_of_the_Environment/18.03%3A_Ozone_in_the_Upper_Atmostphere.txt |
A lot of chemistry happens in Earth's atmosphere; there are many different kinds of chemicals in the air. Those chemicals often combine with each other in chemical reactions, making new and different chemicals. This is called "atmospheric chemistry".
Earth's atmosphere has different layers. The lowest layer is called the troposphere. We live in the troposphere. This page explains about atmospheric chemistry in the troposphere.
Most of the gas in our atmosphere is nitrogen. About 4/5ths of the air is nitrogen. What about the other 1/5th? Almost all of it is oxygen, the stuff in the air we need to breathe. There are also very small amounts of a bunch of other chemicals.
Have you heard of greenhouse gases? They are kinds of gases that trap the heat from sunlight in our atmosphere. Earth would be very cold if we didn't have any greenhouse gases. Carbon dioxide and methane are two very important greenhouse gases.
Some of the chemicals in the air come from pollution. When we burn coal in a factory or gasoline in our cars, we make air pollution. Coal and oil have sulfur in them. When they burn, they make chemicals called sulfur oxides. These can turn into sulfuric acid when they mix with water droplets in the air. These droplets of acid can fall to the ground as acid rain. Cars and trucks also give off chemicals called nitrogen oxides. Nitrogen oxides combine with other chemicals to make smog. They also help make nitric acid, which is another acid in acid rain.
Nature also does things to change the chemistry of the troposphere. Volcanoes, lightning, and wildfires all add chemicals to the air or change the ones that are already there. Energy from sunlight can make chemical reactions happen, changing one gas into another. Some chemicals move in cycles between the atmosphere, living creatures, and the oceans. The Carbon Cycle and the Nitrogen Cycles are two important cycles that change the chemistry of the atmosphere.
This table (below) describes some of the chemicals in the troposphere, and some of the chemical reactions that happen in the air:
Chemicals in the troposphere, and some of the chemical reactions that happen in the air
Chemical Formula Role in Tropospheric Chemistry
Carbon dioxide CO2 Carbon dioxide is a kind of greenhouse gas. When we breathe, we take in oxygen and breathe out carbon dioxide. Plants and some kinds of microbes use carbon dioxide during photosynthesis to make food. Burning fuels also puts carbon dioxide into the atmosphere.
Carbon monoxide CO When things burn, they mostly make carbon dioxide. Sometimes they make carbon monoxide, too. Carbon monoxide is a poisonous gas. Volcanoes and car enginesmake carbon monoxide.
Hydrocarbons CxOy Hydrocarbons are chemicals made up of hydrogen and carbon atoms. When fuel burns, it puts some hydrocarbons into the air. Hydrocarbons help to make smog, a kind of air pollution.
Methane CH4 Methane is a kind of greenhouse gas.
Nitrogen N2 Most of the gas in Earth's atmosphere is nitrogen. About 4/5ths of the air is nitrogen. The nitrogen cycle explains how nitrogen moves around in the environment. When fuel burns hot, like it does in the engine of a car, nitrogen combines with oxygen to make nitrogen oxides.
Nitrogen Oxides NO & NO2 Nitrogen oxides are a kind of pollution. Burning fuels like gasoline in air makes nitrogen oxides. Most nitrogen oxides come from cars and trucks. They help to make smog. They also mix with water droplets in the air to make nitric acid. Nitric acid is a part of acid rain.
Nitric Acid HNO3 Nitric acid is part of acid rain. Nitric acid forms when nitrogen oxides mix with water droplets in the air. Nitrogen oxides are a kind of pollution that comes from the engines of cars and trucks.
Oxygen & Ozone O2 & O3 About 1/5th of the gas in the atmosphere is oxygen. When you breathe, your body uses the oxygen to keep you alive. Ozone is a special kind of oxygen that has three atoms instead of two.
PAN(Peroxyacytyl nitrate) C2H3O5N PAN is a kind of air pollution. Smog has PAN in it. PAN forms when nitrogen dioxide, oxygen, and Volatile Organic Compounds (VOCs) get together.
Smog - Smog is a mixture of smoke and fog. Photochemical smog is a kind of air pollution. It has nitrogen oxides, ozone, VOCs, and PAN in it.
Photodissociation - When a photon of sunlight breaks apart a molecule.
Sulfur Oxides SO2 & SO3 Sulfur dioxide and sulfur trioxide are types of pollution. People make them when we burn coal and oil. Volcanoes also give off sulfur oxides. Sulfur dioxide mixes with water droplets in the air to make sulfuric acid. Sulfuric acid is in acid rain.
Sulfuric Acid H2SO4 Sulfuric acid is in acid rain. Sulfuric acid in the air is made when sulfur dioxide gasmixes with water droplets. The sulfur dioxide gas comes from volcanoes and from coal and oil that people burn for fuel.
www.windows2universe.org/ear...oposphere.html | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/18%3A_Chemistry_of_the_Environment/18.04%3A_Chemistry_of_the_Troposphere.txt |
Discussion Questions
• What is hard water?
• What are the differences between temporary and permanent hard water?
• How can hard water be converted to soft water?
• How to produce deionized water?
Water is the most important resource. Without water life is not possible. From a chemical point of view, water, H2O, is a pure compound, but in reality, you seldom drink, see, touch or use pure water. Water from various sources contains dissolved gases, minerals, organic and inorganic substances.
The Hydrosphere
The total water system surrounding the planet Earth is called the hydrosphere. It includes freshwater systems, oceans, atmosphere vapour, and biological waters. The Arctic, Atlantic, Indian, and Pacific oceans cover 71% of the Earth surface, and contain 97% of all water. Less than 1% is fresh water, and 2-3 % is ice caps and glaciers. The Antarctic Ice Sheet is almost the size of North America continent. These waters dominate our weather and climate, directly and indirectly affecting our daily lives. They cover 3.35x108 km2. The four oceans have a total volume of 1.35x109 km3.
• The sunlight dims by 1/10 for every 75 m in the ocean, and humans barely see light below 500 m. The temperature of almost all of the deep ocean is 4°C (277 K).
• The average ocean depth is 4 km, and the deepest point at the Mariana Trench is 10,912 m (35,802 ft), which compares to the height of 8.8 km for Mount Everest.
Hydrospheric processes are steps by which water cycles on the planet Earth. These processes include sublimation of ice, evaporation of liquid, transportation of moisture by air, rain, snow, river, lake, and ocean currents. All these processes are related to the physical and chemical properties of water, and many government agencies are set up to study and record phenomena related to them. The study of these processes is called hydrology
Among the planets, Earth is the only one in which there are solid, liquid and gaseous waters. These conditions are just right for life, for which water is a vital part. Water is the most abundant substance in the biosphere of Earth. Groundwater is an important part of the water system. When vapor is cooled, clouds and rain develop. Some of the rain percolate through the soil and into the underlying rocks. The water in the rocks is groundwater, which moves slowly.
A body of rock, which contains appreciable quantities of water, is called an aquifier. Below the water table, the aquifier is filled (or saturated) with water. Above the water table is the unsaturated zone. Some regions have two or more water tables. These zones are usually separated by water-impermeable material such as boulder and clay. Groundwater can be brought to the surface by drilling below the water table, and pumped out. The amount of water that can be pumped out depends on the structure of the aquifer. Little water is stored in tight granite layers, but large quantities of water are stored in limestone aquifier layers. In some areas, there are under ground rivers.
Table $1$: Ions in sea water
Species $\ce{Cl^{-}}$ $\ce{Na^{+}}$ $\ce{SO4^{^-}}$ $\ce{Mg^{2+}}$ $\ce{Ca^{2+}}$ $\ce{K^{+}}$ $\ce{HCO3^{-}}$ $\ce{Br^{-}}$ $\ce{Sr^{2+}}$ $\ce{BO4^{3-}}$ $\ce{F^{-}}$ $\ce{H4SiO4}$ $\ce{H^{+}}$
mg/Kg 10,760 2,710 2,710 1,290 411 399 142 67 8 4.5 1.3 0.5-10 $10^{-8.35}$
Common Ions Present in Natural Water
Hydrology is also the study of how solids and solute interact in, and with, water. In this link, the compositions of seawater, composition of the atmosphere, compositions of rain and snow, and compositions of river waters and lake waters are given in details. Table $1$ list the major ions present in seawater. The composition does vary, depending on region, depth, latitude, and water temperature. Waters at the river mouths contain less salt. If the ions are utilized by living organism, its contents vary according to the populations of organisms.
Dust particles and ions present in the air are nucleation center of water drops. Thus, waters from rain and snow also contain such ions: Ca2+, Mg2+, Na+, K+, NH4+. These cations are balanced by anions, HCO3-, SO4-, NO2-, Cl-, and NO3-. The pH of rain is between 5.5 and 5.6. Rain and snow waters eventually become river or lake waters. When the rain or snow waters fall, they interact with vegetation, top soil, bed rock, river bed and lake bed, dissolving whatever is soluble. Bacteria, algae, and water insects also thrive. Solubilities of inorganic salts are governed by the kinetics and equilibria of dissolution. The most common ions in lake and river waters are the same as those present in rainwater, but at higher concentrations. The pH of these waters depends on the river bed and lake bed. Natural waters contain dissolved minerals. Waters containing Ca2+ and Mg2+ ions are usually called hard water.
Hard Water
Minerals usually dissolve in natural water bodies such as lakes, rivers, springs, and underground waterways (ground waters). Calcium carbonate, CaCO3, is one of the most common inorganic compounds in the Earth crust. It is the ingredient for both calcite and aragonite. These two minerals have different crystal structures and appearance. This photograph shows crystals of typical Calcite.
Calcium-carbonate minerals dissolve in water, with a solubility product as shown below.
$CaCO_3 \rightleftharpoons Ca^{2+} + CO_3^{2-} \;\;\; K_{sp} = 5 \times 10^{-9}$
From the solubility product, we can (see example 1) evaluate the molar solubility to be 7.1x10-5 M or 7.1 mg/L (7.1 ppm of CaCO3 in water). The solubility increases as the pH decrease (increase acidity). This is compounded when the water is saturated with carbon dioxide, CO2. Saturated CO2 solution contains carbonic acid, which help the dissolution due to the reaction:
$H_2O + CO_2 \rightleftharpoons H_2CO_3$
$CaCO_3 + H_2CO_3 \rightleftharpoons Ca^{2+} + 2 HCO_3^-$
Because of these reactions, some natural waters contain more than 300 ppm calcium carbonates or its equivalents.
The carbon dioxide in natural water creates an interesting phenomenon. Rainwater is saturated with CO2, and it dissolves limestones. When CO2 is lost due to temperature changes or escaping from water drops, the reverse reaction takes place. The solid formed, however, may be a less stable phase called aragonite, which has the same chemical formula as, but a different crystal structure than that of calcite.
The rain dissolves calcium carbonate by the two reactions shown above. The water carries the ions with it, sips through the crack of the rocks. When it reached the ceiling of a cave, the drop dangles there for a long time before fallen. During this time, the carbon dioxide escapes and the pH of the water increases. Calcium carbonate crystals begin to appear. Calcite, aragonite, stalactite, and stalagmite are four common solids found in the formation of caves.
Natural waters contain metal ions. Water containing calcium, magnesium and their counter anions are called hard waters. Hard waters need to be treated for the following applications.
• Heat transfer carrier in boilers and in cooling systems
• Solvents and reagents in industrial chemical applications
• Domestic water for washing and cleaning
Temporary vs. Permanent Hard Water
Due to the reversibility of the reaction,
$CaCO_{3(s)} + H_2CO_3 \rightleftharpoons Ca^{2+} + 2 HCO_3^-$
water containing Ca2+, Mg2+ and CO32- ions is called temporary hard water, because the hardness can be removed by boiling. Boiling drives the reverse reaction, causing deposit in pipes and scales in boilers. The deposits lower the efficiency of heat transfer in boilers, and diminish flow rates of water in pipes. Thus, temporary hard water has to be softened before it enters the boiler, hot-water tank, or a cooling system. The amount of metal ions that can be removed by boiling is called temporary hardness
After boiling, metal ions remain due to presence of chloride ions, sulfate ions, nitrate ions, and a rather high solubility of MgCO3. Amount of metal ions that can not be removed by boiling is called permanent hardness. Total hardness is the sum of temporary hardness and permanent hardness. Hardness is often expressed as equivalence of amount of calcium ions in the solution. Thus, water conditioning is an important topic. The value of water treatment market has been estimated to be worth \$30 billion.
Lime-soda Softening
Lime-soda softening is the removal of temporary hardness by adding a calculated amount of hydrated lime, Ca(OH)2:
$Ca^{2+} + 2 HCO_3^- + Ca(OH)_{2(s)} \rightarrow 2 CaCO_{3(s)} + 2 H_2O$
Adding more lime causes the pH of water to increase, and as a result, magnesium ions are removed by the reaction:
$Mg^{2+} + Ca(OH)_{2(s)} \rightarrow Mg(OH)_{2(s)} + Ca^{2+}$
The extra calcium ions can be removed by the addition of sodium carbonate.
$Na_2CO_3 \rightarrow 2 Na^+ + CO_3^{2-}$
$Ca^{2+} + CO_3^{2-} + \rightarrow CaCO_{3(s)}$
In this treatment, the amount of Ca(OH)3 required is equivalent to the temporary hardness plus the magnesium hardness. The amount of sodium carbonate required is equivalent to the permanent hardness. Thus, lime-soda softening is effective if both the temporary and total hardness have been determined. The sodium ion will remain in the water after the treatment. The pH of the water is also rather high depending on the amount of lime and sodium carbonates used.
Complexation Treatment
Addition of complexing reagent to form soluble complexes with Ca2+ and Mg2+ prevents the formation of solid. One of the complexing agents is sodium triphosphate Na3PO4, which is marketed as Calgon, etc. The phosphate is the complexing agent. Other complexing agents such as Na2H2EDTA can also be used, but the complexing agent EDTA4- forms strong complexes with transition metals. This causes corrosion problem, unless the pipes of the system are made of stainless steel.
Ion Exchange
Today, most water softeners are using zeolites and employing ion exchange technique to soften hard water. Zeolites are a group of hydrated crystalline aluminosilicates found in certain volcanic rocks. The tetrahedrally coordinated aluminum and silicon atoms form AlO4 and SiO4 tetrahedral groups. They interconnect to each other sharing oxygen atoms forming cage-type structures as shown on the right. This diagram and the next structural diagram are taken from an introduction to zeolites There are many kinds of zeolites, some newly synthesized.
Whatever kind, the crystal structure of zeolites contains large cages. The cages are connected to each other forming a framework with many cavities and channels. Both positive and negative ions can be trapped in these cavities and channels as shown below.
For each oxygen that is not shared in the AlO4 and SiO4 tetrahedral groups, a negative charge is left on the group. These negative charges are balanced by trapping alkali metal and alkaline earth metal ions. When more cations are trapped, hydroxide and chloride ions will remain in the cavities and channels of the zeolites.
To prepare a zeolite for water treatment, they are soaked in concentrated NaCl solution. The cavities trap as many sodium ions as they can accommodate. After the treatment, the zeolite is designated as Na-zeolite. Then the salt solution is drained, and the zeolite is washed with water to eliminate the extra salt. When hard water flow through them, calcium and magnesium ions will be trapped by the Na-zeolite. For every Ca2+ or Mg2+ trapped, two Na+ ions are released. The treated water contains a rather high concentration of Na+ ions, but low concentrations of Mg2+ and Ca2+. Thus, zeolite ion exchange convert hard water into soft water.
Pure Water by Ion Exchange
In most cases, the resins are polystyrene with functional -SO3H groups attached to the polymer chain for cation exchange resin, and with functional group -N(CH3)3+ attached to the chain for anion exchange resin. To prepare the resin for making pure or deionized water, the cation resin is regenerated with HCl so that the groups are really -SO3H. The anion resin is regenerated with NaOH, so that the functional groups are -N(CH3)3(OH). When water containing any metal ion M+ and anion A- passes through the ion exchange resins in two stages, the following reactions take place,
$\ce{M+ + -SO3H \rightarrow H+ + -SO3M}$
$\ce{A^{-} + -N(CH3)3(OH) \rightarrow OH^{-} + -N(CH3)3A}$
$\ce{H^{+} + OH- <=> H2O}$
Thus, ion exchange provides pure water to meet laboratory requirement.
Reverse Osmosis Water Filter System
This method can also be used to prepare water for domestic and laboratory applications. This method has been discussed in Wastewater treatment
Magnetic Water Treatment
The following is a list of companies selling magnetic devices for magnetic water treatment. All devices are based on results of some ressearch indicating that when water runs through a magnetic field, the calcium carbonate will precipitate as aragonite rather than the usual calcite. For example, K.J. Kronenberg has published an article in IEEE Transactions on Magnetics, (Vol. Mag-21, No. 5, September 1985, pages 2059-2061). and stated the following:
The crystallization mode of the water's mineral content was found to change from a dendritic, substrate-bound solidification habit to the form of separate disc-shaped crystals after the water had moved through a number of magnetic fields. The former scarcity of crystallization nucleii in the water had been turned into an abundance of nucleation centers in the water. The reduction of the number of the substrate-bound crystals has been used as a quantitative measure of the magnetic effect.
Many companies have made various devices for magnetic conditioning of water, and they claim that their devices will clean up the pipes and boilers at little or no cost. I have yet to test one of these devices for its claim, but my preliminary tests shows that permanent magnet has little effect on the calcium carbonate deposit of temporary hard water. The cleaning effect they have claimed is probably much overstated.
Example $1$
From the solubility product shown for the dissolution of calcium carbonate,
$CaCO_3 \rightleftharpoons Ca^{2+} + CO_3^{2-} \;\;\; K_{sp} = 5 \times 10^{-9}$
Evaluate the molar solubility of Ca2+ in saturated solution.
Solution
From the definition of solubility product, we have
$[Ca^{2+}] [CO_3^{2-}] = 5 \times 10^{-9}$
Thus,
$[Ca^{2+}] = [CO_3^{2-}] = 7.1 \times 10^{-5}\; M$
The concentration of 7.1x10-5 M is equivalent to 7.1 mg/L (7.1 ppm of CaCO3 in water).
DISCUSSION
There may be other ions present in the system and other equilibria conditions in addition to the equilibrium mentioned here. Problems are more complexed in the real world.
Exercise $1$
Boiling of 1.0 L of water produced 10 mg of CaCO3 solid. What is the temporary hardness of the water?
Answer
10 ppm | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/18%3A_Chemistry_of_the_Environment/18.05%3A_The_World_Ocean.txt |
Discussion Questions
• Is water the source of life, why?
• What is life?
• Is life a symbiotic system of many lives?
• What properties of water make water the prime ingredient of life?
• What are the essential substances of life?
• Other than material, what else is essential for life?
• What is the role of water in supporting life?
• Is there a life form involving no water?
• Is there a method to determine the total water content?
Not finished yet, but the major ideas are here now!
Water Biology
Chemical and physical properties of water discussed in other pages are essential considerations for water biology. Natural water also contains biological matters as well as living creatures. In the discusssion of biology and water, you must feel pleasant, because they make each other grow more lovely as this picture from water lily cottage. The water pages of National Wildlife Federation, offers some interesting reading too.
All living things have a cycle of life. A cycle involves all or some of these process: birth, growth, mature, reproduction, matamorphosis, and death. There are millions of living organisms on Earth, ranging from single-cell amebas, bacteria, to the complicated homosapiens. There are also viruses which are fragment of DNA or RNA that depend on host cells for their reproduction. They are not cells.
Living things usually have cells that isolate their systems so that the cells contain unique materials to sustain the lives of cells. Cells regulate their contents (homeostatic), and carry out their metabolsims. They divide or making copies of themselves. Many reproduction process involve two individuals and future populations are subject to a greater diversity. Mutation is a fact of life, and many adopt to their changing environment.
How life started? Let the research and debate continue by not giving any clusive statements here. A physical geography course suggests that the marine invertebrates began their life 600 million years ago, and they are followed by fish, land plants, amphibians, reptiles, mommals, and then flowering plants, in this order. All these started more than a hundred million years ago, and hominid (primate) line began its evolution 20 to 15 million years ago.
There are strong evidences that life on earth appeared in a body of water. Only the planet Earth has three states of water, and it offers a suitable environment for life to began, among all nine solar planets. Since all life forms involve water. Water is seen as the source, matrix, and mother of life. Water is important, because water is required for life, and some people even consider water as life blood.
Since water supports life, living organisms also modify their environment, changing the nature of the water in which they live. Biology of water pollution, lists the syllabus on a course including a laboratory section. Water and biology interweave into an entangled maze waiting for explorers and curious minds.
Water dissolves or emulsifies other life-supporting substances and transport them to intercellular and intracellular fluids. It is also a medium in which reactions take place. Reactions provide energy (non-matter) for living. Energy causes changes, and manifestation of changes is at least related to, if not the whole, life. An organized and systematized set of reactions is essential in each life.
Balancing Water in Bio-systems
Many living organisms live their lives entirely in water as shown here in this photo from a job center talking about work in marine biology. Aquatic living organisms extract neutrients from water, yet maintaining a balance of electrolyte and nurrishment concentration in their cells. For living things not living in water, they extract water from their environment by whatever mechanism they can. Cells in their body are surrounded by body fluid, and all cells maintain constant concentrations of electrolytes, neutrints, and metabolites. The process of maintaining constant concentrations is called homeostasis. Certainly, some active transport mechanisms are involved in this balance.
The rooting of every type of plants is unique. Generally speaking, plants having extensive roots are able to extract water under harsh conditions. On the other hand, some plants such as cactus, jade and juniper have little roots, but their leaves have a layer of wax that prevents water from evaporation. Water conserving plants tolerate draught, and they survive under harsh conditions. The picture shown here is a jade plant from the above link.
Lately, some pumpkin growers harvested squash weighing almost 500 kg. At the peak of the growing season, the squash grows almost 0.5 kg a day. That is equivalent to 25 moles of water collected by the roots, discounting the water evaporated through the leaves. The growth is particularly good during a hot and wet day, but during a hot sunny after noon, the temperature of the leaves and fruits get very hot.
Essential Electrolytes for Life Support
In addition to water, many inorganic substances or minerals are essential to life. These substances ionize in water to form ions and their solutions conduct electricity. Therefore, they are called electrolytes. Because most of these substances are already dissoved in natural water, we list ions instead of the mineral they come from.
When ions dissove, they form complexes with water molecules. For most metals, the first sphere of coordination usually involve 6 water molecules. For example, when sodium chloride dissolve, we have
\[\ce{NaCl + 12 H2O <=> Na(H2O)6+ + Cl(H2O)6^{-}}\]
\[\ce{FeCl2 + 18 H2O <=> Fe(H2O)62+ + 2 Cl(H2O)6^{-}}\]
Formation of complexes are due to the high dipole moment of water, and the dissolution can be attribute to the high dielectric constant (80). However, in most publications, we ignore the water molecules in the complexes, and simply consider them as ions.
In the following, we describe some essential ions or salts as electrolytes.
• Sodium chloride, Na+ and Cl-
NaCl is readily dissolved and absorbed in extracellular fluid. The two ions help to balance water, acid/base, osmotic pressure, carbon dioxide transport, and excreted in human urine and sweat. Lack of sodium chloride shows symptoms of dehydration.
• Potassium, K+
Good sources of potassium ions are vegetables, fruits, grains, meat, milk, and legumes. It is readily absorbed, and actively transported into the intracellular fluid. Its function is similar to that of sodium ions, but cells prefer potassium ions over sodium. Lack of potassium leads to cardiac arrest.
• Calcium, Ca2+
Divalent calcium ions are usually poorly absorbed by human, but the are essential for the bones, teeth, blood clotting. Lack of calcium hinders growth and osteoporosis in old age.
• Phosphates, PO43-
Calcium phosphate is essential for bones, teeth, etc. However, phosphates are also responsible for many life reactions. ATP, NAD, FAD etc are metabolic intermediates, and they involve phosphate. Phospholipids and phosphoproteins are some other phosphate containing species.
• Magnesium, Mg2+
Magnesium ions are essential in chlorophyll. These ions are absorbed readily, and compete with calcium at times. Magnesium and calcium ions are present in hard water, and this link alerts the lack of magnesium leading to cardiovascular disease.
• Ferrous or ferric ions, Fe2+ or Fe3+
Usually known as iron, but iron are present either as divalent or as trivalent ions. Iron is absorbed according to body need; aided by HCl, ascorbic acid (vitamin C), and regulated by apoferritin. In mammals, iron is stored in liver as ferritin and hemosiderin. Iron deficiency leads to anemia. Good food sources of iron are liver, meats, egg yolk, green vegetables, and whole grains.
• Zinc ions, Zn2+
Zinc ions are important ingredients for many enzymes. They are present in insulin, carbonic anhydrase, carboxypeptidase, lactic dehydrogenase, alcohol dehydrogenase, alkaline phosphatase etc. Like iron, zinc deficiency leads to anemia and poor growth.
• Copper ions, Cu2+
Copper ions help iron utilization, and this metal is present in may enzymes.
• Cobalt ions, Co2+
Cobalt ions are centers of vitamin B12, and deficiency of which leads to anemia.
• Iondine ions, I-
Iondine is a constituent of thyroxin, which regulates cellular oxidation.
• Fluoride ions, F-
Fluoridation of drinking water is often a controversal issue. Childen's teeth are less seceptible to decay. Once they began to bruch their teeth, the fluoride in tooth paste is sufficient.
The eletrolytes listed above are present in significant amount in water, or fluids of organisms. There are some metals present in very minute quantities in biological systems, and these are not listed above.
Metal ions also interact with proteins. An enzyme is usually a very large protein molecule, and it folds into a kidney shape enclosing one or more metal ions forming a complex. The metal is usually responsible for the enzyme activity. Cobalt, copper, iron, molebdenium, nickel, and zinc have groups of enzymes each, and further discussion can be found in The Prosthetic groups and Metal Ions in Protein Active Sites Database (PROMISE). A general discussion is called bioinorganic chemistry and this site has an extensive General references on Bioinorganic Chemistry.
Balancing Electrolytes
Ions Extracellular Intracellular Interstitial
Na+ 140 10 150
K+ 5 150 4
Ca2+ 10 4 6
Mg2+ 6 80 4
Total 161 244 164
Cl- 103 2 120
HCO3- 30 10 30
HPO42- 4 177 4
SO42- 2 10 2
Organic acid 6 5 6
Protein 16 40 2
Total 161 244 164
Electrolyte balance are maintained by passive transport or diffusion and slective active transport mechanisms. Diffusion process tends to make the concentration all the same throughout the entire fluid, but active or selective transport moves ions to special compartment. For example, the active transport of sodium and potasium by an enzyme called sodium-potassium ATPase is usually known as sodium-potassium pump. This process pumps potassium ions inside a cell while removing sodium ions from the cells. Thus, a high concentration of potassium is maintained inside cells. Energy is required in active transport, and cellular metabolism provides the energy and the necessary molecular motions to facilitate the process.
Hormons are produced by special cells, and they are responsible for the communication between various part of the body. Some complicate harmon actions regulates the rate of transport and balance the ion concentrations depending on the portion of the tissue and the need. This is generally called the hormonal effects following the suggestion of human biochemistry.
Gibbs-Donnan effect considers the equilibrium in compartments that are separated by memberances or cell walls. There will be no net change when the products of concentrations of say [Na+]1, [Cl-]1 are the same for compartments 1 and 2.
[Na+]1 [Cl-]1 = [Na+]2 [Cl-]2
the subscript 1 and 2 refer to the two compartments. When no other components are present, we have
[Na+]1 = [Cl-]1 = [Na+]2 = [Cl-]2
But if compartment 2 has a sodium salt with other anions, this salt ionize to give Na+ too. The above condition will not be maintained, in this case. In other words, thermodynamic will be a force to adjust the concentrations.
In general, the cations should be balanced with anions. Otherwise, the solution will be charged.
Water in Human Biology
In human, water in the tissue and body fluid is mostly free, but some fraction may be bounded in pockets of hydrophilic compartments. Body fluids have many electrolytes and neutrients dissolve in them.
Intracellular fluid 70%
Interstitial fluid (lymph) 20%
Blood plasma 7%
Intestinal lumen etc. 3%
Human Biochemistry by J.M. Orten and O.W. Neuhaus (1982), 10th Ed. suggests that about 70% of human body weight is water, most found in three major compartments: 70% intracellular fluid, 20% interstitial fluid, and 7% blood plasma, and only 3 % in intestinal lumen, cerebrospinal fluid and other compartments.
However, Human Biochemistry also suggests that blood makes up about 8% of the total body weight.
Example 1
For a person weighing 50 kg (110 lb), what is the weight of the blood?
Solution
From the distribution given above,
Amount of blood = 50 kg * 0.08
= 4 kg.
This is a lot of blood, and donating 0.5 L of blood will not affect the normal function of the blood.
Input Output
Drinking 400 g Skin 500 g
Beverage 580 Expired air 350
Preformed water
in solid food
720 Urine 1100
Metabolic water 320 Feces 150
Total 2020 Total 2100
Balance -80 g?
Water in human comes from ingestion. Aside from drinking water, there is other beverages. Much of the food also contain water. When food is oxidized in the cells, all hydrogen in food converts to water, which is called metabolic water. Water is excreted via urine, feces, skin, and expiration. A typical daily water balance is shown in a table here. Water balance is maintained between cells and fluid, and the output depends on kidney functions and body insensible perspiration (Expired air from the lung is saturated with water vapor, and evaporation from the skin).
Drinking Water
Drinking water affects health. An Excite search using the phrase "drinking water" came up with 57890 documents. Drinking Water Resources gives annotated links to web sites that provide information about the drinking water.
A rather recent book Chemistry of Water Treatment by S.D. Faust and O.M. Aly, 2nd Ed. (1998) [TD433 F38 1998] addresses the standards for drinking water in the first Chapter. The standards have changed over the years, as we better understand the science.
Safe drinking water is a suitable combination of minerals and electrolytes. Usually, one should not drink water softened by water softeners. Using distilled water for beverages and cooking may not reach your set goals. Hard water with calcium and magnesium ions is good for drinking.
Usually, a government set up a non-profit organization to provide rules for safe drinking water. This organization has an infrastructure to monitor drinking water systems, and it shall also carry out research to improve the quality of drinking water.
Regarding making rules, reliable tests should be developed to determine the electrolyte we have mentioned, plus others such as lead ions, Pb2+, mercury Hg2+, methylmercury, arsenic, radioactivity, etc. Bacteria tests should be carried out regularly. This organization should also have a communication channel to release relavant message.
The Environmental Protection Agency of the U.S. gives a list of comtaminants. The list has suggested limits, and it divides the contaminants into
• Inorganic substnaces
• Organic substances
• Radioactivities (alpha and beta rays, radium)
• Micro-organisms
Among inorganic substances, limits are given to contents of antimony, arsenic, asbestos, barium, beryllum, cadmium, chromium, copper, mercury, nitrate, nitrite, selenium, and thallium.
More than 50 organic compounds are on the list, and some familar ones are: acrylamide, benzene, carbon tetrachloride, chlorobenzene, 2 4 D, dichlorobenzen, dioxin, polychlorinated biphenyls (PCBs), toluene, and vinyl chloride. Many of these have a zero limit.
In terms of microorganisms, Giardia lamblia, and Legionella, are checked. Furthermore, viruses, turbidity, total coliforms, and heterotrophic plate should be checked.
Contaminant Standard
Aluminum 0.05 to 0.2 mg/L
Chloride 250 mg/L
Color 15 (color units)
Copper 1.0 mg/L
Corrosivity noncorrosive
Fluoride 2.0 mg/L
Foaming Agents 0.5 mg/L
Iron 0.3 mg/L
Manganese 0.05 mg/L
Odor 3 threshold odor number
pH 6.5-8.5
Silver 0.10 mg/L
Sulfate 250 mg/L
Total Dissolved Solids 500 mg/L
Zinc 5 mg/L
The secondary standard lists most electrolytes as shown in this table on the right.
The Secondary Drinking Water Standards are non-enforceable guidelines regulating contaminants that may cause cosmetic effects (such as skin or tooth discoloration) or aesthetic effects (such as taste, odor, or color) in drinking water.
There are many brands of bottled drinking water, they have been very popular only in recent years. Do we know the bottle procedure? Is the industry regulated? Is the water quality reliable? Are all bottled water the same? Is there a brand bottled water for real good health? Do we know what should be in healthy drinking water? There is an opinion expressed in Ontario Clean Water Agency (OCWA). Check it out.
The magnesium web site gave the following news release Oct. 4, 1999. According to the U.S. National Academy of Sciences (1977) there have been more than 50 studies, in nine countries, that have indicated an inverse relationship between water hardness and mortality from cardiovascular disease. That is, people who drink water that is deficient in magnesium and calcium generally appear more susceptible to this disease. The U.S. National Academy of Sciences has estimated that a nation-wide initiative to add calcium and magnesium to soft water might reduce the annual cardiovascular death rate by 150,000 in the United States. This is a good summary from the report.
Sports Drinks
Sports talk gives information about Sports Drinks. This is science, art, testing, and myth. However, some fundamentals should be considered.
Our bodies are mostly water, about 70%. The body fluid has many different things dissolved in it, particularly salt. the salinity - varies somewhat with where you take the sample of water to measure. Don't worry about that). I recall the concentration as being 0.5%, but I'm not sure. Doctors call this "normal saline".
Now if you put a human, or other animal, cell in saltwater that is the same concentration as the saltwater inside the cell, the cell pretty much just sits there. If you put it in distilled water, the cell absorbs the water through its cell membrane - called diffusion - until it eventually pops. If you put the cell in concentrated saltwater, the cell looses water. The water diffuses out of the cell through the membrane, leaving a small, shriveled up cell.
What does this have to do with sports drinks? If you give someone distilled water, it seems like they would absorb the water faster because of what I just described. On the other hand, during sweating, you're loosing sodium, potassium and small quantities of other electrolytes. If you're exercising particularly long or hard, you need to replace those electrolytes. Researchers found that adding some salt to water replaced the salt lost through sweating and helped the body to get water to the cells. If you look at a label on a Gatorade or other drink, you'll find that the main electrolyte is simple salt. But if you put too much of the electrolytes in the water, the cells shrivel up just like the way I described above.
I hope that helps you understand what happens. This stuff on how cells swell up or shrivel up is in high school biology books, and maybe even in something you can find in your school's library.
Taste and Order of Drinks
Taste and order are sensations, and thus hard to quantify and systematize. Often, the Weber Fechner law is used. This law expresses the taste or order sensations S as being proportional to the logarithm of the stimulus R, with the proportional constant K,
S = K log R
For some common substances, the minimum amounts detected by expert nose or expert taster gives the sensation as a threshold, below which no taste or order was detected.
However, orderous sensation of water may be reported as threshold order number (TON). If A mL of ordorous sample is diluted with B mL of order-free water to be "just detectable" to the expert nose, the TON is defined as
A + B
TON = -------------
A
Similarly, a flavour threshold number (FTN) can be defined in the same manner.
A + B
FTN = -------------
A
except that A and B are volumes of samples and taste free water used.
These formulations show a method of defining some quantities that are, otherwise, very diffcult to quantify. There are other methods of reporting orders and taste, and some bottling companies have their standard methods of comparison.
Of course, the source of order and taste comes are organic, inorganic compounds as well as bacteria, algae. For example, mercaptans such as C2H2SH and ammonia offers disagreeable smell and taste.
Example 2
A sample of water was tested by 10 expert noses, and only 5 of them can detect an order. Thus, this is "just detected". What is the TON for this sample?
Solution
Since no dilution was used,
TON = A / A = 1.
DISCUSSION
If equal amount of orderless water is required to dilute it so that the order is "just detected", then the test order number is (1+1)/1 = 2.
18.07: Green Chemistry
Green chemistry, also called sustainable chemistry, is an area of chemistry and chemical engineering focused on the design of products and processes that minimize the use and generation of hazardous substances. Whereas environmental chemistry focuses on the effects of polluting chemicals on nature, green chemistry focuses on technological approaches to preventing pollution and reducing consumption of nonrenewable resources.
Green chemistry overlaps with all subdisciplines of chemistry but with a particular focus on chemical synthesis, process chemistry, and chemical engineering, in industrial applications. To a lesser extent, the principles of green chemistry also affect laboratory practices. The overarching goals of green chemistry—namely, more resource-efficient and inherently safer design of molecules, materials, products, and processes—can be pursued in a wide range of contexts.
• Wikipedia | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/18%3A_Chemistry_of_the_Environment/18.06%3A_Fresh_Water.txt |
Our goal in this chapter is to extend the concepts of thermochemistry to an exploration of thermodynamics (from the Greek thermo and dynamic, meaning “heat” and “power,” respectively), the study of the interrelationships among heat, work, and the energy content of a system at equilibrium. Thermodynamics tells chemists whether a particular reaction is energetically possible in the direction in which it is written, and it gives the composition of the reaction system at equilibrium. It does not, however, say anything about whether an energetically feasible reaction will actually occur as written, and it tells us nothing about the reaction rate or the pathway by which it will occur (described by chemical kinetics). Chemical thermodynamics provides a bridge between the macroscopic properties of a substance and the individual properties of its constituent molecules and atoms. As you will see, thermodynamics explains why graphite can be converted to diamond; how chemical energy stored in molecules can be used to perform work; and why certain processes, such as iron rusting and organisms aging and dying, proceed spontaneously in only one direction, requiring no net input of energy to occur.
• 19.1: Spontaneous Processes
Chemical and physical processes have a natural tendency to occur in one direction under certain conditions. A spontaneous process occurs without the need for a continual input of energy from some external source, while a nonspontaneous process requires such. Systems undergoing a spontaneous process may or may not experience a gain or loss of energy, but they will experience a change in the way matter and/or energy is distributed within the system.
• 19.2: Entropy and the Second Law of Thermodynamics
Entropy (S) is a state function whose value increases with an increase in the number of available microstates.For a given system, the greater the number of microstates, the higher the entropy. During a spontaneous process, the entropy of the universe increases.
• 19.3: The Molecular Interpretation of Entropy
These forms of motion are ways in which the molecule can store energy. The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy. A perfectly ordered system with only a single microstate available to it would have an entropy of zero. The only system that meets this criterion is a perfect crystal at a temperature of absolute zero (0 K), in which each component atom, molecule, or ion is fixed in place within a perfect crystal lattice.
• 19.4: Entropy Changes in Chemical Reactions
Changes in internal energy, that are not accompanied by a temperature change, might reflect changes in the entropy of the system.
• 19.5: Gibbs Free Energy
We can predict whether a reaction will occur spontaneously by combining the entropy, enthalpy, and temperature of a system in a new state function called Gibbs free energy (G). The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is spontaneous, ΔG < 0. If the process is not spontaneous as written.
• 19.6: Free Energy and Temperature
We can predict if a reaction will occur spontaneously by combining the entropy, enthalpy, and temperature of a system in a new state function called Gibbs free energy (G). The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is spontaneous, ΔG < 0. If the process is spontaneous in the reverse direction, ΔG > 0.
• 19.7: Free Energy and the Equilibrium Constant
For a reversible process (with no external work), the change in free energy can be expressed in terms of volume, pressure, entropy, and temperature. If the products and reactants are in their standard states and ΔG° < 0, then K > 1, and products are favored over reactants at equilibrium. If ΔG° > 0, then K < 1, and reactants are favored over products at equilibrium. If ΔG° = 0, then K=1, and neither reactants nor products are favored at equilibrium. We can use the measured equilibrium constant
• 19.E: Chemical Thermodynamics (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al.
19: Chemical Thermodynamics
Learning Objectives
• Distinguish between spontaneous and nonspontaneous processes
• Describe the dispersal of matter and energy that accompanies certain spontaneous processes
In this section, consider the differences between two types of changes in a system: Those that occur spontaneously and those that occur only with the continuous input of energy. In doing so, we’ll gain an understanding as to why some systems are naturally inclined to change in one direction under certain conditions. We’ll also gain insight into how the spontaneity of a process affects the distribution of energy and matter within the system.
Spontaneous and Nonspontaneous Processes
Processes have a natural tendency to occur in one direction under a given set of conditions. Water will naturally flow downhill, but uphill flow requires outside intervention such as the use of a pump. A spontaneous process is one that occurs naturally under certain conditions. A nonspontaneous process, on the other hand, will not take place unless it is “driven” by the continual input of energy from an external source. A process that is spontaneous in one direction under a particular set of conditions is nonspontaneous in the reverse direction. At room temperature and typical atmospheric pressure, for example, ice will spontaneously melt, but water will not spontaneously freeze.
The spontaneity of a process is not correlated to the speed of the process. A spontaneous change may be so rapid that it is essentially instantaneous or so slow that it cannot be observed over any practical period of time. To illustrate this concept, consider the decay of radioactive isotopes, a topic more thoroughly treated in the chapter on nuclear chemistry. Radioactive decay is by definition a spontaneous process in which the nuclei of unstable isotopes emit radiation as they are converted to more stable nuclei. All the decay processes occur spontaneously, but the rates at which different isotopes decay vary widely. Technetium-99m is a popular radioisotope for medical imaging studies that undergoes relatively rapid decay and exhibits a half-life of about six hours. Uranium-238 is the most abundant isotope of uranium, and its decay occurs much more slowly, exhibiting a half-life of more than four billion years (Figure $1$).
As another example, consider the conversion of diamond into graphite (Figure $2$).
$\ce{C(s, diamond)}⟶\ce{C(s, graphite)} \label{Eq1}$
The phase diagram for carbon indicates that graphite is the stable form of this element under ambient atmospheric pressure, while diamond is the stable allotrope at very high pressures, such as those present during its geologic formation. Thermodynamic calculations of the sort described in the last section of this chapter indicate that the conversion of diamond to graphite at ambient pressure occurs spontaneously, yet diamonds are observed to exist, and persist, under these conditions. Though the process is spontaneous under typical ambient conditions, its rate is extremely slow, and so for all practical purposes diamonds are indeed “forever.” Situations such as these emphasize the important distinction between the thermodynamic and the kinetic aspects of a process. In this particular case, diamonds are said to be thermodynamically unstable but kinetically stable under ambient conditions.
Dispersal of Matter and Energy
As we extend our discussion of thermodynamic concepts toward the objective of predicting spontaneity, consider now an isolated system consisting of two flasks connected with a closed valve. Initially there is an ideal gas on the left and a vacuum on the right (Figure $3$). When the valve is opened, the gas spontaneously expands to fill both flasks. Recalling the definition of pressure-volume work from the chapter on thermochemistry, note that no work has been done because the pressure in a vacuum is zero.
\begin{align} w&=−PΔV \[4pt]&=0 \,\,\, \mathrm{(P=0\: in\: a\: vaccum)} \label{Eq2} \end{align}
Note as well that since the system is isolated, no heat has been exchanged with the surroundings (q = 0). The first law of thermodynamics confirms that there has been no change in the system’s internal energy as a result of this process.
\begin{align} ΔU&=q+w \tag{First Law of Thermodynamics} \[4pt] &=0+0=0 \label{Eq3}\end{align}
The spontaneity of this process is therefore not a consequence of any change in energy that accompanies the process. Instead, the movement of the gas appears to be related to the greater, more uniform dispersal of matter that results when the gas is allowed to expand. Initially, the system was comprised of one flask containing matter and another flask containing nothing. After the spontaneous process took place, the matter was distributed both more widely (occupying twice its original volume) and more uniformly (present in equal amounts in each flask).
Now consider two objects at different temperatures: object X at temperature TX and object Y at temperature TY, with TX > TY (Figure $4$). When these objects come into contact, heat spontaneously flows from the hotter object (X) to the colder one (Y). This corresponds to a loss of thermal energy by X and a gain of thermal energy by Y.
$q_\ce{X}<0 \hspace{20px} \ce{and} \hspace{20px} q_\ce{Y}=−q_\ce{X}>0 \label{Eq4}$
From the perspective of this two-object system, there was no net gain or loss of thermal energy, rather the available thermal energy was redistributed among the two objects. This spontaneous process resulted in a more uniform dispersal of energy.
As illustrated by the two processes described, an important factor in determining the spontaneity of a process is the extent to which it changes the dispersal or distribution of matter and/or energy. In each case, a spontaneous process took place that resulted in a more uniform distribution of matter or energy.
Example $1$: Redistribution of Matter during a Spontaneous Process
Describe how matter and energy are redistributed when the following spontaneous processes take place:
1. A solid sublimes.
2. A gas condenses.
3. A drop of food coloring added to a glass of water forms a solution with uniform color.
Solution
1. Sublimation is the conversion of a solid (relatively high density) to a gas (much lesser density). This process yields a much greater dispersal of matter, since the molecules will occupy a much greater volume after the solid-to-gas transition. However, an input of energy from the surroundings ss required for the molecules to leave the solid phase and enter the gas phase.
2. Condensation is the conversion of a gas (relatively low density) to a liquid (much greater density). This process yields a much lesser dispersal of matter, since the molecules will occupy a much lesser volume after the gas-to-liquid transition. As the gas molecules move together to form the droplets of liquid, they form intermolecular forces and thus release energy to the surroundings.
3. The process in question is dilution. The food dye molecules initially occupy a much smaller volume (the drop of dye solution) than they occupy once the process is complete (in the full glass of water). The process therefore entails a greater dispersal of matter. The process may also yield a more uniform dispersal of matter, since the initial state of the system involves two regions of different dye concentrations (high in the drop, zero in the water), and the final state of the system contains a single dye concentration throughout. This process can occur with out a change in energy because the molecules have kinetic energy relative to the temperature of the water, and so will be constantly in motion.
Exercise $1$
Describe how matter and energy are redistributed when you empty a canister of compressed air into a room.
Answer
This process entails both a greater and more uniform dispersal of matter as the compressed air in the canister is permitted to expand into the lower-pressure air of the room. The process also requires an input of energy to disrupt the intermolecular forces between the closely-spaced gas molecules that are originally compressed into the container. If you were to touch the nozzle of the canister, you would notice that it is cold because the exiting molecules are taking energy away from their surroundings, and the canister is part of the surroundings.
Summary
Chemical and physical processes have a natural tendency to occur in one direction under certain conditions. A spontaneous process occurs without the need for a continual input of energy from some external source, while a nonspontaneous process requires such. Systems undergoing a spontaneous process may or may not experience a gain or loss of energy, but they will experience a change in the way matter and/or energy is distributed within the system. In this section we have only discussed nuclear decay, physical changes of pure substances, and macroscopic events such as water flowing downhill. In the following sections we will discuss mixtures and chemical reactions, situations in which the description of sponteneity becomes more challenging.
Glossary
nonspontaneous process
process that requires continual input of energy from an external source
spontaneous change
process that takes place without a continuous input of energy from an external source | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/19%3A_Chemical_Thermodynamics/19.01%3A_Spontaneous_Processes.txt |
Learning Objectives
• To understand the relationship between internal energy and entropy.
The first law of thermodynamics governs changes in the state function we have called internal energy ($U$). Changes in the internal energy (ΔU) are closely related to changes in the enthalpy (ΔH), which is a measure of the heat flow between a system and its surroundings at constant pressure. You also learned previously that the enthalpy change for a chemical reaction can be calculated using tabulated values of enthalpies of formation. This information, however, does not tell us whether a particular process or reaction will occur spontaneously.
Let’s consider a familiar example of spontaneous change. If a hot frying pan that has just been removed from the stove is allowed to come into contact with a cooler object, such as cold water in a sink, heat will flow from the hotter object to the cooler one, in this case usually releasing steam. Eventually both objects will reach the same temperature, at a value between the initial temperatures of the two objects. This transfer of heat from a hot object to a cooler one obeys the first law of thermodynamics: energy is conserved.
Now consider the same process in reverse. Suppose that a hot frying pan in a sink of cold water were to become hotter while the water became cooler. As long as the same amount of thermal energy was gained by the frying pan and lost by the water, the first law of thermodynamics would be satisfied. Yet we all know that such a process cannot occur: heat always flows from a hot object to a cold one, never in the reverse direction. That is, by itself the magnitude of the heat flow associated with a process does not predict whether the process will occur spontaneously.
For many years, chemists and physicists tried to identify a single measurable quantity that would enable them to predict whether a particular process or reaction would occur spontaneously. Initially, many of them focused on enthalpy changes and hypothesized that an exothermic process would always be spontaneous. But although it is true that many, if not most, spontaneous processes are exothermic, there are also many spontaneous processes that are not exothermic. For example, at a pressure of 1 atm, ice melts spontaneously at temperatures greater than 0°C, yet this is an endothermic process because heat is absorbed. Similarly, many salts (such as NH4NO3, NaCl, and KBr) dissolve spontaneously in water even though they absorb heat from the surroundings as they dissolve (i.e., ΔHsoln > 0). Reactions can also be both spontaneous and highly endothermic, like the reaction of barium hydroxide with ammonium thiocyanate shown in Figure $1$.
Thus enthalpy is not the only factor that determines whether a process is spontaneous. For example, after a cube of sugar has dissolved in a glass of water so that the sucrose molecules are uniformly dispersed in a dilute solution, they never spontaneously come back together in solution to form a sugar cube. Moreover, the molecules of a gas remain evenly distributed throughout the entire volume of a glass bulb and never spontaneously assemble in only one portion of the available volume. To help explain why these phenomena proceed spontaneously in only one direction requires an additional state function called entropy (S), a thermodynamic property of all substances that is proportional to their degree of "disorder". In Chapter 13, we introduced the concept of entropy in relation to solution formation. Here we further explore the nature of this state function and define it mathematically.
Entropy
Chemical and physical changes in a system may be accompanied by either an increase or a decrease in the disorder of the system, corresponding to an increase in entropy (ΔS > 0) or a decrease in entropy (ΔS < 0), respectively. As with any other state function, the change in entropy is defined as the difference between the entropies of the final and initial states: ΔS = Sf − Si.
When a gas expands into a vacuum, its entropy increases because the increased volume allows for greater atomic or molecular disorder. The greater the number of atoms or molecules in the gas, the greater the disorder. The magnitude of the entropy of a system depends on the number of microscopic states, or microstates, associated with it (in this case, the number of atoms or molecules); that is, the greater the number of microstates, the greater the entropy.
We can illustrate the concepts of microstates and entropy using a deck of playing cards, as shown in Figure $2$. In any new deck, the 52 cards are arranged by four suits, with each suit arranged in descending order. If the cards are shuffled, however, there are approximately 1068 different ways they might be arranged, which corresponds to 1068 different microscopic states. The entropy of an ordered new deck of cards is therefore low, whereas the entropy of a randomly shuffled deck is high. Card games assign a higher value to a hand that has a low degree of disorder. In games such as five-card poker, only 4 of the 2,598,960 different possible hands, or microstates, contain the highly ordered and valued arrangement of cards called a royal flush, almost 1.1 million hands contain one pair, and more than 1.3 million hands are completely disordered and therefore have no value. Because the last two arrangements are far more probable than the first, the value of a poker hand is inversely proportional to its entropy.
We can see how to calculate these kinds of probabilities for a chemical system by considering the possible arrangements of a sample of four gas molecules in a two-bulb container (Figure $3$). There are five possible arrangements: all four molecules in the left bulb (I); three molecules in the left bulb and one in the right bulb (II); two molecules in each bulb (III); one molecule in the left bulb and three molecules in the right bulb (IV); and four molecules in the right bulb (V). If we assign a different color to each molecule to keep track of it for this discussion (remember, however, that in reality the molecules are indistinguishable from one another), we can see that there are 16 different ways the four molecules can be distributed in the bulbs, each corresponding to a particular microstate. As shown in Figure $3$, arrangement I is associated with a single microstate, as is arrangement V, so each arrangement has a probability of 1/16. Arrangements II and IV each have a probability of 4/16 because each can exist in four microstates. Similarly, six different microstates can occur as arrangement III, making the probability of this arrangement 6/16. Thus the arrangement that we would expect to encounter, with half the gas molecules in each bulb, is the most probable arrangement. The others are not impossible but simply less likely.
There are 16 different ways to distribute four gas molecules between the bulbs, with each distribution corresponding to a particular microstate. Arrangements I and V each produce a single microstate with a probability of 1/16. This particular arrangement is so improbable that it is likely not observed. Arrangements II and IV each produce four microstates, with a probability of 4/16. Arrangement III, with half the gas molecules in each bulb, has a probability of 6/16. It is the one encompassing the most microstates, so it is the most probable.
Instead of four molecules of gas, let’s now consider 1 L of an ideal gas at standard temperature and pressure (STP), which contains 2.69 × 1022 molecules (6.022 × 1023 molecules/22.4 L). If we allow the sample of gas to expand into a second 1 L container, the probability of finding all 2.69 × 1022 molecules in one container and none in the other at any given time is extremely small, approximately $\frac{2}{2.69 \times 10^{22}}$. The probability of such an occurrence is effectively zero. Although nothing prevents the molecules in the gas sample from occupying only one of the two bulbs, that particular arrangement is so improbable that it is never actually observed. The probability of arrangements with essentially equal numbers of molecules in each bulb is quite high, however, because there are many equivalent microstates in which the molecules are distributed equally. Hence a macroscopic sample of a gas occupies all of the space available to it, simply because this is the most probable arrangement.
A disordered system has a greater number of possible microstates than does an ordered system, so it has a higher entropy. This is most clearly seen in the entropy changes that accompany phase transitions, such as solid to liquid or liquid to gas. As you know, a crystalline solid is composed of an ordered array of molecules, ions, or atoms that occupy fixed positions in a lattice, whereas the molecules in a liquid are free to move and tumble within the volume of the liquid; molecules in a gas have even more freedom to move than those in a liquid. Each degree of motion increases the number of available microstates, resulting in a higher entropy. Thus the entropy of a system must increase during melting (ΔSfus > 0). Similarly, when a liquid is converted to a vapor, the greater freedom of motion of the molecules in the gas phase means that ΔSvap > 0. Conversely, the reverse processes (condensing a vapor to form a liquid or freezing a liquid to form a solid) must be accompanied by a decrease in the entropy of the system: ΔS < 0.
Entropy (S) is a thermodynamic property of all substances that is proportional to their degree of disorder. The greater the number of possible microstates for a system, the greater the disorder and the higher the entropy.
Experiments show that the magnitude of ΔSvap is 80–90 J/(mol•K) for a wide variety of liquids with different boiling points. However, liquids that have highly ordered structures due to hydrogen bonding or other intermolecular interactions tend to have significantly higher values of ΔSvap. For instance, ΔSvap for water is 102 J/(mol•K). Another process that is accompanied by entropy changes is the formation of a solution. As illustrated in Figure $4$, the formation of a liquid solution from a crystalline solid (the solute) and a liquid solvent is expected to result in an increase in the number of available microstates of the system and hence its entropy. Indeed, dissolving a substance such as NaCl in water disrupts both the ordered crystal lattice of NaCl and the ordered hydrogen-bonded structure of water, leading to an increase in the entropy of the system. At the same time, however, each dissolved Na+ ion becomes hydrated by an ordered arrangement of at least six water molecules, and the Cl ions also cause the water to adopt a particular local structure. Both of these effects increase the order of the system, leading to a decrease in entropy. The overall entropy change for the formation of a solution therefore depends on the relative magnitudes of these opposing factors. In the case of an NaCl solution, disruption of the crystalline NaCl structure and the hydrogen-bonded interactions in water is quantitatively more important, so ΔSsoln > 0.
Dissolving NaCl in water results in an increase in the entropy of the system. Each hydrated ion, however, forms an ordered arrangement with water molecules, which decreases the entropy of the system. The magnitude of the increase is greater than the magnitude of the decrease, so the overall entropy change for the formation of an NaCl solution is positive.
Example $1$
Predict which substance in each pair has the higher entropy and justify your answer.
1. 1 mol of NH3(g) or 1 mol of He(g), both at 25°C
2. 1 mol of Pb(s) at 25°C or 1 mol of Pb(l) at 800°C
Given: amounts of substances and temperature
Asked for: higher entropy
Strategy:
From the number of atoms present and the phase of each substance, predict which has the greater number of available microstates and hence the higher entropy.
Solution:
1. Both substances are gases at 25°C, but one consists of He atoms and the other consists of NH3 molecules. With four atoms instead of one, the NH3 molecules have more motions available, leading to a greater number of microstates. Hence we predict that the NH3 sample will have the higher entropy.
2. The nature of the atomic species is the same in both cases, but the phase is different: one sample is a solid, and one is a liquid. Based on the greater freedom of motion available to atoms in a liquid, we predict that the liquid sample will have the higher entropy.
Exercise $1$
Predict which substance in each pair has the higher entropy and justify your answer.
1. 1 mol of He(g) at 10 K and 1 atm pressure or 1 mol of He(g) at 250°C and 0.2 atm
2. a mixture of 3 mol of H2(g) and 1 mol of N2(g) at 25°C and 1 atm or a sample of 2 mol of NH3(g) at 25°C and 1 atm
Answer a
1 mol of He(g) at 250°C and 0.2 atm (higher temperature and lower pressure indicate greater volume and more microstates)
Answer a
a mixture of 3 mol of H2(g) and 1 mol of N2(g) at 25°C and 1 atm (more molecules of gas are present)
Reversible and Irreversible Changes
Changes in entropy (ΔS), together with changes in enthalpy (ΔH), enable us to predict in which direction a chemical or physical change will occur spontaneously. Before discussing how to do so, however, we must understand the difference between a reversible process and an irreversible one. In a reversible process, every intermediate state between the extremes is an equilibrium state, regardless of the direction of the change. In contrast, an irreversible process is one in which the intermediate states are not equilibrium states, so change occurs spontaneously in only one direction. As a result, a reversible process can change direction at any time, whereas an irreversible process cannot. When a gas expands reversibly against an external pressure such as a piston, for example, the expansion can be reversed at any time by reversing the motion of the piston; once the gas is compressed, it can be allowed to expand again, and the process can continue indefinitely. In contrast, the expansion of a gas into a vacuum (Pext = 0) is irreversible because the external pressure is measurably less than the internal pressure of the gas. No equilibrium states exist, and the gas expands irreversibly. When gas escapes from a microscopic hole in a balloon into a vacuum, for example, the process is irreversible; the direction of airflow cannot change.
Because work done during the expansion of a gas depends on the opposing external pressure (w = - PextΔV), work done in a reversible process is always equal to or greater than work done in a corresponding irreversible process: wrev ≥ wirrev. Whether a process is reversible or irreversible, ΔU = q + w. Because U is a state function, the magnitude of ΔU does not depend on reversibility and is independent of the path taken. So
$ΔU = q_{rev} + w_{rev} = q_{irrev} + w_{irrev} \label{Eq1}$
Work done in a reversible process is always equal to or greater than work done in a corresponding irreversible process: wrev ≥ wirrev.
In other words, ΔU for a process is the same whether that process is carried out in a reversible manner or an irreversible one. We now return to our earlier definition of entropy, using the magnitude of the heat flow for a reversible process (qrev) to define entropy quantitatively.
The Relationship between Internal Energy and Entropy
Because the quantity of heat transferred (qrev) is directly proportional to the absolute temperature of an object (T) (qrev ∝ T), the hotter the object, the greater the amount of heat transferred. Moreover, adding heat to a system increases the kinetic energy of the component atoms and molecules and hence their disorder (ΔS ∝ qrev). Combining these relationships for any reversible process,
$q_{\textrm{rev}}=T\Delta S\;\textrm{ and }\;\Delta S=\dfrac{q_{\textrm{rev}}}{T} \label{Eq2}$
Because the numerator (qrev) is expressed in units of energy (joules), the units of ΔS are joules/kelvin (J/K). Recognizing that the work done in a reversible process at constant pressure is wrev = −PΔV, we can express Equation $\ref{Eq1}$ as follows:
\begin{align} ΔU &= q_{rev} + w_{rev} \[4pt] &= TΔS − PΔV \label{Eq3} \end{align}
Thus the change in the internal energy of the system is related to the change in entropy, the absolute temperature, and the $PV$ work done.
To illustrate the use of Equation $\ref{Eq2}$ and Equation $\ref{Eq3}$, we consider two reversible processes before turning to an irreversible process. When a sample of an ideal gas is allowed to expand reversibly at constant temperature, heat must be added to the gas during expansion to keep its $T$ constant (Figure $5$). The internal energy of the gas does not change because the temperature of the gas does not change; that is, $ΔU = 0$ and $q_{rev} = −w_{rev}$. During expansion, ΔV > 0, so the gas performs work on its surroundings:
$w_{rev} = −PΔV < 0. \nonumber$
According to Equation $\ref{Eq3}$, this means that qrev must increase during expansion; that is, the gas must absorb heat from the surroundings during expansion, and the surroundings must give up that same amount of heat. The entropy change of the system is therefore ΔSsys = +qrev/T, and the entropy change of the surroundings is
$ΔS_{surr} = −\dfrac{q_{rev}}{T}. \nonumber$
The corresponding change in entropy of the universe is then as follows:
\begin{align*} \Delta S_{\textrm{univ}} &=\Delta S_{\textrm{sys}}+\Delta S_{\textrm{surr}} \[4pt] &= \dfrac{q_{\textrm{rev}}}{T}+\left(-\dfrac{q_\textrm{rev}}{T}\right) \[4pt] &= 0 \label{Eq4} \end{align*}
Thus no change in ΔSuniv has occurred.
In the initial state (top), the temperatures of a gas and the surroundings are the same. During the reversible expansion of the gas, heat must be added to the gas to maintain a constant temperature. Thus the internal energy of the gas does not change, but work is performed on the surroundings. In the final state (bottom), the temperature of the surroundings is lower because the gas has absorbed heat from the surroundings during expansion.
Now consider the reversible melting of a sample of ice at 0°C and 1 atm. The enthalpy of fusion of ice is 6.01 kJ/mol, which means that 6.01 kJ of heat are absorbed reversibly from the surroundings when 1 mol of ice melts at 0°C, as illustrated in Figure $6$. The surroundings constitute a sample of low-density carbon foam that is thermally conductive, and the system is the ice cube that has been placed on it. The direction of heat flow along the resulting temperature gradient is indicated with an arrow. From Equation $\ref{Eq2}$, we see that the entropy of fusion of ice can be written as follows:
$\Delta S_{\textrm{fus}}=\dfrac{q_{\textrm{rev}}}{T}=\dfrac{\Delta H_{\textrm{fus}}}{T} \label{Eq5}$
By convention, a thermogram shows cold regions in blue, warm regions in red, and thermally intermediate regions in green. When an ice cube (the system, dark blue) is placed on the corner of a square sample of low-density carbon foam with very high thermal conductivity, the temperature of the foam is lowered (going from red to green). As the ice melts, a temperature gradient appears, ranging from warm to very cold. An arrow indicates the direction of heat flow from the surroundings (red and green) to the ice cube. The amount of heat lost by the surroundings is the same as the amount gained by the ice, so the entropy of the universe does not change.
In this case, ΔSfus = (6.01 kJ/mol)/(273 K) = 22.0 J/(mol•K) = ΔSsys. The amount of heat lost by the surroundings is the same as the amount gained by the ice, so ΔSsurr = qrev/T = −(6.01 kJ/mol)/(273 K) = −22.0 J/(mol•K). Once again, we see that the entropy of the universe does not change:
ΔSuniv = ΔSsys + ΔSsurr = 22.0 J/(mol•K) − 22.0 J/(mol•K) = 0
In these two examples of reversible processes, the entropy of the universe is unchanged. This is true of all reversible processes and constitutes part of the second law of thermodynamics: the entropy of the universe remains constant in a reversible process, whereas the entropy of the universe increases in an irreversible (spontaneous) process.
The Second Law of Thermodynamics
The entropy of the universe increases during a spontaneous process. It also increases during an observable non-spontaneous process.
As an example of an irreversible process, consider the entropy changes that accompany the spontaneous and irreversible transfer of heat from a hot object to a cold one, as occurs when lava spewed from a volcano flows into cold ocean water. The cold substance, the water, gains heat (q > 0), so the change in the entropy of the water can be written as ΔScold = q/Tcold. Similarly, the hot substance, the lava, loses heat (q < 0), so its entropy change can be written as ΔShot = −q/Thot, where Tcold and Thot are the temperatures of the cold and hot substances, respectively. The total entropy change of the universe accompanying this process is therefore
$\Delta S_{\textrm{univ}}=\Delta S_{\textrm{cold}}+\Delta S_{\textrm{hot}}=\dfrac{q}{T_{\textrm{cold}}}+\left(-\dfrac{q}{T_{\textrm{hot}}}\right) \label{Eq6}$
The numerators on the right side of Equation $\ref{Eq6}$ are the same in magnitude but opposite in sign. Whether ΔSuniv is positive or negative depends on the relative magnitudes of the denominators. By definition, Thot > Tcold, so −q/Thot must be less than q/Tcold, and ΔSuniv must be positive. As predicted by the second law of thermodynamics, the entropy of the universe increases during this irreversible process. Any process for which ΔSuniv is positive is, by definition, a spontaneous one that will occur as written. Conversely, any process for which ΔSuniv is negative will not occur as written but will occur spontaneously in the reverse direction. We see, therefore, that heat is spontaneously transferred from a hot substance, the lava, to a cold substance, the ocean water. In fact, if the lava is hot enough (e.g., if it is molten), so much heat can be transferred that the water is converted to steam (Figure $7$).
Example $2$: Tin Pest
Tin has two allotropes with different structures. Gray tin (α-tin) has a structure similar to that of diamond, whereas white tin (β-tin) is denser, with a unit cell structure that is based on a rectangular prism. At temperatures greater than 13.2°C, white tin is the more stable phase, but below that temperature, it slowly converts reversibly to the less dense, powdery gray phase. This phenomenon was argued to have plagued Napoleon’s army during his ill-fated invasion of Russia in 1812: the buttons on his soldiers’ uniforms were made of tin and may have disintegrated during the Russian winter, adversely affecting the soldiers’ health (and morale). The conversion of white tin to gray tin is exothermic, with ΔH = −2.1 kJ/mol at 13.2°C.
1. What is ΔS for this process?
2. Which is the more highly ordered form of tin—white or gray?
Given: ΔH and temperature
Asked for: ΔS and relative degree of order
Strategy:
Use Equation $\ref{Eq2}$ to calculate the change in entropy for the reversible phase transition. From the calculated value of ΔS, predict which allotrope has the more highly ordered structure.
Solution
1. We know from Equation $\ref{Eq2}$ that the entropy change for any reversible process is the heat transferred (in joules) divided by the temperature at which the process occurs. Because the conversion occurs at constant pressure, and ΔH and ΔU are essentially equal for reactions that involve only solids, we can calculate the change in entropy for the reversible phase transition where qrev = ΔH. Substituting the given values for ΔH and temperature in kelvins (in this case, T = 13.2°C = 286.4 K),
$\Delta S=\dfrac{q_{\textrm{rev}}}{T}=\dfrac{(-2.1\;\mathrm{kJ/mol})(1000\;\mathrm{J/kJ})}{\textrm{286.4 K}}=-7.3\;\mathrm{J/(mol\cdot K)}$
1. The fact that ΔS < 0 means that entropy decreases when white tin is converted to gray tin. Thus gray tin must be the more highly ordered structure.
Video $1$: Time lapse tin pest reaction.
Note: Whether failing buttons were indeed a contributing factor in the failure of the invasion remains disputed; critics of the theory point out that the tin used would have been quite impure and thus more tolerant of low temperatures. Laboratory tests provide evidence that the time required for unalloyed tin to develop significant tin pest damage at lowered temperatures is about 18 months, which is more than twice the length of Napoleon's Russian campaign. It is clear though that some of the regiments employed in the campaign had tin buttons and that the temperature reached sufficiently low values (at least -40 °C)
Exercise $2$
Elemental sulfur exists in two forms: an orthorhombic form (Sα), which is stable below 95.3°C, and a monoclinic form (Sβ), which is stable above 95.3°C. The conversion of orthorhombic sulfur to monoclinic sulfur is endothermic, with ΔH = 0.401 kJ/mol at 1 atm.
1. What is ΔS for this process?
2. Which is the more highly ordered form of sulfur—Sα or Sβ?
Answer a
1.09 J/(mol•K)
Answer b
Sα
Entropy: Entropy(opens in new window) [youtu.be]
Summary
For a given system, the greater the number of microstates, the higher the entropy. During a spontaneous process, the entropy of the universe increases. $\Delta S=\frac{q_{\textrm{rev}}}{T} \nonumber$
A measure of the disorder of a system is its entropy (S), a state function whose value increases with an increase in the number of available microstates. A reversible process is one for which all intermediate states between extremes are equilibrium states; it can change direction at any time. In contrast, an irreversible process occurs in one direction only. The change in entropy of the system or the surroundings is the quantity of heat transferred divided by the temperature. The second law of thermodynamics states that in a reversible process, the entropy of the universe is constant, whereas in an irreversible process, such as the transfer of heat from a hot object to a cold object, the entropy of the universe increases. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/19%3A_Chemical_Thermodynamics/19.02%3A_Entropy_and_the_Second_Law_of_Thermodynamics.txt |
Learning Objectives
• To understand entropy from a molecular interpretation
As discussed previously, the second law of thermodynamics argues that all processes must increase the total entropy of the universe. However, the universe if often separated into the System and the Surroundings differing changes of entropy can be observed from a system level perspective. Many processes results in an increase in a system's entropy $\Delta S > 0$:
• Increasing the volume that a gas can occupy will increase the disorder of a gas
• Dissolving a solute into a solution will increase the entropy of the solute - typically resulting in an increase in the entropy of the system. (Note: the solvation of a solute can sometimes result in a significant decrease in the solvent entropy - leading to a net decrease in entropy of the system)
• Phase changes from solid to liquid, or liquid to gas, lead to an increase in the entropy of the system
Some processes result in a decrease in the entropy of a system $\Delta S < 0$:
• A gas molecule dissolved in a liquid is much more confined by neighboring molecules than when its in the gaseous state. Thus, the entropy of the gas molecule will decrease when it is dissolved in a liquid
• A phase change from a liquid to a solid (i.e. freezing), or from a gas to a liquid (i.e. condensation) results in an decrease in the disorder of the substance, and a decrease in the entropy
• A chemical reaction between gas molecules that results in a net decrease in the overall number of gas molecules will decrease the disorder of the system, and result in a decrease in the entropy
$2NO_{(g)} + O_{2(g)} \rightarrow 2NO_{2(g)} \;\;\; \Delta S < 0 \label{19.3.1}$
What is the molecular basis for the above observations for the change in entropy? Let's first consider the last example, the decrease in entropy associated with a decrease in the number of gas molecules for a chemical reaction
The product of this reaction ($NO_2$) involves the formation of a new N-O bond and the O atoms, originally in a separate $O_2$ molecule, are now connected to the $NO$ molecule via a new $N-O$ bond.
• Since they are now physically bonded to the other molecule (forming a new, larger, single molecule) the O atoms have less freedom to move around
• The reaction has resulted in a loss of freedom of the atoms (O atoms)
• There is a reduction in the disorder of the system (i.e. due to the reduction in the degrees of freedom, the system is more ordered after the reaction). $\Delta S < 0$.
Molecular Degrees of Freedom
The atoms, molecules, or ions that compose a chemical system can undergo several types of molecular motion, including translation, rotation, and vibration (Figure $2$).
• Translational motion. The entire molecule can move in some direction in three dimensions
• Rotational motion. The entire molecule can rotate around any axis, (even though it may not actually change its position translationally)
• Vibrational motion. The atoms within a molecule have certain freedom of movement relative to each other; this displacement can be periodic motion like the vibration of a tuning fork
These forms of molecular motion are ways in which molecules can store energy. The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy.
The Third Law of Thermodynamics
These forms of motion are ways in which the molecule can store energy. The greater the molecular motion of a system, the greater the number of possible microstates and the higher the entropy. A perfectly ordered system with only a single microstate available to it would have an entropy of zero. The only system that meets this criterion is a perfect crystal at a temperature of absolute zero (0 K), in which each component atom, molecule, or ion is fixed in place within a crystal lattice and exhibits no motion (ignoring quantum effects). Such a state of perfect order (or, conversely, zero disorder) corresponds to zero entropy. In practice, absolute zero is an ideal temperature that is unobtainable, and a perfect single crystal is also an ideal that cannot be achieved. Nonetheless, the combination of these two ideals constitutes the basis for the third law of thermodynamics: the entropy of any perfectly ordered, crystalline substance at absolute zero is zero.
The Third Law of Thermodynamics
The entropy of a pure crystalline substance at absolute zero (i.e. 0 Kelvin) is 0.
Since S = 0 corresponds to perfect order. The position of the atoms or molecules in the crystal would be perfectly defined
• As the temperature increases, the entropy of the atoms in the lattice increase
• Vibrational motions cause the atoms and molecules in the lattice to be less well ordered
The third law of thermodynamics has two important consequences: it defines the sign of the entropy of any substance at temperatures above absolute zero as positive, and it provides a fixed reference point that allows us to measure the absolute entropy of any substance at any temperature.In practice, chemists determine the absolute entropy of a substance by measuring the molar heat capacity (Cp) as a function of temperature and then plotting the quantity Cp/T versus T. The area under the curve between 0 K and any temperature T is the absolute entropy of the substance at T. In contrast, other thermodynamic properties, such as internal energy and enthalpy, can be evaluated in only relative terms, not absolute terms. In this section, we examine two different ways to calculate ΔS for a reaction or a physical change. The first, based on the definition of absolute entropy provided by the third law of thermodynamics, uses tabulated values of absolute entropies of substances. The second, based on the fact that entropy is a state function, uses a thermodynamic cycle similar to those discussed previously.
Motion never stops at absolute zero!
From classical kinetic theory, all motions cease at absolute zero. However things are more complicated from the more advanced (and more accurate) quantum mechanics. The Heisenberg Uncertainly Principle of quantum mechanics argues that molecules, even at absolute zero, always always have motion. Nonetheless, this motion is often ignored in the introduction of the third law of thermodynamics (which is incorrect of course).
Continued heating of a Solid Lattice
One way of calculating ΔS for a reaction is to use tabulated values of the standard molar entropy (S°), which is the entropy of 1 mol of a substance at a standard temperature of 298 K; the units of S° are J/(mol·K). Unlike enthalpy or internal energy, it is possible to obtain absolute entropy values by measuring the entropy change that occurs between the reference point of 0 K [corresponding to S = 0 J/(mol·K)] and 298 K.
As shown in Table $1$, for substances with approximately the same molar mass and number of atoms, S° values fall in the order S°(gas) > S°(liquid) > S°(solid). For instance, S° for liquid water is 70.0 J/(mol·K), whereas S° for water vapor is 188.8 J/(mol·K). Likewise, S° is 260.7 J/(mol·K) for gaseous I2 and 116.1 J/(mol·K) for solid I2. This order makes qualitative sense based on the kinds and extents of motion available to atoms and molecules in the three phases. The correlation between physical state and absolute entropy is illustrated in Figure $2$, which is a generalized plot of the entropy of a substance versus temperature.
Table $1$: Standard Molar Entropy Values of Selected Substances at 25°C
Gases Liquids Solids
Substance S° [J/(mol·K)] Substance S° [J/(mol·K)] Substance S° [J/(mol·K)]
He 126.2 H2O 70.0 C (diamond) 2.4
H2 130.7 CH3OH 126.8 C (graphite) 5.7
Ne 146.3 Br2 152.2 LiF 35.7
Ar 154.8 CH3CH2OH 160.7 SiO2 (quartz) 41.5
Kr 164.1 C6H6 173.4 Ca 41.6
Xe 169.7 CH3COCl 200.8 Na 51.3
H2O 188.8 C6H12 (cyclohexane) 204.4 MgF2 57.2
N2 191.6 C8H18 (isooctane) 329.3 K 64.7
O2 205.2 NaCl 72.1
CO2 213.8 KCl 82.6
I2 260.7 I2 116.1
Entropy increases with softer, less rigid solids, solids that contain larger atoms, and solids with complex molecular structures.
A closer examination of Table $1$ also reveals that substances with similar molecular structures tend to have similar S° values. Among crystalline materials, those with the lowest entropies tend to be rigid crystals composed of small atoms linked by strong, highly directional bonds, such as diamond [S° = 2.4 J/(mol·K)]. In contrast, graphite, the softer, less rigid allotrope of carbon, has a higher S° [5.7 J/(mol·K)] due to more disorder in the crystal. Soft crystalline substances and those with larger atoms tend to have higher entropies because of increased molecular motion and disorder. Similarly, the absolute entropy of a substance tends to increase with increasing molecular complexity because the number of available microstates increases with molecular complexity. For example, compare the S° values for CH3OH(l) and CH3CH2OH(l). Finally, substances with strong hydrogen bonds have lower values of S°, which reflects a more ordered structure.
Entropy: Entropy(opens in new window) [youtu.be]
Summary
In general, the entropy is expected to increase for the following types of processes:
1. The melting of a solid to form a liquid
2. The vaporization of a liquid (or solid) to produce a gas
3. Chemical reactions that involve phase changes of $\text{solid} \rightarrow \text{liquid/gas}$, or $\text{liquid} \rightarrow \text{gas}$
4. Chemical reactions that result in an increase in the number of gaseous molecules
5. Any time the temperature of a substance is increased | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/19%3A_Chemical_Thermodynamics/19.03%3A_The_Molecular_Interpretation_of_Entropy.txt |
Learning Objectives
• To calculate entropy changes for a chemical reaction
We have seen that the energy given off (or absorbed) by a reaction, and monitored by noting the change in temperature of the surroundings, can be used to determine the enthalpy of a reaction (e.g. by using a calorimeter). Tragically, there is no comparable easy way to experimentally measure the change in entropy for a reaction. Suppose we know that energy is going into a system (or coming out of it), and yet we do not observe any change in temperature. What is going on in such a situation? Changes in internal energy, that are not accompanied by a temperature change, might reflect changes in the entropy of the system.
For example, consider water at °0C at 1 atm pressure
• This is the temperature and pressure condition where liquid and solid phases of water are in equilibrium (also known as the melting point of ice)
$\ce{H2O(s) \rightarrow H2O(l)} \label{19.4.1}$
• At such a temperature and pressure we have a situation (by definition) where we have some ice and some liquid water
• If a small amount of energy is input into the system the equilibrium will shift slightly to the right (i.e. in favor of the liquid state)
• Likewise if a small amount of energy is withdrawn from the system, the equilibrium will shift to the left (more ice)
However, in both of the above situations, the energy change is not accompanied by a change in temperature (the temperature will not change until we no longer have an equilibrium condition; i.e. all the ice has melted or all the liquid has frozen)
Since the quantitative term that relates the amount of heat energy input vs. the rise in temperature is the heat capacity, it would seem that in some way, information about the heat capacity (and how it changes with temperature) would allow us to determine the entropy change in a system. In fact, values for the "standard molar entropy" of a substance have units of J/mol K, the same units as for molar heat capacity.
Standard Molar Entropy, S0
The entropy of a substance has an absolute value of 0 entropy at 0 K.
• Standard molar entropies are listed for a reference temperature (like 298 K) and 1 atm pressure (i.e. the entropy of a pure substance at 298 K and 1 atm pressure). A table of standard molar entropies at 0K would be pretty useless because it would be 0 for every substance (duh!) Standard molar entropy values are listed for a variety of substances in Table T2.
• When comparing standard molar entropies for a substance that is either a solid, liquid or gas at 298 K and 1 atm pressure, the gas will have more entropy than the liquid, and the liquid will have more entropy than the solid
• Unlike enthalpies of formation, standard molar entropies of elements are not 0.
The entropy change in a chemical reaction is given by the sum of the entropies of the products minus the sum of the entropies of the reactants. As with other calculations related to balanced equations, the coefficients of each component must be taken into account in the entropy calculation (the n, and m, terms below are there to indicate that the coefficients must be accounted for):
$\Delta S^0 = \sum_n nS^0(products) - \sum_m mS^0(reactants) \nonumber$
Example $1$: Haber Process
Calculate the change in entropy associated with the Haber process for the production of ammonia from nitrogen and hydrogen gas.
$\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)} \nonumber$
At 298K as a standard temperature:
• S0(NH3) = 192.5 J/mol K
• S0(H2) = 130.6 J/mol K
• S0(N2) = 191.5 J/mol K
Solution
From the balanced equation we can write the equation for ΔS0 (the change in the standard molar entropy for the reaction):
ΔS0 = 2*S0(NH3) - [S0(N2) + (3*S0(H2))]
ΔS0 = 2*192.5 - [191.5 + (3*130.6)]
ΔS0 = -198.3 J/mol K
It would appear that the process results in a decrease in entropy - i.e. a decrease in disorder. This is expected because we are decreasing the number of gas molecules. In other words the N2(g) used to float around independently of the H2 gas molecules. After the reaction, the two are bonded together and can't float around freely from one another. (I guess you can consider marriage as a negative entropy process!)
To calculate ΔS° for a chemical reaction from standard molar entropies, we use the familiar “products minus reactants” rule, in which the absolute entropy of each reactant and product is multiplied by its stoichiometric coefficient in the balanced chemical equation. Example $2$ illustrates this procedure for the combustion of the liquid hydrocarbon isooctane (C8H18; 2,2,4-trimethylpentane).
ΔS° for a reaction can be calculated from absolute entropy values using the same “products minus reactants” rule used to calculate ΔH°.
Example $2$: Combustion of Octane
Use the data in Table T2 to calculate ΔS° for the combustion reaction of liquid isooctane with O2(g) to give CO2(g) and H2O(g) at 298 K.
Given: standard molar entropies, reactants, and products
Asked for: ΔS°
Strategy:
Write the balanced chemical equation for the reaction and identify the appropriate quantities in Table T2. Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each multiplied by their appropriate stoichiometric coefficients, to obtain ΔS° for the reaction.
Solution:
The balanced chemical equation for the complete combustion of isooctane (C8H18) is as follows:
$\ce{C8H_{18}(l) + 25/2 O2(g) \rightarrow 8CO2(g) + 9H2O(g)} \nonumber$
We calculate ΔS° for the reaction using the “products minus reactants” rule, where m and n are the stoichiometric coefficients of each product and each reactant:
\begin{align*}\Delta S^\circ_{\textrm{rxn}}&=\sum mS^\circ(\textrm{products})-\sum nS^\circ(\textrm{reactants})
\ &=[8S^\circ(\mathrm{CO_2})+9S^\circ(\mathrm{H_2O})]-[S^\circ(\mathrm{C_8H_{18}})+\dfrac{25}{2}S^\circ(\mathrm{O_2})]
\ &=\left \{ [8\textrm{ mol }\mathrm{CO_2}\times213.8\;\mathrm{J/(mol\cdot K)}]+[9\textrm{ mol }\mathrm{H_2O}\times188.8\;\mathrm{J/(mol\cdot K)}] \right \}
\ &-\left \{[1\textrm{ mol }\mathrm{C_8H_{18}}\times329.3\;\mathrm{J/(mol\cdot K)}]+\left [\dfrac{25}{2}\textrm{ mol }\mathrm{O_2}\times205.2\textrm{ J}/(\mathrm{mol\cdot K})\right ] \right \} \ &=515.3\;\mathrm{J/K}\end{align*}
ΔS° is positive, as expected for a combustion reaction in which one large hydrocarbon molecule is converted to many molecules of gaseous products.
Exercise $2$
Use the data in Table T2 to calculate ΔS° for the reaction of H2(g) with liquid benzene (C6H6) to give cyclohexane (C6H12).
Answer
−361.1 J/K
Calculating the Entropy of Reaction using S: Calculating the Entropy of Reaction using S(opens in new window) [youtu.be] | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/19%3A_Chemical_Thermodynamics/19.04%3A_Entropy_Changes_in_Chemical_Reactions.txt |
Learning Objectives
• To understand the relationship between Gibbs free energy and work.
One of the major goals of chemical thermodynamics is to establish criteria for predicting whether a particular reaction or process will occur spontaneously. We have developed one such criterion, the change in entropy of the universe: if ΔSuniv > 0 for a process or a reaction, then the process will occur spontaneously as written. Conversely, if ΔSuniv < 0, a process cannot occur spontaneously; if ΔSuniv = 0, the system is at equilibrium. The sign of ΔSuniv is a universally applicable and infallible indicator of the spontaneity of a reaction. Unfortunately, using ΔSuniv requires that we calculate ΔS for both a system and its surroundings. This is not particularly useful for two reasons: we are normally much more interested in the system than in the surroundings, and it is difficult to make quantitative measurements of the surroundings (i.e., the rest of the universe). A criterion of spontaneity that is based solely on the state functions of a system would be much more convenient and is provided by a new state function: the Gibbs free energy.
Gibbs Free Energy and the Direction of Spontaneous Reactions
The Gibbs free energy ($G$), often called simply free energy, was named in honor of J. Willard Gibbs (1838–1903), an American physicist who first developed the concept. It is defined in terms of three other state functions with which you are already familiar: enthalpy, temperature, and entropy:
$G = H- TS$
Because it is a combination of state functions, $G$ is also a state function.
J. Willard Gibbs (1839–1903)
Born in Connecticut, Josiah Willard Gibbs attended Yale, as did his father, a professor of sacred literature at Yale, who was involved in the Amistad trial. In 1863, Gibbs was awarded the first engineering doctorate granted in the United States. He was appointed professor of mathematical physics at Yale in 1871, the first such professorship in the United States. His series of papers entitled “On the Equilibrium of Heterogeneous Substances” was the foundation of the field of physical chemistry and is considered one of the great achievements of the 19th century. Gibbs, whose work was translated into French by Le Chatelier, lived with his sister and brother-in-law until his death in 1903, shortly before the inauguration of the Nobel Prizes.
The criterion for predicting spontaneity is based on ($ΔG$), the change in $G$, at constant temperature and pressure. Although very few chemical reactions actually occur under conditions of constant temperature and pressure, most systems can be brought back to the initial temperature and pressure without significantly affecting the value of thermodynamic state functions such as $G$. At constant temperature and pressure,
$ΔG = ΔH − TΔS \label{Eq2}$
where all thermodynamic quantities are those of the system. Recall that at constant pressure, $ΔH = q$, whether a process is reversible or irreversible, and TΔS = qrev. Using these expressions, we can reduce Equation $\ref{Eq2}$ to $ΔG = q − q_{rev}$. Thus ΔG is the difference between the heat released during a process (via a reversible or an irreversible path) and the heat released for the same process occurring in a reversible manner. Under the special condition in which a process occurs reversibly, q = qrev and ΔG = 0. As we shall soon see, if $ΔG$ is zero, the system is at equilibrium, and there will be no net change.
What about processes for which ΔG ≠ 0? To understand how the sign of ΔG for a system determines the direction in which change is spontaneous, we can rewrite the relationship between $\Delta S$ and $q_{rev}$, discussed earlier.
$\Delta S= \dfrac{q_{rev}}{T} \nonumber$
with the definition of $\Delta H$ in terms of $q_{rev}$
$q_{rev} = ΔH \nonumber$
to obtain
$\Delta S_{\textrm{surr}}=-\dfrac{\Delta H_{\textrm{sys}}}{T} \label{Eq3}$
Thus the entropy change of the surroundings is related to the enthalpy change of the system. We have stated that for a spontaneous reaction, $ΔS_{univ} > 0$, so substituting we obtain
\begin{align} \Delta S_{\textrm{univ}}&=\Delta S_{\textrm{sys}}+\Delta S_{\textrm{surr}}>0 \[4pt] &=\Delta S_{\textrm{sys}}-\dfrac{\Delta H_{\textrm{sys}}}{T}>0\end{align} \nonumber
Multiplying both sides of the inequality by −T reverses the sign of the inequality; rearranging,
$ΔH_{sys}−TΔS_{sys}<0 \nonumber$
which is equal to $ΔG$ (Equation $\ref{Eq2}$). We can therefore see that for a spontaneous process, $ΔG < 0$.
The relationship between the entropy change of the surroundings and the heat gained or lost by the system provides the key connection between the thermodynamic properties of the system and the change in entropy of the universe. The relationship shown in Equation $\ref{Eq2}$ allows us to predict spontaneity by focusing exclusively on the thermodynamic properties and temperature of the system. We predict that highly exothermic processes ($ΔH \ll 0$) that increase the disorder of a system ($ΔS_{sys} \gg 0$) would therefore occur spontaneously. An example of such a process is the decomposition of ammonium nitrate fertilizer. Ammonium nitrate was also used to destroy the Murrah Federal Building in Oklahoma City, Oklahoma, in 1995. For a system at constant temperature and pressure, we can summarize the following results:
• If $ΔG < 0$, the process occurs spontaneously.
• If $ΔG = 0$, the system is at equilibrium.
• If $ΔG > 0$, the process is not spontaneous as written but occurs spontaneously in the reverse direction.
To further understand how the various components of ΔG dictate whether a process occurs spontaneously, we now look at a simple and familiar physical change: the conversion of liquid water to water vapor. If this process is carried out at 1 atm and the normal boiling point of 100.00°C (373.15 K), we can calculate ΔG from the experimentally measured value of ΔHvap (40.657 kJ/mol). For vaporizing 1 mol of water, $ΔH = 40,657; J$, so the process is highly endothermic. From the definition of ΔS (Equation $\ref{Eq3}$), we know that for 1 mol of water,
\begin{align*} \Delta S_{\textrm{vap}}&=\dfrac{\Delta H_{\textrm{vap}}}{T_\textrm b} \[4pt] &=\dfrac{\textrm{40,657 J}}{\textrm{373.15 K}} \[4pt] &=\textrm{108.96 J/K} \end{align*} \nonumber
Hence there is an increase in the disorder of the system. At the normal boiling point of water,
\begin{align*}\Delta G_{100^\circ\textrm C}&=\Delta H_{100^\circ\textrm C}-T\Delta S_{100^\circ\textrm C} \[4pt] &=\textrm{40,657 J}-[(\textrm{373.15 K})(\textrm{108.96 J/K})] \[4pt] &=\textrm{0 J}\end{align*} \nonumber
The energy required for vaporization offsets the increase in disorder of the system. Thus ΔG = 0, and the liquid and vapor are in equilibrium, as is true of any liquid at its boiling point under standard conditions.
Now suppose we were to superheat 1 mol of liquid water to 110°C. The value of ΔG for the vaporization of 1 mol of water at 110°C, assuming that ΔH and ΔS do not change significantly with temperature, becomes
\begin{align*}\Delta G_{110^\circ\textrm C}&=\Delta H-T\Delta S \[4pt] &=\textrm{40,657 J}-[(\textrm{383.15 K})(\textrm{108.96 J/K})] \[4pt] &=-\textrm{1091 J}\end{align*} \nonumber
At 110°C, $ΔG < 0$, and vaporization is predicted to occur spontaneously and irreversibly.
We can also calculate $ΔG$ for the vaporization of 1 mol of water at a temperature below its normal boiling point—for example, 90°C—making the same assumptions:
\begin{align*}\Delta G_{90^\circ\textrm C}&=\Delta H-T\Delta S \[4pt] &=\textrm{40,657 J}-[(\textrm{363.15 K})(\textrm{108.96 J/K})] \[4pt] &=\textrm{1088 J}\end{align*} \nonumber
At 90°C, ΔG > 0, and water does not spontaneously convert to water vapor. When using all the digits in the calculator display in carrying out our calculations, ΔG110°C = 1090 J = −ΔG90°C, as we would predict.
Relating Enthalpy and Entropy changes under Equilibrium Conditions
$ΔG = 0$ only if $ΔH = TΔS$.
We can also calculate the temperature at which liquid water is in equilibrium with water vapor. Inserting the values of ΔH and ΔS into the definition of ΔG (Equation $\ref{Eq2}$), setting $ΔG = 0$, and solving for $T$,
0 J=40,657 J−T(108.96 J/K)
T=373.15 K
Thus $ΔG = 0$ at T = 373.15 K and 1 atm, which indicates that liquid water and water vapor are in equilibrium; this temperature is called the normal boiling point of water. At temperatures greater than 373.15 K, $ΔG$ is negative, and water evaporates spontaneously and irreversibly. Below 373.15 K, $ΔG$ is positive, and water does not evaporate spontaneously. Instead, water vapor at a temperature less than 373.15 K and 1 atm will spontaneously and irreversibly condense to liquid water. Figure $1$ shows how the $ΔH$ and $TΔS$ terms vary with temperature for the vaporization of water. When the two lines cross, $ΔG = 0$, and $ΔH = TΔS$.
A similar situation arises in the conversion of liquid egg white to a solid when an egg is boiled. The major component of egg white is a protein called albumin, which is held in a compact, ordered structure by a large number of hydrogen bonds. Breaking them requires an input of energy (ΔH > 0), which converts the albumin to a highly disordered structure in which the molecules aggregate as a disorganized solid (ΔS > 0). At temperatures greater than 373 K, the TΔS term dominates, and ΔG < 0, so the conversion of a raw egg to a hard-boiled egg is an irreversible and spontaneous process above 373 K.
The Definition of Gibbs Free Energy: The Definition of Gibbs Free Energy (opens in new window) [youtu.be]
The Relationship between ΔG and Work
In the previous subsection, we learned that the value of ΔG allows us to predict the spontaneity of a physical or a chemical change. In addition, the magnitude of ΔG for a process provides other important information. The change in free energy (ΔG) is equal to the maximum amount of work that a system can perform on the surroundings while undergoing a spontaneous change (at constant temperature and pressure): ΔG = wmax. To see why this is true, let’s look again at the relationships among free energy, enthalpy, and entropy expressed in Equation $\ref{Eq2}$. We can rearrange this equation as follows:
$ΔH = ΔG + TΔS \label{Eq4}$
This equation tells us that when energy is released during an exothermic process (ΔH < 0), such as during the combustion of a fuel, some of that energy can be used to do work (ΔG < 0), while some is used to increase the entropy of the universe (TΔS > 0). Only if the process occurs infinitely slowly in a perfectly reversible manner will the entropy of the universe be unchanged. (For more information on entropy and reversibility, see the previous section). Because no real system is perfectly reversible, the entropy of the universe increases during all processes that produce energy. As a result, no process that uses stored energy can ever be 100% efficient; that is, ΔH will never equal ΔG because ΔS has a positive value.
One of the major challenges facing engineers is to maximize the efficiency of converting stored energy to useful work or converting one form of energy to another. As indicated in Table $1$, the efficiencies of various energy-converting devices vary widely. For example, an internal combustion engine typically uses only 25%–30% of the energy stored in the hydrocarbon fuel to perform work; the rest of the stored energy is released in an unusable form as heat. In contrast, gas–electric hybrid engines, now used in several models of automobiles, deliver approximately 50% greater fuel efficiency. A large electrical generator is highly efficient (approximately 99%) in converting mechanical to electrical energy, but a typical incandescent light bulb is one of the least efficient devices known (only approximately 5% of the electrical energy is converted to light). In contrast, a mammalian liver cell is a relatively efficient machine and can use fuels such as glucose with an efficiency of 30%–50%.
Table $1$: Approximate Thermodynamic Efficiencies of Various Devices
Device Energy Conversion Approximate Efficiency (%)
large electrical generator mechanical → electrical 99
chemical battery chemical → electrical 90
home furnace chemical → heat 65
small electric tool electrical → mechanical 60
space shuttle engine chemical → mechanical 50
mammalian liver cell chemical → chemical 30–50
spinach leaf cell light → chemical 30
internal combustion engine chemical → mechanical 25–30
fluorescent light electrical → light 20
solar cell light → electricity 10-20
incandescent light bulb electricity → light 5
yeast cell chemical → chemical 2–4
Standard Free-Energy Change
We have seen that there is no way to measure absolute enthalpies, although we can measure changes in enthalpy (ΔH) during a chemical reaction. Because enthalpy is one of the components of Gibbs free energy, we are consequently unable to measure absolute free energies; we can measure only changes in free energy. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free-energy change can be calculated from the definition of free energy, if the standard enthalpy and entropy changes are known, using Equation $\ref{Eq5}$:
$ΔG° = ΔH° − TΔS° \label{Eq5}$
If ΔS° and ΔH° for a reaction have the same sign, then the sign of ΔG° depends on the relative magnitudes of the ΔH° and TΔS° terms. It is important to recognize that a positive value of ΔG° for a reaction does not mean that no products will form if the reactants in their standard states are mixed; it means only that at equilibrium the concentrations of the products will be less than the concentrations of the reactants.
A positive ΔG° means that the equilibrium constant is less than 1.
Example $1$
Calculate the standard free-energy change ($ΔG^o$) at 25°C for the reaction
$\ce{ H2(g) + O2(g) \rightleftharpoons H2O2(l)}\nonumber$
At 25°C, the standard enthalpy change (ΔH°) is −187.78 kJ/mol, and the absolute entropies of the products and reactants are:
• S°(H2O2) = 109.6 J/(mol•K),
• S°(O2) = 205.2 J/(mol•K), and
• S°(H2) = 130.7 J/(mol•K).
Is the reaction spontaneous as written?
Given: balanced chemical equation, ΔH° and S° for reactants and products
Asked for: spontaneity of reaction as written
Strategy:
1. Calculate ΔS° from the absolute molar entropy values given.
2. Use Equation $\ref{Eq5}$, the calculated value of ΔS°, and other data given to calculate ΔG° for the reaction. Use the value of ΔG° to determine whether the reaction is spontaneous as written.
Solution
A To calculate ΔG° for the reaction, we need to know $ΔH^o$, $ΔS^o$, and $T$. We are given $ΔH^o$, and we know that T = 298.15 K. We can calculate ΔS° from the absolute molar entropy values provided using the “products minus reactants” rule:
\begin{align*}\Delta S^\circ &=S^\circ(\mathrm{H_2O_2})-[S^\circ(\mathrm{O_2})+S^\circ(\mathrm{H_2})] \nonumber \[4pt] &=[\mathrm{1\;mol\;H_2O_2}\times109.6\;\mathrm{J/(mol\cdot K})] \nonumber \[4pt] &-\left \{ [\textrm{1 mol H}_2\times130.7\;\mathrm{J/(mol\cdot K)}]+[\textrm{1 mol O}_2\times205.2\;\mathrm{J/(mol\cdot K)}] \right \} \nonumber \[4pt]&=-226.3\textrm{ J/K }(\textrm{per mole of }\mathrm{H_2O_2}) \end{align*} \nonumber
As we might expect for a reaction in which 2 mol of gas is converted to 1 mol of a much more ordered liquid, $ΔS^o$ is very negative for this reaction.
B Substituting the appropriate quantities into Equation $\ref{Eq5}$,
\begin{align*}\Delta G^\circ &=\Delta H^\circ -T\Delta S^\circ \[4pt] &=-187.78\textrm{ kJ/mol}-(\textrm{298.15 K}) [-226.3\;\mathrm{J/(mol\cdot K)}\times\textrm{1 kJ/1000 J}]\nonumber \[4pt] &=-187.78\textrm{ kJ/mol}+67.47\textrm{ kJ/mol} \[4pt] &=-120.31\textrm{ kJ/mol}\nonumber \end{align*} \nonumber
The negative value of $ΔG^o$ indicates that the reaction is spontaneous as written. Because $ΔS^o$ and $ΔH^o$ for this reaction have the same sign, the sign of $ΔG^o$ depends on the relative magnitudes of the $ΔH^o$ and $TΔS^o$ terms. In this particular case, the enthalpy term dominates, indicating that the strength of the bonds formed in the product more than compensates for the unfavorable $ΔS^o$ term and for the energy needed to break bonds in the reactants.
Exercise $1$
Calculate the standard free-energy change ($ΔG^o$) at 25°C for the reaction
$2H_2(g)+N_2(g) \rightleftharpoons N_2H_4(l)\nonumber . \nonumber$
Is the reaction spontaneous as written at 25°C?
Hint
At 25°C, the standard enthalpy change ($ΔH^o$) is 50.6 kJ/mol, and the absolute entropies of the products and reactants are
• S°(N2H4) = 121.2 J/(mol•K),
• S°(N2) = 191.6 J/(mol•K), and
• S°(H2) = 130.7 J/(mol•K).
Answer
149.5 kJ/mol
no, not spontaneous
Video Solution
Determining if a Reaction is Spontaneous: Determining if a Reaction is Spontaneous(opens in new window) [youtu.be] (opens in new window)
Tabulated values of standard free energies of formation allow chemists to calculate the values of ΔG° for a wide variety of chemical reactions rather than having to measure them in the laboratory. The standard free energy of formation ($ΔG^∘_f$)of a compound is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. By definition, the standard free energy of formation of an element in its standard state is zero at 298.15 K. One mole of Cl2 gas at 298.15 K, for example, has $\Delta G^∘_f = 0$. The standard free energy of formation of a compound can be calculated from the standard enthalpy of formation (ΔHf) and the standard entropy of formation (ΔSf) using the definition of free energy:
$\Delta G^o_{f} =ΔH^o_{f} −TΔS^o_{f} \label{Eq6}$
Using standard free energies of formation to calculate the standard free energy of a reaction is analogous to calculating standard enthalpy changes from standard enthalpies of formation using the familiar “products minus reactants” rule:
$ΔG^o_{rxn}=\sum mΔG^o_{f} (products)− \sum nΔ^o_{f} (reactants) \label{Eq7a}$
where m and n are the stoichiometric coefficients of each product and reactant in the balanced chemical equation. A very large negative ΔG° indicates a strong tendency for products to form spontaneously from reactants; it does not, however, necessarily indicate that the reaction will occur rapidly. To make this determination, we need to evaluate the kinetics of the reaction.
The "Products minus Reactants" Rule
The $ΔG^o$ of a reaction can be calculated from tabulated ΔGf values (Table T1) using the “products minus reactants” rule.
Example $2$
Calculate ΔG° for the reaction of isooctane with oxygen gas to give carbon dioxide and water (described in Example 7). Use the following data:
• ΔG°f(isooctane) = −353.2 kJ/mol,
• ΔG°f(CO2) = −394.4 kJ/mol, and
• ΔG°f(H2O) = −237.1 kJ/mol. Is the reaction spontaneous as written?
Given: balanced chemical equation and values of ΔG°f for isooctane, CO2, and H2O
Asked for: spontaneity of reaction as written
Strategy:
Use the “products minus reactants” rule to obtain ΔGrxn, remembering that ΔG°f for an element in its standard state is zero. From the calculated value, determine whether the reaction is spontaneous as written.
Solution
The balanced chemical equation for the reaction is as follows:
$\ce{C8H_{18}(l) + 25/2 O2 (g) \rightarrow 8CO2(g) + 9H2O(l)}\nonumber$
We are given ΔGf values for all the products and reactants except O2(g). Because oxygen gas is an element in its standard state, ΔGf (O2) is zero. Using the “products minus reactants” rule,
\begin{align*} \Delta G^\circ &=[8\Delta G^\circ_\textrm f(\mathrm{CO_2})+9\Delta G^\circ_\textrm f(\mathrm{H_2O})]-\left[1\Delta G^\circ_\textrm f(\mathrm{C_8H_{18}})+\dfrac{25}{2}\Delta G^\circ_\textrm f(\mathrm{O_2})\right] \nonumber \[4pt] &=[(\textrm{8 mol})(-394.4\textrm{ kJ/mol})+(\textrm{9 mol})(-237.1\textrm{ kJ/mol})] \nonumber\[4pt]&-\left [(\textrm{1 mol})(-353.2\textrm{ kJ/mol})+\left(\dfrac{25}{2}\;\textrm{mol}\right)(0 \textrm{ kJ/mol}) \right ] \nonumber \[4pt] &=-4935.9\textrm{ kJ }(\textrm{per mol of }\mathrm{C_8H_{18}})\nonumber \end{align*} \nonumber
Because ΔG° is a large negative number, there is a strong tendency for the spontaneous formation of products from reactants (though not necessarily at a rapid rate). Also notice that the magnitude of ΔG° is largely determined by the ΔGf of the stable products: water and carbon dioxide.
Exercise $2$
Calculate ΔG° for the reaction of benzene with hydrogen gas to give cyclohexane using the following data
• ΔGf(benzene) = 124.5 kJ/mol
• ΔGf (cyclohexane) = 217.3 kJ/mol.
Is the reaction spontaneous as written?
Answer
92.8 kJ; no
Video Solution
Calculating Grxn using Gf: Calculating Grxn using Gf(opens in new window) [youtu.be]
Calculated values of ΔG° are extremely useful in predicting whether a reaction will occur spontaneously if the reactants and products are mixed under standard conditions. We should note, however, that very few reactions are actually carried out under standard conditions, and calculated values of ΔG° may not tell us whether a given reaction will occur spontaneously under nonstandard conditions. What determines whether a reaction will occur spontaneously is the free-energy change (ΔG) under the actual experimental conditions, which are usually different from ΔG°. If the ΔH and TΔS terms for a reaction have the same sign, for example, then it may be possible to reverse the sign of ΔG by changing the temperature, thereby converting a reaction that is not thermodynamically spontaneous, having Keq < 1, to one that is, having a Keq > 1, or vice versa. Because ΔH and ΔS usually do not vary greatly with temperature in the absence of a phase change, we can use tabulated values of ΔH° and ΔS° to calculate ΔG° at various temperatures, as long as no phase change occurs over the temperature range being considered.
In the absence of a phase change, neither $ΔH$ nor $ΔS$ vary greatly with temperature.
Example $3$
Calculate (a) ΔG° and (b) ΔG300°C for the reaction N2(g)+3H2(g)⇌2NH3(g), assuming that ΔH and ΔS do not change between 25°C and 300°C. Use these data:
• S°(N2) = 191.6 J/(mol•K),
• S°(H2) = 130.7 J/(mol•K),
• S°(NH3) = 192.8 J/(mol•K), and
• ΔHf (NH3) = −45.9 kJ/mol.
Given: balanced chemical equation, temperatures, S° values, and ΔHf for NH3
Asked for: ΔG° and ΔG at 300°C
Strategy:
1. Convert each temperature to kelvins. Then calculate ΔS° for the reaction. Calculate ΔH° for the reaction, recalling that ΔHf for any element in its standard state is zero.
2. Substitute the appropriate values into Equation $\ref{Eq5}$ to obtain ΔG° for the reaction.
3. Assuming that ΔH and ΔS are independent of temperature, substitute values into Equation $\ref{Eq2}$ to obtain ΔG for the reaction at 300°C.
Solution
A To calculate ΔG° for the reaction using Equation $\ref{Eq5}$, we must know the temperature as well as the values of ΔS° and ΔH°. At standard conditions, the temperature is 25°C, or 298 K. We can calculate ΔS° for the reaction from the absolute molar entropy values given for the reactants and the products using the “products minus reactants” rule:
\begin{align}\Delta S^\circ_{\textrm{rxn}}&=2S^\circ(\mathrm{NH_3})-[S^\circ(\mathrm{N_2})+3S^\circ(\mathrm{H_2})] \nonumber\ &=[\textrm{2 mol NH}_3\times192.8\;\mathrm{J/(mol\cdot K)}] \nonumber\ &-\left \{[\textrm{1 mol N}_2\times191.6\;\mathrm{J/(mol\cdot K)}]+[\textrm{3 mol H}_2\times130.7\;\mathrm{J/(mol\cdot K)}]\right \}\nonumber\ &=-198.1\textrm{ J/K (per mole of N}_2)\end{align}\nonumber
We can also calculate ΔH° for the reaction using the “products minus reactants” rule. The value of ΔHf (NH3) is given, and ΔHf is zero for both N2 and H2:
\begin{align}\Delta H^\circ_{\textrm{rxn}}&=2\Delta H^\circ_\textrm f(\mathrm{NH_3})-[\Delta H^\circ_\textrm f(\mathrm{N_2})+3\Delta H^\circ_\textrm f(\mathrm{H_2})]\nonumber \ &=[2\times(-45.9\textrm{ kJ/mol})]-[(1\times0\textrm{ kJ/mol})+(3\times0 \textrm{ kJ/mol})]\nonumber \ &=-91.8\textrm{ kJ(per mole of N}_2)\nonumber\end{align}\nonumber
B Inserting the appropriate values into Equation $\ref{Eq5}$
$\Delta G^\circ_{\textrm{rxn}}=\Delta H^\circ-T\Delta S^\circ=(-\textrm{91.8 kJ})-(\textrm{298 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J})=-\textrm{32.7 kJ (per mole of N}_2)\nonumber$
C To calculate ΔG for this reaction at 300°C, we assume that ΔH and ΔS are independent of temperature (i.e., ΔH300°C = H° and ΔS300°C = ΔS°) and insert the appropriate temperature (573 K) into Equation $\ref{Eq2}$:
\begin{align*}\Delta G_{300^\circ\textrm C}&=\Delta H_{300^\circ\textrm C}-(\textrm{573 K})(\Delta S_{300^\circ\textrm C}) \[4pt] &=\Delta H^\circ -(\textrm{573 K})\Delta S^\circ\nonumber \[4pt] &=(-\textrm{91.8 kJ})-(\textrm{573 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J}) \[4pt]&=21.7\textrm{ kJ (per mole of N}_2) \end{align*} \nonumber
In this example, changing the temperature has a major effect on the thermodynamic spontaneity of the reaction. Under standard conditions, the reaction of nitrogen and hydrogen gas to produce ammonia is thermodynamically spontaneous, but in practice, it is too slow to be useful industrially. Increasing the temperature in an attempt to make this reaction occur more rapidly also changes the thermodynamics by causing the −TΔS° term to dominate, and the reaction is no longer spontaneous at high temperatures; that is, its Keq is less than one. This is a classic example of the conflict encountered in real systems between thermodynamics and kinetics, which is often unavoidable.
Exercise $3$
Calculate
1. $ΔG°$ and
2. $ΔG_{750°C}$
for the following reaction
$\ce{ 2NO(g) + O2 (g) \rightleftharpoons 2NO2 (g)}\nonumber$
which is important in the formation of urban smog. Assume that $ΔH$ and $ΔS$ do not change between 25.0°C and 750°C and use these data:
• S°(NO) = 210.8 J/(mol•K),
• S°(O2) = 205.2 J/(mol•K),
• S°(NO2) = 240.1 J/(mol•K),
• ΔHf(NO2) = 33.2 kJ/mol, and
• ΔHf (NO) = 91.3 kJ/mol.
Answer a
−72.5 kJ/mol of $O_2$
Answer b
33.8 kJ/mol of $O_2$
The effect of temperature on the spontaneity of a reaction, which is an important factor in the design of an experiment or an industrial process, depends on the sign and magnitude of both ΔH° and ΔS°. The temperature at which a given reaction is at equilibrium can be calculated by setting ΔG° = 0 in Equation $\ref{Eq5}$, as illustrated in Example $4$.
Example $4$
As you saw in Example $3$, the reaction of nitrogen and hydrogen gas to produce ammonia is one in which ΔH° and ΔS° are both negative. Such reactions are predicted to be thermodynamically spontaneous at low temperatures but nonspontaneous at high temperatures. Use the data in Example 9.5.3 to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous, assuming that ΔH° and ΔS° are independent of temperature.
Given: ΔH° and ΔS°
Asked for: temperature at which reaction changes from spontaneous to nonspontaneous
Strategy:
Set ΔG° equal to zero in Equation $\ref{Eq5}$ and solve for T, the temperature at which the reaction becomes nonspontaneous.
Solution
In Example $3$, we calculated that ΔH° is −91.8 kJ/mol of N2 and ΔS° is −198.1 J/K per mole of N2, corresponding to ΔG° = −32.7 kJ/mol of N2 at 25°C. Thus the reaction is indeed spontaneous at low temperatures, as expected based on the signs of ΔH° and ΔS°. The temperature at which the reaction becomes nonspontaneous is found by setting ΔG° equal to zero and rearranging Equation $\ref{Eq5}$ to solve for T:
\begin{align*}\Delta G^\circ &=\Delta H^\circ - T\Delta S^\circ=0 \[4pt] \Delta H^\circ &=T\Delta S^\circ \[4pt] T=\dfrac{\Delta H^\circ}{\Delta S^\circ}&=\dfrac{(-\textrm{91.8 kJ})(\textrm{1000 J/kJ})}{-\textrm{198.1 J/K}}=\textrm{463 K}\end{align*} \nonumber
This is a case in which a chemical engineer is severely limited by thermodynamics. Any attempt to increase the rate of reaction of nitrogen with hydrogen by increasing the temperature will cause reactants to be favored over products above 463 K.
Exercise $4$
As you found in the exercise in Example $3$, ΔH° and ΔS° are both negative for the reaction of nitric oxide and oxygen to form nitrogen dioxide. Use those data to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous.
Answer
792.6 K
Video Solution
Summary
• The change in Gibbs free energy, which is based solely on changes in state functions, is the criterion for predicting the spontaneity of a reaction.
• Free-energy change:
$ΔG = ΔH − TΔS\nonumber$
• Standard free-energy change:
$ΔG° = ΔH° − TΔS°\nonumber$
We can predict whether a reaction will occur spontaneously by combining the entropy, enthalpy, and temperature of a system in a new state function called Gibbs free energy (G). The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is spontaneous, ΔG < 0. If the process is not spontaneous as written but is spontaneous in the reverse direction, ΔG > 0. At constant temperature and pressure, ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free energy of formation (ΔGf), is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. Tabulated values of standard free energies of formation are used to calculate ΔG° for a reaction. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/19%3A_Chemical_Thermodynamics/19.05%3A_Gibbs_Free_Energy.txt |
Learning Objectives
• To understand the relationship between Gibbs free energy and work.
One of the major goals of chemical thermodynamics is to establish criteria for predicting whether a particular reaction or process will occur spontaneously. We have developed one such criterion, the change in entropy of the universe: if ΔSuniv > 0 for a process or a reaction, then the process will occur spontaneously as written. Conversely, if ΔSuniv < 0, a process cannot occur spontaneously; if ΔSuniv = 0, the system is at equilibrium. The sign of ΔSuniv is a universally applicable and infallible indicator of the spontaneity of a reaction. Unfortunately, using ΔSuniv requires that we calculate ΔS for both a system and its surroundings. This is not particularly useful for two reasons: we are normally much more interested in the system than in the surroundings, and it is difficult to make quantitative measurements of the surroundings (i.e., the rest of the universe). A criterion of spontaneity that is based solely on the state functions of a system would be much more convenient and is provided by a new state function: the Gibbs free energy.
Gibbs Energy and Spontaneity
The criterion for predicting spontaneity is based on ($ΔG$), the change in $G$, at constant temperature and pressure. Although very few chemical reactions actually occur under conditions of constant temperature and pressure, most systems can be brought back to the initial temperature and pressure without significantly affecting the value of thermodynamic state functions such as $G$. At constant temperature and pressure,
$\color{red} ΔG = ΔH − TΔS \label{Eq2}$
where all thermodynamic quantities are those of the system. Recall that at constant pressure, $ΔH = q$, whether a process is reversible or irreversible, and TΔS = qrev. Using these expressions, we can reduce Equation $\ref{Eq2}$ to $ΔG = q − q_{rev}$. Thus ΔG is the difference between the heat released during a process (via a reversible or an irreversible path) and the heat released for the same process occurring in a reversible manner. Under the special condition in which a process occurs reversibly, q = qrev and ΔG = 0. As we shall soon see, if $ΔG$ is zero, the system is at equilibrium, and there will be no net change.
What about processes for which ΔG ≠ 0? To understand how the sign of ΔG for a system determines the direction in which change is spontaneous, we can rewrite the relationship between $\Delta S$ and $q_{rev}$, discussed earlier.
$\Delta S= \dfrac{q_{rev}}{T}\nonumber$
with the definition of $\Delta H$ in terms of $q_{rev}$
$q_{rev} = ΔH\nonumber$
to obtain
$\Delta S_{\textrm{surr}}=-\dfrac{\Delta H_{\textrm{sys}}}{T} \label{Eq3}$
Thus the entropy change of the surroundings is related to the enthalpy change of the system. We have stated that for a spontaneous reaction, $ΔS_{univ} > 0$, so substituting we obtain
\begin{align*} \Delta S_{\textrm{univ}}&=\Delta S_{\textrm{sys}}+\Delta S_{\textrm{surr}}>0 \ &=\Delta S_{\textrm{sys}}-\dfrac{\Delta H_{\textrm{sys}}}{T}>0\end{align*} \nonumber
Multiplying both sides of the inequality by −T reverses the sign of the inequality; rearranging,
$ΔH_{sys}−TΔS_{sys}<0 \nonumber$
which is equal to $ΔG$ (Equation $\ref{Eq2}$). We can therefore see that for a spontaneous process, $ΔG < 0$.
The relationship between the entropy change of the surroundings and the heat gained or lost by the system provides the key connection between the thermodynamic properties of the system and the change in entropy of the universe. The relationship shown in Equation $\ref{Eq2}$ allows us to predict spontaneity by focusing exclusively on the thermodynamic properties and temperature of the system. We predict that highly exothermic processes ($ΔH \ll 0$) that increase the disorder of a system ($ΔS_{sys} \gg 0$) would therefore occur spontaneously. An example of such a process is the decomposition of ammonium nitrate fertilizer. Ammonium nitrate was also used to destroy the Murrah Federal Building in Oklahoma City, Oklahoma, in 1995. For a system at constant temperature and pressure, we can summarize the following results:
• If $ΔG < 0$, the process occurs spontaneously.
• If $ΔG = 0$, the system is at equilibrium.
• If $ΔG > 0$, the process is not spontaneous as written but occurs spontaneously in the reverse direction.
To further understand how the various components of ΔG dictate whether a process occurs spontaneously, we now look at a simple and familiar physical change: the conversion of liquid water to water vapor. If this process is carried out at 1 atm and the normal boiling point of 100.00°C (373.15 K), we can calculate ΔG from the experimentally measured value of ΔHvap (40.657 kJ/mol). For vaporizing 1 mol of water, $ΔH = 40,657 J$, so the process is highly endothermic. From the definition of ΔS (Equation $\ref{Eq3}$), we know that for 1 mol of water,
$\Delta S_{\textrm{vap}}=\dfrac{\Delta H_{\textrm{vap}}}{T_\textrm b}=\dfrac{\textrm{40,657 J}}{\textrm{373.15 K}}=\textrm{108.96 J/K}$
Hence there is an increase in the disorder of the system. At the normal boiling point of water,
$\Delta G_{100^\circ\textrm C}=\Delta H_{100^\circ\textrm C}-T\Delta S_{100^\circ\textrm C}=\textrm{40,657 J}-[(\textrm{373.15 K})(\textrm{108.96 J/K})]=\textrm{0 J} \nonumber$
The energy required for vaporization offsets the increase in disorder of the system. Thus ΔG = 0, and the liquid and vapor are in equilibrium, as is true of any liquid at its boiling point under standard conditions.
Now suppose we were to superheat 1 mol of liquid water to 110°C. The value of ΔG for the vaporization of 1 mol of water at 110°C, assuming that ΔH and ΔS do not change significantly with temperature, becomes
$\Delta G_{110^\circ\textrm C}=\Delta H-T\Delta S=\textrm{40,657 J}-[(\textrm{383.15 K})(\textrm{108.96 J/K})]=-\textrm{1091 J} \nonumber$
At 110°C, ΔG < 0, and vaporization is predicted to occur spontaneously and irreversibly.
We can also calculate ΔG for the vaporization of 1 mol of water at a temperature below its normal boiling point—for example, 90°C—making the same assumptions:
$\Delta G_{90^\circ\textrm C}=\Delta H-T\Delta S=\textrm{40,657 J}-[(\textrm{363.15 K})(\textrm{108.96 J/K})]=\textrm{1088 J} \nonumber$
At 90°C, ΔG > 0, and water does not spontaneously convert to water vapor. When using all the digits in the calculator display in carrying out our calculations, ΔG110°C = 1090 J = −ΔG90°C, as we would predict.
Equilibrium Conditions
ΔG = 0 only if ΔH = TΔS.
We can also calculate the temperature at which liquid water is in equilibrium with water vapor. Inserting the values of ΔH and ΔS into the definition of ΔG (Equation $\ref{Eq2}$), setting ΔG = 0, and solving for T,
0 J=40,657 J−T(108.96 J/K)
T=373.15 K
Thus ΔG = 0 at T = 373.15 K and 1 atm, which indicates that liquid water and water vapor are in equilibrium; this temperature is called the normal boiling point of water. At temperatures greater than 373.15 K, ΔG is negative, and water evaporates spontaneously and irreversibly. Below 373.15 K, ΔG is positive, and water does not evaporate spontaneously. Instead, water vapor at a temperature less than 373.15 K and 1 atm will spontaneously and irreversibly condense to liquid water. Figure $1$ shows how the ΔH and TΔS terms vary with temperature for the vaporization of water. When the two lines cross, ΔG = 0, and ΔH = TΔS.
A similar situation arises in the conversion of liquid egg white to a solid when an egg is boiled. The major component of egg white is a protein called albumin, which is held in a compact, ordered structure by a large number of hydrogen bonds. Breaking them requires an input of energy (ΔH > 0), which converts the albumin to a highly disordered structure in which the molecules aggregate as a disorganized solid (ΔS > 0). At temperatures greater than 373 K, the TΔS term dominates, and ΔG < 0, so the conversion of a raw egg to a hard-boiled egg is an irreversible and spontaneous process above 373 K.
Figure $1$:
$\Delta{H}$ $\Delta{S}$ $-T\Delta S$ $\Delta{G}$
-
(exothermic)
+
(products more disordered)
-
(favors spontaneity)
-
(spontaneous at all T)
-
(exothermic)
-
(products less disordered)
+
(opposes spontaneity)
-
(spontaneous) at low T
+
(non-spontaneous) at high T "Enthalpically-driven process"
+
(endothermic)
+
(products more disordered)
-
(favors spontaneity)
+
(non-spontaneous) at low T
-
(spontaneous) at high T "Entropically-driven process"
+
(endothermic)
-
(products less disordered)
+
(opposes spontaneity)
+
(non-spontaneous at all T)
Determining if a Reaction is Spontaneous: Determining if a Reaction is Spontaneous(opens in new window) [youtu.be]
Enthalpically vs. Entropically Driving Reactions
Consider the following possible states for two different types of molecules with some attractive force:
There would appear to be greater entropy on the left (state 1) than on the right (state 2). Thus the entropic change for the reaction as written (i.e. going to the right) would be (-) in magnitude, and the energetic contribution to the free energy change would be (+) (i.e. unfavorable) for the reaction as written.
In going to the right, there is an attractive force and the molecules adjacent to each other is a lower energy state (heat energy, q, is liberated). To go to the left, we have to overcome this attractive force (input heat energy) and the left direction is unfavorable with regard to heat energy q. The change in enthalpy is (-) in going to the right (q released), and this enthalpy change is negative (-) in going to the right (and (+) in going to the left). This reaction as written, is therefore, enthalpically favorable and entropically unfavorable. Hence, it is an enthalpically driven reaction.
From Table $1$, it would appear that we might be able to get the reaction to go to the right at low temperatures (lower temperature would minimize the energetic contribution of the entropic change). Looking at the same process from an opposite direction:
This reaction as written, is entropically favorable, and enthalpically unfavorable. It is an entropically driven reaction. From Table $1$, it would appear that we might be able to get the reaction to go to the right at high temperatures (high temperature would increase the energetic contribution of the entropic change).
The Relationship between ΔG and Work
In the previous subsection, we learned that the value of ΔG allows us to predict the spontaneity of a physical or a chemical change. In addition, the magnitude of ΔG for a process provides other important information. The change in free energy (ΔG) is equal to the maximum amount of work that a system can perform on the surroundings while undergoing a spontaneous change (at constant temperature and pressure): ΔG = wmax. To see why this is true, let’s look again at the relationships among free energy, enthalpy, and entropy expressed in Equation $\ref{Eq2}$. We can rearrange this equation as follows:
$ΔH = ΔG + TΔS \label{Eq4}$
This equation tells us that when energy is released during an exothermic process (ΔH < 0), such as during the combustion of a fuel, some of that energy can be used to do work (ΔG < 0), while some is used to increase the entropy of the universe (TΔS > 0). Only if the process occurs infinitely slowly in a perfectly reversible manner will the entropy of the universe be unchanged. Because no real system is perfectly reversible, the entropy of the universe increases during all processes that produce energy. As a result, no process that uses stored energy can ever be 100% efficient; that is, ΔH will never equal ΔG because ΔS has a positive value.
One of the major challenges facing engineers is to maximize the efficiency of converting stored energy to useful work or converting one form of energy to another. As indicated in Table $2$, the efficiencies of various energy-converting devices vary widely. For example, an internal combustion engine typically uses only 25%–30% of the energy stored in the hydrocarbon fuel to perform work; the rest of the stored energy is released in an unusable form as heat. In contrast, gas–electric hybrid engines, now used in several models of automobiles, deliver approximately 50% greater fuel efficiency. A large electrical generator is highly efficient (approximately 99%) in converting mechanical to electrical energy, but a typical incandescent light bulb is one of the least efficient devices known (only approximately 5% of the electrical energy is converted to light). In contrast, a mammalian liver cell is a relatively efficient machine and can use fuels such as glucose with an efficiency of 30%–50%.
Table $2$: Approximate Thermodynamic Efficiencies of Various Devices
Device Energy Conversion Approximate Efficiency (%)
large electrical generator mechanical → electrical 99
chemical battery chemical → electrical 90
home furnace chemical → heat 65
small electric tool electrical → mechanical 60
space shuttle engine chemical → mechanical 50
mammalian liver cell chemical → chemical 30–50
spinach cell light → chemical 30
internal combustion engine chemical → mechanical 25–30
fluorescent light electrical → light 20
solar cell light → electricity 10
incandescent light bulb electricity → light 5
yeast cell chemical → chemical 2–4
Standard Free-Energy Change
We have seen that there is no way to measure absolute enthalpies, although we can measure changes in enthalpy (ΔH) during a chemical reaction. Because enthalpy is one of the components of Gibbs free energy, we are consequently unable to measure absolute free energies; we can measure only changes in free energy. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free-energy change can be calculated from the definition of free energy, if the standard enthalpy and entropy changes are known, using Equation $\ref{Eq5}$:
$ΔG° = ΔH° − TΔS° \label{Eq5}$
If ΔS° and ΔH° for a reaction have the same sign, then the sign of ΔG° depends on the relative magnitudes of the ΔH° and TΔS° terms. It is important to recognize that a positive value of ΔG° for a reaction does not mean that no products will form if the reactants in their standard states are mixed; it means only that at equilibrium the concentrations of the products will be less than the concentrations of the reactants.
A positive ΔG° means that the equilibrium constant is less than 1.
Example $1$
Calculate the standard free-energy change (ΔG°) at 25°C for the reaction
$\ce{H2(g) + O2(g) <=> H_2O2(l)} \nonumber$
At 25°C, the standard enthalpy change (ΔH°) is −187.78 kJ/mol, and the absolute entropies of the products and reactants are:
• S°(H2O2) = 109.6 J/(mol•K),
• S°(O2) = 205.2 J/(mol•K), and
• S°(H2) = 130.7 J/(mol•K).
Is the reaction spontaneous as written?
Given: balanced chemical equation, ΔH° and S° for reactants and products
Asked for: spontaneity of reaction as written
Strategy:
1. Calculate ΔS° from the absolute molar entropy values given.
2. Use Equation $\ref{Eq5}$, the calculated value of ΔS°, and other data given to calculate ΔG° for the reaction. Use the value of ΔG° to determine whether the reaction is spontaneous as written.
Solution
A To calculate ΔG° for the reaction, we need to know ΔH°, ΔS°, and T. We are given ΔH°, and we know that T = 298.15 K. We can calculate ΔS° from the absolute molar entropy values provided using the “products minus reactants” rule:
\begin{align*}\Delta S^\circ &=S^\circ(\mathrm{H_2O_2})-[S^\circ(\mathrm{O_2})+S^\circ(\mathrm{H_2})] \nonumber \ &=[\mathrm{1\;mol\;H_2O_2}\times109.6\;\mathrm{J/(mol\cdot K})] \nonumber \ &-\left \{ [\textrm{1 mol H}_2\times130.7\;\mathrm{J/(mol\cdot K)}]+[\textrm{1 mol O}_2\times205.2\;\mathrm{J/(mol\cdot K)}] \right \} \nonumber \&=-226.3\textrm{ J/K }(\textrm{per mole of }\mathrm{H_2O_2}) \end{align*}
As we might expect for a reaction in which 2 mol of gas is converted to 1 mol of a much more ordered liquid, ΔS° is very negative for this reaction.
B Substituting the appropriate quantities into Equation $\ref{Eq5}$,
\begin{align*}\Delta G^\circ=\Delta H^\circ -T\Delta S^\circ &=-187.78\textrm{ kJ/mol}-(\textrm{298.15 K}) [-226.3\;\mathrm{J/(mol\cdot K)}\times\textrm{1 kJ/1000 J}]\nonumber \ &=-187.78\textrm{ kJ/mol}+67.47\textrm{ kJ/mol}=-120.31\textrm{ kJ/mol}\nonumber \end{align*} \nonumber
The negative value of ΔG° indicates that the reaction is spontaneous as written. Because ΔS° and ΔH° for this reaction have the same sign, the sign of ΔG° depends on the relative magnitudes of the ΔH° and TΔS° terms. In this particular case, the enthalpy term dominates, indicating that the strength of the bonds formed in the product more than compensates for the unfavorable ΔS° term and for the energy needed to break bonds in the reactants.
Exercise $1$
Calculate the standard free-energy change (ΔG°) at 25°C for the reaction
$2H_2(g)+N_2(g) \rightleftharpoons N_2H_4(l)\nonumber$
At 25°C, the standard enthalpy change (ΔH°) is 50.6 kJ/mol, and the absolute entropies of the products and reactants are S°(N2H4) = 121.2 J/(mol•K), S°(N2) = 191.6 J/(mol•K), and S°(H2) = 130.7 J/(mol•K). Is the reaction spontaneous as written?
Answer
149.5 kJ/mol; no
Video Solution
Tabulated values of standard free energies of formation allow chemists to calculate the values of ΔG° for a wide variety of chemical reactions rather than having to measure them in the laboratory. The standard free energy of formation ($ΔG^∘_f$)of a compound is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. By definition, the standard free energy of formation of an element in its standard state is zero at 298.15 K. One mole of Cl2 gas at 298.15 K, for example, has $ΔG^∘_f = 0$. The standard free energy of formation of a compound can be calculated from the standard enthalpy of formation (ΔHf) and the standard entropy of formation (ΔSf) using the definition of free energy:
$Δ^o_{f} =ΔH^o_{f} −TΔS^o_{f} \label{Eq6}$
Using standard free energies of formation to calculate the standard free energy of a reaction is analogous to calculating standard enthalpy changes from standard enthalpies of formation using the familiar “products minus reactants” rule:
$ΔG^o_{rxn}=\sum mΔG^o_{f} (products)− \sum nΔ^o_{f} (reactants) \label{Eq7a}$
where m and n are the stoichiometric coefficients of each product and reactant in the balanced chemical equation. A very large negative ΔG° indicates a strong tendency for products to form spontaneously from reactants; it does not, however, necessarily indicate that the reaction will occur rapidly. To make this determination, we need to evaluate the kinetics of the reaction.
The "Products minus Reactants" Rule
The $ΔG^o$ of a reaction can be calculated from tabulated ΔGf values (Table T1) using the “products minus reactants” rule.
Example $2$
Calculate ΔG° for the reaction of isooctane with oxygen gas to give carbon dioxide and water. Use the following data:
• ΔG°f(isooctane) = −353.2 kJ/mol,
• ΔG°f(CO2) = −394.4 kJ/mol, and
• ΔG°f(H2O) = −237.1 kJ/mol. Is the reaction spontaneous as written?
Given: balanced chemical equation and values of ΔG°f for isooctane, CO2, and H2O
Asked for: spontaneity of reaction as written
Strategy:
Use the “products minus reactants” rule to obtain ΔGrxn, remembering that ΔG°f for an element in its standard state is zero. From the calculated value, determine whether the reaction is spontaneous as written.
Solution
The balanced chemical equation for the reaction is as follows:
$\mathrm{C_8H_{18}(l)}+\frac{25}{2}\mathrm{O_2(g)}\rightarrow\mathrm{8CO_2(g)}+\mathrm{9H_2O(l)}\nonumber$
We are given ΔGf values for all the products and reactants except O2(g). Because oxygen gas is an element in its standard state, ΔGf (O2) is zero. Using the “products minus reactants” rule,
\begin{align*} \Delta G^\circ &=[8\Delta G^\circ_\textrm f(\mathrm{CO_2})+9\Delta G^\circ_\textrm f(\mathrm{H_2O})]-\left[1\Delta G^\circ_\textrm f(\mathrm{C_8H_{18}})+\dfrac{25}{2}\Delta G^\circ_\textrm f(\mathrm{O_2})\right] \nonumber \ &=[(\textrm{8 mol})(-394.4\textrm{ kJ/mol})+(\textrm{9 mol})(-237.1\textrm{ kJ/mol})] \nonumber\&-\left [(\textrm{1 mol})(-353.2\textrm{ kJ/mol})+\left(\dfrac{25}{2}\;\textrm{mol}\right)(0 \textrm{ kJ/mol}) \right ] \nonumber \ &=-4935.9\textrm{ kJ }(\textrm{per mol of }\mathrm{C_8H_{18}})\nonumber \end{align*}
Because ΔG° is a large negative number, there is a strong tendency for the spontaneous formation of products from reactants (though not necessarily at a rapid rate). Also notice that the magnitude of ΔG° is largely determined by the ΔGf of the stable products: water and carbon dioxide.
Exercise $2$
Calculate ΔG° for the reaction of benzene with hydrogen gas to give cyclohexane using the following data
• ΔGf(benzene) = 124.5 kJ/mol
• ΔGf (cyclohexane) = 217.3 kJ/mol.
Is the reaction spontaneous as written?
Answer
92.8 kJ; no
Video Solution
Calculated values of ΔG° are extremely useful in predicting whether a reaction will occur spontaneously if the reactants and products are mixed under standard conditions. We should note, however, that very few reactions are actually carried out under standard conditions, and calculated values of ΔG° may not tell us whether a given reaction will occur spontaneously under nonstandard conditions. What determines whether a reaction will occur spontaneously is the free-energy change (ΔG) under the actual experimental conditions, which are usually different from ΔG°. If the ΔH and TΔS terms for a reaction have the same sign, for example, then it may be possible to reverse the sign of ΔG by changing the temperature, thereby converting a reaction that is not thermodynamically spontaneous, having Keq < 1, to one that is, having a Keq > 1, or vice versa. Because ΔH and ΔS usually do not vary greatly with temperature in the absence of a phase change, we can use tabulated values of ΔH° and ΔS° to calculate ΔG° at various temperatures, as long as no phase change occurs over the temperature range being considered.
In the absence of a phase change, neither $ΔH$ nor $ΔS$ vary greatly with temperature.
Example $3$
Calculate (a) ΔG° and (b) ΔG300°C for the reaction N2(g)+3H2(g)⇌2NH3(g), assuming that ΔH and ΔS do not change between 25°C and 300°C. Use these data:
• S°(N2) = 191.6 J/(mol•K),
• S°(H2) = 130.7 J/(mol•K),
• S°(NH3) = 192.8 J/(mol•K), and
• ΔHf (NH3) = −45.9 kJ/mol.
Given: balanced chemical equation, temperatures, S° values, and ΔHf for NH3
Asked for: ΔG° and ΔG at 300°C
Strategy:
1. Convert each temperature to kelvins. Then calculate ΔS° for the reaction. Calculate ΔH° for the reaction, recalling that ΔHf for any element in its standard state is zero.
2. Substitute the appropriate values into Equation $\ref{Eq5}$ to obtain ΔG° for the reaction.
3. Assuming that ΔH and ΔS are independent of temperature, substitute values into Equation $\ref{Eq2}$ to obtain ΔG for the reaction at 300°C.
Solution
A To calculate ΔG° for the reaction using Equation $\ref{Eq5}$, we must know the temperature as well as the values of ΔS° and ΔH°. At standard conditions, the temperature is 25°C, or 298 K. We can calculate ΔS° for the reaction from the absolute molar entropy values given for the reactants and the products using the “products minus reactants” rule:
\begin{align*}\Delta S^\circ_{\textrm{rxn}}&=2S^\circ(\mathrm{NH_3})-[S^\circ(\mathrm{N_2})+3S^\circ(\mathrm{H_2})] \nonumber\ &=[\textrm{2 mol NH}_3\times192.8\;\mathrm{J/(mol\cdot K)}] \nonumber\ &-\left \{[\textrm{1 mol N}_2\times191.6\;\mathrm{J/(mol\cdot K)}]+[\textrm{3 mol H}_2\times130.7\;\mathrm{J/(mol\cdot K)}]\right \}\nonumber\ &=-198.1\textrm{ J/K (per mole of N}_2)\end{align*}\nonumber
We can also calculate ΔH° for the reaction using the “products minus reactants” rule. The value of ΔHf (NH3) is given, and ΔHf is zero for both N2 and H2:
\begin{align*}\Delta H^\circ_{\textrm{rxn}}&=2\Delta H^\circ_\textrm f(\mathrm{NH_3})-[\Delta H^\circ_\textrm f(\mathrm{N_2})+3\Delta H^\circ_\textrm f(\mathrm{H_2})]\nonumber \ &=[2\times(-45.9\textrm{ kJ/mol})]-[(1\times0\textrm{ kJ/mol})+(3\times0 \textrm{ kJ/mol})]\nonumber \ &=-91.8\textrm{ kJ(per mole of N}_2)\nonumber\end{align*}\nonumber
B Inserting the appropriate values into Equation $\ref{Eq5}$
$\Delta G^\circ_{\textrm{rxn}}=\Delta H^\circ-T\Delta S^\circ=(-\textrm{91.8 kJ})-(\textrm{298 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J})=-\textrm{32.7 kJ (per mole of N}_2)\nonumber$
C To calculate ΔG for this reaction at 300°C, we assume that ΔH and ΔS are independent of temperature (i.e., ΔH300°C = H° and ΔS300°C = ΔS°) and insert the appropriate temperature (573 K) into Equation $\ref{Eq2}$:
\begin{align*}\Delta G_{300^\circ\textrm C}&=\Delta H_{300^\circ\textrm C}-(\textrm{573 K})(\Delta S_{300^\circ\textrm C})=\Delta H^\circ -(\textrm{573 K})\Delta S^\circ\nonumber \ &=(-\textrm{91.8 kJ})-(\textrm{573 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J})=21.7\textrm{ kJ (per mole of N}_2)\nonumber \end{align*}\nonumber
In this example, changing the temperature has a major effect on the thermodynamic spontaneity of the reaction. Under standard conditions, the reaction of nitrogen and hydrogen gas to produce ammonia is thermodynamically spontaneous, but in practice, it is too slow to be useful industrially. Increasing the temperature in an attempt to make this reaction occur more rapidly also changes the thermodynamics by causing the −TΔS° term to dominate, and the reaction is no longer spontaneous at high temperatures; that is, its Keq is less than one. This is a classic example of the conflict encountered in real systems between thermodynamics and kinetics, which is often unavoidable.
Exercise $3$
Calculate
1. $ΔG°$ and
2. $ΔG_{750°C}$
for the following reaction
$2NO(g) + O_2(g) <=> 2NO_2 (g) \nonumber$
which is important in the formation of urban smog. Assume that $ΔH$ and $ΔS$ do not change between 25.0°C and 750°C and use these data:
• S°(NO) = 210.8 J/(mol•K),
• S°(O2) = 205.2 J/(mol•K),
• S°(NO2) = 240.1 J/(mol•K),
• ΔHf(NO2) = 33.2 kJ/mol, and
• ΔHf (NO) = 91.3 kJ/mol.
Answer a
−72.5 kJ/mol of $O_2$
Answer b
33.8 kJ/mol of $O_2$
Video Solution
The effect of temperature on the spontaneity of a reaction, which is an important factor in the design of an experiment or an industrial process, depends on the sign and magnitude of both ΔH° and ΔS°. The temperature at which a given reaction is at equilibrium can be calculated by setting ΔG° = 0 in Equation $\ref{Eq5}$, as illustrated in Example $4$.
Example $4$
As you saw in Example $3$, the reaction of nitrogen and hydrogen gas to produce ammonia is one in which ΔH° and ΔS° are both negative. Such reactions are predicted to be thermodynamically spontaneous at low temperatures but nonspontaneous at high temperatures. Use the data in Example 9.5.3 to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous, assuming that ΔH° and ΔS° are independent of temperature.
Given: ΔH° and ΔS°
Asked for: temperature at which reaction changes from spontaneous to nonspontaneous
Strategy:
Set ΔG° equal to zero in Equation $\ref{Eq5}$ and solve for T, the temperature at which the reaction becomes nonspontaneous.
Solution
In Example $3$, we calculated that ΔH° is −91.8 kJ/mol of N2 and ΔS° is −198.1 J/K per mole of N2, corresponding to ΔG° = −32.7 kJ/mol of N2 at 25°C. Thus the reaction is indeed spontaneous at low temperatures, as expected based on the signs of ΔH° and ΔS°. The temperature at which the reaction becomes nonspontaneous is found by setting ΔG° equal to zero and rearranging Equation $\ref{Eq5}$ to solve for T:
\begin{align*}\Delta G^\circ &=\Delta H^\circ - T\Delta S^\circ=0 \ \Delta H^\circ &=T\Delta S^\circ \ T=\dfrac{\Delta H^\circ}{\Delta S^\circ}&=\dfrac{(-\textrm{91.8 kJ})(\textrm{1000 J/kJ})}{-\textrm{198.1 J/K}}=\textrm{463 K}\end{align*} \nonumber
This is a case in which a chemical engineer is severely limited by thermodynamics. Any attempt to increase the rate of reaction of nitrogen with hydrogen by increasing the temperature will cause reactants to be favored over products above 463 K.
Exercise $4$
As you found in the exercise in Example 9.5.3, ΔH° and ΔS° are both negative for the reaction of nitric oxide and oxygen to form nitrogen dioxide. Use those data to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous.
Answer
792.6 K
Summary
• The change in Gibbs free energy, which is based solely on changes in state functions, is the criterion for predicting the spontaneity of a reaction.
• Free-energy change
$ΔG = ΔH − TΔS\nonumber$
• Standard free-energy change
$ΔG° = ΔH° − TΔS°\nonumber$
We can predict whether a reaction will occur spontaneously by combining the entropy, enthalpy, and temperature of a system in a new state function called Gibbs free energy (G). The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is spontaneous, ΔG < 0. If the process is not spontaneous as written but is spontaneous in the reverse direction, ΔG > 0. At constant temperature and pressure, ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free energy of formation (ΔGf), is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. Tabulated values of standard free energies of formation are used to calculate ΔG° for a reaction.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/19%3A_Chemical_Thermodynamics/19.06%3A_Free_Energy_and_Temperature.txt |
Learning Objectives
• To know the relationship between free energy and the equilibrium constant.
We have identified three criteria for whether a given reaction will occur spontaneously:
1. $ΔS_{univ} > 0$,
2. $ΔG_{sys} < 0$ (applicable under constant temperature and constant pressure conditions), and
3. the relative magnitude of the reaction quotient $Q$ versus the equilibrium constant $K$.
Recall that if $Q < K$, then the reaction proceeds spontaneously to the right as written, resulting in the net conversion of reactants to products. Conversely, if $Q > K$, then the reaction proceeds spontaneously to the left as written, resulting in the net conversion of products to reactants. If $Q = K$, then the system is at equilibrium, and no net reaction occurs. Table $1$ summarizes these criteria and their relative values for spontaneous, nonspontaneous, and equilibrium processes.
Table $1$: Criteria for the Spontaneity of a Process as Written
Spontaneous Equilibrium Nonspontaneous*
*Spontaneous in the reverse direction.
ΔSuniv > 0 ΔSuniv = 0 ΔSuniv < 0
ΔGsys < 0 ΔGsys = 0 ΔGsys > 0
Q < K Q = K Q > K
Because all three criteria are assessing the same thing—the spontaneity of the process—it would be most surprising indeed if they were not related. In this section, we explore the relationship between the standard free energy of reaction ($ΔG^o$) and the equilibrium constant ($K$).
Free Energy and the Equilibrium Constant
Because $ΔH^o$ and $ΔS^o$ determine the magnitude and sign of $ΔG^o$ and also because $K$ is a measure of the ratio of the concentrations of products to the concentrations of reactants, we should be able to express K in terms of $ΔG^o$ and vice versa. "Free Energy", $ΔG$ is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature, thereby eliminating $ΔH$ from the equation for $ΔG$. The general relationship can be shown as follows (derivation not shown):
$\Delta G = V \Delta P − S \Delta T \label{18.29}$
If a reaction is carried out at constant temperature ($ΔT = 0$), then Equation $\ref{18.29}$ simplifies to
$\Delta{G} = V\Delta{P} \label{18.30}$
Under normal conditions, the pressure dependence of free energy is not important for solids and liquids because of their small molar volumes. For reactions that involve gases, however, the effect of pressure on free energy is very important.
Assuming ideal gas behavior, we can replace the $V$ in Equation $\ref{18.30}$ by $nRT/P$ (where $n$ is the number of moles of gas and $R$ is the ideal gas constant) and express $\Delta{G}$ in terms of the initial and final pressures ($P_i$ and $P_f$, respectively):
\begin{align} \Delta G &=\left(\dfrac{nRT}{P}\right)\Delta P \[4pt] &=nRT\dfrac{\Delta P}{P} \[4pt] &=nRT\ln\left(\dfrac{P_\textrm f}{P_\textrm i}\right) \label{18.31}\end{align}
If the initial state is the standard state with $P_i = 1 \,atm$, then the change in free energy of a substance when going from the standard state to any other state with a pressure $P$ can be written as follows:
$G − G^° = nRT\ln{P} \nonumber$
This can be rearranged as follows:
$G = G^° + nRT\ln {P} \label{18.32}$
As you will soon discover, Equation $\ref{18.32}$ allows us to relate $ΔG^o$ and $K_p$. Any relationship that is true for $K_p$ must also be true for $K$ because $K_p$ and $K$ are simply different ways of expressing the equilibrium constant using different units.
Let’s consider the following hypothetical reaction, in which all the reactants and the products are ideal gases and the lowercase letters correspond to the stoichiometric coefficients for the various species:
$aA+bB \rightleftharpoons cC+dD \label{18.33}$
Because the free-energy change for a reaction is the difference between the sum of the free energies of the products and the reactants, we can write the following expression for $ΔG$:
\begin{align} \Delta{G} &=\sum_m G_{products}−\sum_n G_{reactants} \[4pt] &=(cG_C+dG_D)−(aG_A+bG_B) \label{18.34} \end{align}
Substituting Equation $\ref{18.32}$ for each term into Equation $\ref{18.34}$,
$\Delta G = [(cG^o_C+cRT \ln P_C)+(dG^o_D+dRT\ln P_D)]−[(aG^o_A+aRT\ln P_A)+(bG^o_B+bRT\ln P_B)] \nonumber$
Combining terms gives the following relationship between $ΔG$ and the reaction quotient $Q$:
\begin{align} \Delta G &=\Delta G^\circ+RT \ln\left(\dfrac{P^c_\textrm CP^d_\textrm D}{P^a_\textrm AP^b_\textrm B}\right) \[4pt] &=\Delta G^\circ+RT\ln Q \label{18.35} \end{align}
where $ΔG^o$ indicates that all reactants and products are in their standard states. For gases at equilibrium ($Q = K_p$), and as you’ve learned in this chapter, $ΔG = 0$ for a system at equilibrium. Therefore, we can describe the relationship between $ΔG^o$ and $K_p$ for gases as follows:
\begin{align} 0 &= ΔG^o + RT\ln K_p \label{18.36a} \[4pt] ΔG^o &= −RT\ln K_p \label{18.36b} \end{align}
If the products and reactants are in their standard states and $ΔG^o < 0$, then $K_p > 1$, and products are favored over reactants when the reaction is at equilibrium. Conversely, if $ΔG^o > 0$, then $K_p < 1$, and reactants are favored over products when the reaction is at equilibrium. If $ΔG^o = 0$, then $K_p = 1$, and neither reactants nor products are favored when the reaction is at equilibrium.
For a spontaneous process under standard conditions, $K_{eq}$ and $K_p$ are greater than 1.
Example $1$
$ΔG^o$ is −32.7 kJ/mol of N2 for the reaction
$\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \nonumber$
This calculation was for the reaction under standard conditions—that is, with all gases present at a partial pressure of 1 atm and a temperature of 25°C. Calculate $ΔG$ for the same reaction under the following nonstandard conditions:
• $P_{\textrm N_2}$ = 2.00 atm,
• $P_{\textrm H_2}$ = 7.00 atm,
• $P_{\textrm{NH}_3}$ = 0.021 atm, and
• $T = 100 ^oC$.
Does the reaction proceed to the right, as written, or to the left to reach equilibrium?
Given: balanced chemical equation, partial pressure of each species, temperature, and ΔG°
Asked for: whether the reaction proceeds to the right or to the left to reach equilibrium
Strategy:
1. Using the values given and Equation $\ref{18.35}$, calculate $Q$.
2. Determine if $Q$ is >, <, or = to $K$
3. Substitute the values of $ΔG^o$ and $Q$ into Equation $\ref{18.35}$ to obtain $ΔG$ for the reaction under nonstandard conditions.
Solution:
A The relationship between $ΔG^o$ and $ΔG$ under nonstandard conditions is given in Equation $\ref{18.35}$. Substituting the partial pressures given, we can calculate $Q$:
\begin{align*} Q &=\dfrac{P^2_{\textrm{NH}_3}}{P_{\textrm N_2}P^3_{\textrm H_2}} \[4pt] &=\dfrac{(0.021)^2}{(2.00)(7.00)^3}=6.4 \times 10^{-7} \end{align*} \nonumber
B Because $ΔG^o$ is −, K must be a number greater than 1
C Substituting the values of $ΔG^o$ and $Q$ into Equation $\ref{18.35}$,
\begin{align*} \Delta G &=\Delta G^\circ+RT\ln Q \[4pt] &=-32.7\textrm{ kJ}+\left[(\textrm{8.314 J/K})(\textrm{373 K})\left(\dfrac{\textrm{1 kJ}}{\textrm{1000 J}}\right)\ln(6.4\times10^{-7})\right] \[4pt] &=-32.7\textrm{ kJ}+(-44\textrm{ kJ}) \[4pt] &=-77\textrm{ kJ/mol of N}_2 \end{align*} \nonumber
Because $ΔG < 0$ and $Q < K$ (because $Q < 1$), the reaction proceeds spontaneously to the right, as written, in order to reach equilibrium.
Exercise $1$
Calculate $ΔG$ for the reaction of nitric oxide with oxygen to give nitrogen dioxide under these conditions: T = 50°C, PNO = 0.0100 atm, $P_{\ce{O_2}}$ = 0.200 atm, and $P_{\ce{NO_2}} = 1.00 × 10^{−4} atm$. The value of $ΔG^o$ for this reaction is −72.5 kJ/mol of O2. Are products or reactants favored?
Answer
−92.9 kJ/mol of $\ce{O2}$; the reaction is spontaneous to the right as written. The reaction will proceed in the forward direction to reach equilibrium.
Example $2$
Calculate $K_p$ for the reaction of $\ce{H_2}$ with $\ce{N2}$ to give $\ce{NH3}$ at 25°C. $ΔG^o$ for this reaction is −32.7 kJ/mol of $\ce{N2}$.
Given: balanced chemical equation from Example $1$, $ΔG^o$, and temperature
Asked for: $K_p$
Strategy:
Substitute values for $ΔG^o$ and T (in kelvin) into Equation $\ref{18.36b}$ to calculate $K_p$, the equilibrium constant for the formation of ammonia.
Solution
In Example $1$, we used tabulated values of ΔGf to calculate $ΔG^o$ for this reaction (−32.7 kJ/mol of N2). For equilibrium conditions, rearranging Equation $\ref{18.36b}$,
\begin{align*} \Delta G^\circ &=-RT\ln K_\textrm p \[4pt] \dfrac{-\Delta G^\circ}{RT} &=\ln K_\textrm p \end{align*} \nonumber
Inserting the value of $ΔG^o$ and the temperature (25°C = 298 K) into this equation,
\begin{align*}\ln K_\textrm p &=-\dfrac{(-\textrm{32.7 kJ})(\textrm{1000 J/kJ})}{(\textrm{8.314 J/K})(\textrm{298 K})}=13.2 \[4pt] K_\textrm p &=5.4\times10^5\end{align*} \nonumber
Thus the equilibrium constant for the formation of ammonia at room temperature is product-favored. However, the rate at which the reaction occurs at room temperature is too slow to be useful.
Exercise $3$
Calculate Kp for the reaction of NO with O2 to give NO2 at 25°C. $ΔG^o$ for this reaction is −70.5 kJ/mol of $\ce{O2}$.
Answer
2.3 × 1012
Although $K_p$ is defined in terms of the partial pressures of the reactants and the products, the equilibrium constant $K$ is defined in terms of the concentrations of the reactants and the products. The numerical magnitude of $K_p$ and $K$ are related:
$K_p = K(RT)^{Δn} \label{18.37}$
where $Δn$ is the number of moles of gaseous product minus the number of moles of gaseous reactant. For reactions that involve only solutions, liquids, and solids, $Δn = 0$, so $K_p = K$. For all reactions that do not involve a change in the number of moles of gas present, the relationship in Equation $\ref{18.36b}$ can be written in a more general form:
$ΔG° = −RT \ln K \label{18.38}$
Only when a reaction results in a net production or consumption of gases is it necessary to correct Equation $\ref{18.38}$ for the difference between $K_p$ and $K$.
Non-Ideal Behavior
Although we typically use concentrations or pressures in our equilibrium calculations, recall that equilibrium constants are generally expressed as unitless numbers because of the use of activities or fugacities in precise thermodynamic work. Systems that contain gases at high pressures or concentrated solutions that deviate substantially from ideal behavior require the use of fugacities or activities, respectively.
Combining Equation $\ref{18.38}$ with $ΔG^o = ΔH^o − TΔS^o$ provides insight into how the components of $ΔG^o$ influence the magnitude of the equilibrium constant:
\begin{align} ΔG° &= ΔH° − TΔS° \[4pt] &= −RT \ln K \label{18.39} \end{align}
Notice that $K$ becomes larger as $ΔS^o$ becomes more positive, indicating that the magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder. Moreover, $K$ increases as $ΔH^o$ decreases. Thus the magnitude of the equilibrium constant is also directly influenced by the tendency of a system to seek the lowest energy state possible.
The magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum entropy and seek the lowest energy state possible.
Relating Grxn and Kp: Relating Grxn and Kp(opens in new window) [youtu.be]
Temperature Dependence of the Equilibrium Constant
The fact that $ΔG^o$ and $K$ are related provides us with another explanation of why equilibrium constants are temperature dependent. This relationship is shown explicitly in Equation $\ref{18.39}$, which can be rearranged as follows:
$\ln K=-\dfrac{\Delta H^\circ}{RT}+\dfrac{\Delta S^\circ}{R} \label{18.40}$
Assuming $ΔH^o$ and $ΔS^o$ are temperature independent, for an exothermic reaction ($ΔH^o < 0$), the magnitude of $K$ decreases with increasing temperature, whereas for an endothermic reaction (ΔH° > 0), the magnitude of $K$ increases with increasing temperature. The quantitative relationship expressed in Equation $\ref{18.40}$ agrees with the qualitative predictions made by applying Le Chatelier’s principle. Because heat is produced in an exothermic reaction, adding heat (by increasing the temperature) will shift the equilibrium to the left, favoring the reactants and decreasing the magnitude of $K$. Conversely, because heat is consumed in an endothermic reaction, adding heat will shift the equilibrium to the right, favoring the products and increasing the magnitude of $K$. Equation $\ref{18.40}$ also shows that the magnitude of $ΔH^o$ dictates how rapidly $K$ changes as a function of temperature. In contrast, the magnitude and sign of $ΔS^o$ affect the magnitude of $K$ but not its temperature dependence.
If we know the value of $K$ at a given temperature and the value of $ΔH^o$ for a reaction, we can estimate the value of $K$ at any other temperature, even in the absence of information on $ΔS^o$. Suppose, for example, that $K_1$ and $K_2$ are the equilibrium constants for a reaction at temperatures $T_1$ and $T_2$, respectively. Applying Equation $\ref{18.40}$ gives the following relationship at each temperature:
$\ln K_1 =\dfrac{-\Delta H^\circ}{RT_1}+\dfrac{\Delta S^\circ}{R} \nonumber$
and
$\ln K_2 =\dfrac{-\Delta H^\circ}{RT_2}+\dfrac{\Delta S^\circ}{R} \nonumber$
Subtracting $\ln K_1$ from $\ln K_2$,
$\ln K_2-\ln K_1=\ln\dfrac{K_2}{K_1}=\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \label{18.41}$
Thus calculating $ΔH^o$ from tabulated enthalpies of formation and measuring the equilibrium constant at one temperature ($K_1$) allow us to calculate the value of the equilibrium constant at any other temperature ($K_2$), assuming that $ΔH^o$ and $ΔS^o$ are independent of temperature.
Example $4$
The equilibrium constant for the formation of $\ce{NH3}$ from $\ce{H2}$ and $\ce{N2}$ at 25°C was calculated to be Kp = 5.4 × 105 in Example $3$. What is $K_p$ at 500°C? (Use the data from Example $1$.)
Given: balanced chemical equation, $ΔH^o°$, initial and final $T$, and $K_p$ at 25°C
Asked for: $K_p$ at 500°C
Strategy:
Convert the initial and final temperatures to kelvin. Then substitute appropriate values into Equation $\ref{18.41}$ to obtain $K_2$, the equilibrium constant at the final temperature.
Solution:
The value of $ΔH^o$ for the reaction obtained using Hess’s law is −91.8 kJ/mol of N2. If we set T1 = 25°C = 298.K and T2 = 500°C = 773 K, then from Equation $\ref{18.41}$ we obtain the following:
\begin{align*}\ln\dfrac{K_2}{K_1}&=\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \[4pt] &=\dfrac{(-\textrm{91.8 kJ})(\textrm{1000 J/kJ})}{\textrm{8.314 J/K}}\left(\dfrac{1}{\textrm{298 K}}-\dfrac{1}{\textrm{773 K}}\right)=-22.8 \[4pt] \dfrac{K_2}{K_1}&=1.3\times10^{-10} \[4pt] K_2&=(5.4\times10^5)(1.3\times10^{-10})=7.0\times10^{-5}\end{align*} \nonumber
Thus at 500°C, the equilibrium strongly favors the reactants over the products.
Exercise $4$
In the exercise in Example $3$, you calculated Kp = 2.2 × 1012 for the reaction of NO with O2 to give NO2 at 25°C. Use the $ΔH^o_f$ values in the exercise in Example $1$ to calculate $K_p$ for this reaction at 1000°C.
Answer
5.6 × 10−4
The Van't Hoff Equation: The Van't Hoff Equation (opens in new window) [youtu.be]
Summary
For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature. If we assume ideal gas behavior, the ideal gas law allows us to express ΔG in terms of the partial pressures of the reactants and products, which gives us a relationship between ΔG and Kp, the equilibrium constant of a reaction involving gases, or K, the equilibrium constant expressed in terms of concentrations. If $ΔG^o$ < 0, then K > 1, and products are favored over reactants at equilibrium. Conversely, if $ΔG^o$ > 0, then K < 1, and reactants are favored over products at equilibrium. If $ΔG^o$ = 0, then K=1, and neither reactants nor products are favored at equilibrium. We can use the measured equilibrium constant K at one temperature and ΔH° to estimate the equilibrium constant for a reaction at any other temperature.
Contributors and Attributions
• Modified by Tom Neils (Grand Rapids Community College) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/19%3A_Chemical_Thermodynamics/19.07%3A_Free_Energy_and_the_Equilibrium_Constant.txt |
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
Conceptual Problems
1. A Russian space vehicle developed a leak, which resulted in an internal pressure drop from 1 atm to 0.85 atm. Is this an example of a reversible expansion? Has work been done?
2. Which member of each pair do you expect to have a higher entropy? Why?
1. solid phenol or liquid phenol
2. 1-butanol or butane
3. cyclohexane or cyclohexanol
4. 1 mol of N2 mixed with 2 mol of O2 or 2 mol of NO2
5. 1 mol of O2 or 1 mol of O3
6. 1 mol of propane at 1 atm or 1 mol of propane at 2 atm
1. Determine whether each process is reversible or irreversible.
1. ice melting at 0°C
2. salt crystallizing from a saline solution
3. evaporation of a liquid in equilibrium with its vapor in a sealed flask
4. a neutralization reaction
1. Determine whether each process is reversible or irreversible.
1. cooking spaghetti
2. the reaction between sodium metal and water
3. oxygen uptake by hemoglobin
4. evaporation of water at its boiling point
1. Explain why increasing the temperature of a gas increases its entropy. What effect does this have on the internal energy of the gas?
1. For a series of related compounds, does ΔSvap increase or decrease with an increase in the strength of intermolecular interactions in the liquid state? Why?
1. Is the change in the enthalpy of reaction or the change in entropy of reaction more sensitive to changes in temperature? Explain your reasoning.
1. Solid potassium chloride has a highly ordered lattice structure. Do you expect ΔSsoln to be greater or less than zero? Why? What opposing factors must be considered in making your prediction?
1. Aniline (C6H5NH2) is an oily liquid at 25°C that darkens on exposure to air and light. It is used in dying fabrics and in staining wood black. One gram of aniline dissolves in 28.6 mL of water, but aniline is completely miscible with ethanol. Do you expect ΔSsoln in H2O to be greater than, less than, or equal to ΔSsoln in CH3CH2OH? Why?
Conceptual Answers
1. No, it is irreversible; no work is done because the external pressure is effectively zero.
1. reversible
2. irreversible
3. reversible
4. irreversible
1. Water has a highly ordered, hydrogen-bonded structure that must reorganize to accommodate hydrophobic solutes like aniline. In contrast, we expect that aniline will be able to disperse randomly throughout ethanol, which has a significantly less ordered structure. We therefore predict that ΔSsoln in ethanol will be more positive than ΔSsoln in water.
Numerical Problems
1. Liquid nitrogen, which has a boiling point of −195.79°C, is used as a coolant and as a preservative for biological tissues. Is the entropy of nitrogen higher or lower at −200°C than at −190°C? Explain your answer. Liquid nitrogen freezes to a white solid at −210.00°C, with an enthalpy of fusion of 0.71 kJ/mol. What is its entropy of fusion? Is freezing biological tissue in liquid nitrogen an example of a reversible process or an irreversible process?
2. Using the second law of thermodynamics, explain why heat flows from a hot body to a cold body but not from a cold body to a hot body.
3. One test of the spontaneity of a reaction is whether the entropy of the universe increases: ΔSuniv > 0. Using an entropic argument, show that the following reaction is spontaneous at 25°C:
4Fe(s) + 3O2(g) → 2Fe2O3(s)
Why does the entropy of the universe increase in this reaction even though gaseous molecules, which have a high entropy, are consumed?
1. Calculate the missing data in the following table.
Compound ΔHfus (kJ/mol) ΔSfus [J/(mol·K)] Melting Point (°C)
acetic acid 11.7 16.6
CH3CN 8.2 35.9
CH4 0.94 −182.5
CH3OH 18.2 −97.7
formic acid 12.7 45.1
Based on this table, can you conclude that entropy is related to the nature of functional groups? Explain your reasoning.
1. Calculate the missing data in the following table.
Compound ΔHvap (kJ/mol) ΔSvap [J/(mol·K)] Boiling Point (°C)
hexanoic acid 71.1 105.7
hexane 28.9 85.5
formic acid 60.7 100.8
1-hexanol 44.5 157.5
The text states that the magnitude of ΔSvap tends to be similar for a wide variety of compounds. Based on the values in the table, do you agree?
Conceptual Problems
1. How does each example illustrate the fact that no process is 100% efficient?
1. burning a log to stay warm
2. the respiration of glucose to provide energy
3. burning a candle to provide light
1. Neither the change in enthalpy nor the change in entropy is, by itself, sufficient to determine whether a reaction will occur spontaneously. Why?
1. If a system is at equilibrium, what must be the relationship between ΔH and ΔS?
1. The equilibrium 2AB⇌A2B2 is exothermic in the forward direction. Which has the higher entropy—the products or the reactants? Why? Which is favored at high temperatures?
1. Is ΔG a state function that describes a system or its surroundings? Do its components—ΔH and ΔS—describe a system or its surroundings?
1. How can you use ΔG to determine the temperature of a phase transition, such as the boiling point of a liquid or the melting point of a solid?
1. Occasionally, an inventor claims to have invented a “perpetual motion” machine, which requires no additional input of energy once the machine has been put into motion. Using your knowledge of thermodynamics, how would you respond to such a claim? Justify your arguments.
1. Must the entropy of the universe increase in a spontaneous process? If not, why is no process 100% efficient?
1. The reaction of methyl chloride with water produces methanol and hydrogen chloride gas at room temperature, despite the fact that ΔHrxn = 7.3 kcal/mol. Using thermodynamic arguments, propose an explanation as to why methanol forms.
Conceptual Answers
1. In order for the reaction to occur spontaneously, ΔG for the reaction must be less than zero. In this case, ΔS must be positive, and the TΔS term outweighs the positive value of ΔH.
Numerical Problems
1. Use the tables in the text to determine whether each reaction is spontaneous under standard conditions. If a reaction is not spontaneous, write the corresponding spontaneous reaction.
1. $\mathrm{H_2(g)}+\frac{1}{2}\mathrm{O_2(g)}\rightarrow\mathrm{H_2O(l)}$
2. 2H2(g) + C2H2(g) → C2H6(g)
3. (CH3)2O(g) + H2O(g) → 2CH3OH(l)
4. CH4(g) + H2O(g) → CO(g) + 3H2(g)
1. Use the tables in the text to determine whether each reaction is spontaneous under standard conditions. If a reaction is not spontaneous, write the corresponding spontaneous reaction.
1. K2O2(s) → 2K(s) + O2(g)
2. PbCO3(s) → PbO(s) + CO2(g)
3. P4(s) + 6H2(g) → 4PH3(g)
4. 2AgCl(s) + H2S(g) → Ag2S(s) + 2HCl(g)
1. Nitrogen fixation is the process by which nitrogen in the atmosphere is reduced to NH3 for use by organisms. Several reactions are associated with this process; three are listed in the following table. Which of these are spontaneous at 25°C? If a reaction is not spontaneous, at what temperature does it become spontaneous?
Reaction ΔH298 (kcal/mol) ΔS298 [cal/(°·mol)]
(a) $\frac{1}{2}\mathrm{N_2}+\mathrm{O_2}\rightarrow\mathrm{NO_2}$ 8.0 −14.4
(b) $\frac{1}{2}\mathrm{N_2}+\frac{1}{2}\mathrm{O_2}\rightarrow\mathrm{NO}$ 21.6 2.9
(c) $\frac{1}{2}\mathrm{N_2}+\frac{3}{2}\mathrm{H_2}\rightarrow\mathrm{NH_3}$ −11.0 −23.7
1. A student was asked to propose three reactions for the oxidation of carbon or a carbon compound to CO or CO2. The reactions are listed in the following table. Are any of these reactions spontaneous at 25°C? If a reaction does not occur spontaneously at 25°C, at what temperature does it become spontaneous?
Reaction ΔH298 (kcal/mol) ΔS298 [cal/(°·mol)]
C(s) + H2O(g) → CO(g) + H2(g) 42 32
CO(g) + H2O(g) → CO2(g) + H2(g) −9.8 −10.1
CH4(g) + H2O(g) → CO(g) + 3H2(g) 49.3 51.3
1. Tungsten trioxide (WO3) is a dense yellow powder that, because of its bright color, is used as a pigment in oil paints and water colors (although cadmium yellow is more commonly used in artists’ paints). Tungsten metal can be isolated by the reaction of WO3 with H2 at 1100°C according to the equation WO3(s) + 3H2(g) → W(s) + 3H2O(g). What is the lowest temperature at which the reaction occurs spontaneously? ΔH° = 27.4 kJ/mol and ΔS° = 29.8 J/K.
1. Sulfur trioxide (SO3) is produced in large quantities in the industrial synthesis of sulfuric acid. Sulfur dioxide is converted to sulfur trioxide by reaction with oxygen gas.
1. Write a balanced chemical equation for the reaction of SO2 with O2(g) and determine its ΔG°.
2. What is the value of the equilibrium constant at 600°C?
3. If you had to rely on the equilibrium concentrations alone, would you obtain a higher yield of product at 400°C or at 600°C?
1. Calculate ΔG° for the general reaction MCO3(s) → MO(s) + CO2(g) at 25°C, where M is Mg or Ba. At what temperature does each of these reactions become spontaneous?
Compound ΔHf (kJ/mol) S° [J/(mol·K)]
MCO3
Mg −1111 65.85
Ba −1213.0 112.1
MO
Mg −601.6 27.0
Ba −548.0 72.1
CO2 −393.5 213.8
1. The reaction of aqueous solutions of barium nitrate with sodium iodide is described by the following equation:
Ba(NO3)2(aq) + 2NaI(aq) → BaI2(aq) + 2NaNO3(aq)
You want to determine the absolute entropy of BaI2, but that information is not listed in your tables. However, you have been able to obtain the following information:
Ba(NO3)2 NaI BaI2 NaNO3
ΔHf (kJ/mol) −952.36 −295.31 −605.4 −447.5
S° [J/(mol·K)] 302.5 170.3 205.4
You know that ΔG° for the reaction at 25°C is 22.64 kJ/mol. What is ΔH° for this reaction? What is S° for BaI2?
Numerical Answers
1. −237.1 kJ/mol; spontaneous as written
2. −241.9 kJ/mol; spontaneous as written
3. 8.0 kJ/mol; spontaneous in reverse direction.
4. 141.9 kJ/mol; spontaneous in reverse direction.
1. Not spontaneous at any T
2. Not spontaneous at 25°C; spontaneous above 7400 K
3. Spontaneous at 25°C
1. 919 K
1. MgCO3: ΔG° = 63 kJ/mol, spontaneous above 663 K; BaCO3: ΔG° = 220 kJ/mol, spontaneous above 1562 K
Conceptual Problems
1. Do you expect products or reactants to dominate at equilibrium in a reaction for which ΔG° is equal to
1. 1.4 kJ/mol?
2. 105 kJ/mol?
3. −34 kJ/mol?
1. The change in free energy enables us to determine whether a reaction will proceed spontaneously. How is this related to the extent to which a reaction proceeds?
1. What happens to the change in free energy of the reaction N2(g) + 3F2(g) → 2NF3(g) if the pressure is increased while the temperature remains constant? if the temperature is increased at constant pressure? Why are these effects not so important for reactions that involve liquids and solids?
1. Compare the expressions for the relationship between the change in free energy of a reaction and its equilibrium constant where the reactants are gases versus liquids. What are the differences between these expressions?
Numerical Problems
1. Carbon monoxide, a toxic product from the incomplete combustion of fossil fuels, reacts with water to form CO2 and H2, as shown in the equation CO(g)+H2O(g)⇌CO2(g)+H2(g), for which ΔH° = −41.0 kJ/mol and ΔS° = −42.3 J cal/(mol·K) at 25°C and 1 atm.
1. What is ΔG° for this reaction?
2. What is ΔG if the gases have the following partial pressures: PCO = 1.3 atm, $P_{\mathrm{H_2O}}$ = 0.8 atm, $P_{\mathrm{CO_2}}$ = 2.0 atm, and $P_{\mathrm{H_2}}$ = 1.3 atm?
3. What is ΔG if the temperature is increased to 150°C assuming no change in pressure?
1. Methane and water react to form carbon monoxide and hydrogen according to the equation CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g).
1. What is the standard free energy change for this reaction?
2. What is Kp for this reaction?
3. What is the carbon monoxide pressure if 1.3 atm of methane reacts with 0.8 atm of water, producing 1.8 atm of hydrogen gas?
4. What is the hydrogen gas pressure if 2.0 atm of methane is allowed to react with 1.1 atm of water?
5. At what temperature does the reaction become spontaneous?
1. Calculate the equilibrium constant at 25°C for each equilibrium reaction and comment on the extent of the reaction.
1. CCl4(g)+6H2O(l)⇌CO2(g)+4HCl(aq); ΔG° = −377 kJ/mol
2. Xe(g)+2F2(g)⇌XeF4(s); ΔH° = −66.3 kJ/mol, ΔS° = −102.3 J/(mol·K)
3. PCl3(g)+S⇌PSCl3(l); ΔGf(PCl3) = −272.4 kJ/mol, ΔGf (PSCl3) = −363.2 kJ/mol
1. Calculate the equilibrium constant at 25°C for each equilibrium reaction and comment on the extent of the reaction.
1. 2KClO3(s)⇌2KCl(s)+3O2(g); ΔG° = −225.8 kJ/mol
2. CoCl2(s)+6H2O(g)⇌CoCl2⋅6H2O(s); ΔHrxn = −352 kJ/mol, ΔSrxn = −899 J/(mol·K)
3. 2PCl3(g)+O2(g)⇌2POCl3(g); ΔGf(PCl3) = −272.4 kJ/mol, ΔGf (POCl3) = −558.5 kJ/mol
1. The gas-phase decomposition of N2O4 to NO2 is an equilibrium reaction with Kp = 4.66 × 10−3. Calculate the standard free-energy change for the equilibrium reaction between N2O4 and NO2.
1. The standard free-energy change for the dissolution K4Fe(CN)6⋅H2O(s)⇌4K+(aq)+Fe(CN)64−(aq)+H2O(l) is 26.1 kJ/mol. What is the equilibrium constant for this process at 25°C?
1. Ammonia reacts with water in liquid ammonia solution (am) according to the equation NH3(g) + H2O(am) ⇌ NH4+(am) + OH(am). The change in enthalpy for this reaction is 21 kJ/mol, and ΔS° = −303 J/(mol·K). What is the equilibrium constant for the reaction at the boiling point of liquid ammonia (−31°C)?
1. At 25°C, a saturated solution of barium carbonate is found to have a concentration of [Ba2+] = [CO32−] = 5.08 × 10−5 M. Determine ΔG° for the dissolution of BaCO3.
1. Lead phosphates are believed to play a major role in controlling the overall solubility of lead in acidic soils. One of the dissolution reactions is Pb3(PO4)2(s)+4H+(aq)⇌3Pb2+(aq)+2H2PO4(aq), for which log K = −1.80. What is ΔG° for this reaction?
1. The conversion of butane to 2-methylpropane is an equilibrium process with ΔH° = −2.05 kcal/mol and ΔG° = −0.89 kcal/mol.
1. What is the change in entropy for this conversion?
2. Based on structural arguments, are the sign and magnitude of the entropy change what you would expect? Why?
3. What is the equilibrium constant for this reaction?
1. The reaction of CaCO3(s) to produce CaO(s) and CO2(g) has an equilibrium constant at 25°C of 2 × 10−23. Values of ΔHf are as follows: CaCO3, −1207.6 kJ/mol; CaO, −634.9 kJ/mol; and CO2, −393.5 kJ/mol.
1. What is ΔG° for this reaction?
2. What is the equilibrium constant at 900°C?
3. What is the partial pressure of CO2(g) in equilibrium with CaO and CaCO3 at this temperature?
4. Are reactants or products favored at the lower temperature? at the higher temperature?
1. In acidic soils, dissolved Al3+ undergoes a complex formation reaction with SO42− to form [AlSO4+]. The equilibrium constant at 25°C for the reaction Al3+(aq)+SO42−(aq)⇌AlSO4+(aq) is 1585.
1. What is ΔG° for this reaction?
2. How does this value compare with ΔG° for the reaction Al3+(aq)+F(aq)⇌AlF2+(aq), for which K = 107 at 25°C?
3. Which is the better ligand to use to trap Al3+ from the soil?
Numerical Answers
1. −28.4 kJ/mol
2. −26.1 kJ/mol
3. −19.9 kJ/mol
1. 1.21 × 1066; equilibrium lies far to the right.
2. 1.89 × 106; equilibrium lies to the right.
3. 5.28 × 1016; equilibrium lies far to the right.
1. 13.3 kJ/mol
1. 5.1 × 10−21
1. 10.3 kJ/mol
1. 129.5 kJ/mol
2. 6
3. 6.0 atm
4. Products are favored at high T; reactants are favored at low T. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/19%3A_Chemical_Thermodynamics/19.E%3A_Chemical_Thermodynamics_%28Exercises%29.txt |
In oxidation–reduction (redox) reactions, electrons are transferred from one species (the reductant) to another (the oxidant). This transfer of electrons provides a means for converting chemical energy to electrical energy or vice versa. The study of the relationship between electricity and chemical reactions is called electrochemistry. In this chapter, we describe electrochemical reactions in more depth and explore some of their applications. We start by describing how they can be used to generate an electrical potential, or voltage; and discuss factors that affect the magnitude of the potential. We then explore the relationships among the electrical potential, the change in free energy, and the equilibrium constant for a redox reaction, which are all measures of the thermodynamic driving force for a reaction. Finally, we examine two kinds of applications of electrochemical principles: (1) those in which a spontaneous reaction is used to provide electricity and (2) those in which electrical energy is used to drive a thermodynamically nonspontaneous reaction.
By the end of this chapter, you will understand why different kinds of batteries are used in cars, flashlights, cameras, and portable computers; how rechargeable batteries operate; and why corrosion occurs and how to slow—if not prevent—it. You will also discover how metal objects can be plated with silver or chromium for protection; how silver polish removes tarnish; and how to calculate the amount of electricity needed to produce aluminum, chlorine, copper, and sodium on an industrial scale.
• 20.1: Oxidation States and Redox Reactions
Oxidation state is a useful tool for keeping track of electron transfers. It is most commonly used in dealing with metals and especially with transition metals. Oxidation signifies a loss of electrons and reduction signifies a gain of electrons. Balancing redox reactions is an important step that changes in neutral, basic, and acidic solutions.
• 20.2: Balanced Oxidation-Reduction Equations
Oxidation–reduction reactions are balanced by separating the overall chemical equation into an oxidation equation and a reduction equation. In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reduction reactions in solution using the oxidation state method, in which the overall reaction is separated into an oxidation equation and a reduction equation.
• 20.3: Voltaic Cells
A galvanic (voltaic) cell uses the energy released during a spontaneous redox reaction to generate electricity, whereas an electrolytic cell consumes electrical energy from an external source to force a reaction to occur. Electrochemistry is the study of the relationship between electricity and chemical reactions. The oxidation–reduction reaction that occurs during an electrochemical process consists of two half-reactions, one representing the oxidation process and one the reduction process.
• 20.4: Cell Potential Under Standard Conditions
Redox reactions can be balanced using the half-reaction method. The standard cell potential is a measure of the driving force for the reaction. \(E°_{cell} = E°_{cathode} − E°_{anode} \] The flow of electrons in an electrochemical cell depends on the identity of the reacting substances, the difference in the potential energy of their valence electrons, and their concentrations. The potential of the cell under standard conditions is called the standard cell potential (E°cell).
• 20.5: Gibbs Energy and Redox Reactions
A coulomb (C) relates electrical potential, expressed in volts, and energy, expressed in joules. The faraday (F) is Avogadro’s number multiplied by the charge on an electron and corresponds to the charge on 1 mol of electrons. Spontaneous redox reactions have a negative ΔG and therefore a positive Ecell. Because the equilibrium constant K is related to ΔG, E°cell and K are also related. Large equilibrium constants correspond to large positive values of E°.
• 20.6: Cell Potential Under Nonstandard Conditions
The Nernst equation allows us to determine the spontaneous direction of any redox reaction under any reaction conditions from values of the relevant standard electrode potentials. Concentration cells consist of anode and cathode compartments that are identical except for the concentrations of the reactant. Because ΔG = 0 at equilibrium, the measured potential of a concentration cell is zero at equilibrium (the concentrations are equal).
• 20.7: Batteries and Fuel Cells
Commercial batteries are galvanic cells that use solids or pastes as reactants to maximize the electrical output per unit mass. A battery is a contained unit that produces electricity, whereas a fuel cell is a galvanic cell that requires a constant external supply of one or more reactants to generate electricity. One type of battery is the Leclanché dry cell, which contains an electrolyte in an acidic water-based paste.
• 20.8: Corrosion
Corrosion is a galvanic process that can be prevented using cathodic protection. The deterioration of metals through oxidation is a galvanic process called corrosion. Protective coatings consist of a second metal that is more difficult to oxidize than the metal being protected. Alternatively, a more easily oxidized metal can be applied to a metal surface, thus providing cathodic protection of the surface. A thin layer of zinc protects galvanized steel. Sacrificial electrodes can also be attached
• 20.9: Electrolysis
In electrolysis, an external voltage is applied to drive a nonspontaneous reaction. Electrolysis can also be used to produce \(H_2\) and \(O_2\) from water. Electroplating is the process by which a second metal is deposited on a metal surface. The amount of material consumed or produced in a reaction can be calculated from the stoichiometry of an electrolysis reaction, the amount of current passed, and the duration of the electrolytic reaction.
• 20.E: Electrochemistry (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al.
Thumbnail: Schematic of Zn-Cu galvanic cell. (CC BY-SA 3.0; Ohiostandard).
20: Electrochemistry
Electron transfer is one of the most basic processes that can happen in chemistry. It simply involves the movement of an electron from one atom to another. Many important biological processes rely on electron transfer, as do key industrial transformations used to make valuable products. In biology, for example, electron transfer plays a central role in respiration and the harvesting of energy from glucose, as well as the storage of energy during photosynthesis. In society, electron transfer has been used to obtain metals from ores since the dawn of civilization.
Oxidation States (Numbers)
Oxidation state is a useful tool for keeping track of electron transfers. It is most commonly used in dealing with metals and especially with transition metals. Unlike metals from the first two columns of the periodic table, such as sodium or magnesium, transition metals can often transfer different numbers of electrons, leading to different metal ions (e.g., sodium is generally found as $\ce{Na^{+}}$ and magnesium is almost always $\ce{Mg^{2+}}$, but manganese could be $\ce{Mn^{2+}}$, $\ce{Mn^{3+}}$, and so on, as far as $\ce{Mn^{7+}}$). Oxidation state is a number assigned to an element in a compound according to some rules. This number enables us to describe oxidation-reduction reactions, and balancing redox chemical reactions. When a covalent bond forms between two atoms with different electronegativities the shared electrons in the bond lie closer to the more electronegative atom:
The oxidation state of an atom is the charge that results when the electrons in a covalent bond are assigned to the more electronegative atom and is the charge an atom would possess if the bonding were ionic. In $\ce{HCl}$ (above) the oxidation number for the hydrogen would be +1 and that of the $\ce{Cl}$ would be -1.
Example $1$
Determine which element is oxidized and which element is reduced in the following reactions (be sure to include the Oxidation State of each):
1. $\ce{Zn + 2H^+ → Zn^{2+} + H2}$
2. $\ce{2Al + 3Cu^{2+}→2Al^{3+} +3Cu}$
3. $\ce{CO3^{2-} + 2H^+→ CO2 + H2O}$
Solutions
1. $\ce{Zn}$ is oxidized (Oxidation number: 0 → +2); $\ce{H^{+}}$ is reduced (Oxidation number: +1 → 0)
2. $\ce{Al}$ is oxidized (Oxidation number: 0 → +3); $\ce{Cu^{2+}}$ is reduced (+2 → 0)
3. This is not a redox reaction because each element has the same oxidation number in both reactants and products: O= -2, H= +1, C= +4.
An atom is oxidized if its oxidation number increases, and an atom is reduced if its oxidation number decreases. The atom that is oxidized is the reducing agent, and the atom that is reduced is the oxidizing agent. (Note: the oxidizing and reducing agents can be the same element or compound).
Oxidation Numbers and Nomenclature
Compounds of the alkali (oxidation number +1) and alkaline earth metals (oxidation number +2) are typically ionic in nature. Compounds of metals with higher oxidation numbers (e.g., tin +4) tend to form molecular compounds
• In ionic and covalent molecular compounds usually the less electronegative element is given first.
• In ionic compounds the names are given which refer to the oxidation (ionic) state
• In molecular compounds the names are given which refer to the number of molecules present in the compound
Figure $1$: Example of nomenclature based on oxidation states.
Ionic Molecular
MgH2 magnesium hydride H2S dihydrogen sulfide
FeF2 iron(II) fluoride OF2 oxygen difluoride
Mn2O3 manganese(III) oxide Cl2O3 dichlorine trioxide
An oxidation-reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between two species. An oxidation-reduction reaction is any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron. Redox reactions are common and vital to some of the basic functions of life, including photosynthesis, respiration, combustion, and corrosion or rusting.
Oxidation-Reduction Reaction Examples
Redox reactions are comprised of two parts, a reduced half and an oxidized half, that always occur together. The reduced half gains electrons and the oxidation number decreases, while the oxidized half loses electrons and the oxidation number increases. Simple ways to remember this include the mnemonic devices OIL RIG, meaning "oxidation is loss" and "reduction is gain," and LEO says GER, meaning "loss of e- = oxidation" and "gain of e- = reduced." There is no net change in the number of electrons in a redox reaction. Those given off in the oxidation half reaction are taken up by another species in the reduction half reaction.
The two species that exchange electrons in a redox reaction are given special names. The ion or molecule that accepts electrons is called the oxidizing agent; by accepting electrons it causes the oxidation of another species. Conversely, the species that donates electrons is called the reducing agent; when the reaction occurs, it reduces the other species. In other words, what is oxidized is the reducing agent and what is reduced is the oxidizing agent. (Note: the oxidizing and reducing agents can be the same element or compound, as in disproportionation reactions).
A good example of a redox reaction is the thermite reaction, in which iron atoms in ferric oxide lose (or give up) $\ce{O}$ atoms to $\ce{Al}$ atoms, producing $\ce{Al2O3}$ (Figure $1$).
$\ce{Fe2O3(s) + 2Al(s) -> Al2O3(s) + 2Fe(l)} \nonumber$
Another example of the redox reaction (although less dangerous) is the reaction between zinc and copper sulfate.
$\ce{Zn + CuSO4 -> ZnSO4 + Cu} \nonumber$
Example $2$: Identifying Oxidized and Reduced Elements
Determine what is oxidized and what is reduced in the following reaction.
$\ce{Zn + 2H^+ \rightarrow Zn^{2+} + H_2} \nonumber$
Solution
The oxidation state of $\ce{H^{+}}$ changes from +1 to 0, and the oxidation state of $\ce{Zn}$ changes from 0 to +2. Hence, $\ce{Zn}$ is oxidized and acts as the reducing agent.
The oxidation state of $\ce{H^{+}}$ changes from +1 to 0, and the oxidation state of $\ce{Zn}$ changes from 0 to +2. Hence, $\ce{H^{+}}$ ion is reduced and acts as the oxidizing agent.
Combination Reactions
Combination reactions are among the simplest redox reactions and, as the name suggests, involves "combining" elements to form a chemical compound. As usual, oxidation and reduction occur together. The general equation for a combination reaction is given below:
$\ce{A + B \rightarrow AB} \nonumber$
Example $3$: Combination Reaction
Equation: $\ce{H2 + O2 → H2O} \nonumber$
Calculation: 0 + 0 → (2)(+1) + (-2) = 0
Explanation:
In this equation both $\ce{H2}$ and $\ce{O2}$ are free elements; following Rule #1, their Oxidation States are 0. The product is $\ce{H2O}$, which has a total Oxidation State of 0. According to Rule #6, the Oxidation State of oxygen is usually -2. Therefore, the Oxidation State of $\ce{H}$ in $\ce{H2O}$ must be +1.
Decomposition Reactions
A decomposition reaction is the reverse of a combination reaction, the breakdown of a chemical compound into individual elements:
$AB \rightarrow A + B \nonumber$
Example $4$: Decomposition Reaction
Identify the oxidation state of the products and reactant in the decomposition of water:
$\ce{H_2O \rightarrow H_2 + O_2}\nonumber$
Calculation
$(2)(+1) + (-2) = 0 → 0 + 0 \nonumber$
In this reaction, water is "decomposed" into hydrogen and oxygen. As in the previous example the H2O has a total Oxidation State of 0; thus, according to Rule #6 the Oxidation State of oxygen is usually -2, so the Oxidation State of hydrogen in H2O must be +1.
Single Replacement Reactions
A single replacement reaction involves the "replacing" of an element in the reactants with another element in the products:
$\ce{A + BC \rightarrow AB + C} \nonumber$
Example $5$: Single Replacement Reaction
Chlorine gas is a great oxidizing agent and will replace bromide ions from sodium bromide salt.
$\ce{Cl2(g) + Na \underline{Br}(s) -> Na\underline{Cl}(s) + Br2(l)} \nonumber$
Explanation
In this equation, $\ce{Br}$ is replaced with $\ce{Cl}$, and the $\ce{Cl}$ atoms in $\ce{Cl2}$ are reduced, while the $\ce{Br}$ ion in $\ce{NaBr}$ is oxidized.
Double Replacement Reactions
A double replacement reaction is similar to a single replacement reaction, but involves "replacing" two elements in the reactants, with two in the products:
$\ce{AB + CD \rightarrow AD + CB} \nonumber$
Example $6$: Double Replacement Reaction
The reaction of gaseous hydrogen chloride and iron oxide is a double replacement reaction. Write the expected reaction for this chemistry equation.
Solution
$\ce{Fe2O3 + 6HCl → 2FeCl3 + 3H2O} \nonumber$
In this equation, $\ce{Fe}$ and $\ce{H}$ trade places, and oxygen and chlorine trade places.
Combustion Reactions
Combustion reactions almost always involve oxygen in the form of $\ce{O2}$, and are almost always exothermic, meaning they produce heat. Chemical reactions that give off light and heat and light are colloquially referred to as "burning."
$\ce{C_{x}H_{y} + O_2 \rightarrow CO_2 + H_2O} \nonumber$
Although combustion reactions typically involve redox reactions with a chemical being oxidized by oxygen, many chemicals "burn" in other environments. For example, both titanium and magnesium burn in nitrogen as well:
$\ce{2Ti(s) + N2(g) -> 2TiN(s)} \nonumber$
$\ce{ 3 Mg(s) + N2(g) -> Mg3N2(s)} \nonumber$
Moreover, chemicals can be oxidized by other chemicals than oxygen, such as $\ce{Cl2}$ or $\ce{F2}$; these processes are also considered combustion reactions
Disproportionation Reactions
A single substance can be both oxidized and reduced in some redox reactions. These are known as disproportionation reactions, with the following general equation:
$\ce{2A -> A^{+n} + A^{-n}} \nonumber$
where $n$ is the number of electrons transferred. Disproportionation reactions do not need begin with neutral molecules, and can involve more than two species with differing oxidation states (but rarely).
Example $7$: Disproportionation Reaction
Disproportionation reactions have some practical significance in everyday life, including the reaction of hydrogen peroxide, $\ce{H2O2}$ poured over a cut. This is a decomposition reaction of hydrogen peroxide (catalyzed by the catalase enzyme) that produces oxygen and water. Oxygen is present in all parts of the chemical equation and as a result it is both oxidized and reduced. The reaction is as follows:
$\ce{2H2O2(aq) \rightarrow 2H2O(l) + O2(g)} \nonumber$
Explanation
On the reactant side, $\ce{H}$ has an Oxidation State of +1 and $\ce{O}$ has an Oxidation State of -1, which changes to -2 for the product $\ce{H2O}$ (oxygen is reduced), and 0 in the product $\ce{O2}$ (oxygen is oxidized).
Redox Reactions: Redox Reactions(opens in new window) [youtu.be]
Summary
Oxidation signifies a loss of electrons and reduction signifies a gain of electrons. Balancing redox reactions is an important step that changes in neutral, basic, and acidic solutions. The types of redox reactions: Combination and decomposition, Displacement reactions (single and double), Combustion, Disproportionation. The oxidizing agent undergoes reduction and the reducing agent undergoes oxidation.
Contributors and Attributions
• Christopher Spohrer (UCD), Christina Breitenbuecher (UCD), Luvleen Brar (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/20%3A_Electrochemistry/20.01%3A_Oxidation_States_and_Redox_Reactions.txt |
Learning Objectives
• To identify oxidation–reduction reactions in solution.
We described the defining characteristics of oxidation–reduction, or redox, reactions. Most of the reactions we considered there were relatively simple, and balancing them was straightforward. When oxidation–reduction reactions occur in aqueous solution, however, the equations are more complex and can be more difficult to balance by inspection. Because a balanced chemical equation is the most important prerequisite for solving any stoichiometry problem, we need a method for balancing oxidation–reduction reactions in aqueous solution that is generally applicable. One such method uses oxidation states, and a second is referred to as the half-reaction method.
Balancing Redox Equations Using Oxidation States
To balance a redox equation using the oxidation state method, we conceptually separate the overall reaction into two parts: an oxidation—in which the atoms of one element lose electrons—and a reduction—in which the atoms of one element gain electrons. Consider, for example, the reaction of Cr2+(aq) with manganese dioxide (MnO2) in the presence of dilute acid. Equation $\ref{20.2.1}$ is the net ionic equation for this reaction before balancing; the oxidation state of each element in each species has been assigned using the procedure described previously (in red above each element):
$\overset{\color{red}{+2}}{\ce{Cr^{2+}} ( aq }) + \overset{\color{red}{+4}}{\ce{Mn}} \overset{\color{red}{-2}}{\ce{O_2} ( aq )} + \overset{\color{red}{+1}} {\ce{H^{+}} ( aq )} \rightarrow \overset{\color{red}{+3}}{\ce{Cr^{3+}} ( aq )} + \overset{\color{red}{+2}}{\ce{Mn^{2+}}( aq )} + \overset{\color{red}{+1}} {\ce{H_2}} \overset{\color{red}{-2}} {\ce{O} (l)} \label{20.2.1}$
Notice that chromium is oxidized from the +2 to the +3 oxidation state, while manganese is reduced from the +4 to the +2 oxidation state. We can write an equation for this reaction that shows only the atoms that are oxidized and reduced (ignoring the oxygen and hydrogen atoms):
$\ce{Cr^{2+} + Mn^{4+} -> Cr^{3+} + Mn^{2+}} \label{20.2.2}$
The oxidation can be written as
$\underbrace{\ce{Cr^{2+} -> Cr^{3+} + e^{-}}}_{\text{oxidation with 1 electron lost}} \label{20.2.3}$
and the reduction as
$\underbrace{\ce{Mn^{4+} + 2e^{-} \rightarrow Mn^{2+}}}_{\text{reduction with 2 electrons gained}} \label{20.2.4}$
For the overall chemical equation to be balanced, the number of electrons lost by the reductant must equal the number gained by the oxidant. We must therefore multiply the oxidation and the reduction equations by appropriate coefficients to give us the same number of electrons in both. In this example, we must multiply the oxidation (Equation \ref{20.2.3}) by 2 to give
$\ce{2Cr^{2+} -> 2Cr^{3+} + 2e^{-}} \label{20.2.5}$
The number of electrons lost in the oxidation now equals the number of electrons gained in the reduction (Equation \ref{20.2.4}):
\begin{align*} \ce{2Cr^{2+}} &\rightarrow \ce{2Cr^{3+}} + \ce{2e^{-}} \label{20.2.6} \[8pt] \ce{Mn^{4+}} + \ce{2e^{-}} &\rightarrow \ce{Mn^{2+}} \end{align*}
We then add the equations for the oxidation and the reduction and cancel the electrons on both sides of the equation, using the actual chemical forms of the reactants and products:
\begin{align*} \ce{2Cr^{2+}} &\rightarrow \ce{2Cr^{3+}} + \cancel{\ce{2e^{-}}} \[8pt] \ce{Mn^{4+}} + \cancel{\ce{2e^{-}}} &\rightarrow \ce{Mn^{2+}} \end{align*} \nonumber
to result in the balanced redox reaction (metals only)
$\ce{ Mn^{4+} +2Cr^{2+} \rightarrow 2Cr^{3+} + Mn^{2+}} \label{20.2.7}$
now we can add the non-redox active atoms back into the equation (ignoring water and hydronium for now)
$\ce{MnO2(aq) + 2Cr^{2+}(aq) -> 2Cr^{3+}(aq) + Mn^{2+}(aq)} \label{20.2.7b}$
In a balanced redox reaction, the number of electrons lost by the reductant equals the number of electrons gained by the oxidant.
Although the electrons cancel and the metal atoms are balanced, the total charge on the left side of Equation \ref{20.2.7b} (+4) does not equal the charge on the right side (+8). Because the reaction is carried out in the presence of aqueous acid, we can add $\ce{H^{+}}$ as necessary to either side of the equation to balance the charge. By the same token, if the reaction were carried out in the presence of aqueous base, we could balance the charge by adding $\ce{OH^{−}}$ as necessary to either side of the equation to balance the charges.
In this case, adding four $\ce{H^{+}}$ ions to the left side of Equation \ref{20.2.7b} to give
$\ce{ MnO2(s) + 2Cr^{2+}(aq) + 4H^{+}(aq) -> 2Cr^{3+}(aq) + Mn^{2+}(aq)} \label{20.2.8}$
Although the charges are now balanced in Equation \ref{20.2.8}, we have two oxygen atoms on the left side of the equation and none on the right. We can balance the oxygen atoms without affecting the overall charge balance by adding $\ce{H2O}$ as necessary to either side of the equation. Here, we need to add two $\ce{H2O}$ molecules to the right side of Equation \ref{20.2.8}:
$\ce{MnO2(s) + 2Cr^{2+}(aq) + 4H^{+}(aq) -> 2Cr^{3+}(aq) + Mn^{2+}(aq) + 2H_2O(l)} \label{20.2.9}$
Although we did not explicitly balance the hydrogen atoms, we can see by inspection that the overall chemical equation is now balanced with respect to all atoms and charge. All that remains is to check to make sure that we have not made a mistake. This procedure for balancing reactions is summarized below and illustrated in Example $1$ below.
Procedure for Balancing Oxidation–Reduction Reactions by the Oxidation State Method
1. Write the unbalanced chemical equation for the reaction, showing the reactants and the products.
2. Assign oxidation states to all atoms in the reactants and the products and determine which atoms change oxidation state.
3. Write separate equations for oxidation and reduction, showing (a) the atom(s) that is (are) oxidized and reduced and (b) the number of electrons accepted or donated by each.
4. Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number of electrons.
5. Write the oxidation and reduction equations showing the actual chemical forms of the reactants and the products, adjusting the coefficients as necessary to give the numbers of atoms in step 4.
6. Add the two equations and cancel the electrons.
7. Balance the charge by adding $\ce{H^{+}}$ or $\ce{OH^{−}}$ ions as necessary for reactions in acidic or basic solution, respectively.
8. Balance the oxygen atoms by adding $\ce{H2O}$ molecules to one side of the equation.
9. Check to make sure that the equation is balanced in both atoms and total charges.
Example $1$: Balancing in Acid Solutions
Arsenic acid ($\ce{H3AsO4}$) is a highly poisonous substance that was once used as a pesticide. The reaction of elemental zinc with arsenic acid in acidic solution yields arsine ($\ce{AsH3}$, a highly toxic and unstable gas) and $\ce{Zn^{2+}(aq)}$. Balance the equation for this reaction using oxidation states:
$\ce{H3AsO4(aq) + Zn(s) -> AsH3(g) + Zn^{2+}(aq)} \nonumber$
Given: reactants and products in acidic solution
Asked for: balanced chemical equation using oxidation states
Strategy:
Follow the procedure given above for balancing a redox equation using oxidation states. When you are done, be certain to check that the equation is balanced.
Solution:
1. Write a chemical equation showing the reactants and the products. Because we are given this information, we can skip this step.
2. Assign oxidation states and determine which atoms change oxidation state. The oxidation state of arsenic in arsenic acid is +5, and the oxidation state of arsenic in arsine is −3. Conversely, the oxidation state of zinc in elemental zinc is 0, and the oxidation state of zinc in $Zn^{2+}(aq)$ is +2: $H_3\overset{\color{red}{+5}}{As}O_4(aq) + \overset{\color{red}{0}}{Zn}(s) \rightarrow \overset{\color{red}{-3}}{As}H_3(g) + \overset{\color{red}{+2}}{Zn^{2+}}(aq) \nonumber$
3. Write separate equations for oxidation and reduction. The arsenic atom in H3AsO4 is reduced from the +5 to the −3 oxidation state, which requires the addition of eight electrons:
$\underbrace{ \overset{\color{red}{+5}}{As} + 8e^- \rightarrow \overset{\color{red}{-3}}{As}}_{\text{Reduction with gain of 8 electrons}} \nonumber$
Each zinc atom in elemental zinc is oxidized from 0 to +2, which requires the loss of two electrons per zinc atom:
$\underbrace{ \overset{\color{red}{0}} {Zn} \rightarrow \overset{\color{red}{+2}} {Zn^{2+}} + 2e^- }_{\text{Oxidation with loss of 2 electrons}}\nonumber$
4. Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number of electrons. The reduction equation has eight electrons, and the oxidation equation has two electrons, so we need to multiply the oxidation equation by 4 to obtain \begin{align*} \overset{\color{red}{+5}}{\ce{As}} + \ce{8e^{-}} & \rightarrow \overset{\color{red}{-3}}{\ce{As}} \nonumber \ \overset{\color{red}{0}} {\ce{4Zn}} & \rightarrow \overset{\color{red}{+2}} {\ce{4Zn^{2+}}} + \ce{8e^{-}} \end{align*} \nonumber
5. Write the oxidation and reduction equations showing the actual chemical forms of the reactants and the products, adjusting coefficients as necessary to give the numbers of atoms shown in step 4. Inserting the actual chemical forms of arsenic and zinc and adjusting the coefficients gives
• Reduction: $\ce{H3AsO4(aq) + 8e^{-} \rightarrow AsH3(g)} \nonumber$
• Oxidation: $\ce{4Zn(s) -> 4Zn^{2+}(aq) + 8e^{-}} \nonumber$
6. Add the two equations and cancel the electrons. The sum of the two equations in step 5 is $\ce{H3AsO4(aq) + 4Zn(s)} + \cancel{\ce{8e^{-}}} \rightarrow \ce{AsH3(g)} + \ce{4Zn^{2+}(aq)} + \cancel{\ce{8e^{-}}} \nonumber$ which then yields after canceling electrons $\ce{H3AsO4(aq) + 4Zn(s) \rightarrow AsH3(g) + 4Zn^{2+}(aq)} \nonumber$
7. Balance the charge by adding $\ce{H^{+}}$ or $\ce{OH^{−}}$ ions as necessary for reactions in acidic or basic solution, respectively. Because the reaction is carried out in acidic solution, we can add H+ ions to whichever side of the equation requires them to balance the charge. The overall charge on the left side is zero, and the total charge on the right side is 4 × (+2) = +8. Adding eight H+ ions to the left side gives a charge of +8 on both sides of the equation: $\ce{H3AsO4(aq) + 4Zn(s) + 8H^{+}(aq) \rightarrow AsH3(g) + 4Zn^{2+}(aq)} \nonumber$
8. Balance the oxygen atoms by adding $\ce{H2O}$ molecules to one side of the equation. There are 4 $\ce{O}$ atoms on the left side of the equation. Adding 4 $\ce{H2O}$ molecules to the right side balances the $\ce{O}$ atoms: $\ce{H3AsO4(aq) + 4Zn(s) + 8H^{+}(aq) \rightarrow AsH3(g) + 4Zn^{2+}(aq) + 4H2O(l)} \nonumber$ Although we have not explicitly balanced $\ce{H}$ atoms, each side of the equation has 11 $\ce{H}$ atoms.
9. Check to make sure that the equation is balanced in both atoms and total charges. To guard against careless errors, it is important to check that both the total number of atoms of each element and the total charges are the same on both sides of the equation:
• Atoms: $\ce{1As + 4Zn + 4O + 11H} \overset{\checkmark}{=} \ce{1As + 4Zn + 4O + 11H} \nonumber$
• Charge: $8(+1) \overset{\checkmark}{=} 4(+2) \nonumber$
The balanced chemical equation (both for charge and for atoms) for this reaction is therefore:
$\ce{H3AsO4(aq) + 4Zn(s) + 8H^{+}(aq) -> AsH3(g) + 4Zn^{2+}(aq) + 4H2O(l)} \nonumber$
Exercise $1$: Oxidizing Copper
Copper commonly occurs as the sulfide mineral $\ce{CuS}$. The first step in extracting copper from $\ce{CuS}$ is to dissolve the mineral in nitric acid, which oxidizes the sulfide to sulfate and reduces nitric acid to $\ce{NO}$. Balance the equation for this reaction using oxidation states:
$\ce{CuS(s) + H^{+}(aq) + NO^{-}3(aq) -> Cu^{2+}(aq) + NO(g) + SO^{2-}4(aq)} \nonumber$
Answer
$\ce{3CuS(s) + 8H^{+}(aq) + 8NO^{-}3(aq) \rightarrow 3Cu^{2+}(aq) + 8NO(g) + 3SO^{2-}4(aq) + 4H2O(l)} \nonumber$
Reactions in basic solutions are balanced in exactly the same manner. To make sure you understand the procedure, consider Example $2$.
Example $2$: Balancing in Basic Solution
The commercial solid drain cleaner, Drano, contains a mixture of sodium hydroxide and powdered aluminum. The sodium hydroxide dissolves in standing water to form a strongly basic solution, capable of slowly dissolving organic substances, such as hair, that may be clogging the drain. The aluminum dissolves in the strongly basic solution to produce bubbles of hydrogen gas that agitate the solution to help break up the clogs. The reaction is as follows:
$\ce{Al(s) + H2O(l) \rightarrow [Al(OH)4]^{-}(aq) + H2(g)} \nonumber$
Balance this equation using oxidation states.
Given: reactants and products in a basic solution
Asked for: balanced chemical equation
Strategy:
Follow the procedure given above for balancing a redox reaction using oxidation states. When you are done, be certain to check that the equation is balanced.
Solution:
We will apply the same procedure used in Example $1$, but in a more abbreviated form.
1. The equation for the reaction is given, so we can skip this step.
2. The oxidation state of aluminum changes from 0 in metallic $Al$ to +3 in $\ce{[Al(OH)4]^{−}}$. The oxidation state of hydrogen changes from +1 in $\ce{H_2O}$ to 0 in $\ce{H2}$. Aluminum is oxidized, while hydrogen is reduced: $\overset{\color{red}{0}}{Al}_{(s)} + \overset{\color{red}{+1}}{H}_2 O_{(aq)} \rightarrow [ \overset{\color{red}{+3}}{Al} (OH)_4 ]^- _{(aq)} + \overset{\color{red}{0}}{H_2}_{(g)} \nonumber$
3. Write separate equations for oxidation and reduction.
• Reduction: $\overset{\color{red}{+1}}{H} + e^- \rightarrow \overset{\color{red}{0}}{H} \: (in\: H_2 ) \nonumber$
• Oxidation: $\overset{\color{red}{0}}{Al} \rightarrow \overset{\color{red}{+3}}{Al} + 3e^- \nonumber$
4. Multiply the reduction equation by 3 to obtain an equation with the same number of electrons as the oxidation equation:
• Reduction: $\ce{3H^{+} + 3e^{-} -> 3H^0}\: (in\: \ce{H2}) \nonumber$
• Oxidation: $\ce{Al^0 -> Al^{3+} + 3e^{-}} \nonumber$
5. Insert the actual chemical forms of the reactants and products, adjusting the coefficients as necessary to obtain the correct numbers of atoms as in step 4. Because a molecule of $\ce{H2O}$ contains two protons, in this case, $\ce{3H^{+}}$ corresponds to $\ce{3/2 H2O}$. Similarly, each molecule of hydrogen gas contains two H atoms, so $\ce{3H}$ corresponds to $\ce{3/2H2}$.
• Reduction: $\ce{3/2 H2O + 3e^{-} -> 3/2 H2} \nonumber$
• Oxidation: $\ce{Al -> [Al(OH)4]^{-} + 3e^{-}} \nonumber$
6. Adding the equations and canceling the electrons gives $\ce{Al} + \ce{3/2 H2O} + \cancel{\ce{3e^{-}}} \ce{->} \ce{[Al(OH)4]^{-}} + \ce{3/2 H2} + \cancel{\ce{3e^{-}}} \nonumber$ $\ce{Al} + \ce{3/2 H2O} \ce{->} \ce{[Al(OH)4]^{-}} + \ce{3/2 H2} \nonumber$ To remove the fractional coefficients, multiply both sides of the equation by 2: $\ce{2Al + 3H2O \rightarrow 2[Al(OH)4]^{-} + 3H2} \nonumber$
7. The right side of the equation has a total charge of −2, whereas the left side has a total charge of 0. Because the reaction is carried out in basic solution, we can balance the charge by adding two $\ce{OH^{−}}$ ions to the left side: $\ce{2Al + 2OH^{-} + 3H2O -> 2[Al(OH)4]^{-} + 3H2} \nonumber$
8. The left side of the equation contains five O atoms, and the right side contains eight O atoms. We can balance the O atoms by adding three H2O molecules to the left side: $\ce{2Al + 2OH^{-} + 6H2O -> 2[Al(OH)4]^{-} + 3H2} \nonumber$
9. Be sure the equation is balanced:
1. Atoms: $\ce{2Al + 8O + 14H} \overset{\checkmark}{=} \ce{2Al + 8O + 14H} \nonumber$
2. Charge: $(2)(0) + (2)(-1) + (6)(0) \overset{\checkmark}{=} (2)(-1) + (3)(0) \nonumber$
The balanced chemical equation is therefore
$\ce{ 2Al(s) + 2OH^{-}(aq) + 6H2O(l) \rightarrow 2[Al(OH)4]^{-}(aq) + 3H2(g)} \nonumber$
Thus 3 mol of $\ce{H2}$ gas are produced for every 2 mol of $\ce{Al}$ consumed.
Exercise $2$: Reducing Manganese in permanganate
The permanganate ion reacts with nitrite ion in basic solution to produce manganese (IV) oxide and nitrate ion. Write a balanced chemical equation for the reaction.
Answer
$\ce{2MnO4^{-}(aq) + 3NO2^{-}(aq) + H2O(l) -> 2MnO2(s) + 3NO3^{-}(aq) + 2OH^{-}(aq)} \nonumber$
As suggested in Examples $1$ and $2$, a wide variety of redox reactions are possible in aqueous solutions. The identity of the products obtained from a given set of reactants often depends on both the ratio of oxidant to reductant and whether the reaction is carried out in acidic or basic solution, which is one reason it can be difficult to predict the outcome of a reaction. Because oxidation–reduction reactions in solution are so common and so important, however, chemists have developed two general guidelines for predicting whether a redox reaction will occur and the identity of the products:
1. Compounds of elements in high oxidation states (such as $\ce{ClO4^{−}}$, $\ce{NO3^{−}}$, $\ce{MnO4^{−}}$, $\ce{Cr2O7^{2−}}$, and $\ce{UF6}$) tend to act as oxidants and become reduced in chemical reactions.
2. Compounds of elements in low oxidation states (such as $\ce{CH4}$, $\ce{NH3}$, $\ce{H2S}$, and $\ce{HI}$) tend to act as reductants and become oxidized in chemical reactions.
When an aqueous solution of a compound that contains an element in a high oxidation state is mixed with an aqueous solution of a compound that contains an element in a low oxidation state, an oxidation–reduction reaction is likely to occur.
Species in high oxidation states act as oxidants, whereas species in low oxidation states act as reductants.
Balancing a Redox Reaction in Acidic Conditions: Balancing a Redox Reaction in Acidic Conditions (opens in new window) [youtu.be]
Summary
Oxidation–reduction reactions are balanced by separating the overall chemical equation into an oxidation equation and a reduction equation. In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reduction reactions in solution using the oxidation state method, in which the overall reaction is separated into an oxidation equation and a reduction equation. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/20%3A_Electrochemistry/20.02%3A_Balanced_Oxidation-Reduction_Equations.txt |
Learning Objectives
• To understand the basics of voltaic cells
• To connect voltage from a voltaic cell to underlying redox chemistry
In any electrochemical process, electrons flow from one chemical substance to another, driven by an oxidation–reduction (redox) reaction. A redox reaction occurs when electrons are transferred from a substance that is oxidized to one that is being reduced. The reductant is the substance that loses electrons and is oxidized in the process; the oxidant is the species that gains electrons and is reduced in the process. The associated potential energy is determined by the potential difference between the valence electrons in atoms of different elements.
Because it is impossible to have a reduction without an oxidation and vice versa, a redox reaction can be described as two half-reactions, one representing the oxidation process and one the reduction process. For the reaction of zinc with bromine, the overall chemical reaction is as follows:
$\ce{Zn(s) + Br2(aq) \rightarrow Zn^{2+} (aq) + 2Br^{−} (aq)} \nonumber$
The half-reactions are as follows:
reduction half-reaction:
$\ce{Br2 (aq) + 2e^{−} \rightarrow 2Br^{−} (aq)} \nonumber$
oxidation half-reaction:
$\ce{Zn (s) \rightarrow Zn^{2+} (aq) + 2e^{−} }\nonumber$
Each half-reaction is written to show what is actually occurring in the system; $\ce{Zn}$ is the reductant in this reaction (it loses electrons), and $\ce{Br2}$ is the oxidant (it gains electrons). Adding the two half-reactions gives the overall chemical reaction (Equation $1$). A redox reaction is balanced when the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. Like any balanced chemical equation, the overall process is electrically neutral; that is, the net charge is the same on both sides of the equation.
In any redox reaction, the number of electrons lost by the oxidation reaction(s) equals the number of electrons gained by the reduction reaction(s).
In most of our discussions of chemical reactions, we have assumed that the reactants are in intimate physical contact with one another. Acid–base reactions, for example, are usually carried out with the acid and the base dispersed in a single phase, such as a liquid solution. With redox reactions, however, it is possible to physically separate the oxidation and reduction half-reactions in space, as long as there is a complete circuit, including an external electrical connection, such as a wire, between the two half-reactions. As the reaction progresses, the electrons flow from the reductant to the oxidant over this electrical connection, producing an electric current that can be used to do work. An apparatus that is used to generate electricity from a spontaneous redox reaction or, conversely, that uses electricity to drive a nonspontaneous redox reaction is called an electrochemical cell.
There are two types of electrochemical cells: galvanic cells and electrolytic cells. Galvanic cells are named for the Italian physicist and physician Luigi Galvani (1737–1798), who observed that dissected frog leg muscles twitched when a small electric shock was applied, demonstrating the electrical nature of nerve impulses. A galvanic (voltaic) cell uses the energy released during a spontaneous redox reaction ($ΔG < 0$) to generate electricity. This type of electrochemical cell is often called a voltaic cell after its inventor, the Italian physicist Alessandro Volta (1745–1827). In contrast, an electrolytic cell consumes electrical energy from an external source, using it to cause a nonspontaneous redox reaction to occur ($ΔG > 0$). Both types contain two electrodes, which are solid metals connected to an external circuit that provides an electrical connection between the two parts of the system (Figure $1$). The oxidation half-reaction occurs at one electrode (the anode), and the reduction half-reaction occurs at the other (the cathode). When the circuit is closed, electrons flow from the anode to the cathode. The electrodes are also connected by an electrolyte, an ionic substance or solution that allows ions to transfer between the electrode compartments, thereby maintaining the system’s electrical neutrality. In this section, we focus on reactions that occur in galvanic cells.
Voltaic (Galvanic) Cells
To illustrate the basic principles of a galvanic cell, let’s consider the reaction of metallic zinc with cupric ion (Cu2+) to give copper metal and Zn2+ ion. The balanced chemical equation is as follows:
$\ce{Zn (s) + Cu^{2+} (aq) \rightarrow Zn^{2+} (aq) + Cu(s)} \label{20.3.4}$
We can cause this reaction to occur by inserting a zinc rod into an aqueous solution of copper(II) sulfate. As the reaction proceeds, the zinc rod dissolves, and a mass of metallic copper forms. These changes occur spontaneously, but all the energy released is in the form of heat rather than in a form that can be used to do work.
This same reaction can be carried out using the galvanic cell illustrated in Figure $\PageIndex{3a}$. To assemble the cell, a copper strip is inserted into a beaker that contains a 1 M solution of $\ce{Cu^{2+}}$ ions, and a zinc strip is inserted into a different beaker that contains a 1 M solution of $\ce{Zn^{2+}}$ ions. The two metal strips, which serve as electrodes, are connected by a wire, and the compartments are connected by a salt bridge, a U-shaped tube inserted into both solutions that contains a concentrated liquid or gelled electrolyte. The ions in the salt bridge are selected so that they do not interfere with the electrochemical reaction by being oxidized or reduced themselves or by forming a precipitate or complex; commonly used cations and anions are $\ce{Na^{+}}$ or $\ce{K^{+}}$ and $\ce{NO3^{−}}$ or $\ce{SO4^{2−}}$, respectively. (The ions in the salt bridge do not have to be the same as those in the redox couple in either compartment.) When the circuit is closed, a spontaneous reaction occurs: zinc metal is oxidized to $\ce{Zn^{2+}}$ ions at the zinc electrode (the anode), and $\ce{Cu^{2+}}$ ions are reduced to $\ce{Cu}$ metal at the copper electrode (the cathode). As the reaction progresses, the zinc strip dissolves, and the concentration of $\ce{Zn^{2+}}$ ions in the solution increases; simultaneously, the copper strip gains mass, and the concentration of $\ce{Cu^{2+}}$ ions in the solution decreases (Figure $\PageIndex{3b}$). Thus we have carried out the same reaction as we did using a single beaker, but this time the oxidative and reductive half-reactions are physically separated from each other. The electrons that are released at the anode flow through the wire, producing an electric current. Galvanic cells therefore transform chemical energy into electrical energy that can then be used to do work.
The electrolyte in the salt bridge serves two purposes: it completes the circuit by carrying electrical charge and maintains electrical neutrality in both solutions by allowing ions to migrate between them. The identity of the salt in a salt bridge is unimportant, as long as the component ions do not react or undergo a redox reaction under the operating conditions of the cell. Without such a connection, the total positive charge in the $\ce{Zn^{2+}}$ solution would increase as the zinc metal dissolves, and the total positive charge in the $\ce{Cu^{2+}}$ solution would decrease. The salt bridge allows charges to be neutralized by a flow of anions into the $\ce{Zn^{2+}}$ solution and a flow of cations into the $\ce{Cu^{2+}}$ solution. In the absence of a salt bridge or some other similar connection, the reaction would rapidly cease because electrical neutrality could not be maintained.
A voltmeter can be used to measure the difference in electrical potential between the two compartments. Opening the switch that connects the wires to the anode and the cathode prevents a current from flowing, so no chemical reaction occurs. With the switch closed, however, the external circuit is closed, and an electric current can flow from the anode to the cathode. The potential ($E_{cell}$) of the cell, measured in volts, is the difference in electrical potential between the two half-reactions and is related to the energy needed to move a charged particle in an electric field. In the cell we have described, the voltmeter indicates a potential of 1.10 V (Figure $\PageIndex{3a}$). Because electrons from the oxidation half-reaction are released at the anode, the anode in a galvanic cell is negatively charged. The cathode, which attracts electrons, is positively charged.
Not all electrodes undergo a chemical transformation during a redox reaction. The electrode can be made from an inert, highly conducting metal such as platinum to prevent it from reacting during a redox process, where it does not appear in the overall electrochemical reaction. This phenomenon is illustrated in Example $1$.
A galvanic (voltaic) cell converts the energy released by a spontaneous chemical reaction to electrical energy. An electrolytic cell consumes electrical energy from an external source to drive a nonspontaneous chemical reaction.
Example $1$
A chemist has constructed a galvanic cell consisting of two beakers. One beaker contains a strip of tin immersed in aqueous sulfuric acid, and the other contains a platinum electrode immersed in aqueous nitric acid. The two solutions are connected by a salt bridge, and the electrodes are connected by a wire. Current begins to flow, and bubbles of a gas appear at the platinum electrode. The spontaneous redox reaction that occurs is described by the following balanced chemical equation:
$\ce{3Sn(s) + 2NO3^{-}(aq) + 8H^{+} (aq) \rightarrow 3Sn^{2+} (aq) + 2NO (g) + 4H2O (l)} \nonumber$
For this galvanic cell,
1. write the half-reaction that occurs at each electrode.
2. indicate which electrode is the cathode and which is the anode.
3. indicate which electrode is the positive electrode and which is the negative electrode.
Given: galvanic cell and redox reaction
Asked for: half-reactions, identity of anode and cathode, and electrode assignment as positive or negative
Strategy:
1. Identify the oxidation half-reaction and the reduction half-reaction. Then identify the anode and cathode from the half-reaction that occurs at each electrode.
2. From the direction of electron flow, assign each electrode as either positive or negative.
Solution
A In the reduction half-reaction, nitrate is reduced to nitric oxide. (The nitric oxide would then react with oxygen in the air to form NO2, with its characteristic red-brown color.) In the oxidation half-reaction, metallic tin is oxidized. The half-reactions corresponding to the actual reactions that occur in the system are as follows:
reduction: $\ce{NO3^{−} (aq) + 4H^{+}(aq) + 3e^{−} → NO(g) + 2H2O(l)} \nonumber$
oxidation: $\ce{Sn(s) → Sn^{2+}(aq) + 2e^{−}} \nonumber$
Thus nitrate is reduced to NO, while the tin electrode is oxidized to Sn2+.
Because the reduction reaction occurs at the Pt electrode, it is the cathode. Conversely, the oxidation reaction occurs at the tin electrode, so it is the anode.
B Electrons flow from the tin electrode through the wire to the platinum electrode, where they transfer to nitrate. The electric circuit is completed by the salt bridge, which permits the diffusion of cations toward the cathode and anions toward the anode. Because electrons flow from the tin electrode, it must be electrically negative. In contrast, electrons flow toward the Pt electrode, so that electrode must be electrically positive.
Exercise $1$
Consider a simple galvanic cell consisting of two beakers connected by a salt bridge. One beaker contains a solution of $\ce{MnO_4^{−}}$ in dilute sulfuric acid and has a Pt electrode. The other beaker contains a solution of $\ce{Sn^{2+}}$ in dilute sulfuric acid, also with a Pt electrode. When the two electrodes are connected by a wire, current flows and a spontaneous reaction occurs that is described by the following balanced chemical equation:
$\ce{2MnO^{−}4(aq) + 5Sn^{2+}(aq) + 16H^{+}(aq) \rightarrow 2Mn^{2+}(aq) + 5Sn^{4+}(aq) + 8H2O(l)} \nonumber$
For this galvanic cell,
1. write the half-reaction that occurs at each electrode.
2. indicate which electrode is the cathode and which is the anode.
3. indicate which electrode is positive and which is negative.
Answer a
\begin{align*} \ce{MnO4^{−}(aq) + 8H^{+}(aq) + 5e^{−}} &→ \ce{Mn^{2+}(aq) + 4H2O(l)} \[4pt] \ce{Sn^{2+}(aq)} &→ \ce{Sn^{4+}(aq) + 2e^{−}} \end{align*} \nonumber
Answer b
The Pt electrode in the permanganate solution is the cathode; the one in the tin solution is the anode.
Answer c
The cathode (electrode in beaker that contains the permanganate solution) is positive, and the anode (electrode in beaker that contains the tin solution) is negative.
Electrochemical Cells: Electrochemical Cells(opens in new window) [youtu.be]
Constructing Cell Diagrams (Cell Notation)
Because it is somewhat cumbersome to describe any given galvanic cell in words, a more convenient notation has been developed. In this line notation, called a cell diagram, the identity of the electrodes and the chemical contents of the compartments are indicated by their chemical formulas, with the anode written on the far left and the cathode on the far right. Phase boundaries are shown by single vertical lines, and the salt bridge, which has two phase boundaries, by a double vertical line. Thus the cell diagram for the $\ce{Zn/Cu}$ cell shown in Figure $\PageIndex{3a}$ is written as follows:
Galvanic cells can have arrangements other than the examples we have seen so far. For example, the voltage produced by a redox reaction can be measured more accurately using two electrodes immersed in a single beaker containing an electrolyte that completes the circuit. This arrangement reduces errors caused by resistance to the flow of charge at a boundary, called the junction potential. One example of this type of galvanic cell is as follows:
$\ce{Pt(s)\, | \, H2(g) | HCl(aq, \, 1\,M)\,|\, AgCl(s) \,Ag(s)} \nonumber$
This cell diagram does not include a double vertical line representing a salt bridge because there is no salt bridge providing a junction between two dissimilar solutions. Moreover, solution concentrations have not been specified, so they are not included in the cell diagram. The half-reactions and the overall reaction for this cell are as follows:
cathode reaction:
$\ce{AgCl (s) + e^{−} \rightarrow Ag(s) + Cl^{−}(aq)} \nonumber$
anode reaction:
$\ce{ 1/2 H2(g) -> H^{+}(aq) + e^{-}} \nonumber$
overall:
$\ce{ AgCl(s) + 1/2H2(g) -> Ag(s) + Cl^{-} + H^{+}(aq)} \nonumber$
A single-compartment galvanic cell will initially exhibit the same voltage as a galvanic cell constructed using separate compartments, but it will discharge rapidly because of the direct reaction of the reactant at the anode with the oxidized member of the cathodic redox couple. Consequently, cells of this type are not particularly useful for producing electricity.
Example $2$
Draw a cell diagram for the galvanic cell described in Example $1$. The balanced chemical reaction is as follows:
$\ce{3Sn(s) + 2NO^{−}3(aq) + 8H^{+}(aq) \rightarrow 3Sn^{2+}(aq) + 2NO(g) + 4H2O(l)} \nonumber$
Given: galvanic cell and redox reaction
Asked for: cell diagram
Strategy:
Using the symbols described, write the cell diagram beginning with the oxidation half-reaction on the left.
Solution
The anode is the tin strip, and the cathode is the $\ce{Pt}$ electrode. Beginning on the left with the anode, we indicate the phase boundary between the electrode and the tin solution by a vertical bar. The anode compartment is thus $\ce{Sn(s)∣Sn^{2+}(aq)}$. We could include $\ce{H2SO4(aq)}$ with the contents of the anode compartment, but the sulfate ion (as $\ce{HSO4^{−}}$) does not participate in the overall reaction, so it does not need to be specifically indicated. The cathode compartment contains aqueous nitric acid, which does participate in the overall reaction, together with the product of the reaction ($\ce{NO}$) and the $\ce{Pt}$ electrode. These are written as $\ce{HNO3(aq)∣NO(g)∣Pt(s)}$, with single vertical bars indicating the phase boundaries. Combining the two compartments and using a double vertical bar to indicate the salt bridge,
$\ce{Sn(s)\,|\,Sn^{2+}(aq)\,||\,HNO3(aq)\,|\,NO(g)\,|\,Pt_(s)} \nonumber$
The solution concentrations were not specified, so they are not included in this cell diagram.
Exercise $2$
Draw the cell diagram for the following reaction, assuming the concentration of $\ce{Ag^{+}}$ and $\ce{Mg^{2+}}$ are each 1 M:
$\ce{Mg(s) + 2Ag^{+}(aq) \rightarrow Mg^{2+}(aq) + 2Ag(s)} \nonumber$
Answer
$\ce{ Mg(s) \,|\,Mg^{2+}(aq, \;1 \,M )\,||\,Ag^+(aq, \;1\, M)\,|\,Ag(s)} \nonumber$
Cell Diagrams: Cell Diagrams(opens in new window) [youtu.be]
Summary
A galvanic (voltaic) cell uses the energy released during a spontaneous redox reaction to generate electricity, whereas an electrolytic cell consumes electrical energy from an external source to force a reaction to occur. Electrochemistry is the study of the relationship between electricity and chemical reactions. The oxidation–reduction reaction that occurs during an electrochemical process consists of two half-reactions, one representing the oxidation process and one the reduction process. The sum of the half-reactions gives the overall chemical reaction. The overall redox reaction is balanced when the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. An electric current is produced from the flow of electrons from the reductant to the oxidant. An electrochemical cell can either generate electricity from a spontaneous redox reaction or consume electricity to drive a nonspontaneous reaction. In a galvanic (voltaic) cell, the energy from a spontaneous reaction generates electricity, whereas in an electrolytic cell, electrical energy is consumed to drive a nonspontaneous redox reaction. Both types of cells use two electrodes that provide an electrical connection between systems that are separated in space. The oxidative half-reaction occurs at the anode, and the reductive half-reaction occurs at the cathode. A salt bridge connects the separated solutions, allowing ions to migrate to either solution to ensure the system’s electrical neutrality. A voltmeter is a device that measures the flow of electric current between two half-reactions. The potential of a cell, measured in volts, is the energy needed to move a charged particle in an electric field. An electrochemical cell can be described using line notation called a cell diagram, in which vertical lines indicate phase boundaries and the location of the salt bridge. Resistance to the flow of charge at a boundary is called the junction potential. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/20%3A_Electrochemistry/20.03%3A_Voltaic_Cells.txt |
Learning Objectives
• To use redox potentials to predict whether a reaction is spontaneous.
• To balance redox reactions using half-reactions.
In a galvanic cell, current is produced when electrons flow externally through the circuit from the anode to the cathode because of a difference in potential energy between the two electrodes in the electrochemical cell. In the Zn/Cu system, the valence electrons in zinc have a substantially higher potential energy than the valence electrons in copper because of shielding of the s electrons of zinc by the electrons in filled d orbitals. Hence electrons flow spontaneously from zinc to copper(II) ions, forming zinc(II) ions and metallic copper. Just like water flowing spontaneously downhill, which can be made to do work by forcing a waterwheel, the flow of electrons from a higher potential energy to a lower one can also be harnessed to perform work.
Because the potential energy of valence electrons differs greatly from one substance to another, the voltage of a galvanic cell depends partly on the identity of the reacting substances. If we construct a galvanic cell similar to the one in part (a) in Figure $1$ but instead of copper use a strip of cobalt metal and 1 M Co2+ in the cathode compartment, the measured voltage is not 1.10 V but 0.51 V. Thus we can conclude that the difference in potential energy between the valence electrons of cobalt and zinc is less than the difference between the valence electrons of copper and zinc by 0.59 V.
The measured potential of a cell also depends strongly on the concentrations of the reacting species and the temperature of the system. To develop a scale of relative potentials that will allow us to predict the direction of an electrochemical reaction and the magnitude of the driving force for the reaction, the potentials for oxidations and reductions of different substances must be measured under comparable conditions. To do this, chemists use the standard cell potential (E°cell), defined as the potential of a cell measured under standard conditions—that is, with all species in their standard states (1 M for solutions, concentrated solutions of salts (about 1 M) generally do not exhibit ideal behavior, and the actual standard state corresponds to an activity of 1 rather than a concentration of 1 M. Corrections for non ideal behavior are important for precise quantitative work but not for the more qualitative approach that we are taking here. 1 atm for gases, pure solids or pure liquids for other substances) and at a fixed temperature, usually 25°C.
Measured redox potentials depend on the potential energy of valence electrons, the concentrations of the species in the reaction, and the temperature of the system.
Measuring Standard Electrode Potentials
It is physically impossible to measure the potential of a single electrode: only the difference between the potentials of two electrodes can be measured (this is analogous to measuring absolute enthalpies or free energies; recall that only differences in enthalpy and free energy can be measured.) We can, however, compare the standard cell potentials for two different galvanic cells that have one kind of electrode in common. This allows us to measure the potential difference between two dissimilar electrodes. For example, the measured standard cell potential (E°) for the Zn/Cu system is 1.10 V, whereas E° for the corresponding Zn/Co system is 0.51 V. This implies that the potential difference between the Co and Cu electrodes is 1.10 V − 0.51 V = 0.59 V. In fact, that is exactly the potential measured under standard conditions if a cell is constructed with the following cell diagram:
$Co_{(s)} ∣ Co^{2+}(aq, 1 M)∥Cu^{2+}(aq, 1 M) ∣ Cu (s)\;\;\; E°=0.59\; V \label{20.4.1}$
This cell diagram corresponds to the oxidation of a cobalt anode and the reduction of Cu2+ in solution at the copper cathode.
All tabulated values of standard electrode potentials by convention are listed for a reaction written as a reduction, not as an oxidation, to be able to compare standard potentials for different substances (Table P1). The standard cell potential (E°cell) is therefore the difference between the tabulated reduction potentials of the two half-reactions, not their sum:
$E°_{cell} = E°_{cathode} − E°_{anode} \label{20.4.2}$
In contrast, recall that half-reactions are written to show the reduction and oxidation reactions that actually occur in the cell, so the overall cell reaction is written as the sum of the two half-reactions. According to Equation $\ref{20.4.2}$, when we know the standard potential for any single half-reaction, we can obtain the value of the standard potential of many other half-reactions by measuring the standard potential of the corresponding cell.
Although it is impossible to measure the potential of any electrode directly, we can choose a reference electrode whose potential is defined as 0 V under standard conditions. The standard hydrogen electrode (SHE) is universally used for this purpose and is assigned a standard potential of 0 V. It consists of a strip of platinum wire in contact with an aqueous solution containing 1 M H+. The [H+] in solution is in equilibrium with H2 gas at a pressure of 1 atm at the Pt-solution interface (Figure $2$). Protons are reduced or hydrogen molecules are oxidized at the Pt surface according to the following equation:
$2H^+_{(aq)}+2e^− \rightleftharpoons H_{2(g)} \label{20.4.3}$
One especially attractive feature of the SHE is that the Pt metal electrode is not consumed during the reaction.
Figure $3$ shows a galvanic cell that consists of a SHE in one beaker and a Zn strip in another beaker containing a solution of Zn2+ ions. When the circuit is closed, the voltmeter indicates a potential of 0.76 V. The zinc electrode begins to dissolve to form Zn2+, and H+ ions are reduced to H2 in the other compartment. Thus the hydrogen electrode is the cathode, and the zinc electrode is the anode. The diagram for this galvanic cell is as follows:
$Zn_{(s)}∣Zn^{2+}_{(aq)}∥H^+(aq, 1 M)∣H_2(g, 1 atm)∣Pt_{(s)} \label{20.4.4}$
The half-reactions that actually occur in the cell and their corresponding electrode potentials are as follows:
• cathode: $2H^+_{(aq)} + 2e^− \rightarrow H_{2(g)}\;\;\; E°_{cathode}=0 V \label{20.4.5}$
• anode: $Zn_{(s)} \rightarrow Zn^{2+}_{(aq)}+2e^−\;\;\; E°_{anode}=−0.76\; V \label{20.4.6}$
• overall: $Zn_{(s)}+2H^+_{(aq)} \rightarrow Zn^{2+}_{(aq)}+H_{2(g)} \label{20.4.7}$
We then use Equation \ref{20.4.2} to calculate the cell potential
\begin{align*} E°_{cell} &=E°_{cathode}−E°_{anode}\[4pt] &=0.76\; V \end{align*} \nonumber
Although the reaction at the anode is an oxidation, by convention its tabulated E° value is reported as a reduction potential. The potential of a half-reaction measured against the SHE under standard conditions is called the standard electrode potential for that half-reaction. In this example, the standard reduction potential for Zn2+(aq) + 2e → Zn(s) is −0.76 V, which means that the standard electrode potential for the reaction that occurs at the anode, the oxidation of Zn to Zn2+, often called the Zn/Zn2+ redox couple, or the Zn/Zn2+ couple, is −(−0.76 V) = 0.76 V. We must therefore subtract E°anode from E°cathode to obtain
$E°_{cell}: 0 \,V − (−0.76\, V) = 0.76\, V \nonumber$
Because electrical potential is the energy needed to move a charged particle in an electric field, standard electrode potentials for half-reactions are intensive properties and do not depend on the amount of substance involved. Consequently, E° values are independent of the stoichiometric coefficients for the half-reaction, and, most important, the coefficients used to produce a balanced overall reaction do not affect the value of the cell potential.
E° values do NOT depend on the stoichiometric coefficients for a half-reaction, because it is an intensive property.
The Standard Hydrogen Electrode (SHE): The Standard Hydrogen Electrode (SHE)(opens in new window) [youtu.be]
Standard Electrode Potentials
To measure the potential of the Cu/Cu2+ couple, we can construct a galvanic cell analogous to the one shown in Figure $3$ but containing a Cu/Cu2+ couple in the sample compartment instead of Zn/Zn2+. When we close the circuit this time, the measured potential for the cell is negative (−0.34 V) rather than positive. The negative value of $E°_{cell}$ indicates that the direction of spontaneous electron flow is the opposite of that for the Zn/Zn2+ couple. Hence the reactions that occur spontaneously, indicated by a positive $E°_{cell}$, are the reduction of Cu2+ to Cu at the copper electrode. The copper electrode gains mass as the reaction proceeds, and H2 is oxidized to H+ at the platinum electrode. In this cell, the copper strip is the cathode, and the hydrogen electrode is the anode. The cell diagram therefore is written with the SHE on the left and the Cu2+/Cu couple on the right:
$Pt_{(s)}∣H_2(g, 1 atm)∣H^+(aq, 1\; M)∥Cu^{2+}(aq, 1 M)∣Cu_{(s)} \label{20.4.8}$
The half-cell reactions and potentials of the spontaneous reaction are as follows:
• Cathode: $Cu^{2+}{(aq)} + 2e^− \rightarrow Cu_{(g)}\;\;\; E°_{cathode} = 0.34\; V \label{20.4.9}$
• Anode: $H_{2(g)} \rightarrow 2H^+_{(aq)} + 2e^−\;\;\; E°_{anode} = 0\; V \label{20.4.10}$
• Overall: $H_{2(g)} + Cu^{2+}_{(aq)} \rightarrow 2H^+_{(aq)} + Cu_{(s)} \label{20.4.11}$
We then use Equation \ref{20.4.2} to calculate the cell potential
\begin{align*} E°_{cell} &= E°_{cathode}− E°_{anode} \[4pt] &= 0.34\; V \end{align*} \nonumber
Thus the standard electrode potential for the Cu2+/Cu couple is 0.34 V.
Electrode Potentials and ECell: Electrode and Potentials and Ecell(opens in new window) [youtu.be]
Balancing Redox Reactions Using the Half-Reaction Method
In Section 4.4, we described a method for balancing redox reactions using oxidation numbers. Oxidation numbers were assigned to each atom in a redox reaction to identify any changes in the oxidation states. Here we present an alternative approach to balancing redox reactions, the half-reaction method, in which the overall redox reaction is divided into an oxidation half-reaction and a reduction half-reaction, each balanced for mass and charge. This method more closely reflects the events that take place in an electrochemical cell, where the two half-reactions may be physically separated from each other.
We can illustrate how to balance a redox reaction using half-reactions with the reaction that occurs when Drano, a commercial solid drain cleaner, is poured into a clogged drain. Drano contains a mixture of sodium hydroxide and powdered aluminum, which in solution reacts to produce hydrogen gas:
$Al_{(s)} + OH^−_{(aq)} \rightarrow Al(OH)^−_{4(aq)} + H_{2(g)} \label{20.4.12}$
In this reaction, $Al_{(s)}$ is oxidized to Al3+, and H+ in water is reduced to H2 gas, which bubbles through the solution, agitating it and breaking up the clogs.
The overall redox reaction is composed of a reduction half-reaction and an oxidation half-reaction. From the standard electrode potentials listed Table P1, we find the corresponding half-reactions that describe the reduction of H+ ions in water to H2and the oxidation of Al to Al3+ in basic solution:
• reduction: $2H_2O_{(l)} + 2e^− \rightarrow 2OH^−_{(aq)} + H_{2(g)} \label{20.4.13}$
• oxidation: $Al_{(s)} + 4OH^−_{(aq)} \rightarrow Al(OH)^−_{4(aq)} + 3e^− \label{20.4.14}$
The half-reactions chosen must exactly reflect the reaction conditions, such as the basic conditions shown here. Moreover, the physical states of the reactants and the products must be identical to those given in the overall reaction, whether gaseous, liquid, solid, or in solution.
In Equation $\ref{20.4.13}$, two H+ ions gain one electron each in the reduction; in Equation $\ref{20.4.14}$, the aluminum atom loses three electrons in the oxidation. The charges are balanced by multiplying the reduction half-reaction (Equation $\ref{20.4.13}$) by 3 and the oxidation half-reaction (Equation $\ref{20.4.14}$) by 2 to give the same number of electrons in both half-reactions:
• reduction:
$6H_2O_{(l)} + 6e^− \rightarrow 6OH^−_{(aq)} + 3H_{2(g)} \label{20.4.15}$
• oxidation:
$2Al_{(s)} + 8OH^−_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 6e^− \label{20.4.16}$
Adding the two half-reactions,
$6H_2O_{(l)} + 2Al_{(s)} + 8OH^−_{(aq)} \rightarrow 2Al(OH)^−{4(aq)} + 3H_{2(g)} + 6OH^−_{(aq)} \label{20.4.17}$
Simplifying by canceling substances that appear on both sides of the equation,
$6H_2O_{(l)} + 2Al_{(s)} + 2OH^−_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 3H_{2(g)} \label{20.4.18}$
We have a −2 charge on the left side of the equation and a −2 charge on the right side. Thus the charges are balanced, but we must also check that atoms are balanced:
$2Al + 8O + 14H = 2Al + 8O + 14H \label{20.4.19}$
The atoms also balance, so Equation $\ref{20.4.18}$ is a balanced chemical equation for the redox reaction depicted in Equation $\ref{20.4.12}$.
The half-reaction method requires that half-reactions exactly reflect reaction conditions, and the physical states of the reactants and the products must be identical to those in the overall reaction.
We can also balance a redox reaction by first balancing the atoms in each half-reaction and then balancing the charges. With this alternative method, we do not need to use the half-reactions listed in Table P1, but instead focus on the atoms whose oxidation states change, as illustrated in the following steps:
Step 1: Write the reduction half-reaction and the oxidation half-reaction.
For the reaction shown in Equation $\ref{20.4.12}$, hydrogen is reduced from H+ in OH to H2, and aluminum is oxidized from Al° to Al3+:
• reduction:
$OH^−_{(aq)} \rightarrow H_{2(g)} \label{20.4.20}$
• oxidation:
$Al_{(s)} \rightarrow Al(OH)^−_{4(aq)} \label{20.4.21}$
Step 2: Balance the atoms by balancing elements other than O and H. Then balance O atoms by adding H2O and balance H atoms by adding H+.
Elements other than O and H in the previous two equations are balanced as written, so we proceed with balancing the O atoms. We can do this by adding water to the appropriate side of each half-reaction:
• reduction:
$OH^−_{(aq)} \rightarrow H_{2(g)} + H_2O_{(l)} \label{20.4.22}$
• oxidation:
$Al_{(s)} + 4H_2O_{(l)} \rightarrow Al(OH)^−_{4(aq)} \label{20.4.23}$
Balancing H atoms by adding H+, we obtain the following:
• reduction:
$OH^−_{(aq)} + 3H^+_{(aq)} \rightarrow H_{2(g)} + H_2O_{(l)} \label{20.4.24}$
• oxidation:
$Al_{(s)} + 4H_2O_{(l)} \rightarrow Al(OH)^−_{4(aq)} + 4H^+_{(aq)} \label{20.4.25}$
We have now balanced the atoms in each half-reaction, but the charges are not balanced.
Step 3: Balance the charges in each half-reaction by adding electrons.
Two electrons are gained in the reduction of H+ ions to H2, and three electrons are lost during the oxidation of Al° to Al3+:
• reduction:
$OH^−_{(aq)} + 3H^+_{(aq)} + 2e^− \rightarrow H_{2(g)} + H_2O_{(l)} \label{20.4.26}$
• oxidation:
$Al_{(s)} + 4H_2O_{(l)} \rightarrow Al(OH)^−_{4(aq)} + 4H^+_{(aq)} + 3e^− \label{20.4.27}$
Step 4: Multiply the reductive and oxidative half-reactions by appropriate integers to obtain the same number of electrons in both half-reactions.
In this case, we multiply Equation $\ref{20.4.26}$ (the reductive half-reaction) by 3 and Equation $\ref{20.4.27}$ (the oxidative half-reaction) by 2 to obtain the same number of electrons in both half-reactions:
• reduction:
$3OH^−_{(aq)} + 9H^+_{(aq)} + 6e^− \rightarrow 3H_{2(g)} + 3H_2O_{(l)} \label{20.4.28}$
• oxidation:
$2Al_{(s)} + 8H_2O_{(l)} \rightarrow 2Al(OH)^−_{4(aq)} + 8H^+_{(aq)} + 6e^− \label{20.4.29}$
Step 5: Add the two half-reactions and cancel substances that appear on both sides of the equation.
Adding and, in this case, canceling 8H+, 3H2O, and 6e,
$2Al_{(s)} + 5H_2O_{(l)} + 3OH^−_{(aq)} + H^+_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 3H_{2(g)} \label{20.4.30}$
We have three OH and one H+ on the left side. Neutralizing the H+ gives us a total of 5H2O + H2O = 6H2O and leaves 2OH on the left side:
$2Al_{(s)} + 6H_2O_{(l)} + 2OH^−_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 3H_{2(g)} \label{20.4.31}$
Step 6: Check to make sure that all atoms and charges are balanced.
Equation $\ref{20.4.31}$ is identical to Equation $\ref{20.4.18}$, obtained using the first method, so the charges and numbers of atoms on each side of the equation balance.
Example $1$
In acidic solution, the redox reaction of dichromate ion ($\ce{Cr2O7^{2−}}$) and iodide ($\ce{I^{−}}$) can be monitored visually. The yellow dichromate solution reacts with the colorless iodide solution to produce a solution that is deep amber due to the presence of a green $\ce{Cr^{3+}(aq)}$ complex and brown $\ce{I2(aq)}$ ions (Figure $4$):
$\ce{Cr2O7^{2−}(aq) + I^{−}(aq) -> Cr^{3+}(aq) + I2(aq)} \nonumber$
Balance this equation using half-reactions.
Given: redox reaction and Table P1
Asked for: balanced chemical equation using half-reactions
Strategy:
Follow the steps to balance the redox reaction using the half-reaction method.
Solution
From the standard electrode potentials listed in Table P1, we find the half-reactions corresponding to the overall reaction:
• reduction: $\ce{Cr2O7^{2−}(aq) + 14H^{+}(aq) + 6e^{−} -> 2Cr^{3+}(aq) + 7H2O(l)} \nonumber$
• oxidation: $\ce{2I^{−}(aq) -> I2(aq) + 2e^{−}} \nonumber$
Balancing the number of electrons by multiplying the oxidation reaction by 3,
• oxidation: $\ce{6I^{−}(aq) -> 3I2(aq) + 6e^{−}} \nonumber$
Adding the two half-reactions and canceling electrons,
$\ce{Cr2O^{2−}7(aq) + 14H^{+}(aq) + 6I^{−}(aq) -> 2Cr^{3+}(aq) + 7H2O(l) + 3I2(aq)} \nonumber$
We must now check to make sure the charges and atoms on each side of the equation balance:
\begin{align*} (−2) + 14 + (−6) &= +6 \[4pt] +6 &\overset{\checkmark}{=} +6 \end{align*} \nonumber
and atoms
$\ce{2Cr + 7O + 14H + 6I} \overset{\checkmark}{=} \ce{2Cr + 7O + 14H + 6I} \nonumber$
Both the charges and atoms balance, so our equation is balanced.
We can also use the alternative procedure, which does not require the half-reactions listed in Table P1.
Step 1: Chromium is reduced from $\ce{Cr^{6+}}$ in $\ce{Cr2O7^{2−}}$ to $\ce{Cr^{3+}}$, and $\ce{I^{−}}$ ions are oxidized to $\ce{I2}$. Dividing the reaction into two half-reactions,
• reduction: $Cr_2O^{2−}_{7(aq)} \rightarrow Cr^{3+}_{(aq)} \nonumber$
• oxidation: $I^−_{(aq)} \rightarrow I_{2(aq)} \nonumber$
Step 2: Balancing the atoms other than oxygen and hydrogen,
• reduction: $Cr_2O^{2−}_{7(aq)} \rightarrow 2Cr^{3+}_{(aq)} \nonumber$
• oxidation: $2I^−_{(aq)} \rightarrow I_{2(aq)} \nonumber$
We now balance the O atoms by adding H2O—in this case, to the right side of the reduction half-reaction. Because the oxidation half-reaction does not contain oxygen, it can be ignored in this step.
• reduction: $Cr_2O^{2−}_{7(aq)} \rightarrow 2Cr^{3+}_{(aq)} + 7H_2O_{(l)} \nonumber$
Next we balance the H atoms by adding H+ to the left side of the reduction half-reaction. Again, we can ignore the oxidation half-reaction.
• reduction: $Cr_2O^{2−}_{7(aq)} + 14H^+_{(aq)} \rightarrow 2Cr^{3+}_{(aq)} + 7H_2O_{(l)} \nonumber$
Step 3: We must now add electrons to balance the charges. The reduction half-reaction (2Cr+6 to 2Cr+3) has a +12 charge on the left and a +6 charge on the right, so six electrons are needed to balance the charge. The oxidation half-reaction (2I to I2) has a −2 charge on the left side and a 0 charge on the right, so it needs two electrons to balance the charge:
• reduction: Cr2O72(aq) + 14H+(aq) + 6e → 2Cr3+(aq) + 7H2O(l)
• oxidation: 2I(aq) → I2(aq) + 2e
Step 4: To have the same number of electrons in both half-reactions, we must multiply the oxidation half-reaction by 3:
• oxidation: 6I(aq) → 3I2(s) + 6e
Step 5: Adding the two half-reactions and canceling substances that appear in both reactions,
$\ce{Cr2O7^{2−}(aq) + 14H^{+}(aq) + 6I^{−}(aq) → 2Cr^{3+}(aq) + 7H2O(l) + 3I2(aq)} \nonumber$
Step 6: This is the same equation we obtained using the first method. Thus the charges and atoms on each side of the equation balance.
Exercise $1$
Copper is found as the mineral covellite ($\ce{CuS}$). The first step in extracting the copper is to dissolve the mineral in nitric acid ($\ce{HNO3}$), which oxidizes sulfide to sulfate and reduces nitric acid to $\ce{NO}$:
$\ce{CuS(s) + HNO3(aq) \rightarrow NO(g) + CuSO4(aq)} \nonumber$
Balance this equation using the half-reaction method.
Answer
$\ce{3CuS(s) + 8HNO3(aq) -> 8NO(g) + 3CuSO4(aq) + 4H2O(l)} \nonumber$
Calculating Standard Cell Potentials
The standard cell potential for a redox reaction (E°cell) is a measure of the tendency of reactants in their standard states to form products in their standard states; consequently, it is a measure of the driving force for the reaction, which earlier we called voltage. We can use the two standard electrode potentials we found earlier to calculate the standard potential for the Zn/Cu cell represented by the following cell diagram:
$Zn{(s)}∣Zn^{2+}(aq, 1 M)∥Cu^{2+}(aq, 1 M)∣Cu_{(s)} \label{20.4.32}$
We know the values of E°anode for the reduction of Zn2+ and E°cathode for the reduction of Cu2+, so we can calculate $E°_{cell}$:
• cathode: $Cu^{2+}_{(aq)} + 2e^− \rightarrow Cu_{(s)} \;\;\; E°_{cathode} = 0.34\; V \label{20.4.33}$
• anode: $Zn_{(s)} \rightarrow Zn^{2+}(aq, 1 M) + 2e^−\;\;\; E°_{anode} = −0.76\; V \label{20.4.34}$
• overall: $Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)} \label{20.4.35}$
$E°_{cell} = E°_{cathode} − E°_{anode} = 1.10\; V \nonumber$
This is the same value that is observed experimentally. If the value of $E°_{cell}$ is positive, the reaction will occur spontaneously as written. If the value of $E°_{cell}$ is negative, then the reaction is not spontaneous, and it will not occur as written under standard conditions; it will, however, proceed spontaneously in the opposite direction. As we shall see in Section 20.9, this does not mean that the reaction cannot be made to occur at all under standard conditions. With a sufficient input of electrical energy, virtually any reaction can be forced to occur. Example $2$ and its corresponding exercise illustrate how we can use measured cell potentials to calculate standard potentials for redox couples.
A positive $E°_{cell}$ means that the reaction will occur spontaneously as written. A negative $E°_{cell}$ means that the reaction will proceed spontaneously in the opposite direction.
Example $2$
A galvanic cell with a measured standard cell potential of 0.27 V is constructed using two beakers connected by a salt bridge. One beaker contains a strip of gallium metal immersed in a 1 M solution of GaCl3, and the other contains a piece of nickel immersed in a 1 M solution of NiCl2. The half-reactions that occur when the compartments are connected are as follows:
cathode: Ni2+(aq) + 2e → Ni(s)
anode: Ga(s) → Ga3+(aq) + 3e
If the potential for the oxidation of Ga to Ga3+ is 0.55 V under standard conditions, what is the potential for the oxidation of Ni to Ni2+?
Given: galvanic cell, half-reactions, standard cell potential, and potential for the oxidation half-reaction under standard conditions
Asked for: standard electrode potential of reaction occurring at the cathode
Strategy:
1. Write the equation for the half-reaction that occurs at the anode along with the value of the standard electrode potential for the half-reaction.
2. Use Equation $\ref{20.4.2}$ to calculate the standard electrode potential for the half-reaction that occurs at the cathode. Then reverse the sign to obtain the potential for the corresponding oxidation half-reaction under standard conditions.
Solution
A We have been given the potential for the oxidation of Ga to Ga3+ under standard conditions, but to report the standard electrode potential, we must reverse the sign. For the reduction reaction Ga3+(aq) + 3e → Ga(s), E°anode = −0.55 V.
B Using the value given for $E°_{cell}$ and the calculated value of E°anode, we can calculate the standard potential for the reduction of Ni2+ to Ni from Equation $\ref{20.4.2}$:
\begin{align*} E°_{cell} &= E°_{cathode} − E°_{anode} \[4pt] 0.27\, V &= E^o°_{cathhode} − (−0.55\, V) \[4pt] E^°_{cathode} &= −0.28 \,V \end{align*} \nonumber
This is the standard electrode potential for the reaction Ni2+(aq) + 2e → Ni(s). Because we are asked for the potential for the oxidation of Ni to Ni2+ under standard conditions, we must reverse the sign of E°cathode. Thus E° = −(−0.28 V) = 0.28 V for the oxidation. With three electrons consumed in the reduction and two produced in the oxidation, the overall reaction is not balanced. Recall, however, that standard potentials are independent of stoichiometry.
Exercise $2$
A galvanic cell is constructed with one compartment that contains a mercury electrode immersed in a 1 M aqueous solution of mercuric acetate $\ce{Hg(CH_3CO_2)_2}$ and one compartment that contains a strip of magnesium immersed in a 1 M aqueous solution of $\ce{MgCl2}$. When the compartments are connected, a potential of 3.22 V is measured and the following half-reactions occur:
• cathode: $\ce{Hg^{2+} (aq) + 2e^{−} → Hg(l)}$
• anode: $\ce{Mg(s) → Mg^{2+}(aq) + 2e^{−}}$
If the potential for the oxidation of $\ce{Mg}$ to $\ce{Mg^{2+}}$ is 2.37 V under standard conditions, what is the standard electrode potential for the reaction that occurs at the cathode?
Answer
0.85 V
Reference Electrodes and Measuring Concentrations
When using a galvanic cell to measure the concentration of a substance, we are generally interested in the potential of only one of the electrodes of the cell, the so-called indicator electrode, whose potential is related to the concentration of the substance being measured. To ensure that any change in the measured potential of the cell is due to only the substance being analyzed, the potential of the other electrode, the reference electrode, must be constant. You are already familiar with one example of a reference electrode: the SHE. The potential of a reference electrode must be unaffected by the properties of the solution, and if possible, it should be physically isolated from the solution of interest. To measure the potential of a solution, we select a reference electrode and an appropriate indicator electrode. Whether reduction or oxidation of the substance being analyzed occurs depends on the potential of the half-reaction for the substance of interest (the sample) and the potential of the reference electrode.
The potential of any reference electrode should not be affected by the properties of the solution to be analyzed, and it should also be physically isolated.
There are many possible choices of reference electrode other than the SHE. The SHE requires a constant flow of highly flammable hydrogen gas, which makes it inconvenient to use. Consequently, two other electrodes are commonly chosen as reference electrodes. One is the silver–silver chloride electrode, which consists of a silver wire coated with a very thin layer of AgCl that is dipped into a chloride ion solution with a fixed concentration. The cell diagram and reduction half-reaction are as follows:
$Cl^−_{(aq)}∣AgCl_{(s)}∣Ag_{(s)} \label{20.4.36}$
$AgCl_{(s)}+e^− \rightarrow Ag_{(s)} + Cl^−_{(aq)} \nonumber$
If a saturated solution of KCl is used as the chloride solution, the potential of the silver–silver chloride electrode is 0.197 V versus the SHE. That is, 0.197 V must be subtracted from the measured value to obtain the standard electrode potential measured against the SHE.
A second common reference electrode is the saturated calomel electrode (SCE), which has the same general form as the silver–silver chloride electrode. The SCE consists of a platinum wire inserted into a moist paste of liquid mercury (Hg2Cl2; called calomel in the old chemical literature) and KCl. This interior cell is surrounded by an aqueous KCl solution, which acts as a salt bridge between the interior cell and the exterior solution (part (a) in Figure $5$. Although it sounds and looks complex, this cell is actually easy to prepare and maintain, and its potential is highly reproducible. The SCE cell diagram and corresponding half-reaction are as follows:
$Pt_{(s)} ∣ Hg_2Cl_{2(s)}∣KCl_{(aq, sat)} \label{20.4.37}$
$Hg_2Cl_{2(s)} + 2e^− \rightarrow 2Hg_{(l)} + 2Cl^−{(aq)} \label{20.4.38}$
At 25°C, the potential of the SCE is 0.2415 V versus the SHE, which means that 0.2415 V must be subtracted from the potential versus an SCE to obtain the standard electrode potential.
One of the most common uses of electrochemistry is to measure the H+ ion concentration of a solution. A glass electrode is generally used for this purpose, in which an internal Ag/AgCl electrode is immersed in a 0.10 M HCl solution that is separated from the solution by a very thin glass membrane (part (b) in Figure $5$. The glass membrane absorbs protons, which affects the measured potential. The extent of the adsorption on the inner side is fixed because [H+] is fixed inside the electrode, but the adsorption of protons on the outer surface depends on the pH of the solution. The potential of the glass electrode depends on [H+] as follows (recall that pH = −log[H+]):
$E_{glass} = E′ + (0.0591\; V \times \log[H^+]) = E′ − 0.0591\; V \times pH \label{20.4.39}$
The voltage E′ is a constant that depends on the exact construction of the electrode. Although it can be measured, in practice, a glass electrode is calibrated; that is, it is inserted into a solution of known pH, and the display on the pH meter is adjusted to the known value. Once the electrode is properly calibrated, it can be placed in a solution and used to determine an unknown pH.
Ion-selective electrodes are used to measure the concentration of a particular species in solution; they are designed so that their potential depends on only the concentration of the desired species (part (c) in Figure $5$). These electrodes usually contain an internal reference electrode that is connected by a solution of an electrolyte to a crystalline inorganic material or a membrane, which acts as the sensor. For example, one type of ion-selective electrode uses a single crystal of Eu-doped $LaF_3$ as the inorganic material. When fluoride ions in solution diffuse to the surface of the solid, the potential of the electrode changes, resulting in a so-called fluoride electrode. Similar electrodes are used to measure the concentrations of other species in solution. Some of the species whose concentrations can be determined in aqueous solution using ion-selective electrodes and similar devices are listed in Table $1$.
Table $1$: Some Species Whose Aqueous Concentrations Can Be Measured Using Electrochemical Methods
Species Type of Sample
H+ laboratory samples, blood, soil, and ground and surface water
NH3/NH4+ wastewater and runoff water
K+ blood, wine, and soil
CO2/HCO3 blood and groundwater
F groundwater, drinking water, and soil
Br grains and plant extracts
I milk and pharmaceuticals
NO3 groundwater, drinking water, soil, and fertilizer
Summary
Redox reactions can be balanced using the half-reaction method. The standard cell potential is a measure of the driving force for the reaction. $E°_{cell} = E°_{cathode} − E°_{anode} \nonumber \] The flow of electrons in an electrochemical cell depends on the identity of the reacting substances, the difference in the potential energy of their valence electrons, and their concentrations. The potential of the cell under standard conditions (1 M for solutions, 1 atm for gases, pure solids or liquids for other substances) and at a fixed temperature (25°C) is called the standard cell potential (E°cell). Only the difference between the potentials of two electrodes can be measured. By convention, all tabulated values of standard electrode potentials are listed as standard reduction potentials. The overall cell potential is the reduction potential of the reductive half-reaction minus the reduction potential of the oxidative half-reaction (E°cell = E°cathode − E°anode). The potential of the standard hydrogen electrode (SHE) is defined as 0 V under standard conditions. The potential of a half-reaction measured against the SHE under standard conditions is called its standard electrode potential. The standard cell potential is a measure of the driving force for a given redox reaction. All E° values are independent of the stoichiometric coefficients for the half-reaction. Redox reactions can be balanced using the half-reaction method, in which the overall redox reaction is divided into an oxidation half-reaction and a reduction half-reaction, each balanced for mass and charge. The half-reactions selected from tabulated lists must exactly reflect reaction conditions. In an alternative method, the atoms in each half-reaction are balanced, and then the charges are balanced. Whenever a half-reaction is reversed, the sign of E° corresponding to that reaction must also be reversed. If \(E°_{cell}$ is positive, the reaction will occur spontaneously under standard conditions. If $E°_{cell}$ is negative, then the reaction is not spontaneous under standard conditions, although it will proceed spontaneously in the opposite direction. The potential of an indicator electrode is related to the concentration of the substance being measured, whereas the potential of the reference electrode is held constant. Whether reduction or oxidation occurs depends on the potential of the sample versus the potential of the reference electrode. In addition to the SHE, other reference electrodes are the silver–silver chloride electrode; the saturated calomel electrode (SCE); the glass electrode, which is commonly used to measure pH; and ion-selective electrodes, which depend on the concentration of a single ionic species in solution. Differences in potential between the SHE and other reference electrodes must be included when calculating values for E°. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/20%3A_Electrochemistry/20.04%3A_Cell_Potential_Under_Standard_Conditions.txt |
Learning Objectives
• To understand the relationship between cell potential and the equilibrium constant.
• To use cell potentials to calculate solution concentrations.
Changes in reaction conditions can have a tremendous effect on the course of a redox reaction. For example, under standard conditions, the reaction of $\ce{Co(s)}$ with $\ce{Ni^{2+}(aq)}$ to form $\ce{Ni(s)}$ and $\ce{Co^{2+}(aq)}$ occurs spontaneously, but if we reduce the concentration of $\ce{Ni^{2+}}$ by a factor of 100, so that $\ce{[Ni^{2+}]}$ is 0.01 M, then the reverse reaction occurs spontaneously instead. The relationship between voltage and concentration is one of the factors that must be understood to predict whether a reaction will be spontaneous.
The Relationship between Cell Potential & Gibbs Energy
Electrochemical cells convert chemical energy to electrical energy and vice versa. The total amount of energy produced by an electrochemical cell, and thus the amount of energy available to do electrical work, depends on both the cell potential and the total number of electrons that are transferred from the reductant to the oxidant during the course of a reaction. The resulting electric current is measured in coulombs (C), an SI unit that measures the number of electrons passing a given point in 1 s. A coulomb relates energy (in joules) to electrical potential (in volts). Electric current is measured in amperes (A); 1 A is defined as the flow of 1 C/s past a given point (1 C = 1 A·s):
$\dfrac{\textrm{1 J}}{\textrm{1 V}}=\textrm{1 C}=\mathrm{A\cdot s} \label{20.5.1}$
In chemical reactions, however, we need to relate the coulomb to the charge on a mole of electrons. Multiplying the charge on the electron by Avogadro’s number gives us the charge on 1 mol of electrons, which is called the faraday (F), named after the English physicist and chemist Michael Faraday (1791–1867):
\begin{align}F &=(1.60218\times10^{-19}\textrm{ C})\left(\dfrac{6.02214 \times 10^{23}\, J}{\textrm{1 mol e}^-}\right) \[4pt] &=9.64833212\times10^4\textrm{ C/mol e}^- \[4pt] &\simeq 96,485\, J/(\mathrm{V\cdot mol\;e^-})\end{align} \label{20.5.2}
The total charge transferred from the reductant to the oxidant is therefore $nF$, where $n$ is the number of moles of electrons.
Michael Faraday (1791–1867)
Faraday was a British physicist and chemist who was arguably one of the greatest experimental scientists in history. The son of a blacksmith, Faraday was self-educated and became an apprentice bookbinder at age 14 before turning to science. His experiments in electricity and magnetism made electricity a routine tool in science and led to both the electric motor and the electric generator. He discovered the phenomenon of electrolysis and laid the foundations of electrochemistry. In fact, most of the specialized terms introduced in this chapter (electrode, anode, cathode, and so forth) are due to Faraday. In addition, he discovered benzene and invented the system of oxidation state numbers that we use today. Faraday is probably best known for “The Chemical History of a Candle,” a series of public lectures on the chemistry and physics of flames.
The maximum amount of work that can be produced by an electrochemical cell ($w_{max}$) is equal to the product of the cell potential ($E^°_{cell}$) and the total charge transferred during the reaction ($nF$):
$w_{max} = nFE_{cell} \label{20.5.3}$
Work is expressed as a negative number because work is being done by a system (an electrochemical cell with a positive potential) on its surroundings.
The change in free energy ($\Delta{G}$) is also a measure of the maximum amount of work that can be performed during a chemical process ($ΔG = w_{max}$). Consequently, there must be a relationship between the potential of an electrochemical cell and $\Delta{G}$; this relationship is as follows:
$\Delta{G} = −nFE_{cell} \label{20.5.4}$
A spontaneous redox reaction is therefore characterized by a negative value of $\Delta{G}$ and a positive value of $E^°_{cell}$, consistent with our earlier discussions. When both reactants and products are in their standard states, the relationship between ΔG° and $E^°_{cell}$ is as follows:
$\Delta{G^°} = −nFE^°_{cell} \label{20.5.5}$
A spontaneous redox reaction is characterized by a negative value of ΔG°, which corresponds to a positive value of E°cell.
Example $1$
Suppose you want to prepare elemental bromine from bromide using the dichromate ion as an oxidant. Using the data in Table P2, calculate the free-energy change (ΔG°) for this redox reaction under standard conditions. Is the reaction spontaneous?
Given: redox reaction
Asked for: $ΔG^o$ for the reaction and spontaneity
Strategy:
1. From the relevant half-reactions and the corresponding values of $E^o$, write the overall reaction and calculate $E^°_{cell}$.
2. Determine the number of electrons transferred in the overall reaction. Then use Equation \ref{20.5.5} to calculate $ΔG^o$. If $ΔG^o$ is negative, then the reaction is spontaneous.
A
As always, the first step is to write the relevant half-reactions and use them to obtain the overall reaction and the magnitude of $E^o$. From Table P2, we can find the reduction and oxidation half-reactions and corresponding $E^o$ values:
\begin{align*} & \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cathode}} =\textrm{1.36 V} \ & \textrm{anode:} &\quad & \mathrm{2Br^{-}(aq)} \rightarrow \mathrm{Br_2(aq)} +\mathrm{2e^-} &\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V} \end{align*} \nonumber
To obtain the overall balanced chemical equation, we must multiply both sides of the oxidation half-reaction by 3 to obtain the same number of electrons as in the reduction half-reaction, remembering that the magnitude of $E^o$ is not affected:
\begin{align*} & \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cathode}} =\textrm{1.36 V} \[4pt] & \textrm{anode:} &\quad & \mathrm{6Br^{-}(aq)} \rightarrow \mathrm{3Br_2(aq)} +\mathrm{6e^-} &\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V} \[4pt] \hline & \textrm{overall:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{6Br^{-}(aq)} + \mathrm{14H^+(aq)} \rightarrow \mathrm{2Cr^{3+}(aq)} + \mathrm{3Br_2(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cell}} =\textrm{0.27 V} \end{align*} \nonumber
B
We can now calculate ΔG° using Equation $\ref{20.5.5}$. Because six electrons are transferred in the overall reaction, the value of $n$ is 6:
\begin{align*}\Delta G^\circ &=-(n)(F)(E^\circ_{\textrm{cell}}) \[4pt] & =-(\textrm{6 mole})[96,485\;\mathrm{J/(V\cdot mol})(\textrm{0.27 V})] \& =-15.6 \times 10^4\textrm{ J} \ & =-156\;\mathrm{kJ/mol\;Cr_2O_7^{2-}} \end{align*} \nonumber
Thus $ΔG^o$ is −168 kJ/mol for the reaction as written, and the reaction is spontaneous.
Exercise $1$
Use the data in Table P2 to calculate $ΔG^o$ for the reduction of ferric ion by iodide:
$\ce{2Fe^{3+}(aq) + 2I^{−}(aq) → 2Fe^{2+}(aq) + I2(s)}\nonumber$
Is the reaction spontaneous?
Answer
−44 kJ/mol I2; yes
Relating G and Ecell: Relating G and Ecell(opens in new window) [youtu.be]
Potentials for the Sums of Half-Reactions
Although Table P2 list several half-reactions, many more are known. When the standard potential for a half-reaction is not available, we can use relationships between standard potentials and free energy to obtain the potential of any other half-reaction that can be written as the sum of two or more half-reactions whose standard potentials are available. For example, the potential for the reduction of $\ce{Fe^{3+}(aq)}$ to $\ce{Fe(s)}$ is not listed in the table, but two related reductions are given:
$\ce{Fe^{3+}(aq) + e^{−} -> Fe^{2+}(aq)} \;\;\;E^° = +0.77 V \label{20.5.6}$
$\ce{Fe^{2+}(aq) + 2e^{−} -> Fe(s)} \;\;\;E^° = −0.45 V \label{20.5.7}$
Although the sum of these two half-reactions gives the desired half-reaction, we cannot simply add the potentials of two reductive half-reactions to obtain the potential of a third reductive half-reaction because $E^o$ is not a state function. However, because $ΔG^o$ is a state function, the sum of the $ΔG^o$ values for the individual reactions gives us $ΔG^o$ for the overall reaction, which is proportional to both the potential and the number of electrons ($n$) transferred. To obtain the value of $E^o$ for the overall half-reaction, we first must add the values of $ΔG^o (= −nFE^o)$ for each individual half-reaction to obtain $ΔG^o$ for the overall half-reaction:
\begin{align*}\ce{Fe^{3+}(aq)} + \ce{e^-} &\rightarrow \ce\mathrm{Fe^{2+}(aq)} &\quad \Delta G^\circ &=-(1)(F)(\textrm{0.77 V})\[4pt] \ce{Fe^{2+}(aq)}+\ce{2e^-} &\rightarrow\ce{Fe(s)} &\quad\Delta G^\circ &=-(2)(F)(-\textrm{0.45 V})\[4pt] \ce{Fe^{3+}(aq)}+\ce{3e^-} &\rightarrow \ce{Fe(s)} &\quad\Delta G^\circ & =[-(1)(F)(\textrm{0.77 V})]+[-(2)(F)(-\textrm{0.45 V})] \end{align*} \nonumber
Solving the last expression for ΔG° for the overall half-reaction,
$\Delta{G^°} = F[(−0.77 V) + (−2)(−0.45 V)] = F(0.13 V) \label{20.5.9}$
Three electrons ($n = 3$) are transferred in the overall reaction, so substituting into Equation $\ref{20.5.5}$ and solving for $E^o$ gives the following:
\begin{align*}\Delta G^\circ & =-nFE^\circ_{\textrm{cell}} \[4pt] F(\textrm{0.13 V}) & =-(3)(F)(E^\circ_{\textrm{cell}}) \[4pt] E^\circ & =-\dfrac{0.13\textrm{ V}}{3}=-0.043\textrm{ V}\end{align*} \nonumber
This value of $E^o$ is very different from the value that is obtained by simply adding the potentials for the two half-reactions (0.32 V) and even has the opposite sign.
Values of $E^o$ for half-reactions cannot be added to give $E^o$ for the sum of the half-reactions; only values of $ΔG^o = −nFE^°_{cell}$ for half-reactions can be added.
The Relationship between Cell Potential & the Equilibrium Constant
We can use the relationship between $\Delta{G^°}$ and the equilibrium constant $K$, to obtain a relationship between $E^°_{cell}$ and $K$. Recall that for a general reaction of the type $aA + bB \rightarrow cC + dD$, the standard free-energy change and the equilibrium constant are related by the following equation:
$\Delta{G°} = −RT \ln K \label{20.5.10}$
Given the relationship between the standard free-energy change and the standard cell potential (Equation $\ref{20.5.5}$), we can write
$−nFE^°_{cell} = −RT \ln K \label{20.5.12}$
Rearranging this equation,
$E^\circ_{\textrm{cell}}= \left( \dfrac{RT}{nF} \right) \ln K \label{20.5.12B}$
For $T = 298\, K$, Equation $\ref{20.5.12B}$ can be simplified as follows:
\begin{align} E^\circ_{\textrm{cell}} &=\left(\dfrac{RT}{nF}\right)\ln K \[4pt] &=\left[ \dfrac{[8.314\;\mathrm{J/(mol\cdot K})(\textrm{298 K})]}{n[96,485\;\mathrm{J/(V\cdot mol)}]}\right]2.303 \log K \[4pt] &=\left(\dfrac{\textrm{0.0592 V}}{n}\right)\log K \label{20.5.13} \end{align}
Thus $E^°_{cell}$ is directly proportional to the logarithm of the equilibrium constant. This means that large equilibrium constants correspond to large positive values of $E^°_{cell}$ and vice versa.
Example $2$
Use the data in Table P2 to calculate the equilibrium constant for the reaction of metallic lead with PbO2 in the presence of sulfate ions to give PbSO4 under standard conditions. (This reaction occurs when a car battery is discharged.) Report your answer to two significant figures.
Given: redox reaction
Asked for: $K$
Strategy:
1. Write the relevant half-reactions and potentials. From these, obtain the overall reaction and $E^o_{cell}$.
2. Determine the number of electrons transferred in the overall reaction. Use Equation $\ref{20.5.13}$ to solve for $\log K$ and then $K$.
Solution
A The relevant half-reactions and potentials from Table P2 are as follows:
\begin{align*} & \textrm {cathode:} & & \mathrm{PbO_2(s)}+\mathrm{SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}+\mathrm{2e^-}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2H_2O(l)} & & E^\circ_\textrm{cathode}=\textrm{1.69 V} \[4pt] & \textrm{anode:} & & \mathrm{Pb(s)}+\mathrm{SO_4^{2-}(aq)}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2e^-} & & E^\circ_\textrm{anode}=-\textrm{0.36 V} \[4pt] \hline & \textrm {overall:} & & \mathrm{Pb(s)}+\mathrm{PbO_2(s)}+\mathrm{2SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}\rightarrow\mathrm{2PbSO_4(s)}+\mathrm{2H_2O(l)} & & E^\circ_\textrm{cell}=\textrm{2.05 V} \end{align*} \nonumber
B Two electrons are transferred in the overall reaction, so $n = 2$. Solving Equation $\ref{20.5.13}$ for log K and inserting the values of $n$ and $E^o$,
\begin{align*}\log K & =\dfrac{nE^\circ}{\textrm{0.0591 V}}=\dfrac{2(\textrm{2.05 V})}{\textrm{0.0591 V}}=69.37 \[4pt] K & =2.3\times10^{69}\end{align*} \nonumber
Thus the equilibrium lies far to the right, favoring a discharged battery (as anyone who has ever tried unsuccessfully to start a car after letting it sit for a long time will know).
Exercise $2$
Use the data in Table P2 to calculate the equilibrium constant for the reaction of $\ce{Sn^{2+}(aq)}$ with oxygen to produce $\ce{Sn^{4+}(aq)}$ and water under standard conditions. Report your answer to two significant figures. The reaction is as follows:
$\ce{2Sn^{2+}(aq) + O2(g) + 4H^{+}(aq) <=> 2Sn^{4+}(aq) + 2H2O(l)} \nonumber$
Answer
$5.7 \times 10^{72}$
Figure $1$ summarizes the relationships that we have developed based on properties of the system—that is, based on the equilibrium constant, standard free-energy change, and standard cell potential—and the criteria for spontaneity (ΔG° < 0). Unfortunately, these criteria apply only to systems in which all reactants and products are present in their standard states, a situation that is seldom encountered in the real world. A more generally useful relationship between cell potential and reactant and product concentrations, as we are about to see, uses the relationship between $\Delta{G}$ and the reaction quotient $Q$.
Electrode Potentials and ECell: Electrode Potentials and Ecell(opens in new window) [youtu.be]
Summary
A coulomb (C) relates electrical potential, expressed in volts, and energy, expressed in joules. The current generated from a redox reaction is measured in amperes (A), where 1 A is defined as the flow of 1 C/s past a given point. The faraday (F) is Avogadro’s number multiplied by the charge on an electron and corresponds to the charge on 1 mol of electrons. The product of the cell potential and the total charge is the maximum amount of energy available to do work, which is related to the change in free energy that occurs during the chemical process. Adding together the ΔG values for the half-reactions gives ΔG for the overall reaction, which is proportional to both the potential and the number of electrons (n) transferred. Spontaneous redox reactions have a negative ΔG and therefore a positive Ecell. Because the equilibrium constant K is related to ΔG, E°cell and K are also related. Large equilibrium constants correspond to large positive values of E°. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/20%3A_Electrochemistry/20.05%3A_Gibbs_Energy_and_Redox_Reactions.txt |
Learning Objectives
• Relate cell potentials to Gibbs energy changes
• Use the Nernst equation to determine cell potentials at nonstandard conditions
• Perform calculations that involve converting between cell potentials, free energy changes, and equilibrium constants
The Nernst Equation enables the determination of cell potential under non-standard conditions. It relates the measured cell potential to the reaction quotient and allows the accurate determination of equilibrium constants (including solubility constants).
The Effect of Concentration on Cell Potential: The Nernst Equation
Recall that the actual free-energy change for a reaction under nonstandard conditions, $\Delta{G}$, is given as follows:
$\Delta{G} = \Delta{G°} + RT \ln Q \label{Eq1}$
We also know that $ΔG = −nFE_{cell}$ (under non-standard conditions) and $ΔG^o = −nFE^o_{cell}$ (under standard conditions). Substituting these expressions into Equation $\ref{Eq1}$, we obtain
$−nFE_{cell} = −nFE^o_{cell} + RT \ln Q \label{Eq2}$
Dividing both sides of this equation by $−nF$,
$E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{RT}{nF}\right)\ln Q \label{Eq3}$
Equation $\ref{Eq3}$ is called the Nernst equation, after the German physicist and chemist Walter Nernst (1864–1941), who first derived it. The Nernst equation is arguably the most important relationship in electrochemistry. When a redox reaction is at equilibrium ($ΔG = 0$), then Equation $\ref{Eq3}$ reduces to Equation $\ref{Eq31}$ and $\ref{Eq32}$ because $Q = K$, and there is no net transfer of electrons (i.e., Ecell = 0).
$E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{RT}{nF}\right)\ln K = 0 \label{Eq31}$
since
$E^\circ_\textrm{cell}= \left(\dfrac{RT}{nF}\right)\ln K \label{Eq32}$
Substituting the values of the constants into Equation $\ref{Eq3}$ with $T = 298\, K$ and converting to base-10 logarithms give the relationship of the actual cell potential (Ecell), the standard cell potential (E°cell), and the reactant and product concentrations at room temperature (contained in $Q$):
$E_{\textrm{cell}}=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \label{Eq4}$
The Power of the Nernst Equation
The Nernst Equation ($\ref{Eq3}$) can be used to determine the value of Ecell, and thus the direction of spontaneous reaction, for any redox reaction under any conditions.
Equation $\ref{Eq4}$ allows us to calculate the potential associated with any electrochemical cell at 298 K for any combination of reactant and product concentrations under any conditions. We can therefore determine the spontaneous direction of any redox reaction under any conditions, as long as we have tabulated values for the relevant standard electrode potentials. Notice in Equation $\ref{Eq4}$ that the cell potential changes by 0.0591/n V for each 10-fold change in the value of $Q$ because log 10 = 1.
Example $1$
The following reaction proceeds spontaneously under standard conditions because E°cell > 0 (which means that ΔG° < 0):
$\ce{2Ce^{4+}(aq) + 2Cl^{–}(aq) -> 2Ce^{3+}(aq) + Cl2(g)}\;\; E^°_{cell} = 0.25\, V \nonumber$
Calculate $E_{cell}$ for this reaction under the following nonstandard conditions and determine whether it will occur spontaneously: [Ce4+] = 0.013 M, [Ce3+] = 0.60 M, [Cl] = 0.0030 M, $P_\mathrm{Cl_2}$ = 1.0 atm, and T = 25°C.
Given: balanced redox reaction, standard cell potential, and nonstandard conditions
Asked for: cell potential
Strategy:
Determine the number of electrons transferred during the redox process. Then use the Nernst equation to find the cell potential under the nonstandard conditions.
Solution
We can use the information given and the Nernst equation to calculate Ecell. Moreover, because the temperature is 25°C (298 K), we can use Equation $\ref{Eq4}$ instead of Equation $\ref{Eq3}$. The overall reaction involves the net transfer of two electrons:
$2Ce^{4+}_{(aq)} + 2e^− \rightarrow 2Ce^{3+}_{(aq)}\nonumber$
$2Cl^−_{(aq)} \rightarrow Cl_{2(g)} + 2e^−\nonumber$
so n = 2. Substituting the concentrations given in the problem, the partial pressure of Cl2, and the value of E°cell into Equation $\ref{Eq4}$,
\begin{align*}E_\textrm{cell} & =E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \ & =\textrm{0.25 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{[\mathrm{Ce^{3+}}]^2P_\mathrm{Cl_2}}{[\mathrm{Ce^{4+}}]^2[\mathrm{Cl^-}]^2}\right) \ & =\textrm{0.25 V}-[(\textrm{0.0296 V})(8.37)]=\textrm{0.00 V}\end{align*} \nonumber
Thus the reaction will not occur spontaneously under these conditions (because E = 0 V and ΔG = 0). The composition specified is that of an equilibrium mixture
Exercise $1$
Molecular oxygen will not oxidize $MnO_2$ to permanganate via the reaction
$\ce{4MnO2(s) + 3O2(g) + 4OH^{−} (aq) -> 4MnO^{−}4(aq) + 2H2O(l)} \;\;\; E°_{cell} = −0.20\; V\nonumber$
Calculate $E_{cell}$ for the reaction under the following nonstandard conditions and decide whether the reaction will occur spontaneously: pH 10, $P_\mathrm{O_2}$= 0.20 atm, [MNO4] = 1.0 × 10−4 M, and T = 25°C.
Answer
Ecell = −0.22 V; the reaction will not occur spontaneously.
Applying the Nernst equation to a simple electrochemical cell such as the Zn/Cu cell allows us to see how the cell voltage varies as the reaction progresses and the concentrations of the dissolved ions change. Recall that the overall reaction for this cell is as follows:
$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)\;\;\;E°cell = 1.10 V \label{Eq5}$
The reaction quotient is therefore $Q = [Zn^{2+}]/[Cu^{2+}]$. Suppose that the cell initially contains 1.0 M Cu2+ and 1.0 × 10−6 M Zn2+. The initial voltage measured when the cell is connected can then be calculated from Equation $\ref{Eq4}$:
\begin{align}E_\textrm{cell} & =E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log\dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}\ & =\textrm{1.10 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{1.0\times10^{-6}}{1.0}\right)=\textrm{1.28 V}\end{align} \label{Eq6}
Thus the initial voltage is greater than E° because $Q<1$. As the reaction proceeds, [Zn2+] in the anode compartment increases as the zinc electrode dissolves, while [Cu2+] in the cathode compartment decreases as metallic copper is deposited on the electrode. During this process, the ratio Q = [Zn2+]/[Cu2+] steadily increases, and the cell voltage therefore steadily decreases. Eventually, [Zn2+] = [Cu2+], so Q = 1 and Ecell = E°cell. Beyond this point, [Zn2+] will continue to increase in the anode compartment, and [Cu2+] will continue to decrease in the cathode compartment. Thus the value of Q will increase further, leading to a further decrease in Ecell. When the concentrations in the two compartments are the opposite of the initial concentrations (i.e., 1.0 M Zn2+ and 1.0 × 10−6 M Cu2+), Q = 1.0 × 106, and the cell potential will be reduced to 0.92 V.
The variation of Ecell with $\log{Q}$ over this range is linear with a slope of −0.0591/n, as illustrated in Figure $1$. As the reaction proceeds still further, $Q$ continues to increase, and Ecell continues to decrease. If neither of the electrodes dissolves completely, thereby breaking the electrical circuit, the cell voltage will eventually reach zero. This is the situation that occurs when a battery is “dead.” The value of $Q$ when Ecell = 0 is calculated as follows:
\begin{align}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q=0 \ E^\circ &=\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \ \log Q &=\dfrac{E^\circ n}{\textrm{0.0591 V}}=\dfrac{(\textrm{1.10 V})(2)}{\textrm{0.0591 V}}=37.23 \ Q &=10^{37.23}=1.7\times10^{37}\end{align} \label{Eq7}
Recall that at equilibrium, $Q = K$. Thus the equilibrium constant for the reaction of Zn metal with Cu2+ to give Cu metal and Zn2+ is 1.7 × 1037 at 25°C.
The Nernst Equation: The Nernst Equation (opens in new window) [youtu.be]
Concentration Cells
A voltage can also be generated by constructing an electrochemical cell in which each compartment contains the same redox active solution but at different concentrations. The voltage is produced as the concentrations equilibrate. Suppose, for example, we have a cell with 0.010 M AgNO3 in one compartment and 1.0 M AgNO3 in the other. The cell diagram and corresponding half-reactions are as follows:
$\ce{Ag(s)\,|\,Ag^{+}}(aq, 0.010 \;M)\,||\,\ce{Ag^{+}}(aq, 1.0 \;M)\,|\,\ce{Ag(s)} \label{Eq8}$
cathode:
$\ce{Ag^{+}} (aq, 1.0\; M) + \ce{e^{−}} \rightarrow \ce{Ag(s)} \label{Eq9}$
anode:
$\ce{Ag(s)} \rightarrow \ce{Ag^{+}}(aq, 0.010\; M) + \ce{e^{−}} \label{Eq10}$
Overall
$\ce{Ag^{+}}(aq, 1.0 \;M) \rightarrow \ce{Ag^{+}}(aq, 0.010\; M) \label{Eq11}$
As the reaction progresses, the concentration of $Ag^+$ will increase in the left (oxidation) compartment as the silver electrode dissolves, while the $Ag^+$ concentration in the right (reduction) compartment decreases as the electrode in that compartment gains mass. The total mass of $Ag(s)$ in the cell will remain constant, however. We can calculate the potential of the cell using the Nernst equation, inserting 0 for E°cell because E°cathode = −E°anode:
\begin{align*} E_\textrm{cell}&=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \[4pt] &=0-\left(\dfrac{\textrm{0.0591 V}}{1}\right)\log\left(\dfrac{0.010}{1.0}\right) \[4pt] &=\textrm{0.12 V} \end{align*} \nonumber
An electrochemical cell of this type, in which the anode and cathode compartments are identical except for the concentration of a reactant, is called a concentration cell. As the reaction proceeds, the difference between the concentrations of Ag+ in the two compartments will decrease, as will Ecell. Finally, when the concentration of Ag+ is the same in both compartments, equilibrium will have been reached, and the measured potential difference between the two compartments will be zero (Ecell = 0).
Example $2$
Calculate the voltage in a galvanic cell that contains a manganese electrode immersed in a 2.0 M solution of MnCl2 as the cathode, and a manganese electrode immersed in a 5.2 × 10−2 M solution of MnSO4 as the anode (T = 25°C).
Given: galvanic cell, identities of the electrodes, and solution concentrations
Asked for: voltage
Strategy:
1. Write the overall reaction that occurs in the cell.
2. Determine the number of electrons transferred. Substitute this value into the Nernst equation to calculate the voltage.
Solution
A This is a concentration cell, in which the electrode compartments contain the same redox active substance but at different concentrations. The anions (Cl and SO42) do not participate in the reaction, so their identity is not important. The overall reaction is as follows:
$\ce{ Mn^{2+}}(aq, 2.0\, M) \rightarrow \ce{Mn^{2+}} (aq, 5.2 \times 10^{−2}\, M)\nonumber$
B For the reduction of Mn2+(aq) to Mn(s), n = 2. We substitute this value and the given Mn2+ concentrations into Equation $\ref{Eq4}$:
\begin{align*} E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \[4pt] &=\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{5.2\times10^{-2}}{2.0}\right) \[4pt] &=\textrm{0.047 V}\end{align*} \nonumber
Thus manganese will dissolve from the electrode in the compartment that contains the more dilute solution and will be deposited on the electrode in the compartment that contains the more concentrated solution.
Exercise $2$
Suppose we construct a galvanic cell by placing two identical platinum electrodes in two beakers that are connected by a salt bridge. One beaker contains 1.0 M HCl, and the other a 0.010 M solution of Na2SO4 at pH 7.00. Both cells are in contact with the atmosphere, with $P_\mathrm{O_2}$ = 0.20 atm. If the relevant electrochemical reaction in both compartments is the four-electron reduction of oxygen to water:
$\ce{O2(g) + 4H^{+}(aq) + 4e^{−} \rightarrow 2H2O(l)} \nonumber$
What will be the potential when the circuit is closed?
Answer
0.41 V
Using Cell Potentials to Measure Solubility Products
Because voltages are relatively easy to measure accurately using a voltmeter, electrochemical methods provide a convenient way to determine the concentrations of very dilute solutions and the solubility products ($K_{sp}$) of sparingly soluble substances. As you learned previously, solubility products can be very small, with values of less than or equal to 10−30. Equilibrium constants of this magnitude are virtually impossible to measure accurately by direct methods, so we must use alternative methods that are more sensitive, such as electrochemical methods.
To understand how an electrochemical cell is used to measure a solubility product, consider the cell shown in Figure $1$, which is designed to measure the solubility product of silver chloride:
$K_{sp} = [\ce{Ag^{+}}][\ce{Cl^{−}}]. \nonumber$
In one compartment, the cell contains a silver wire inserted into a 1.0 M solution of Ag+; the other compartment contains a silver wire inserted into a 1.0 M Cl solution saturated with AgCl. In this system, the Ag+ ion concentration in the first compartment equals Ksp. We can see this by dividing both sides of the equation for Ksp by [Cl] and substituting:
\begin{align*}[\ce{Ag^{+}}] &= \dfrac{K_{sp}}{[\ce{Cl^{−}}]} \[4pt] &= \dfrac{K_{sp}}{1.0} = K_{sp}. \end{align*} \nonumber
The overall cell reaction is as follows:
Ag+(aq, concentrated) → Ag+(aq, dilute)
Thus the voltage of the concentration cell due to the difference in [Ag+] between the two cells is as follows:
\begin{align} E_\textrm{cell} &=\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{1}\right)\log\left(\dfrac{[\mathrm{Ag^+}]_\textrm{dilute}}{[\mathrm{Ag^+}]_\textrm{concentrated}}\right) \nonumber \[4pt] &= -\textrm{0.0591 V } \log\left(\dfrac{K_{\textrm{sp}}}{1.0}\right) \nonumber \[4pt] &=-\textrm{0.0591 V }\log K_{\textrm{sp}} \label{Eq122} \end{align}
By closing the circuit, we can measure the potential caused by the difference in [Ag+] in the two cells. In this case, the experimentally measured voltage of the concentration cell at 25°C is 0.580 V. Solving Equation $\ref{Eq122}$ for $K_{sp}$,
\begin{align*}\log K_\textrm{sp} & =\dfrac{-E_\textrm{cell}}{\textrm{0.0591 V}}=\dfrac{-\textrm{0.580 V}}{\textrm{0.0591 V}}=-9.81 \[4pt] K_\textrm{sp} & =1.5\times10^{-10}\end{align*} \nonumber
Thus a single potential measurement can provide the information we need to determine the value of the solubility product of a sparingly soluble salt.
Example $3$: Solubility of lead(II) sulfate
To measure the solubility product of lead(II) sulfate (PbSO4) at 25°C, you construct a galvanic cell like the one shown in Figure $1$, which contains a 1.0 M solution of a very soluble Pb2+ salt [lead(II) acetate trihydrate] in one compartment that is connected by a salt bridge to a 1.0 M solution of Na2SO4 saturated with PbSO4 in the other. You then insert a Pb electrode into each compartment and close the circuit. Your voltmeter shows a voltage of 230 mV. What is Ksp for PbSO4? Report your answer to two significant figures.
Given: galvanic cell, solution concentrations, electrodes, and voltage
Asked for: Ksp
Strategy:
1. From the information given, write the equation for Ksp. Express this equation in terms of the concentration of Pb2+.
2. Determine the number of electrons transferred in the electrochemical reaction. Substitute the appropriate values into Equation $\ref{Eq12}$ and solve for Ksp.
Solution
A You have constructed a concentration cell, with one compartment containing a 1.0 M solution of $\ce{Pb^{2+}}$ and the other containing a dilute solution of Pb2+ in 1.0 M Na2SO4. As for any concentration cell, the voltage between the two compartments can be calculated using the Nernst equation. The first step is to relate the concentration of Pb2+ in the dilute solution to Ksp:
\begin{align*}[\mathrm{Pb^{2+}}][\mathrm{SO_4^{2-}}] & =K_\textrm{sp} \ [\mathrm{Pb^{2+}}] &=\dfrac{K_\textrm{sp}}{[\mathrm{SO_4^{2-}}]}=\dfrac{K_\textrm{sp}}{\textrm{1.0 M}}=K_\textrm{sp}\end{align*} \nonumber
B The reduction of Pb2+ to Pb is a two-electron process and proceeds according to the following reaction:
Pb2+(aq, concentrated) → Pb2+(aq, dilute)
so
\begin{align*}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{0.0591}{n}\right)\log Q \ \textrm{0.230 V} & =\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{[\mathrm{Pb^{2+}}]_\textrm{dilute}}{[\mathrm{Pb^{2+}}]_\textrm{concentrated}}\right)=-\textrm{0.0296 V}\log\left(\dfrac{K_\textrm{sp}}{1.0}\right) \ -7.77 & =\log K_\textrm{sp} \ 1.7\times10^{-8} & =K_\textrm{sp}\end{align*} \nonumber
Exercise $3$
A concentration cell similar to the one described in Example $3$ contains a 1.0 M solution of lanthanum nitrate [La(NO3)3] in one compartment and a 1.0 M solution of sodium fluoride saturated with LaF3 in the other. A metallic La strip is inserted into each compartment, and the circuit is closed. The measured potential is 0.32 V. What is the Ksp for LaF3? Report your answer to two significant figures.
Answer
5.7 × 10−17
Using Cell Potentials to Measure Concentrations
Another use for the Nernst equation is to calculate the concentration of a species given a measured potential and the concentrations of all the other species. We saw an example of this in Example $3$, in which the experimental conditions were defined in such a way that the concentration of the metal ion was equal to Ksp. Potential measurements can be used to obtain the concentrations of dissolved species under other conditions as well, which explains the widespread use of electrochemical cells in many analytical devices. Perhaps the most common application is in the determination of [H+] using a pH meter, as illustrated below.
Example $4$: Measuring pH
Suppose a galvanic cell is constructed with a standard Zn/Zn2+ couple in one compartment and a modified hydrogen electrode in the second compartment. The pressure of hydrogen gas is 1.0 atm, but [H+] in the second compartment is unknown. The cell diagram is as follows:
$\ce{Zn(s)}|\ce{Zn^{2+}}(aq, 1.0\, M) || \ce{H^{+}} (aq, ?\, M)| \ce{H2} (g, 1.0\, atm)| Pt(s) \nonumber$
What is the pH of the solution in the second compartment if the measured potential in the cell is 0.26 V at 25°C?
Given: galvanic cell, cell diagram, and cell potential
Asked for: pH of the solution
Strategy:
1. Write the overall cell reaction.
2. Substitute appropriate values into the Nernst equation and solve for −log[H+] to obtain the pH.
Solution
A Under standard conditions, the overall reaction that occurs is the reduction of protons by zinc to give H2 (note that Zn lies below H2 in Table P2):
Zn(s) + 2H2+(aq) → Zn2+(aq) + H2(g) E°=0.76 V
B By substituting the given values into the simplified Nernst equation (Equation $\ref{Eq4}$), we can calculate [H+] under nonstandard conditions:
\begin{align*}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}n\right)\log\left(\dfrac{[\mathrm{Zn^{2+}}]P_\mathrm{H_2}}{[\mathrm{H^+}]^2}\right) \ \textrm{0.26 V} &=\textrm{0.76 V}-\left(\dfrac{\textrm{0.0591 V}}2\right)\log\left(\dfrac{(1.0)(1.0)}{[\mathrm{H^+}]^2}\right) \ 16.9 &=\log\left(\dfrac{1}{[\mathrm{H^+}]^2}\right)=\log[\mathrm{H^+}]^{-2}=(-2)\log[\mathrm{H^+}] \ 8.46 &=-\log[\mathrm{H^+}] \ 8.5 &=\mathrm{pH}\end{align*} \nonumber
Thus the potential of a galvanic cell can be used to measure the pH of a solution.
Exercise $4$
Suppose you work for an environmental laboratory and you want to use an electrochemical method to measure the concentration of Pb2+ in groundwater. You construct a galvanic cell using a standard oxygen electrode in one compartment (E°cathode = 1.23 V). The other compartment contains a strip of lead in a sample of groundwater to which you have added sufficient acetic acid, a weak organic acid, to ensure electrical conductivity. The cell diagram is as follows:
$Pb_{(s)} ∣Pb^{2+}(aq, ? M)∥H^+(aq), 1.0 M∣O_2(g, 1.0 atm)∣Pt_{(s)}\nonumber$
When the circuit is closed, the cell has a measured potential of 1.62 V. Use Table P2 to determine the concentration of Pb2+ in the groundwater.
Answer
$1.2 \times 10^{−9}\; M$
Summary
The Nernst equation can be used to determine the direction of spontaneous reaction for any redox reaction in aqueous solution. The Nernst equation allows us to determine the spontaneous direction of any redox reaction under any reaction conditions from values of the relevant standard electrode potentials. Concentration cells consist of anode and cathode compartments that are identical except for the concentrations of the reactant. Because ΔG = 0 at equilibrium, the measured potential of a concentration cell is zero at equilibrium (the concentrations are equal). A galvanic cell can also be used to measure the solubility product of a sparingly soluble substance and calculate the concentration of a species given a measured potential and the concentrations of all the other species. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/20%3A_Electrochemistry/20.06%3A_Cell_Potential_Under_Nonstandard_Conditions.txt |
Because galvanic cells can be self-contained and portable, they can be used as batteries and fuel cells. A battery (storage cell) is a galvanic cell (or a series of galvanic cells) that contains all the reactants needed to produce electricity. In contrast, a fuel cell is a galvanic cell that requires a constant external supply of one or more reactants to generate electricity. In this section, we describe the chemistry behind some of the more common types of batteries and fuel cells.
Batteries
There are two basic kinds of batteries: disposable, or primary, batteries, in which the electrode reactions are effectively irreversible and which cannot be recharged; and rechargeable, or secondary, batteries, which form an insoluble product that adheres to the electrodes. These batteries can be recharged by applying an electrical potential in the reverse direction. The recharging process temporarily converts a rechargeable battery from a galvanic cell to an electrolytic cell.
Batteries are cleverly engineered devices that are based on the same fundamental laws as galvanic cells. The major difference between batteries and the galvanic cells we have previously described is that commercial batteries use solids or pastes rather than solutions as reactants to maximize the electrical output per unit mass. The use of highly concentrated or solid reactants has another beneficial effect: the concentrations of the reactants and the products do not change greatly as the battery is discharged; consequently, the output voltage remains remarkably constant during the discharge process. This behavior is in contrast to that of the Zn/Cu cell, whose output decreases logarithmically as the reaction proceeds (Figure $1$). When a battery consists of more than one galvanic cell, the cells are usually connected in series—that is, with the positive (+) terminal of one cell connected to the negative (−) terminal of the next, and so forth. The overall voltage of the battery is therefore the sum of the voltages of the individual cells.
The major difference between batteries and the galvanic cells is that commercial typically batteries use solids or pastes rather than solutions as reactants to maximize the electrical output per unit mass. An obvious exception is the standard car battery which used solution phase chemistry.
Leclanché Dry Cell
The dry cell, by far the most common type of battery, is used in flashlights, electronic devices such as the Walkman and Game Boy, and many other devices. Although the dry cell was patented in 1866 by the French chemist Georges Leclanché and more than 5 billion such cells are sold every year, the details of its electrode chemistry are still not completely understood. In spite of its name, the Leclanché dry cell is actually a “wet cell”: the electrolyte is an acidic water-based paste containing $MnO_2$, $NH_4Cl$, $ZnCl_2$, graphite, and starch (part (a) in Figure $1$). The half-reactions at the anode and the cathode can be summarized as follows:
• cathode (reduction):
$\ce{2MnO2(s) + 2NH^{+}4(aq) + 2e^{−} -> Mn2O3(s) + 2NH3(aq) + H2O(l)} \nonumber$
• anode (oxidation):
$\ce{Zn(s) -> Zn^{2+}(aq) + 2e^{−}} \nonumber$
The $\ce{Zn^{2+}}$ ions formed by the oxidation of $\ce{Zn(s)}$ at the anode react with $\ce{NH_3}$ formed at the cathode and $\ce{Cl^{−}}$ ions present in solution, so the overall cell reaction is as follows:
• overall reaction:
$\ce{2MnO2(s) + 2NH4Cl(aq) + Zn(s) -> Mn2O3(s) + Zn(NH3)2Cl2(s) + H2O(l)} \label{Eq3}$
The dry cell produces about 1.55 V and is inexpensive to manufacture. It is not, however, very efficient in producing electrical energy because only the relatively small fraction of the $\ce{MnO2}$ that is near the cathode is actually reduced and only a small fraction of the zinc cathode is actually consumed as the cell discharges. In addition, dry cells have a limited shelf life because the $\ce{Zn}$ anode reacts spontaneously with $\ce{NH4Cl}$ in the electrolyte, causing the case to corrode and allowing the contents to leak out.
The alkaline battery is essentially a Leclanché cell adapted to operate under alkaline, or basic, conditions. The half-reactions that occur in an alkaline battery are as follows:
• cathode (reduction)
$\ce{2MnO2(s) + H2O(l) + 2e^{−} -> Mn2O3(s) + 2OH^{−}(aq)} \nonumber$
• anode (oxidation):
$\ce{Zn(s) + 2OH^{−}(aq) -> ZnO(s) + H2O(l) + 2e^{−}} \nonumber$
• overall reaction:
$\ce{Zn(s) + 2MnO2(s) -> ZnO(s) + Mn2O3(s)} \nonumber$
This battery also produces about 1.5 V, but it has a longer shelf life and more constant output voltage as the cell is discharged than the Leclanché dry cell. Although the alkaline battery is more expensive to produce than the Leclanché dry cell, the improved performance makes this battery more cost-effective.
Button Batteries
Although some of the small button batteries used to power watches, calculators, and cameras are miniature alkaline cells, most are based on a completely different chemistry. In these "button" batteries, the anode is a zinc–mercury amalgam rather than pure zinc, and the cathode uses either $\ce{HgO}$ or $\ce{Ag2O}$ as the oxidant rather than $\ce{MnO2}$ in Figure $\PageIndex{1b}$).
The cathode, anode and overall reactions and cell output for these two types of button batteries are as follows (two half-reactions occur at the anode, but the overall oxidation half-reaction is shown):
• cathode (mercury battery): $\ce{HgO(s) + H2O(l) + 2e^{−} -> Hg(l) + 2OH^{−}(aq)} \nonumber$
• Anode (mercury battery): $\ce{Zn + 2OH^{−} -> ZnO + H2O + 2e^{−}} \nonumber$
• overall reaction (mercury battery): $\ce{Zn(s) + 2HgO(s) -> 2Hg(l) + ZnO(s)} \nonumber$ with $E_{cell} = 1.35 \,V$.
• cathode reaction (silver battery): $\ce{Ag2O(s) + H2O(l) + 2e^{−} -> 2Ag(s) + 2OH^{−}(aq)} \nonumber$
• anode (silver battery): $\ce{Zn + 2OH^{−} -> ZnO + H2O + 2e^{−}} \nonumber$
• Overall reaction (silver battery): $\ce{Zn(s) + 2Ag2O(s) -> 2Ag(s) + ZnO(s)} \nonumber$ with $E_{cell} = 1.6 \,V$.
The major advantages of the mercury and silver cells are their reliability and their high output-to-mass ratio. These factors make them ideal for applications where small size is crucial, as in cameras and hearing aids. The disadvantages are the expense and the environmental problems caused by the disposal of heavy metals, such as $\ce{Hg}$ and $\ce{Ag}$.
Lithium–Iodine Battery
None of the batteries described above is actually “dry.” They all contain small amounts of liquid water, which adds significant mass and causes potential corrosion problems. Consequently, substantial effort has been expended to develop water-free batteries. One of the few commercially successful water-free batteries is the lithium–iodine battery. The anode is lithium metal, and the cathode is a solid complex of $I_2$. Separating them is a layer of solid $LiI$, which acts as the electrolyte by allowing the diffusion of Li+ ions. The electrode reactions are as follows:
• cathode (reduction):
$I_{2(s)} + 2e^− \rightarrow {2I^-}_{(LiI)}\label{Eq11}$
• anode (oxidation):
$2Li_{(s)} \rightarrow 2Li^+_{(LiI)} + 2e^− \label{Eq12}$
• overall:
$2Li_{(s)}+ I_{2(s)} \rightarrow 2LiI_{(s)} \label{Eq12a}$
with $E_{cell} = 3.5 \, V$
As shown in part (c) in Figure $1$, a typical lithium–iodine battery consists of two cells separated by a nickel metal mesh that collects charge from the anode. Because of the high internal resistance caused by the solid electrolyte, only a low current can be drawn. Nonetheless, such batteries have proven to be long-lived (up to 10 yr) and reliable. They are therefore used in applications where frequent replacement is difficult or undesirable, such as in cardiac pacemakers and other medical implants and in computers for memory protection. These batteries are also used in security transmitters and smoke alarms. Other batteries based on lithium anodes and solid electrolytes are under development, using $TiS_2$, for example, for the cathode.
Dry cells, button batteries, and lithium–iodine batteries are disposable and cannot be recharged once they are discharged. Rechargeable batteries, in contrast, offer significant economic and environmental advantages because they can be recharged and discharged numerous times. As a result, manufacturing and disposal costs drop dramatically for a given number of hours of battery usage. Two common rechargeable batteries are the nickel–cadmium battery and the lead–acid battery, which we describe next.
Nickel–Cadmium (NiCad) Battery
The nickel–cadmium, or NiCad, battery is used in small electrical appliances and devices like drills, portable vacuum cleaners, and AM/FM digital tuners. It is a water-based cell with a cadmium anode and a highly oxidized nickel cathode that is usually described as the nickel(III) oxo-hydroxide, NiO(OH). As shown in Figure $2$, the design maximizes the surface area of the electrodes and minimizes the distance between them, which decreases internal resistance and makes a rather high discharge current possible.
The electrode reactions during the discharge of a $NiCad$ battery are as follows:
• cathode (reduction):
$2NiO(OH)_{(s)} + 2H_2O_{(l)} + 2e^− \rightarrow 2Ni(OH)_{2(s)} + 2OH^-_{(aq)} \label{Eq13}$
• anode (oxidation):
$Cd_{(s)} + 2OH^-_{(aq)} \rightarrow Cd(OH)_{2(s)} + 2e^- \label{Eq14}$
• overall:
$Cd_{(s)} + 2NiO(OH)_{(s)} + 2H_2O_{(l)} \rightarrow Cd(OH)_{2(s)} + 2Ni(OH)_{2(s)} \label{Eq15}$
$E_{cell} = 1.4 V$
Because the products of the discharge half-reactions are solids that adhere to the electrodes [Cd(OH)2 and 2Ni(OH)2], the overall reaction is readily reversed when the cell is recharged. Although NiCad cells are lightweight, rechargeable, and high capacity, they have certain disadvantages. For example, they tend to lose capacity quickly if not allowed to discharge fully before recharging, they do not store well for long periods when fully charged, and they present significant environmental and disposal problems because of the toxicity of cadmium.
A variation on the NiCad battery is the nickel–metal hydride battery (NiMH) used in hybrid automobiles, wireless communication devices, and mobile computing. The overall chemical equation for this type of battery is as follows:
$NiO(OH)_{(s)} + MH \rightarrow Ni(OH)_{2(s)} + M_{(s)} \label{Eq16}$
The NiMH battery has a 30%–40% improvement in capacity over the NiCad battery; it is more environmentally friendly so storage, transportation, and disposal are not subject to environmental control; and it is not as sensitive to recharging memory. It is, however, subject to a 50% greater self-discharge rate, a limited service life, and higher maintenance, and it is more expensive than the NiCad battery.
Directive 2006/66/EC of the European Union prohibits the placing on the market of portable batteries that contain more than 0.002% of cadmium by weight. The aim of this directive was to improve "the environmental performance of batteries and accumulators"
Lead–Acid (Lead Storage) Battery
The lead–acid battery is used to provide the starting power in virtually every automobile and marine engine on the market. Marine and car batteries typically consist of multiple cells connected in series. The total voltage generated by the battery is the potential per cell (E°cell) times the number of cells.
As shown in Figure $3$, the anode of each cell in a lead storage battery is a plate or grid of spongy lead metal, and the cathode is a similar grid containing powdered lead dioxide ($PbO_2$). The electrolyte is usually an approximately 37% solution (by mass) of sulfuric acid in water, with a density of 1.28 g/mL (about 4.5 M $H_2SO_4$). Because the redox active species are solids, there is no need to separate the electrodes. The electrode reactions in each cell during discharge are as follows:
• cathode (reduction):
$PbO_{2(s)} + HSO^−_{4(aq)} + 3H^+_{(aq)} + 2e^− \rightarrow PbSO_{4(s)} + 2H_2O_{(l)} \label{Eq17}$
with $E^°_{cathode} = 1.685 \; V$
• anode (oxidation):
$Pb_{(s)} + HSO^−_{4(aq)} \rightarrow PbSO_{4(s) }+ H^+_{(aq)} + 2e^−\label{Eq18}$
with $E^°_{anode} = −0.356 \; V$
• overall:
$Pb_{(s)} + PbO_{2(s)} + 2HSO^−_{4(aq)} + 2H^+_{(aq)} \rightarrow 2PbSO_{4(s)} + 2H_2O_{(l)} \label{Eq19}$
and $E^°_{cell} = 2.041 \; V$
As the cell is discharged, a powder of $PbSO_4$ forms on the electrodes. Moreover, sulfuric acid is consumed and water is produced, decreasing the density of the electrolyte and providing a convenient way of monitoring the status of a battery by simply measuring the density of the electrolyte. This is often done with the use of a hydrometer.
A hydrometer can be used to test the specific gravity of each cell as a measure of its state of charge (www.youtube.com/watch?v=SRcOqfL6GqQ).
When an external voltage in excess of 2.04 V per cell is applied to a lead–acid battery, the electrode reactions reverse, and $PbSO_4$ is converted back to metallic lead and $PbO_2$. If the battery is recharged too vigorously, however, electrolysis of water can occur:
$2H_2O_{(l)} \rightarrow 2H_{2(g)} +O_{2 (g)} \label{EqX}$
This results in the evolution of potentially explosive hydrogen gas. The gas bubbles formed in this way can dislodge some of the $PbSO_4$ or $PbO_2$ particles from the grids, allowing them to fall to the bottom of the cell, where they can build up and cause an internal short circuit. Thus the recharging process must be carefully monitored to optimize the life of the battery. With proper care, however, a lead–acid battery can be discharged and recharged thousands of times. In automobiles, the alternator supplies the electric current that causes the discharge reaction to reverse.
Fuel Cells
A fuel cell is a galvanic cell that requires a constant external supply of reactants because the products of the reaction are continuously removed. Unlike a battery, it does not store chemical or electrical energy; a fuel cell allows electrical energy to be extracted directly from a chemical reaction. In principle, this should be a more efficient process than, for example, burning the fuel to drive an internal combustion engine that turns a generator, which is typically less than 40% efficient, and in fact, the efficiency of a fuel cell is generally between 40% and 60%. Unfortunately, significant cost and reliability problems have hindered the wide-scale adoption of fuel cells. In practice, their use has been restricted to applications in which mass may be a significant cost factor, such as US manned space vehicles.
These space vehicles use a hydrogen/oxygen fuel cell that requires a continuous input of H2(g) and O2(g), as illustrated in Figure $4$. The electrode reactions are as follows:
• cathode (reduction):
$O_{2(g)} + 4H^+ + 4e^− \rightarrow 2H_2O_{(g)} \label{Eq20}$
• anode (oxidation):
$2H_{2(g)} \rightarrow 4H^+ + 4e^− \label{Eq21}$
• overall:
$2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(g)} \label{Eq22}$
The overall reaction represents an essentially pollution-free conversion of hydrogen and oxygen to water, which in space vehicles is then collected and used. Although this type of fuel cell should produce 1.23 V under standard conditions, in practice the device achieves only about 0.9 V. One of the major barriers to achieving greater efficiency is the fact that the four-electron reduction of $O_2 (g)$ at the cathode is intrinsically rather slow, which limits current that can be achieved. All major automobile manufacturers have major research programs involving fuel cells: one of the most important goals is the development of a better catalyst for the reduction of $O_2 (g)$.
Summary
Commercial batteries are galvanic cells that use solids or pastes as reactants to maximize the electrical output per unit mass. A battery is a contained unit that produces electricity, whereas a fuel cell is a galvanic cell that requires a constant external supply of one or more reactants to generate electricity. One type of battery is the Leclanché dry cell, which contains an electrolyte in an acidic water-based paste. This battery is called an alkaline battery when adapted to operate under alkaline conditions. Button batteries have a high output-to-mass ratio; lithium–iodine batteries consist of a solid electrolyte; the nickel–cadmium (NiCad) battery is rechargeable; and the lead–acid battery, which is also rechargeable, does not require the electrodes to be in separate compartments. A fuel cell requires an external supply of reactants as the products of the reaction are continuously removed. In a fuel cell, energy is not stored; electrical energy is provided by a chemical reaction. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/20%3A_Electrochemistry/20.07%3A_Batteries_and_Fuel_Cells.txt |
Learning Objectives
• To understand the process of corrosion.
Corrosion is a galvanic process by which metals deteriorate through oxidation—usually but not always to their oxides. For example, when exposed to air, iron rusts, silver tarnishes, and copper and brass acquire a bluish-green surface called a patina. Of the various metals subject to corrosion, iron is by far the most important commercially. An estimated \$100 billion per year is spent in the United States alone to replace iron-containing objects destroyed by corrosion. Consequently, the development of methods for protecting metal surfaces from corrosion constitutes a very active area of industrial research. In this section, we describe some of the chemical and electrochemical processes responsible for corrosion. We also examine the chemical basis for some common methods for preventing corrosion and treating corroded metals.
Corrosion is a REDOX process.
Under ambient conditions, the oxidation of most metals is thermodynamically spontaneous, with the notable exception of gold and platinum. Hence it is actually somewhat surprising that any metals are useful at all in Earth’s moist, oxygen-rich atmosphere. Some metals, however, are resistant to corrosion for kinetic reasons. For example, aluminum in soft-drink cans and airplanes is protected by a thin coating of metal oxide that forms on the surface of the metal and acts as an impenetrable barrier that prevents further destruction. Aluminum cans also have a thin plastic layer to prevent reaction of the oxide with acid in the soft drink. Chromium, magnesium, and nickel also form protective oxide films. Stainless steels are remarkably resistant to corrosion because they usually contain a significant proportion of chromium, nickel, or both.
In contrast to these metals, when iron corrodes, it forms a red-brown hydrated metal oxide ($\ce{Fe2O3 \cdot xH2O}$), commonly known as rust, that does not provide a tight protective film (Figure $1$). Instead, the rust continually flakes off to expose a fresh metal surface vulnerable to reaction with oxygen and water. Because both oxygen and water are required for rust to form, an iron nail immersed in deoxygenated water will not rust—even over a period of several weeks. Similarly, a nail immersed in an organic solvent such as kerosene or mineral oil will not rust because of the absence of water even if the solvent is saturated with oxygen.
In the corrosion process, iron metal acts as the anode in a galvanic cell and is oxidized to Fe2+; oxygen is reduced to water at the cathode. The relevant reactions are as follows:
• at cathode: $\ce{O2(g) + 4H^{+}(aq) + 4e^{−} -> 2H2O(l)} \nonumber$ with $E^o_{SRP}=1.23\; V$.
• at anode: $\ce{Fe(s) -> Fe^{2+}(aq) + 2e^{−}}\nonumber$ with $E^o_{SRP} = −0.45\; V$.
• overall: $\ce{2Fe(s) + O2(g) + 4H^{+}(aq) -> 2Fe^{2+}(aq) + 2H2O(l)} \label{Eq3}$ with $E^o_{cell} = 1.68\; V$.
The $\ce{Fe^{2+}}$ ions produced in the initial reaction are then oxidized by atmospheric oxygen to produce the insoluble hydrated oxide containing $\ce{Fe^{3+}}$, as represented in the following equation:
$\ce{4Fe^{2+}(aq) + O2(g) + (2 + 4x)H2O \rightarrow 2Fe2O3 \cdot xH2O + 4H^{+}(aq)} \label{Eq4}$
The sign and magnitude of $E^o_{cell}$ for the corrosion process (Equation $\ref{Eq3}$) indicate that there is a strong driving force for the oxidation of iron by O2 under standard conditions (1 M H+). Under neutral conditions, the driving force is somewhat less but still appreciable (E = 1.25 V at pH 7.0). Normally, the reaction of atmospheric CO2 with water to form H+ and HCO3 provides a low enough pH to enhance the reaction rate, as does acid rain. Automobile manufacturers spend a great deal of time and money developing paints that adhere tightly to the car’s metal surface to prevent oxygenated water, acid, and salt from coming into contact with the underlying metal. Unfortunately, even the best paint is subject to scratching or denting, and the electrochemical nature of the corrosion process means that two scratches relatively remote from each other can operate together as anode and cathode, leading to sudden mechanical failure (Figure $2$).
Prophylactic Protection
One of the most common techniques used to prevent the corrosion of iron is applying a protective coating of another metal that is more difficult to oxidize. Faucets and some external parts of automobiles, for example, are often coated with a thin layer of chromium using an electrolytic process. With the increased use of polymeric materials in cars, however, the use of chrome-plated steel has diminished in recent years. Similarly, the “tin cans” that hold soups and other foods are actually consist of steel container that is coated with a thin layer of tin. While neither chromium nor tin metals are intrinsically resistant to corrosion, they both form protective oxide coatings that hinder access of oxygen and water to the underlying steel (iron alloy).
As with a protective paint, scratching a protective metal coating will allow corrosion to occur. In this case, however, the presence of the second metal can actually increase the rate of corrosion. The values of the standard electrode potentials for $\ce{Sn^{2+}}$ (E° = −0.14 V) and Fe2+ (E° = −0.45 V) in Table P2 show that $\ce{Fe}$ is more easily oxidized than $\ce{Sn}$. As a result, the more corrosion-resistant metal (in this case, tin) accelerates the corrosion of iron by acting as the cathode and providing a large surface area for the reduction of oxygen (Figure $3$). This process is seen in some older homes where copper and iron pipes have been directly connected to each other. The less easily oxidized copper acts as the cathode, causing iron to dissolve rapidly near the connection and occasionally resulting in a catastrophic plumbing failure.
Cathodic Protection
One way to avoid these problems is to use a more easily oxidized metal to protect iron from corrosion. In this approach, called cathodic protection, a more reactive metal such as $\ce{Zn}$ (E° = −0.76 V for $\ce{Zn^{2+} + 2e^{−} -> Zn}$) becomes the anode, and iron becomes the cathode. This prevents oxidation of the iron and protects the iron object from corrosion. The reactions that occur under these conditions are as follows:
$\underbrace{O_{2(g)} + 4e^− + 4H^+_{(aq)} \rightarrow 2H_2O_{(l)} }_{\text{reduction at cathode}}\label{Eq5}$
$\underbrace{Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^−}_{\text{oxidation at anode}} \label{Eq6}$
$\underbrace{ 2Zn_{(s)} + O_{2(g)} + 4H^+_{(aq)} \rightarrow 2Zn^{2+}_{(aq)} + 2H_2O_{(l)} }_{\text{overall}}\label{Eq7}$
The more reactive metal reacts with oxygen and will eventually dissolve, “sacrificing” itself to protect the iron object. Cathodic protection is the principle underlying galvanized steel, which is steel protected by a thin layer of zinc. Galvanized steel is used in objects ranging from nails to garbage cans.
In a similar strategy, sacrificial electrodes using magnesium, for example, are used to protect underground tanks or pipes (Figure $4$). Replacing the sacrificial electrodes is more cost-effective than replacing the iron objects they are protecting.
Example $1$
Suppose an old wooden sailboat, held together with iron screws, has a bronze propeller (recall that bronze is an alloy of copper containing about 7%–10% tin).
1. If the boat is immersed in seawater, what corrosion reaction will occur? What is $E^o°_{cell}$?
2. How could you prevent this corrosion from occurring?
Given: identity of metals
Asked for: corrosion reaction, $E^o°_{cell}$, and preventive measures
Strategy:
1. Write the reactions that occur at the anode and the cathode. From these, write the overall cell reaction and calculate $E^o°_{cell}$.
2. Based on the relative redox activity of various substances, suggest possible preventive measures.
Solution
1. A According to Table P2, both copper and tin are less active metals than iron (i.e., they have higher positive values of $E^o°_{cell}$ than iron). Thus if tin or copper is brought into electrical contact by seawater with iron in the presence of oxygen, corrosion will occur. We therefore anticipate that the bronze propeller will act as the cathode at which $\ce{O2}$ is reduced, and the iron screws will act as anodes at which iron dissolves:
\begin{align*} & \textrm{cathode:} & & \mathrm{O_2(s)} + \mathrm{4H^+(aq)}+\mathrm{4e^-}\rightarrow \mathrm{2H_2O(l)} & & E^\circ_{\textrm{cathode}} =\textrm{1.23 V} \ & \textrm{anode:} & & \mathrm{Fe(s)} \rightarrow \mathrm{Fe^{2+}} +\mathrm{2e^-} & & E^\circ_{\textrm{anode}} =-\textrm{0.45 V} \ & \textrm{overall:} & & \mathrm{2Fe(s)}+\mathrm{O_2(g)}+\mathrm{4H^+(aq)} \rightarrow \mathrm{2Fe^{2+}(aq)} +\mathrm{2H_2O(l)} & & E^\circ_{\textrm{overall}} =\textrm{1.68 V} \end{align*} \nonumber
Over time, the iron screws will dissolve, and the boat will fall apart.
1. B Possible ways to prevent corrosion, in order of decreasing cost and inconvenience, are as follows: disassembling the boat and rebuilding it with bronze screws; removing the boat from the water and storing it in a dry place; or attaching an inexpensive piece of zinc metal to the propeller shaft to act as a sacrificial electrode and replacing it once or twice a year. Because zinc is a more active metal than iron, it will act as the sacrificial anode in the electrochemical cell and dissolve (Equation $\ref{Eq7}$).
Exercise $1$
Suppose the water pipes leading into your house are made of lead, while the rest of the plumbing in your house is iron. To eliminate the possibility of lead poisoning, you call a plumber to replace the lead pipes. He quotes you a very low price if he can use up his existing supply of copper pipe to do the job.
1. Do you accept his proposal?
2. What else should you have the plumber do while at your home?
Answer a
Not unless you plan to sell the house very soon because the $\ce{Cu/Fe}$ pipe joints will lead to rapid corrosion.
Answer b
Any existing $\ce{Pb/Fe}$ joints should be examined carefully for corrosion of the iron pipes due to the $\ce{Pb–Fe}$ junction; the less active $\ce{Pb}$ will have served as the cathode for the reduction of $\ce{O2}$, promoting oxidation of the more active $\ce{Fe}$ nearby.
Summary
Corrosion is a galvanic process that can be prevented using cathodic protection. The deterioration of metals through oxidation is a galvanic process called corrosion. Protective coatings consist of a second metal that is more difficult to oxidize than the metal being protected. Alternatively, a more easily oxidized metal can be applied to a metal surface, thus providing cathodic protection of the surface. A thin layer of zinc protects galvanized steel. Sacrificial electrodes can also be attached to an object to protect it. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/20%3A_Electrochemistry/20.08%3A_Corrosion.txt |
Learning Objectives
• To understand electrolysis and describe it quantitatively.
In this chapter, we have described various galvanic cells in which a spontaneous chemical reaction is used to generate electrical energy. In an electrolytic cell, however, the opposite process, called electrolysis, occurs: an external voltage is applied to drive a nonspontaneous reaction. In this section, we look at how electrolytic cells are constructed and explore some of their many commercial applications.
Electrolytic Cells
If we construct an electrochemical cell in which one electrode is copper metal immersed in a 1 M Cu2+ solution and the other electrode is cadmium metal immersed in a $\,1\; M\, Cd^{2+}$ solution and then close the circuit, the potential difference between the two compartments will be 0.74 V. The cadmium electrode will begin to dissolve (Cd is oxidized to Cd2+) and is the anode, while metallic copper will be deposited on the copper electrode (Cu2+ is reduced to Cu), which is the cathode (Figure $\PageIndex{1a}$).
The overall reaction is as follows:
$\ce{Cd (s) + Cu^{2+} (aq) \rightarrow Cd^{2+} (aq) + Cu (s)} \nonumber$
with $E°_{cell} = 0.74\; V$
This reaction is thermodynamically spontaneous as written ($ΔG^o < 0$):
\begin{align*} \Delta G^\circ &=-nFE^\circ_\textrm{cell} \[4pt] &=-(\textrm{2 mol e}^-)[\mathrm{96,485\;J/(V\cdot mol)}](\mathrm{0.74\;V}) \[4pt] &=-\textrm{140 kJ (per mole Cd)} \end{align*} \nonumber
In this direction, the system is acting as a galvanic cell.
In an electrolytic cell, an external voltage is applied to drive a nonspontaneous reaction.
The reverse reaction, the reduction of Cd2+ by Cu, is thermodynamically nonspontaneous and will occur only with an input of 140 kJ. We can force the reaction to proceed in the reverse direction by applying an electrical potential greater than 0.74 V from an external power supply. The applied voltage forces electrons through the circuit in the reverse direction, converting a galvanic cell to an electrolytic cell. Thus the copper electrode is now the anode (Cu is oxidized), and the cadmium electrode is now the cathode (Cd2+ is reduced) (Figure $\PageIndex{1b}$). The signs of the cathode and the anode have switched to reflect the flow of electrons in the circuit. The half-reactions that occur at the cathode and the anode are as follows:
• half-reaction at the cathode:
$\ce{Cd^{2+}(aq) + 2e^{−} \rightarrow Cd(s)}\label{20.9.3}$
with $E^°_{cathode} = −0.40 \, V$
• half-reaction at the anode:
$\ce{Cu(s) \rightarrow Cu^{2+}(aq) + 2e^{−}} \label{20.9.4}$
with $E^°_{anode} = 0.34 \, V$
• Overall Reaction:
$\ce{Cd^{2+}(aq) + Cu(s) \rightarrow Cd(s) + Cu^{2+}(aq) } \label{20.9.5}$
with $E^°_{cell} = −0.74 \: V$
Because $E^°_{cell} < 0$, the overall reaction—the reduction of $Cd^{2+}$ by $Cu$—clearly cannot occur spontaneously and proceeds only when sufficient electrical energy is applied. The differences between galvanic and electrolytic cells are summarized in Table $1$.
Table $1$: Comparison of Galvanic and Electrolytic Cells
Property Galvanic Cell Electrolytic Cell
ΔG < 0 > 0
Ecell > 0 < 0
Electrode Process
anode oxidation oxidation
cathode reduction reduction
Sign of Electrode
anode +
cathode +
Electrolytic Reactions
At sufficiently high temperatures, ionic solids melt to form liquids that conduct electricity extremely well due to the high concentrations of ions. If two inert electrodes are inserted into molten $\ce{NaCl}$, for example, and an electrical potential is applied, $\ce{Cl^{-}}$ is oxidized at the anode, and $\ce{Na^{+}}$ is reduced at the cathode. The overall reaction is as follows:
$\ce{ 2NaCl (l) \rightarrow 2Na(l) + Cl2(g)} \label{20.9.6}$
This is the reverse of the formation of $\ce{NaCl}$ from its elements. The product of the reduction reaction is liquid sodium because the melting point of sodium metal is 97.8°C, well below that of $\ce{NaCl}$ (801°C). Approximately 20,000 tons of sodium metal are produced commercially in the United States each year by the electrolysis of molten $\ce{NaCl}$ in a Downs cell (Figure $2$). In this specialized cell, $\ce{CaCl2}$ (melting point = 772°C) is first added to the $\ce{NaCl}$ to lower the melting point of the mixture to about 600°C, thereby lowering operating costs.
Similarly, in the Hall–Heroult process used to produce aluminum commercially, a molten mixture of about 5% aluminum oxide (Al2O3; melting point = 2054°C) and 95% cryolite (Na3AlF6; melting point = 1012°C) is electrolyzed at about 1000°C, producing molten aluminum at the cathode and CO2 gas at the carbon anode. The overall reaction is as follows:
$\ce{2Al2O3(l) + 3C(s) -> 4Al(l) + 3CO2(g)} \label{20.9.7}$
Oxide ions react with oxidized carbon at the anode, producing CO2(g).
There are two important points to make about these two commercial processes and about the electrolysis of molten salts in general.
1. The electrode potentials for molten salts are likely to be very different from the standard cell potentials listed in Table P2, which are compiled for the reduction of the hydrated ions in aqueous solutions under standard conditions.
2. Using a mixed salt system means there is a possibility of competition between different electrolytic reactions. When a mixture of NaCl and CaCl2 is electrolyzed, Cl is oxidized because it is the only anion present, but either Na+ or Ca2+ can be reduced. Conversely, in the Hall–Heroult process, only one cation is present that can be reduced (Al3+), but there are three species that can be oxidized: C, O2−, and F.
In the Hall–Heroult process, C is oxidized instead of O2− or F because oxygen and fluorine are more electronegative than carbon, which means that C is a weaker oxidant than either O2 or F2. Similarly, in the Downs cell, we might expect electrolysis of a NaCl/CaCl2 mixture to produce calcium rather than sodium because Na is slightly less electronegative than Ca (χ = 0.93 versus 1.00, respectively), making Na easier to oxidize and, conversely, Na+ more difficult to reduce. In fact, the reduction of Na+ to Na is the observed reaction. In cases where the electronegativities of two species are similar, other factors, such as the formation of complex ions, become important and may determine the outcome.
Example $1$
If a molten mixture of MgCl2 and KBr is electrolyzed, what products will form at the cathode and the anode, respectively?
Given: identity of salts
Asked for: electrolysis products
Strategy:
1. List all the possible reduction and oxidation products. Based on the electronegativity values shown in Figure 7.5, determine which species will be reduced and which species will be oxidized.
2. Identify the products that will form at each electrode.
Solution
A The possible reduction products are Mg and K, and the possible oxidation products are Cl2 and Br2. Because Mg is more electronegative than K (χ = 1.31 versus 0.82), it is likely that Mg will be reduced rather than K. Because Cl is more electronegative than Br (3.16 versus 2.96), Cl2 is a stronger oxidant than Br2.
B Electrolysis will therefore produce Br2 at the anode and Mg at the cathode.
Exercise $1$
Predict the products if a molten mixture of AlBr3 and LiF is electrolyzed.
Answer
Br2 and Al
Electrolysis can also be used to drive the thermodynamically nonspontaneous decomposition of water into its constituent elements: H2 and O2. However, because pure water is a very poor electrical conductor, a small amount of an ionic solute (such as H2SO4 or Na2SO4) must first be added to increase its electrical conductivity. Inserting inert electrodes into the solution and applying a voltage between them will result in the rapid evolution of bubbles of H2 and O2 (Figure $3$).
The reactions that occur are as follows:
• cathode: $2H^+_{(aq)} + 2e^− \rightarrow H_{2(g)} \;\;\; E^°_{cathode} = 0 V \label{20.9.8}$
• anode: $2H_2O_{(l)} → O_{2(g)} + 4H^+_{(aq)} + 4e^−\;\;\; E^°_{anode} = 1.23\; V \label{20.9.9}$
• overall: $2H_2O_{(l)} → O_{2(g)} + 2H_{2(g)}\;\;\; E^°_{cell} = −1.23 \;V \label{20.9.10}$
For a system that contains an electrolyte such as Na2SO4, which has a negligible effect on the ionization equilibrium of liquid water, the pH of the solution will be 7.00 and [H+] = [OH] = 1.0 × 10−7. Assuming that $P_\mathrm{O_2}$ = $P_\mathrm{H_2}$ = 1 atm, we can use the standard potentials to calculate E for the overall reaction:
\begin{align}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log(P_\mathrm{O_2}P^2_\mathrm{H_2}) \ &=-\textrm{1.23 V}-\left(\dfrac{\textrm{0.0591 V}}{4}\right)\log(1)=-\textrm{1.23 V}\end{align} \label{20.9.11}
Thus Ecell is −1.23 V, which is the value of E°cell if the reaction is carried out in the presence of 1 M H+ rather than at pH 7.0.
In practice, a voltage about 0.4–0.6 V greater than the calculated value is needed to electrolyze water. This added voltage, called an overvoltage, represents the additional driving force required to overcome barriers such as the large activation energy for the formation of a gas at a metal surface. Overvoltages are needed in all electrolytic processes, which explain why, for example, approximately 14 V must be applied to recharge the 12 V battery in your car.
In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation. The p-block metals and most of the transition metals are in this category, but metals in high oxidation states, which form oxoanions, cannot be reduced to the metal by simple electrolysis. Active metals, such as aluminum and those of groups 1 and 2, react so readily with water that they can be prepared only by the electrolysis of molten salts. Similarly, any nonmetallic element that does not readily oxidize water to O2 can be prepared by the electrolytic oxidation of an aqueous solution that contains an appropriate anion. In practice, among the nonmetals, only F2 cannot be prepared using this method. Oxoanions of nonmetals in their highest oxidation states, such as NO3, SO42, PO43, are usually difficult to reduce electrochemically and usually behave like spectator ions that remain in solution during electrolysis.
In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation.
Electroplating
In a process called electroplating, a layer of a second metal is deposited on the metal electrode that acts as the cathode during electrolysis. Electroplating is used to enhance the appearance of metal objects and protect them from corrosion. Examples of electroplating include the chromium layer found on many bathroom fixtures or (in earlier days) on the bumpers and hubcaps of cars, as well as the thin layer of precious metal that coats silver-plated dinnerware or jewelry. In all cases, the basic concept is the same. A schematic view of an apparatus for electroplating silverware and a photograph of a commercial electroplating cell are shown in Figure $4$.
The half-reactions in electroplating a fork, for example, with silver are as follows:
• cathode (fork): $\ce{Ag^{+}(aq) + e^{−} -> Ag(s)} \quad E°_{cathode} = 0.80 V\ \nonumber$
• anode (silver bar): $\ce{Ag(s) -> Ag^{+}(aq) + e^{-}} \quad E°_{anode} = 0.80 V \nonumber$
The overall reaction is the transfer of silver metal from one electrode (a silver bar acting as the anode) to another (a fork acting as the cathode). Because $E^o_{cell} = 0\, V$, it takes only a small applied voltage to drive the electroplating process. In practice, various other substances may be added to the plating solution to control its electrical conductivity and regulate the concentration of free metal ions, thus ensuring a smooth, even coating.
Quantitative Considerations
If we know the stoichiometry of an electrolysis reaction, the amount of current passed, and the length of time, we can calculate the amount of material consumed or produced in a reaction. Conversely, we can use stoichiometry to determine the combination of current and time needed to produce a given amount of material.
The quantity of material that is oxidized or reduced at an electrode during an electrochemical reaction is determined by the stoichiometry of the reaction and the amount of charge that is transferred. For example, in the reaction
$\ce{Ag^{+}(aq) + e^{−} → Ag(s)} \nonumber$
1 mol of electrons reduces 1 mol of $\ce{Ag^{+}}$ to $\ce{Ag}$ metal. In contrast, in the reaction
$\ce{Cu^{2+}(aq) + 2e^{−} → Cu(s)} \nonumber$
1 mol of electrons reduces only 0.5 mol of $\ce{Cu^{2+}}$ to $\ce{Cu}$ metal. Recall that the charge on 1 mol of electrons is 1 faraday (1 F), which is equal to 96,485 C. We can therefore calculate the number of moles of electrons transferred when a known current is passed through a cell for a given period of time. The total charge ($q$ in coulombs) transferred is the product of the current ($I$ in amperes) and the time ($t$, in seconds):
$q = I \times t \label{20.9.14}$
The stoichiometry of the reaction and the total charge transferred enable us to calculate the amount of product formed during an electrolysis reaction or the amount of metal deposited in an electroplating process.
For example, if a current of 0.60 A passes through an aqueous solution of $\ce{CuSO4}$ for 6.0 min, the total number of coulombs of charge that passes through the cell is as follows:
\begin{align*} q &= \textrm{(0.60 A)(6.0 min)(60 s/min)} \[4pt] &=\mathrm{220\;A\cdot s} \[4pt] &=\textrm{220 C} \end{align*} \nonumber
The number of moles of electrons transferred to $\ce{Cu^{2+}}$ is therefore
\begin{align*} \textrm{moles e}^- &=\dfrac{\textrm{220 C}}{\textrm{96,485 C/mol}} \[4pt] &=2.3\times10^{-3}\textrm{ mol e}^- \end{align*} \nonumber
Because two electrons are required to reduce a single Cu2+ ion, the total number of moles of Cu produced is half the number of moles of electrons transferred, or 1.2 × 10−3 mol. This corresponds to 76 mg of Cu. In commercial electrorefining processes, much higher currents (greater than or equal to 50,000 A) are used, corresponding to approximately 0.5 F/s, and reaction times are on the order of 3–4 weeks.
Example $2$
A silver-plated spoon typically contains about 2.00 g of Ag. If 12.0 h are required to achieve the desired thickness of the Ag coating, what is the average current per spoon that must flow during the electroplating process, assuming an efficiency of 100%?
Given: mass of metal, time, and efficiency
Asked for: current required
Strategy:
1. Calculate the number of moles of metal corresponding to the given mass transferred.
2. Write the reaction and determine the number of moles of electrons required for the electroplating process.
3. Use the definition of the faraday to calculate the number of coulombs required. Then convert coulombs to current in amperes.
Solution
A We must first determine the number of moles of Ag corresponding to 2.00 g of Ag:
$\textrm{moles Ag}=\dfrac{\textrm{2.00 g}}{\textrm{107.868 g/mol}}=1.85\times10^{-2}\textrm{ mol Ag}$
B The reduction reaction is Ag+(aq) + e → Ag(s), so 1 mol of electrons produces 1 mol of silver.
C Using the definition of the faraday,
coulombs = (1.85 × 102mol e)(96,485 C/mol e) = 1.78 × 103 C / mole
The current in amperes needed to deliver this amount of charge in 12.0 h is therefore
\begin{align*}\textrm{amperes} &=\dfrac{1.78\times10^3\textrm{ C}}{(\textrm{12.0 h})(\textrm{60 min/h})(\textrm{60 s/min})}\ & =4.12\times10^{-2}\textrm{ C/s}=4.12\times10^{-2}\textrm{ A}\end{align*} \nonumber
Because the electroplating process is usually much less than 100% efficient (typical values are closer to 30%), the actual current necessary is greater than 0.1 A.
Exercise $2$
A typical aluminum soft-drink can weighs about 29 g. How much time is needed to produce this amount of Al(s) in the Hall–Heroult process, using a current of 15 A to reduce a molten Al2O3/Na3AlF6 mixture?
Answer
5.8 h
Electroplating: Electroplating(opens in new window) [youtu.be]
Summary
In electrolysis, an external voltage is applied to drive a nonspontaneous reaction. The quantity of material oxidized or reduced can be calculated from the stoichiometry of the reaction and the amount of charge transferred. Relationship of charge, current and time:
$q = I \times t \nonumber$
In electrolysis, an external voltage is applied to drive a nonspontaneous reaction. Electrolysis can also be used to produce H2 and O2 from water. In practice, an additional voltage, called an overvoltage, must be applied to overcome factors such as a large activation energy and a junction potential. Electroplating is the process by which a second metal is deposited on a metal surface, thereby enhancing an object’s appearance or providing protection from corrosion. The amount of material consumed or produced in a reaction can be calculated from the stoichiometry of an electrolysis reaction, the amount of current passed, and the duration of the electrolytic reaction.
20.E: Electrochemistry (Exercises)
Questions moved to ADAPT. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/20%3A_Electrochemistry/20.09%3A_Electrolysis.txt |
Until now, you have studied chemical processes in which atoms share or transfer electrons to form new compounds, leaving the atomic nuclei largely unaffected. In this chapter, we examine some properties of the atomic nucleus and the changes that can occur in atomic nuclei. Nuclear reactions differ from other chemical processes in one critical way: in a nuclear reaction, the identities of the elements change. In addition, nuclear reactions are often accompanied by the release of enormous amounts of energy, as much as a billion times more than the energy released by chemical reactions. Moreover, the yields and rates of a nuclear reaction are generally unaffected by changes in temperature, pressure, or the presence of a catalyst.
We begin by examining the structure of the atomic nucleus and the factors that determine whether a particular nucleus is stable or decays spontaneously to another element. We then discuss the major kinds of nuclear decay reactions, as well as the properties and uses of the radiation emitted when nuclei decay. You will learn how radioactive emissions can be used to study the mechanisms of chemical reactions and biological processes and how to calculate the amount of energy released during a nuclear reaction. You will also discover why houses are tested for radon gas, how radiation is used to probe organs such as the brain, and how the energy from nuclear reactions can be harnessed to produce electricity. Last, we explore the nuclear chemistry that takes place in stars, and we describe the role that stars play in producing most of the elements in the universe.
• 21.1: Radioactivity
Nuclei can undergo reactions that change their number of protons, number of neutrons, or energy state. Many different particles can be involved and the most common are protons, neutrons, positrons, alpha (α) particles, beta (β) particles (high-energy electrons), and gamma (γ) rays (which compose high-energy electromagnetic radiation). As with chemical reactions, nuclear reactions are always balanced. When a nuclear reaction occurs, the total mass (number) and the total charge remain unchanged.
• 21.2: Patterns of Nuclear Stability
Protons and neutrons are called nucleons and a nuclide is an atom with a specific number nucleons. Unstable nuclei decay spontaneously are radioactive and its emissions are called radioactivity. Nuclei are bound by the strong nuclear force. Stable nuclei generally have even numbers of protons and neutrons with a ratio of at least 1. Nuclei that contain magic numbers of protons and neutrons are often especially stable including superheavy elements, with atomic numbers near 126.
• 21.3: Nuclear Transmutations
Hydrogen and helium are the most abundant elements in the universe. Heavier elements are formed in the interior of stars via multiple neutron-capture events. Successive fusion reactions of helium nuclei at higher temperatures create elements with even numbers of protons and neutrons up to Mg and then up to Ca. Eventually, the elements up to Fe-56 and Ni-58 are formed by exchange processes at even higher temperatures. Heavier elements can only be made by the explosion of a supernova.
• 21.4: Rates of Radioactive Decay
Unstable nuclei undergo spontaneous radioactive decay. The most common types of radioactivity are α decay, β decay, γ emission, positron emission, and electron capture. Nuclear reactions also often involve γ rays, and some nuclei decay by electron capture. Each of these modes of decay leads to the formation of a new stable nuclei sometimes via multiple decays before ending in a stable isotope. All nuclear decay processes follow first-order kinetics and each radioisotope has its own half-life.
• 21.5: Energy Changes in Nuclear Reactions
Unlike a chemical reaction, a nuclear reaction results in a significant change in mass and an associated change of energy, as described by Einstein’s equation. Nuclear reactions are accompanied by large changes in energy, which result in detectable changes in mass. The experimentally determined mass of an atom is always less than the sum of the masses of the component particles (protons, neutrons, and electrons) by an amount called the mass defect that corresponds to the nuclear binding energy.
• 21.6: Nuclear Fission
Many heavier elements with smaller binding energies per nucleon can decompose into more stable elements that have intermediate mass numbers and larger binding energies per nucleon. Sometimes neutrons are also produced. This decomposition is called fission, the breaking of a large nucleus into smaller pieces. The breaking is rather random with the formation of a large number of different products. Fission usually does not occur naturally, but is induced by bombardment with neutrons.
• 21.7: Nuclear Fusion
The process of converting very light nuclei into heavier nuclei is also accompanied by the conversion of mass into large amounts of energy, a process called fusion. The principal source of energy in the sun is a net fusion reaction in which four hydrogen nuclei fuse and produce one helium nucleus and two positrons. This is a net reaction of a more complicated series of events.
• 21.8: Biological Effects of Radiation
The effects of radiation on matter depend on the energy of the radiation. Nonionizing radiation is relatively low in energy, and the energy is transferred to matter in the form of heat. Ionizing radiation is relatively high in energy, and when it collides with an atom, it can completely remove an electron to form a positively charged ion that can damage biological tissues.
• 21.E: Exercises
Problems and select solutions to chapter 21 of the Brown et al. textmap.
• 21.S: Nuclear Chemistry (Summary)
Summary of Chapter 21 of the Brown et al. textmap.
21: Nuclear Chemistry
Learning Objectives
• Write and balance nuclear equations
• To know the different kinds of radioactive decay.
• To balance a nuclear reaction.
Nuclear chemistry is the study of reactions that involve changes in nuclear structure. The chapter on atoms, molecules, and ions introduced the basic idea of nuclear structure, that the nucleus of an atom is composed of protons and, with the exception of $\ce{^1_1H}$, neutrons. Recall that the number of protons in the nucleus is called the atomic number ($Z$) of the element, and the sum of the number of protons and the number of neutrons is the mass number ($A$). Atoms with the same atomic number but different mass numbers are isotopes of the same element. When referring to a single type of nucleus, we often use the term nuclide and identify it by the notation:
$\large \ce{^{A}_{Z}X} \label{Eq1a}$
where
• $X$ is the symbol for the element,
• $A$ is the mass number, and
• $Z$ is the atomic number.
Often a nuclide is referenced by the name of the element followed by a hyphen and the mass number. For example, $\ce{^{14}_6C}$ is called “carbon-14.”
Protons and neutrons, collectively called nucleons, are packed together tightly in a nucleus. With a radius of about 10−15 meters, a nucleus is quite small compared to the radius of the entire atom, which is about 10−10 meters. Nuclei are extremely dense compared to bulk matter, averaging $1.8 \times 10^{14}$ grams per cubic centimeter. For example, water has a density of 1 gram per cubic centimeter, and iridium, one of the densest elements known, has a density of 22.6 g/cm3. If the earth’s density were equal to the average nuclear density, the earth’s radius would be only about 200 meters (earth’s actual radius is approximately $6.4 \times 10^6$ meters, 30,000 times larger).
Changes of nuclei that result in changes in their atomic numbers, mass numbers, or energy states are nuclear reactions. To describe a nuclear reaction, we use an equation that identifies the nuclides involved in the reaction, their mass numbers and atomic numbers, and the other particles involved in the reaction.
Nuclear Equations
A balanced chemical reaction equation reflects the fact that during a chemical reaction, bonds break and form, and atoms are rearranged, but the total numbers of atoms of each element are conserved and do not change. A balanced nuclear reaction equation indicates that there is a rearrangement during a nuclear reaction, but of subatomic particles rather than atoms. Nuclear reactions also follow conservation laws, and they are balanced in two ways:
1. The sum of the mass numbers of the reactants equals the sum of the mass numbers of the products.
2. The sum of the charges of the reactants equals the sum of the charges of the products.
If the atomic number and the mass number of all but one of the particles in a nuclear reaction are known, we can identify the particle by balancing the reaction. For instance, we could determine that $\ce{^{17}_8O}$ is a product of the nuclear reaction of $\ce{^{14}_7N}$ and $\ce{^4_2He}$ if we knew that a proton, $\ce{^1_1H}$, was one of the two products. Example $1$ shows how we can identify a nuclide by balancing the nuclear reaction.
Example $1$: Balancing Equations for Nuclear Reactions
The reaction of an α particle with magnesium-25 $(\ce{^{25}_{12}Mg})$ produces a proton and a nuclide of another element. Identify the new nuclide produced.
Solution
The nuclear reaction can be written as:
$\ce{^{25}_{12}Mg + ^4_2He \rightarrow ^1_1H + ^{A}_{Z}X} \nonumber$
where
• $\ce A$ is the mass number and
• $\ce Z$ is the atomic number of the new nuclide, $\ce X$.
Because the sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products:
$\mathrm{25+4=A+1} \nonumber$
so
$\mathrm{A=28} \nonumber$
Similarly, the charges must balance, so:
$\mathrm{12+2=Z+1} \nonumber$
so
$\mathrm{Z=13} \nonumber$
Check the periodic table: The element with nuclear charge = +13 is aluminum. Thus, the product is $\ce{^{28}_{13}Al}$.
Exercise $1$
The nuclide $\ce{^{125}_{53}I}$ combines with an electron and produces a new nucleus and no other massive particles. What is the equation for this reaction?
Answer
$\ce{^{125}_{53}I + ^0_{−1}e \rightarrow ^{125}_{52}Te} \nonumber$
The two general kinds of nuclear reactions are nuclear decay reactions and nuclear transmutation reactions. In a nuclear decay reaction, also called radioactive decay, an unstable nucleus emits radiation and is transformed into the nucleus of one or more other elements. The resulting daughter nuclei have a lower mass and are lower in energy (more stable) than the parent nucleus that decayed. In contrast, in a nuclear transmutation reaction, a nucleus reacts with a subatomic particle or another nucleus to form a product nucleus that is more massive than the starting material. As we shall see, nuclear decay reactions occur spontaneously under all conditions, but nuclear transmutation reactions occur only under very special conditions, such as the collision of a beam of highly energetic particles with a target nucleus or in the interior of stars. We begin this section by considering the different classes of radioactive nuclei, along with their characteristic nuclear decay reactions and the radiation they emit.
Nuclear decay reactions occur spontaneously under all conditions, whereas nuclear transmutation reactions are induced.
Nuclear Decay Reactions
Just as we use the number and type of atoms present to balance a chemical equation, we can use the number and type of nucleons present to write a balanced nuclear equation for a nuclear decay reaction. This procedure also allows us to predict the identity of either the parent or the daughter nucleus if the identity of only one is known. Regardless of the mode of decay, the total number of nucleons is conserved in all nuclear reactions.
To describe nuclear decay reactions, chemists have extended the $^A _Z \textrm{X}$ notation for nuclides to include radioactive emissions. Table $1$ lists the name and symbol for each type of emitted radiation. The most notable addition is the positron, a particle that has the same mass as an electron but a positive charge rather than a negative charge.
Table $1$: Nuclear Decay Emissions and Their Symbols
Identity Symbol Charge Mass (amu)
helium nucleus $^4_2\alpha$ +2 4.001506
electron $^0_{-1}\beta$ or $\beta ^-$ −1 0.000549
photon $_0^0\gamma$
neutron $^1_0\textrm n$ 0 1.008665
proton $^1_1\textrm p$ +1 1.007276
positron $^0_{+1}\beta$ or $\beta ^+$ +1 0.000549
Like the notation used to indicate isotopes, the upper left superscript in the symbol for a particle gives the mass number, which is the total number of protons and neutrons. For a proton or a neutron, A = 1. Because neither an electron nor a positron contains protons or neutrons, its mass number is 0. The numbers should not be taken literally, however, as meaning that these particles have zero mass; ejection of a beta particle (an electron) simply has a negligible effect on the mass of a nucleus.
Similarly, the lower left subscript gives the charge of the particle. Because protons carry a positive charge, Z = +1 for a proton. In contrast, a neutron contains no protons and is electrically neutral, so Z = 0. In the case of an electron, Z = −1, and for a positron, Z = +1. Because γ rays are high-energy photons, both A and Z are 0. In some cases, two different symbols are used for particles that are identical but produced in different ways. For example, the symbol $^0_{-1}\textrm e$, which is usually simplified to e, represents a free electron or an electron associated with an atom, whereas the symbol $^0_{-1}\beta$, which is often simplified to β, denotes an electron that originates from within the nucleus, which is a β particle. Similarly, $^4_{2}\textrm{He}^{2+}$ refers to the nucleus of a helium atom, and $^4_{2}\alpha$ denotes an identical particle that has been ejected from a heavier nucleus.
There are six fundamentally different kinds of nuclear decay reactions, and each releases a different kind of particle or energy. The essential features of each reaction are shown in Figure $1$. The most common are alpha and beta decay and gamma emission, but the others are essential to an understanding of nuclear decay reactions.
Alpha $\alpha$ Decay
Many nuclei with mass numbers greater than 200 undergo alpha (α) decay, which results in the emission of a helium-4 nucleus as an alpha (α) particle, $^4_{2}\alpha$. The general reaction is as follows:
$\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A-4}_{Z-2} \textrm X'}+\underset{\textrm{alpha}\ \textrm{particle}}{^4_2 \alpha}\label{Eq1}$
The daughter nuclide contains two fewer protons and two fewer neutrons than the parent. Thus α-particle emission produces a daughter nucleus with a mass number A − 4 and a nuclear charge Z − 2 compared to the parent nucleus. Radium-226, for example, undergoes alpha decay to form radon-222:
$^{226}_{88}\textrm{Ra}\rightarrow ^{222}_{86}\textrm{Rn}+^{4}_{2}\alpha\label{Eq2}$
Because nucleons are conserved in this and all other nuclear reactions, the sum of the mass numbers of the products, 222 + 4 = 226, equals the mass number of the parent. Similarly, the sum of the atomic numbers of the products, 86 + 2 = 88, equals the atomic number of the parent. Thus the nuclear equation is balanced.
Just as the total number of atoms is conserved in a chemical reaction, the total number of nucleons is conserved in a nuclear reaction.
Beta $\beta^-$ Decay
Nuclei that contain too many neutrons often undergo beta (β) decay, in which a neutron is converted to a proton and a high-energy electron that is ejected from the nucleus as a β particle:
$\underset{\textrm{unstable} \ \textrm{neutron in} \ \textrm{nucleus}}{^1_0 \textrm n}\rightarrow \underset{\textrm{proton} \ \textrm{retained} \ \textrm{by nucleus}}{^{1}_{1} \textrm p}+\underset{\textrm{beta particle} \ \textrm{emitted by} \ \textrm{nucleus}}{^0_{-1} \beta}\label{Eq3}$
The general reaction for beta decay is therefore
$\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z+1} \textrm X'}+\underset{\textrm{beta particle}}{^0_{-1} \beta}\label{Eq4}$
Although beta decay does not change the mass number of the nucleus, it does result in an increase of +1 in the atomic number because of the addition of a proton in the daughter nucleus. Thus beta decay decreases the neutron-to-proton ratio, moving the nucleus toward the band of stable nuclei. For example, carbon-14 undergoes beta decay to form nitrogen-14:
$^{14}_{6}\textrm{C}\rightarrow ^{14}_{7}\textrm{N}+\,^{0}_{-1}\beta \nonumber$
Once again, the number of nucleons is conserved, and the charges are balanced. The parent and the daughter nuclei have the same mass number, 14, and the sum of the atomic numbers of the products is 6, which is the same as the atomic number of the carbon-14 parent.
Positron $\beta^+$ Emission
Because a positron has the same mass as an electron but opposite charge, positron emission is the opposite of beta decay. Thus positron emission is characteristic of neutron-poor nuclei, which decay by transforming a proton to a neutron and emitting a high-energy positron:
$^{1}_{1}\textrm{p}^+\rightarrow ^{1}_{0}\textrm{n}+\,^{0}_{+1}\beta^+\label{Eq6}$
The general reaction for positron emission is therefore
$\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z-1} \textrm X'}+\underset{\textrm{positron}}{^0_{+1} \beta^+} \nonumber$
Like beta decay, positron emission does not change the mass number of the nucleus. In this case, however, the atomic number of the daughter nucleus is lower by 1 than that of the parent. Thus the neutron-to-proton ratio has increased, again moving the nucleus closer to the band of stable nuclei. For example, carbon-11 undergoes positron emission to form boron-11:
$^{11}_{6}\textrm{C}\rightarrow ^{11}_{5}\textrm{B}+\,^{0}_{+1}\beta^+ \nonumber$
Nucleons are conserved, and the charges balance. The mass number, 11, does not change, and the sum of the atomic numbers of the products is 6, the same as the atomic number of the parent carbon-11 nuclide.
Electron Capture
A neutron-poor nucleus can decay by either positron emission or electron capture (EC), in which an electron in an inner shell reacts with a proton to produce a neutron:
$^{1}_{1}\textrm{p} +\; ^{0}_{-1}\textrm{e}\rightarrow \, ^{1}_{0}\textrm n\label{Eq9}$
When a second electron moves from an outer shell to take the place of the lower-energy electron that was absorbed by the nucleus, an x-ray is emitted. The overall reaction for electron capture is thus
$\underset{\textrm{parent}}{^A_Z \textrm X}+\underset{\textrm{electron}}{^0_{-1} \textrm e}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z-1} \textrm X'}+\textrm{x-ray} \nonumber$
Electron capture does not change the mass number of the nucleus because both the proton that is lost and the neutron that is formed have a mass number of 1. As with positron emission, however, the atomic number of the daughter nucleus is lower by 1 than that of the parent. Once again, the neutron-to-proton ratio has increased, moving the nucleus toward the band of stable nuclei. For example, iron-55 decays by electron capture to form manganese-55, which is often written as follows:
$^{55}_{26}\textrm{Fe}\overset{\textrm{EC}}{\rightarrow}\, ^{55}_{25}\textrm{Mn}+\textrm{x-ray} \nonumber$
The atomic numbers of the parent and daughter nuclides differ in Equation 20.2.11, although the mass numbers are the same. To write a balanced nuclear equation for this reaction, we must explicitly include the captured electron in the equation:
$^{55}_{26}\textrm{Fe}+\,^{0}_{-1}\textrm{e}\rightarrow \, ^{55}_{25}\textrm{Mn}+\textrm{x-ray} \nonumber$
Both positron emission and electron capture are usually observed for nuclides with low neutron-to-proton ratios, but the decay rates for the two processes can be very different.
Gamma $\gamma$ Emission
Many nuclear decay reactions produce daughter nuclei that are in a nuclear excited state, which is similar to an atom in which an electron has been excited to a higher-energy orbital to give an electronic excited state. Just as an electron in an electronic excited state emits energy in the form of a photon when it returns to the ground state, a nucleus in an excited state releases energy in the form of a photon when it returns to the ground state. These high-energy photons are γ rays. Gamma ($\gamma$) emission can occur virtually instantaneously, as it does in the alpha decay of uranium-238 to thorium-234, where the asterisk denotes an excited state:
$^{238}_{92}\textrm{U}\rightarrow \, \underset{\textrm{excited} \ \textrm{nuclear} \ \textrm{state}}{^{234}_{90}\textrm{Th*}} + ^{4}_{2}\alpha\xrightarrow {\textrm{relaxation}\,}\,^{234}_{90}\textrm{Th} + \ce{^0_0\gamma} \nonumber$
If we disregard the decay event that created the excited nucleus, then
$^{234}_{88}\textrm{Th*} \rightarrow\, ^{234}_{88}\textrm{Th} + ^{0}_{0}\gamma \nonumber$
or more generally,
$^{A}_{Z}\textrm{X*} \rightarrow\, ^{A}_{Z}\textrm{X} + ^{0}_{0}\gamma \nonumber$
Gamma emission can also occur after a significant delay. For example, technetium-99m has a half-life of about 6 hours before emitting a $γ$ ray to form technetium-99 (the m is for metastable). Because γ rays are energy, their emission does not affect either the mass number or the atomic number of the daughter nuclide. Gamma-ray emission is therefore the only kind of radiation that does not necessarily involve the conversion of one element to another, although it is almost always observed in conjunction with some other nuclear decay reaction.
Spontaneous Fission
Only very massive nuclei with high neutron-to-proton ratios can undergo spontaneous fission, in which the nucleus breaks into two pieces that have different atomic numbers and atomic masses. This process is most important for the transactinide elements, with Z ≥ 104. Spontaneous fission is invariably accompanied by the release of large amounts of energy, and it is usually accompanied by the emission of several neutrons as well. An example is the spontaneous fission of $^{254}_{98}\textrm{Cf}$, which gives a distribution of fission products; one possible set of products is shown in the following equation:
$^{254}_{98}\textrm{Cf}\rightarrow \,^{118}_{46}\textrm{Pd}+\,^{132}_{52}\textrm{Te}+4^{1}_{0}\textrm{n} \nonumber$
Once again, the number of nucleons is conserved. Thus the sum of the mass numbers of the products (118 + 132 + 4 = 254) equals the mass number of the reactant. Similarly, the sum of the atomic numbers of the products [46 + 52 + (4 × 0) = 98] is the same as the atomic number of the parent nuclide.
Example $2$
Write a balanced nuclear equation to describe each reaction.
1. the beta decay of $^{35}_{16}\textrm{S}$
2. the decay of $^{201}_{80}\textrm{Hg}$ by electron capture
3. the decay of $^{30}_{15}\textrm{P}$ by positron emission
Given: radioactive nuclide and mode of decay
Asked for: balanced nuclear equation
Strategy:
A Identify the reactants and the products from the information given.
B Use the values of A and Z to identify any missing components needed to balance the equation.
Solution
a.
A We know the identities of the reactant and one of the products (a β particle). We can therefore begin by writing an equation that shows the reactant and one of the products and indicates the unknown product as $^{A}_{Z}\textrm{X}$: $^{35}_{16}\textrm{S}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{-1}\beta \nonumber$
B Because both protons and neutrons must be conserved in a nuclear reaction, the unknown product must have a mass number of A = 35 − 0 = 35 and an atomic number of Z = 16 − (−1) = 17. The element with Z = 17 is chlorine, so the balanced nuclear equation is as follows: $^{35}_{16}\textrm{S}\rightarrow\,^{35}_{17}\textrm{Cl}+\,^{0}_{-1}\beta \nonumber$
b.
A We know the identities of both reactants: $^{201}_{80}\textrm{Hg}$ and an inner electron, $^{0}_{-1}\textrm{e}$. The reaction is as follows: $^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{A}_{Z}\textrm{X} \nonumber$
B Both protons and neutrons are conserved, so the mass number of the product must be A = 201 + 0 = 201, and the atomic number of the product must be Z = 80 + (−1) = 79, which corresponds to the element gold. The balanced nuclear equation is thus $^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{201}_{79}\textrm{Au} \nonumber$
c.
A As in part (a), we are given the identities of the reactant and one of the products—in this case, a positron. The unbalanced nuclear equation is therefore $^{30}_{15}\textrm{P}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{+1}\beta \nonumber$
B The mass number of the second product is A = 30 − 0 = 30, and its atomic number is Z = 15 − 1 = 14, which corresponds to silicon. The balanced nuclear equation for the reaction is as follows: $^{30}_{15}\textrm{P}\rightarrow\,^{30}_{14}\textrm{Si}+\,^{0}_{+1}\beta \nonumber$
Exercise $2$
Write a balanced nuclear equation to describe each reaction.
1. $^{11}_{6}\textrm{C}$ by positron emission
2. the beta decay of molybdenum-99
3. the emission of an α particle followed by gamma emission from $^{185}_{74}\textrm{W}$
Answer a
$^{11}_{6}\textrm{C}\rightarrow\,^{11}_{5}\textrm{B}+\,^{0}_{+1}\beta$
Answer d
$^{99}_{42}\textrm{Mo}\rightarrow\,^{99m}_{43}\textrm{Tc}+\,^{0}_{-1}\beta$
Answer c
$^{185}_{74}\textrm{W}\rightarrow\,^{181}_{72}\textrm{Hf}+\,^{4}_{2}\alpha +\,^{0}_{0}\gamma$
Example $3$
Predict the kind of nuclear change each unstable nuclide undergoes when it decays.
1. $^{45}_{22}\textrm{Ti}$
2. $^{242}_{94}\textrm{Pu}$
3. $^{12}_{5}\textrm{B}$
4. $^{256}_{100}\textrm{Fm}$
Given: nuclide
Asked for: type of nuclear decay
Strategy:
Based on the neutron-to-proton ratio and the value of Z, predict the type of nuclear decay reaction that will produce a more stable nuclide.
Solution
1. This nuclide has a neutron-to-proton ratio of only 1.05, which is much less than the requirement for stability for an element with an atomic number in this range. Nuclei that have low neutron-to-proton ratios decay by converting a proton to a neutron. The two possibilities are positron emission, which converts a proton to a neutron and a positron, and electron capture, which converts a proton and a core electron to a neutron. In this case, both are observed, with positron emission occurring about 86% of the time and electron capture about 14% of the time.
2. Nuclei with Z > 83 are too heavy to be stable and usually undergo alpha decay, which decreases both the mass number and the atomic number. Thus $^{242}_{94}\textrm{Pu}$ is expected to decay by alpha emission.
3. This nuclide has a neutron-to-proton ratio of 1.4, which is very high for a light element. Nuclei with high neutron-to-proton ratios decay by converting a neutron to a proton and an electron. The electron is emitted as a β particle, and the proton remains in the nucleus, causing an increase in the atomic number with no change in the mass number. We therefore predict that $^{12}_{5}\textrm{B}$ will undergo beta decay.
4. This is a massive nuclide, with an atomic number of 100 and a mass number much greater than 200. Nuclides with A ≥ 200 tend to decay by alpha emission, and even heavier nuclei tend to undergo spontaneous fission. We therefore predict that $^{256}_{100}\textrm{Fm}$ will decay by either or both of these two processes. In fact, it decays by both spontaneous fission and alpha emission, in a 97:3 ratio.
Exercise $3$
Predict the kind of nuclear change each unstable nuclide undergoes when it decays.
1. $^{32}_{14}\textrm{Si}$
2. $^{43}_{21}\textrm{Sc}$
3. $^{231}_{91}\textrm{Pa}$
Answer a
beta decay
Answer d
positron emission or electron capture
Answer c
alpha decay
Radioactive Decay Series
The nuclei of all elements with atomic numbers greater than 83 are unstable. Thus all isotopes of all elements beyond bismuth in the periodic table are radioactive. Because alpha decay decreases Z by only 2, and positron emission or electron capture decreases Z by only 1, it is impossible for any nuclide with Z > 85 to decay to a stable daughter nuclide in a single step, except via nuclear fission. Consequently, radioactive isotopes with Z > 85 usually decay to a daughter nucleus that is radiaoctive, which in turn decays to a second radioactive daughter nucleus, and so forth, until a stable nucleus finally results. This series of sequential alpha- and beta-decay reactions is called a radioactive decay series. The most common is the uranium-238 decay series, which produces lead-206 in a series of 14 sequential alpha- and beta-decay reactions (Figure $2$). Although a radioactive decay series can be written for almost any isotope with Z > 85, only two others occur naturally: the decay of uranium-235 to lead-207 (in 11 steps) and thorium-232 to lead-208 (in 10 steps). A fourth series, the decay of neptunium-237 to bismuth-209 in 11 steps, is known to have occurred on the primitive Earth. With a half-life of “only” 2.14 million years, all the neptunium-237 present when Earth was formed decayed long ago, and today all the neptunium on Earth is synthetic.
Due to these radioactive decay series, small amounts of very unstable isotopes are found in ores that contain uranium or thorium. These rare, unstable isotopes should have decayed long ago to stable nuclei with a lower atomic number, and they would no longer be found on Earth. Because they are generated continuously by the decay of uranium or thorium, however, their amounts have reached a steady state, in which their rate of formation is equal to their rate of decay. In some cases, the abundance of the daughter isotopes can be used to date a material or identify its origin.
Induced Nuclear Reactions
The discovery of radioactivity in the late 19th century showed that some nuclei spontaneously transform into nuclei with a different number of protons, thereby producing a different element. When scientists realized that these naturally occurring radioactive isotopes decayed by emitting subatomic particles, they realized that—in principle—it should be possible to carry out the reverse reaction, converting a stable nucleus to another more massive nucleus by bombarding it with subatomic particles in a nuclear transmutation reaction.
The first successful nuclear transmutation reaction was carried out in 1919 by Ernest Rutherford, who showed that α particles emitted by radium could react with nitrogen nuclei to form oxygen nuclei. As shown in the following equation, a proton is emitted in the process:
$^{4}_{2}\alpha + \, ^{14}_{7}\textrm{N} \rightarrow \,^{17}_{8}\textrm{O}+\,^{1}_{1}\textrm{p}\label{Eq17}$
Rutherford’s nuclear transmutation experiments led to the discovery of the neutron. He found that bombarding the nucleus of a light target element with an α particle usually converted the target nucleus to a product that had an atomic number higher by 1 and a mass number higher by 3 than the target nucleus. Such behavior is consistent with the emission of a proton after reaction with the α particle. Very light targets such as Li, Be, and B reacted differently, however, emitting a new kind of highly penetrating radiation rather than a proton. Because neither a magnetic field nor an electrical field could deflect these high-energy particles, Rutherford concluded that they were electrically neutral. Other observations suggested that the mass of the neutral particle was similar to the mass of the proton. In 1932, James Chadwick (Nobel Prize in Physics, 1935), who was a student of Rutherford’s at the time, named these neutral particles neutrons and proposed that they were fundamental building blocks of the atom. The reaction that Chadwick initially used to explain the production of neutrons was as follows:
$^{4}_{2}\alpha + \, ^{9}_{4}\textrm{Be} \rightarrow \,^{12}_{6}\textrm{C}+\,^{1}_{0}\textrm{n}\label{Eq18}$
Because α particles and atomic nuclei are both positively charged, electrostatic forces cause them to repel each other. Only α particles with very high kinetic energy can overcome this repulsion and collide with a nucleus (Figure $3$). Neutrons have no electrical charge, however, so they are not repelled by the nucleus. Hence bombardment with neutrons is a much easier way to prepare new isotopes of the lighter elements. In fact, carbon-14 is formed naturally in the atmosphere by bombarding nitrogen-14 with neutrons generated by cosmic rays:
$^{1}_{0}\textrm{n} + \, ^{14}_{7}\textrm{N} \rightarrow \,^{14}_{6}\textrm{C}+\,^{1}_{1}\textrm{p}\label{Eq19}$
Example $4$
In 1933, Frédéric Joliot and Iréne Joliot-Curie (daughter of Marie and Pierre Curie) prepared the first artificial radioactive isotope by bombarding aluminum-27 with α particles. For each 27Al that reacted, one neutron was released. Identify the product nuclide and write a balanced nuclear equation for this transmutation reaction.
Given: reactants in a nuclear transmutation reaction
Asked for: product nuclide and balanced nuclear equation
Strategy:
A Based on the reactants and one product, identify the other product of the reaction. Use conservation of mass and charge to determine the values of Z and A of the product nuclide and thus its identity.
B Write the balanced nuclear equation for the reaction.
Solution
A Bombarding an element with α particles usually produces an element with an atomic number that is 2 greater than the atomic number of the target nucleus. Thus we expect that aluminum (Z = 13) will be converted to phosphorus (Z = 15). With one neutron released, conservation of mass requires that the mass number of the other product be 3 greater than the mass number of the target. In this case, the mass number of the target is 27, so the mass number of the product will be 30. The second product is therefore phosphorus-30, $^{30}_{15}\textrm{P}$.
B The balanced nuclear equation for the reaction is as follows:
$^{27}_{13}\textrm{Al} + \, ^{4}_{2}\alpha \rightarrow \,^{30}_{15}\textrm{P}+\,^{1}_{0}\textrm{n} \nonumber$
Exercise $4$
Because all isotopes of technetium are radioactive and have short half-lives, it does not exist in nature. Technetium can, however, be prepared by nuclear transmutation reactions. For example, bombarding a molybdenum-96 target with deuterium nuclei $(^{2}_{1}\textrm{H})$ produces technetium-97. Identify the other product of the reaction and write a balanced nuclear equation for this transmutation reaction.
Answer
neutron, $^{1}_{0}\textrm{n}$ ; $^{96}_{42}\textrm{Mo} + \, ^{2}_{1}\textrm{H} \rightarrow \,^{97}_{43}\textrm{Tc}+\,^{1}_{0}\textrm{n}$ :
We noted earlier in this section that very heavy nuclides, corresponding to Z ≥ 104, tend to decay by spontaneous fission. Nuclides with slightly lower values of Z, such as the isotopes of uranium (Z = 92) and plutonium (Z = 94), do not undergo spontaneous fission at any significant rate. Some isotopes of these elements, however, such as $^{235}_{92}\textrm{U}$ and $^{239}_{94}\textrm{Pu}$ undergo induced nuclear fission when they are bombarded with relatively low-energy neutrons, as shown in the following equation for uranium-235 and in Figure $4$:
$^{235}_{92}\textrm{U} + \, ^{1}_{0}\textrm{n} \rightarrow \,^{236}_{92}\textrm{U}\rightarrow \,^{141}_{56}\textrm{Ba}+\,^{92}_{36}\textrm{Kr}+3^{1}_{0}\textrm{n}\label{Eq20}$
Any isotope that can undergo a nuclear fission reaction when bombarded with neutrons is called a fissile isotope.
During nuclear fission, the nucleus usually divides asymmetrically rather than into two equal parts, as shown in Figure $4$. Moreover, every fission event of a given nuclide does not give the same products; more than 50 different fission modes have been identified for uranium-235, for example. Consequently, nuclear fission of a fissile nuclide can never be described by a single equation. Instead, as shown in Figure $5$, a distribution of many pairs of fission products with different yields is obtained, but the mass ratio of each pair of fission products produced by a single fission event is always roughly 3:2.
Synthesis of Transuranium Elements
Uranium (Z = 92) is the heaviest naturally occurring element. Consequently, all the elements with Z > 92, the transuranium elements, are artificial and have been prepared by bombarding suitable target nuclei with smaller particles. The first of the transuranium elements to be prepared was neptunium (Z = 93), which was synthesized in 1940 by bombarding a 238U target with neutrons. As shown in Equation 20.21, this reaction occurs in two steps. Initially, a neutron combines with a 238U nucleus to form 239U, which is unstable and undergoes beta decay to produce 239Np:
$^{238}_{92}\textrm{U} + \, ^{1}_{0}\textrm{n} \rightarrow \,^{239}_{92}\textrm{U}\rightarrow \,^{239}_{93}\textrm{Np}+\,^{0}_{-1}\beta\label{Eq21}$
Subsequent beta decay of 239Np produces the second transuranium element, plutonium (Z = 94):
$^{239}_{93}\textrm{Np} \rightarrow \,^{239}_{94}\textrm{Pu}+\,^{0}_{-1}\beta\label{Eq22}$
Bombarding the target with more massive nuclei creates elements that have atomic numbers significantly greater than that of the target nucleus (Table $2$). Such techniques have resulted in the creation of the superheavy elements 114 and 116, both of which lie in or near the “island of stability."
Table $2$: Some Reactions Used to Synthesize Transuranium Elements
$^{239}_{94}\textrm{Pu}+\,^{4}_{2}\alpha \rightarrow \,^{242}_{96}\textrm{Cm}+\,^{1}_{0}\textrm{n}$
$^{239}_{94}\textrm{Pu}+\,^{4}_{2}\alpha \rightarrow \,^{241}_{95}\textrm{Am}+\,^{1}_{1}\textrm{p}+\,^{1}_{0}\textrm{n}$
$^{242}_{96}\textrm{Cm}+\,^{4}_{2}\alpha \rightarrow \,^{243}_{97}\textrm{Bk}+\,^{1}_{1}\textrm{p}+2^{1}_{0}\textrm{n}$
$^{253}_{99}\textrm{Es}+\,^{4}_{2}\alpha \rightarrow \,^{256}_{101}\textrm{Md}+\,^{1}_{0}\textrm{n}$
$^{238}_{92}\textrm{U}+\,^{12}_{6}\textrm{C} \rightarrow \,^{246}_{98}\textrm{Cf}+4^{1}_{0}\textrm{n}$
$^{252}_{98}\textrm{Cf}+\,^{10}_{5}\textrm{B} \rightarrow \,^{256}_{103}\textrm{Lr}+6^{1}_{0}\textrm{n}$
A device called a particle accelerator is used to accelerate positively charged particles to the speeds needed to overcome the electrostatic repulsions between them and the target nuclei by using electrical and magnetic fields. Operationally, the simplest particle accelerator is the linear accelerator (Figure $6$), in which a beam of particles is injected at one end of a long evacuated tube. Rapid alternation of the polarity of the electrodes along the tube causes the particles to be alternately accelerated toward a region of opposite charge and repelled by a region with the same charge, resulting in a tremendous acceleration as the particle travels down the tube. A modern linear accelerator such as the Stanford Linear Accelerator (SLAC) at Stanford University is about 2 miles long.
To achieve the same outcome in less space, a particle accelerator called a cyclotron forces the charged particles to travel in a circular path rather than a linear one. The particles are injected into the center of a ring and accelerated by rapidly alternating the polarity of two large D-shaped electrodes above and below the ring, which accelerates the particles outward along a spiral path toward the target.
The length of a linear accelerator and the size of the D-shaped electrodes in a cyclotron severely limit the kinetic energy that particles can attain in these devices. These limitations can be overcome by using a synchrotron, a hybrid of the two designs. A synchrotron contains an evacuated tube similar to that of a linear accelerator, but the tube is circular and can be more than a mile in diameter. Charged particles are accelerated around the circle by a series of magnets whose polarities rapidly alternate.
Summary and Key Takeaway
• Nuclear decay reactions occur spontaneously under all conditions and produce more stable daughter nuclei, whereas nuclear transmutation reactions are induced and form a product nucleus that is more massive than the starting material.
In nuclear decay reactions (or radioactive decay), the parent nucleus is converted to a more stable daughter nucleus. Nuclei with too many neutrons decay by converting a neutron to a proton, whereas nuclei with too few neutrons decay by converting a proton to a neutron. Very heavy nuclei (with A ≥ 200 and Z > 83) are unstable and tend to decay by emitting an α particle. When an unstable nuclide undergoes radioactive decay, the total number of nucleons is conserved, as is the total positive charge. Six different kinds of nuclear decay reactions are known. Alpha decay results in the emission of an α particle, $^4 _2 \alpha$, and produces a daughter nucleus with a mass number that is lower by 4 and an atomic number that is lower by 2 than the parent nucleus. Beta decay converts a neutron to a proton and emits a high-energy electron, producing a daughter nucleus with the same mass number as the parent and an atomic number that is higher by 1. Positron emission is the opposite of beta decay and converts a proton to a neutron plus a positron. Positron emission does not change the mass number of the nucleus, but the atomic number of the daughter nucleus is lower by 1 than the parent. In electron capture (EC), an electron in an inner shell reacts with a proton to produce a neutron, with emission of an x-ray. The mass number does not change, but the atomic number of the daughter is lower by 1 than the parent. In gamma emission, a daughter nucleus in a nuclear excited state undergoes a transition to a lower-energy state by emitting a γ ray. Very heavy nuclei with high neutron-to-proton ratios can undergo spontaneous fission, in which the nucleus breaks into two pieces that can have different atomic numbers and atomic masses with the release of neutrons. Many very heavy nuclei decay via a radioactive decay series—a succession of some combination of alpha- and beta-decay reactions. In nuclear transmutation reactions, a target nucleus is bombarded with energetic subatomic particles to give a product nucleus that is more massive than the original. All transuranium elements—elements with Z > 92—are artificial and must be prepared by nuclear transmutation reactions. These reactions are carried out in particle accelerators such as linear accelerators, cyclotrons, and synchrotrons.
Key Equations
alpha decay
$^A_Z \textrm X\rightarrow \, ^{A-4}_{Z-2} \textrm X'+\,^4_2 \alpha \nonumber$
beta decay
$^A_Z \textrm X\rightarrow \, ^{A}_{Z+1} \textrm X'+\,^0_{-1} \beta \nonumber$
positron emission
$^A_Z \textrm X\rightarrow \, ^{A}_{Z-1} \textrm X'+\,^0_{+1} \beta \nonumber$
electron capture
$^A_Z \textrm X+\,^{0}_{-1} \textrm e\rightarrow \, ^{A}_{Z-1} \textrm X'+\textrm{x-ray} \nonumber$
gamma emission
$^A_Z \textrm{X*}\rightarrow \, ^{A}_{Z} \textrm X+\,^0_{0} \gamma \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/21%3A_Nuclear_Chemistry/21.01%3A_Radioactivity.txt |
Learning Objectives
• To understand the factors that affect nuclear stability.
Although most of the known elements have at least one isotope whose atomic nucleus is stable indefinitely, all elements have isotopes that are unstable and disintegrate, or decay, at measurable rates by emitting radiation. Some elements have no stable isotopes and eventually decay to other elements. In contrast to the chemical reactions that were the main focus of earlier chapters and are due to changes in the arrangements of the valence electrons of atoms, the process of nuclear decay results in changes inside an atomic nucleus. We begin our discussion of nuclear reactions by reviewing the conventions used to describe the components of the nucleus.
The Atomic Nucleus
Each element can be represented by the notation \(^A_Z \textrm X\), where A, the mass number, is the sum of the number of protons and the number of neutrons, and Z, the atomic number, is the number of protons. The protons and neutrons that make up the nucleus of an atom are called nucleons, and an atom with a particular number of protons and neutrons is called a nuclide. Nuclides with the same number of protons but different numbers of neutrons are called isotopes. Isotopes can also be represented by an alternative notation that uses the name of the element followed by the mass number, such as carbon-12. The stable isotopes of oxygen, for example, can be represented in any of the following ways:
stable isotopes of oxygen represented in different ways
\(^A_Z \textrm X\) \(\ce{^{16}_8 O}\) \(\ce{^{17}_8 O}\) \(\ce{^{18}_8 O}\)
\(^A \textrm X\) \(\ce{^{16} O}\) \(\ce{^{17} O}\) \(\ce{^{18} O}\)
\(\textrm{element-A:}\) \(\textrm{oxygen-16}\) \(\textrm{oxygen-17}\) \(\textrm{oxygen-18}\)
Because the number of neutrons is equal to AZ, we see that the first isotope of oxygen has 8 neutrons, the second isotope 9 neutrons, and the third isotope 10 neutrons. Isotopes of all naturally occurring elements on Earth are present in nearly fixed proportions, with each proportion constituting an isotope’s natural abundance. For example, in a typical terrestrial sample of oxygen, 99.76% of the O atoms is oxygen-16, 0.20% is oxygen-18, and 0.04% is oxygen-17. Any nucleus that is unstable and decays spontaneously is said to be radioactive, emitting subatomic particles and electromagnetic radiation. The emissions are collectively called radioactivity and can be measured. Isotopes that emit radiation are called radioisotopes.
Nuclear Stability
The nucleus of an atom occupies a tiny fraction of the volume of an atom and contains the number of protons and neutrons that is characteristic of a given isotope. Electrostatic repulsions would normally cause the positively charged protons to repel each other, but the nucleus does not fly apart because of the strong nuclear force, an extremely powerful but very short-range attractive force between nucleons (Figure \(1\)). All stable nuclei except the hydrogen-1 nucleus (1H) contain at least one neutron to overcome the electrostatic repulsion between protons. As the number of protons in the nucleus increases, the number of neutrons needed for a stable nucleus increases even more rapidly. Too many protons (or too few neutrons) in the nucleus result in an imbalance between forces, which leads to nuclear instability.
The relationship between the number of protons and the number of neutrons in stable nuclei, arbitrarily defined as having a half-life longer than 10 times the age of Earth, is shown graphically in Figure \(2\). The stable isotopes form a “peninsula of stability” in a “sea of instability.” Only two stable isotopes, 1H and 3He, have a neutron-to-proton ratio less than 1. Several stable isotopes of light atoms have a neutron-to-proton ratio equal to 1 (e.g., \(^4_2 \textrm{He}\), \(^{10}_5 \textrm{B}\), and \(^{40}_{20} \textrm{Ca}\)). All other stable nuclei have a higher neutron-to-proton ratio, which increases steadily to about 1.5 for the heaviest nuclei. Regardless of the number of neutrons, however, all elements with Z > 83 are unstable and radioactive.
As shown in Figure \(3\), more than half of the stable nuclei (166 out of 279) have even numbers of both neutrons and protons; only 6 of the 279 stable nuclei do not have odd numbers of both. Moreover, certain numbers of neutrons or protons result in especially stable nuclei; these are the so-called magic numbers 2, 8, 20, 50, 82, and 126. For example, tin (Z = 50) has 10 stable isotopes, but the elements on either side of tin in the periodic table, indium (Z = 49) and antimony (Z = 51), have only 2 stable isotopes each. Nuclei with magic numbers of both protons and neutrons are said to be “doubly magic” and are even more stable. Examples of elements with doubly magic nuclei are \(^4_2 \textrm{He}\), with 2 protons and 2 neutrons, and \(^{208}_{82} \textrm{Pb}\), with 82 protons and 126 neutrons, which is the heaviest known stable isotope of any element.
Most stable nuclei contain even numbers of both neutrons and protons
The pattern of stability suggested by the magic numbers of nucleons is reminiscent of the stability associated with the closed-shell electron configurations of the noble gases in group 18 and has led to the hypothesis that the nucleus contains shells of nucleons that are in some ways analogous to the shells occupied by electrons in an atom. As shown in Figure \(2\), the “peninsula” of stable isotopes is surrounded by a “reef” of radioactive isotopes, which are stable enough to exist for varying lengths of time before they eventually decay to produce other nuclei.
Origin of the Magic Numbers
Multiple models have been formulated to explain the origin of the magic numbers and two popular ones are the Nuclear Shell Model and the Liquid Drop Model. Unfortuneatly, both require advanced quantum mechanics to fully understand and are beyond the scope of this text.
Example \(1\)
Classify each nuclide as stable or radioactive.
1. \(\ce{_{15}^{30} P}\)
2. \(\ce{_{43}^{98} Tc}\)
3. tin-118
4. \(\ce{_{94}^{239} Pu}\)
Given: mass number and atomic number
Asked for: predicted nuclear stability
Strategy:
Use the number of protons, the neutron-to-proton ratio, and the presence of even or odd numbers of neutrons and protons to predict the stability or radioactivity of each nuclide.
Solution:
a. This isotope of phosphorus has 15 neutrons and 15 protons, giving a neutron-to-proton ratio of 1.0. Although the atomic number, 15, is much less than the value of 83 above which all nuclides are unstable, the neutron-to-proton ratio is less than that expected for stability for an element with this mass. As shown in Figure \(2\), its neutron-to-proton ratio should be greater than 1. Moreover, this isotope has an odd number of both neutrons and protons, which also tends to make a nuclide unstable. Consequently, \(_{15}^{30} \textrm P\) is predicted to be radioactive, and it is.
b. This isotope of technetium has 55 neutrons and 43 protons, giving a neutron-to-proton ratio of 1.28, which places \(_{43}^{98} \textrm{Tc}\) near the edge of the band of stability. The atomic number, 55, is much less than the value of 83 above which all isotopes are unstable. These facts suggest that \(_{43}^{98} \textrm{Tc}\) might be stable. However, \(_{43}^{98} \textrm{Tc}\) has an odd number of both neutrons and protons, a combination that seldom gives a stable nucleus. Consequently, \(_{43}^{98} \textrm{Tc}\) is predicted to be radioactive, and it is.
c. Tin-118 has 68 neutrons and 50 protons, for a neutron-to-proton ratio of 1.36. As in part b, this value and the atomic number both suggest stability. In addition, the isotope has an even number of both neutrons and protons, which tends to increase nuclear stability. Most important, the nucleus has 50 protons, and 50 is one of the magic numbers associated with especially stable nuclei. Thus \(_{50}^{118} \textrm{Sn}\)should be particularly stable.
d. This nuclide has an atomic number of 94. Because all nuclei with Z > 83 are unstable, \(_{94}^{239} \textrm{Pu}\) must be radioactive.
Exercise \(1\)
Classify each nuclide as stable or radioactive.
1. \(\ce{_{90}^{232} Th}\)
2. \(\ce{_{20}^{40} Ca}\)
3. \(\ce{_8^{15} O}\)
4. \(\ce{_{57}^{139} La}\)
Answer a
radioactive
Answer b
stable
Answer c
radioactive
Answer d
stable
Superheavy Elements
In addition to the “peninsula of stability” there is a small “island of stability” that is predicted to exist in the upper right corner. This island corresponds to the superheavy elements, with atomic numbers near the magic number 126. Because the next magic number for neutrons should be 184, it was suggested that an element with 114 protons and 184 neutrons might be stable enough to exist in nature. Although these claims were met with skepticism for many years, since 1999 a few atoms of isotopes with Z = 114 and Z = 116 have been prepared and found to be surprisingly stable. One isotope of element 114 lasts 2.7 seconds before decaying, described as an “eternity” by nuclear chemists. Moreover, there is recent evidence for the existence of a nucleus with A = 292 that was found in 232Th. With an estimated half-life greater than 108 years, the isotope is particularly stable. Its measured mass is consistent with predictions for the mass of an isotope with Z = 122. Thus a number of relatively long-lived nuclei may well be accessible among the superheavy elements.
Summary
Subatomic particles of the nucleus (protons and neutrons) are called nucleons. A nuclide is an atom with a particular number of protons and neutrons. An unstable nucleus that decays spontaneously is radioactive, and its emissions are collectively called radioactivity. Isotopes that emit radiation are called radioisotopes. Each nucleon is attracted to other nucleons by the strong nuclear force. Stable nuclei generally have even numbers of both protons and neutrons and a neutron-to-proton ratio of at least 1. Nuclei that contain magic numbers of protons and neutrons are often especially stable. Superheavy elements, with atomic numbers near 126, may even be stable enough to exist in nature. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/21%3A_Nuclear_Chemistry/21.02%3A_Patterns_of_Nuclear_Stability.txt |
Learning Objectives
• To understand how nuclear transmutation reactions lead to the formation of the elements in stars and how they can be used to synthesize transuranium elements.
The relative abundances of the elements in the known universe vary by more than 12 orders of magnitude. For the most part, these differences in abundance cannot be explained by differences in nuclear stability. Although the 56Fe nucleus is the most stable nucleus known, the most abundant element in the known universe is not iron, but hydrogen (1H), which accounts for about 90% of all atoms. In fact, 1H is the raw material from which all other elements are formed. In this section, we explain why 1H and 2He together account for at least 99% of all the atoms in the known universe. We also describe the nuclear reactions that take place in stars, which transform one nucleus into another and create all the naturally occurring elements.
Relative Abundances of the Elements on Earth and in the Known Universe
The relative abundances of the elements in the known universe and on Earth relative to silicon are shown in Figure $1$. The data are estimates based on the characteristic emission spectra of the elements in stars, the absorption spectra of matter in clouds of interstellar dust, and the approximate composition of Earth as measured by geologists. The data in Figure $1$ illustrate two important points. First, except for hydrogen, the most abundant elements have even atomic numbers. Not only is this consistent with the known trends in nuclear stability, but it also suggests that heavier elements are formed by combining helium nuclei (Z = 2). Second, the relative abundances of the elements in the known universe and on Earth are often very different, as indicated by the data in Table $1$ for some common elements.
Some of these differences are easily explained. For example, nonmetals such as H, He, C, N, O, Ne, and Kr are much less abundant relative to silicon on Earth than they are in the rest of the universe. These elements are either noble gases (He, Ne, and Kr) or elements that form volatile hydrides, such as NH3, CH4, and H2O. Because Earth’s gravity is not strong enough to hold such light substances in the atmosphere, these elements have been slowly diffusing into outer space ever since our planet was formed. Argon is an exception; it is relatively abundant on Earth compared with the other noble gases because it is continuously produced in rocks by the radioactive decay of isotopes such as 40K. In contrast, many metals, such as Al, Na, Fe, Ca, Mg, K, and Ti, are relatively abundant on Earth because they form nonvolatile compounds, such as oxides, that cannot escape into space. Other metals, however, are much less abundant on Earth than in the universe; some examples are Ru and Ir. This section explains some of the reasons for the great differences in abundances of the metallic elements.
Table $1$: Relative Abundances of Elements on Earth and in the Known Universe
Terrestrial/Universal Element Abundance Ratio
H 0.0020
He 2.4 × 10−8
C 0.36
N 0.02
O 46
Ne 1.9 × 10−6
Na 1200
Mg 48
Al 1600
Si 390
S 0.84
K 5000
Ca 710
Ti 2200
Fe 57
All the elements originally present on Earth (and on other planets) were synthesized from hydrogen and helium nuclei in the interiors of stars that have long since exploded and disappeared. Six of the most abundant elements in the universe (C, O, Ne, Mg, Si, and Fe) have nuclei that are integral multiples of the helium-4 nucleus, which suggests that helium-4 is the primary building block for heavier nuclei.
Synthesis of the Elements in Stars
Elements are synthesized in discrete stages during the lifetime of a star, and some steps occur only in the most massive stars known (Figure $2$). Initially, all stars are formed by the aggregation of interstellar “dust,” which is mostly hydrogen. As the cloud of dust slowly contracts due to gravitational attraction, its density eventually reaches about 100 g/cm3, and the temperature increases to about 1.5 × 107 K, forming a dense plasma of ionized hydrogen nuclei. At this point, self-sustaining nuclear reactions begin, and the star “ignites,” creating a yellow star like our sun.
In the first stage of its life, the star is powered by a series of nuclear fusion reactions that convert hydrogen to helium:
$_1^1\textrm H+\,_1^1\textrm H\rightarrow\,_1^2\textrm H+\,_{+1}^0\beta \_1^2\textrm H+\,_1^1\textrm H\rightarrow\,_2^3\textrm{He}+\,_{0}^0\gamma \_2^3\textrm{He}+\,_2^3\textrm{He}\rightarrow\,_2^4\textrm{He}+2_{1}^1\textrm H\label{Eq1}$
The overall reaction is the conversion of four hydrogen nuclei to a helium-4 nucleus, which is accompanied by the release of two positrons, two $\gamma$ rays, and a great deal of energy:
$4_1^1\textrm H\rightarrow\,_2^4\textrm{He}+2_{+1}^0\beta+2_0^0\gamma\label{Eq2}$
These reactions are responsible for most of the enormous amount of energy that is released as sunlight and solar heat. It takes several billion years, depending on the size of the star, to convert about 10% of the hydrogen to helium.
Once large amounts of helium-4 have been formed, they become concentrated in the core of the star, which slowly becomes denser and hotter. At a temperature of about 2 × 108 K, the helium-4 nuclei begin to fuse, producing beryllium-8:
$2_2^4\textrm{He}\rightarrow\,_4^8\textrm{Be}\label{Eq3}$
Although beryllium-8 has both an even mass number and an even atomic number, its also has a low neutron-to-proton ratio (and other factors beyond the scope of this text) that makes it unstable; it decomposes in only about 10−16 s. Nonetheless, this is long enough for it to react with a third helium-4 nucleus to form carbon-12, which is very stable. Sequential reactions of carbon-12 with helium-4 produce the elements with even numbers of protons and neutrons up to magnesium-24:
$_4^8\textrm{Be}\xrightarrow{_2^4\textrm{He}}\,_6^{12}\textrm C\xrightarrow{_2^4\textrm{He}}\,_8^{16}\textrm O\xrightarrow{_2^4\textrm{He}}\,_{10}^{20}\textrm{Ne}\xrightarrow{_2^4\textrm{He}}\,_{12}^{24}\textrm{Mg}\label{Eq4}$
So much energy is released by these reactions that it causes the surrounding mass of hydrogen to expand, producing a red giant that is about 100 times larger than the original yellow star.
As the star expands, heavier nuclei accumulate in its core, which contracts further to a density of about 50,000 g/cm3, so the core becomes even hotter. At a temperature of about 7 × 108 K, carbon and oxygen nuclei undergo nuclear fusion reactions to produce sodium and silicon nuclei:
$_6^{12}\textrm C+\,_6^{12}\textrm C\rightarrow \,_{11}^{23}\textrm{Na}+\,_1^1\textrm H\label{Eq5}$
$_6^{12}\textrm C+\,_8^{16}\textrm O\rightarrow \,_{14}^{28}\textrm{Si}+\,_0^0\gamma\label{Eq6}$
At these temperatures, carbon-12 reacts with helium-4 to initiate a series of reactions that produce more oxygen-16, neon-20, magnesium-24, and silicon-28, as well as heavier nuclides such as sulfur-32, argon-36, and calcium-40:
$_6^{12}\textrm C\xrightarrow{_2^4\textrm{He}}\,_8^{16}\textrm O\xrightarrow{_2^4\textrm{He}}\,_{10}^{20}\textrm{Ne}\xrightarrow{_2^4\textrm{He}}\,_{12}^{24}\textrm{Mg}\xrightarrow{_2^4\textrm{He}}\,_{14}^{28}\textrm{Si}\xrightarrow{_2^4\textrm{He}}\,_{16}^{32}\textrm S\xrightarrow{_2^4\textrm{He}}\,_{18}^{36}\textrm{Ar}\xrightarrow{_2^4\textrm{He}}\,_{20}^{40}\textrm{Ca}\label{Eq7}$
The energy released by these reactions causes a further expansion of the star to form a red supergiant, and the core temperature increases steadily. At a temperature of about 3 × 109 K, the nuclei that have been formed exchange protons and neutrons freely. This equilibration process forms heavier elements up to iron-56 and nickel-58, which have the most stable nuclei known.
The Formation of Heavier Elements in Supernovas
None of the processes described so far produces nuclei with Z > 28. All naturally occurring elements heavier than nickel are formed in the rare but spectacular cataclysmic explosions called supernovas (Figure $2$). When the fuel in the core of a very massive star has been consumed, its gravity causes it to collapse in about 1 s. As the core is compressed, the iron and nickel nuclei within it disintegrate to protons and neutrons, and many of the protons capture electrons to form neutrons. The resulting neutron star is a dark object that is so dense that atoms no longer exist. Simultaneously, the energy released by the collapse of the core causes the supernova to explode in what is arguably the single most violent event in the universe. The force of the explosion blows most of the star’s matter into space, creating a gigantic and rapidly expanding dust cloud, or nebula (Figure $3$). During the extraordinarily short duration of this event, the concentration of neutrons is so great that multiple neutron-capture events occur, leading to the production of the heaviest elements and many of the less stable nuclides. Under these conditions, for example, an iron-56 nucleus can absorb as many as 64 neutrons, briefly forming an extraordinarily unstable iron isotope that can then undergo multiple rapid β-decay processes to produce tin-120:
$_{26}^{56}\textrm{Fe}+64_0^1\textrm n\rightarrow \,_{26}^{120}\textrm{Fe}\rightarrow\,_{50}^{120}\textrm{Sn}+24_{-1}^0\beta\label{Eq8}$
Although a supernova occurs only every few hundred years in a galaxy such as the Milky Way, these rare explosions provide the only conditions under which elements heavier than nickel can be formed. The force of the explosions distributes these elements throughout the galaxy surrounding the supernova, and eventually they are captured in the dust that condenses to form new stars. Based on its elemental composition, our sun is thought to be a second- or third-generation star. It contains a considerable amount of cosmic debris from the explosion of supernovas in the remote past.
Example $1$: Carbon Burning Stars
The reaction of two carbon-12 nuclei in a carbon-burning star can produce elements other than sodium. Write a balanced nuclear equation for the formation of
1. magnesium-24.
2. neon-20 from two carbon-12 nuclei.
Given: reactant and product nuclides
Asked for: balanced nuclear equation
Strategy:
Use conservation of mass and charge to determine the type of nuclear reaction that will convert the reactant to the indicated product. Write the balanced nuclear equation for the reaction.
Solution
1. A magnesium-24 nucleus (Z = 12, A = 24) has the same nucleons as two carbon-12 nuclei (Z = 6, A = 12). The reaction is therefore a fusion of two carbon-12 nuclei, and no other particles are produced: $_{6}^{12}\textrm C+\,_{6}^{12}\textrm C\rightarrow\,_{12}^{24}\textrm{Mg}$.
2. The neon-20 product has Z = 10 and A = 20. The conservation of mass requires that the other product have A = (2 × 12) − 20 = 4; because of conservation of charge, it must have Z = (2 × 6) − 10 = 2. These are the characteristics of an α particle. The reaction is therefore $_{6}^{12}\textrm C+\,_{6}^{12}\textrm C\rightarrow\,_{10}^{20}\textrm{Ne}+\,_2^4\alpha$.
Exercise $1$
How many neutrons must an iron-56 nucleus absorb during a supernova explosion to produce an arsenic-75 nucleus? Write a balanced nuclear equation for the reaction.
Answer
19 neutrons; $_{26}^{56}\textrm{Fe}+19_0^1\textrm n \rightarrow \,_{26}^{75}\textrm{Fe}\rightarrow \,_{33}^{75}\textrm{As}+7_{-1}^0\beta$
Summary
Hydrogen and helium are the most abundant elements in the universe. Heavier elements are formed in the interior of stars via multiple neutron-capture events. By far the most abundant element in the universe is hydrogen. The fusion of hydrogen nuclei to form helium nuclei is the major process that fuels young stars such as the sun. Elements heavier than helium are formed from hydrogen and helium in the interiors of stars. Successive fusion reactions of helium nuclei at higher temperatures create elements with even numbers of protons and neutrons up to magnesium and then up to calcium. Eventually, the elements up to iron-56 and nickel-58 are formed by exchange processes at even higher temperatures. Heavier elements can only be made by a process that involves multiple neutron-capture events, which can occur only during the explosion of a supernova. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/21%3A_Nuclear_Chemistry/21.03%3A_Nuclear_Transmutations.txt |
Learning Objectives
• To know how to use half-lives to describe the rates of first-order reactions
Another approach to describing reaction rates is based on the time required for the concentration of a reactant to decrease to one-half its initial value. This period of time is called the half-life of the reaction, written as t1/2. Thus the half-life of a reaction is the time required for the reactant concentration to decrease from [A]0 to [A]0/2. If two reactions have the same order, the faster reaction will have a shorter half-life, and the slower reaction will have a longer half-life.
The half-life of a first-order reaction under a given set of reaction conditions is a constant. This is not true for zeroth- and second-order reactions. The half-life of a first-order reaction is independent of the concentration of the reactants. This becomes evident when we rearrange the integrated rate law for a first-order reaction to produce the following equation:
$\ln\dfrac{[\textrm A]_0}{[\textrm A]}=kt \label{21.4.1}$
Substituting [A]0/2 for [A] and t1/2 for t (to indicate a half-life) into Equation $\ref{21.4.1}$ gives
$\ln\dfrac{[\textrm A]_0}{[\textrm A]_0/2}=\ln 2=kt_{1/2}$
Substituting $\ln{2} \approx 0.693$ into the equation results in the expression for the half-life of a first-order reaction:
$t_{1/2}=\dfrac{0.693}{k} \label{21.4.2}$
Thus, for a first-order reaction, each successive half-life is the same length of time, as shown in Figure $1$, and is independent of [A].
If we know the rate constant for a first-order reaction, then we can use half-lives to predict how much time is needed for the reaction to reach a certain percent completion.
Number of Half-Lives Percentage of Reactant Remaining
1 $\dfrac{100\%}{2}=50\%$ $\dfrac{1}{2}(100\%)=50\%$
2 $\dfrac{50\%}{2}=25\%$ $\dfrac{1}{2}\left(\dfrac{1}{2}\right)(100\%)=25\%$
3 $\dfrac{25\%}{2}=12.5\%$ $\dfrac{1}{2}\left(\dfrac{1}{2}\right )\left (\dfrac{1}{2}\right)(100\%)=12.5\%$
n $\dfrac{100\%}{2^n}$ $\left(\dfrac{1}{2}\right)^n(100\%)=\left(\dfrac{1}{2}\right)^n\%$
As you can see from this table, the amount of reactant left after n half-lives of a first-order reaction is (1/2)n times the initial concentration.
For a first-order reaction, the concentration of the reactant decreases by a constant with each half-life and is independent of [A].
Example $1$
The anticancer drug cis-platin hydrolyzes in water with a rate constant of 1.5 × 10−3 min−1 at pH 7.0 and 25°C. Calculate the half-life for the hydrolysis reaction under these conditions. If a freshly prepared solution of cis-platin has a concentration of 0.053 M, what will be the concentration of cis-platin after 5 half-lives? after 10 half-lives? What is the percent completion of the reaction after 5 half-lives? after 10 half-lives?
Given: rate constant, initial concentration, and number of half-lives
Asked for: half-life, final concentrations, and percent completion
Strategy:
1. Use Equation $\ref{21.4.2}$ to calculate the half-life of the reaction.
2. Multiply the initial concentration by 1/2 to the power corresponding to the number of half-lives to obtain the remaining concentrations after those half-lives.
3. Subtract the remaining concentration from the initial concentration. Then divide by the initial concentration, multiplying the fraction by 100 to obtain the percent completion.
Solution
A We can calculate the half-life of the reaction using Equation $\ref{21.4.2}$:
$t_{1/2}=\dfrac{0.693}{k}=\dfrac{0.693}{1.5\times10^{-3}\textrm{ min}^{-1}}=4.6\times10^2\textrm{ min}$
Thus it takes almost 8 h for half of the cis-platin to hydrolyze.
B After 5 half-lives (about 38 h), the remaining concentration of cis-platin will be as follows:
$\dfrac{0.053\textrm{ M}}{2^5}=\dfrac{0.053\textrm{ M}}{32}=0.0017\textrm{ M}$
After 10 half-lives (77 h), the remaining concentration of cis-platin will be as follows:
$\dfrac{0.053\textrm{ M}}{2^{10}}=\dfrac{0.053\textrm{ M}}{1024}=5.2\times10^{-5}\textrm{ M}$
C The percent completion after 5 half-lives will be as follows:
$\textrm{percent completion}=\dfrac{(0.053\textrm{ M}-0.0017\textrm{ M})(100)}{0.053}=97\%$
The percent completion after 10 half-lives will be as follows:
$\textrm{percent completion}=\dfrac{(0.053\textrm{ M}-5.2\times10^{-5}\textrm{ M})(100)}{0.053\textrm{ M}}=100\%$
Thus a first-order chemical reaction is 97% complete after 5 half-lives and 100% complete after 10 half-lives.
Exercise $1$
Ethyl chloride decomposes to ethylene and HCl in a first-order reaction that has a rate constant of 1.6 × 10−6 s−1 at 650°C.
1. What is the half-life for the reaction under these conditions?
2. If a flask that originally contains 0.077 M ethyl chloride is heated at 650°C, what is the concentration of ethyl chloride after 4 half-lives?
Answer a
4.3 × 105 s = 120 h = 5.0 days;
Answer b
4.8 × 10−3 M
Radioactive Decay Rates
Radioactivity, or radioactive decay, is the emission of a particle or a photon that results from the spontaneous decomposition of the unstable nucleus of an atom. The rate of radioactive decay is an intrinsic property of each radioactive isotope that is independent of the chemical and physical form of the radioactive isotope. The rate is also independent of temperature. In this section, we will describe radioactive decay rates and how half-lives can be used to monitor radioactive decay processes.
In any sample of a given radioactive substance, the number of atoms of the radioactive isotope must decrease with time as their nuclei decay to nuclei of a more stable isotope. Using N to represent the number of atoms of the radioactive isotope, we can define the rate of decay of the sample, which is also called its activity (A) as the decrease in the number of the radioisotope’s nuclei per unit time:
$A=-\dfrac{\Delta N}{\Delta t} \label{21.4.3}$
Activity is usually measured in disintegrations per second (dps) or disintegrations per minute (dpm).
The activity of a sample is directly proportional to the number of atoms of the radioactive isotope in the sample:
$A = kN \label{21.4.4}$
Here, the symbol k is the radioactive decay constant, which has units of inverse time (e.g., s−1, yr−1) and a characteristic value for each radioactive isotope. If we combine Equation $\ref{21.4.3}$ and Equation $\ref{21.4.4}$, we obtain the relationship between the number of decays per unit time and the number of atoms of the isotope in a sample:
$-\dfrac{\Delta N}{\Delta t}=kN \label{21.4.5}$
Equation $\ref{21.4.5}$ is the same as the equation for the reaction rate of a first-order reaction, except that it uses numbers of atoms instead of concentrations. In fact, radioactive decay is a first-order process and can be described in terms of either the differential rate law (Equation $\ref{21.4.5}$) or the integrated rate law:
$N = N_0e^{−kt}$
$\ln \dfrac{N}{N_0}=-kt \label{21.4.6}$
Because radioactive decay is a first-order process, the time required for half of the nuclei in any sample of a radioactive isotope to decay is a constant, called the half-life of the isotope. The half-life tells us how radioactive an isotope is (the number of decays per unit time); thus it is the most commonly cited property of any radioisotope. For a given number of atoms, isotopes with shorter half-lives decay more rapidly, undergoing a greater number of radioactive decays per unit time than do isotopes with longer half-lives. The half-lives of several isotopes are listed in Table 14.6, along with some of their applications.
Table $2$: Half-Lives and Applications of Some Radioactive Isotopes
Radioactive Isotope Half-Life Typical Uses
*The m denotes metastable, where an excited state nucleus decays to the ground state of the same isotope.
hydrogen-3 (tritium) 12.32 yr biochemical tracer
carbon-11 20.33 min positron emission tomography (biomedical imaging)
carbon-14 5.70 × 103 yr dating of artifacts
sodium-24 14.951 h cardiovascular system tracer
phosphorus-32 14.26 days biochemical tracer
potassium-40 1.248 × 109 yr dating of rocks
iron-59 44.495 days red blood cell lifetime tracer
cobalt-60 5.2712 yr radiation therapy for cancer
technetium-99m* 6.006 h biomedical imaging
iodine-131 8.0207 days thyroid studies tracer
radium-226 1.600 × 103 yr radiation therapy for cancer
uranium-238 4.468 × 109 yr dating of rocks and Earth’s crust
americium-241 432.2 yr smoke detectors
Note
Radioactive decay is a first-order process.
Radioisotope Dating Techniques
In our earlier discussion, we used the half-life of a first-order reaction to calculate how long the reaction had been occurring. Because nuclear decay reactions follow first-order kinetics and have a rate constant that is independent of temperature and the chemical or physical environment, we can perform similar calculations using the half-lives of isotopes to estimate the ages of geological and archaeological artifacts. The techniques that have been developed for this application are known as radioisotope dating techniques.
The most common method for measuring the age of ancient objects is carbon-14 dating. The carbon-14 isotope, created continuously in the upper regions of Earth’s atmosphere, reacts with atmospheric oxygen or ozone to form 14CO2. As a result, the CO2 that plants use as a carbon source for synthesizing organic compounds always includes a certain proportion of 14CO2 molecules as well as nonradioactive 12CO2 and 13CO2. Any animal that eats a plant ingests a mixture of organic compounds that contains approximately the same proportions of carbon isotopes as those in the atmosphere. When the animal or plant dies, the carbon-14 nuclei in its tissues decay to nitrogen-14 nuclei by a radioactive process known as beta decay, which releases low-energy electrons (β particles) that can be detected and measured:
$\ce{^{14}C \rightarrow ^{14}N + \beta^{−}} \label{21.4.7}$
The half-life for this reaction is 5700 ± 30 yr.
The 14C/12C ratio in living organisms is 1.3 × 10−12, with a decay rate of 15 dpm/g of carbon (Figure $2$). Comparing the disintegrations per minute per gram of carbon from an archaeological sample with those from a recently living sample enables scientists to estimate the age of the artifact, as illustrated in Example 11.Using this method implicitly assumes that the 14CO2/12CO2 ratio in the atmosphere is constant, which is not strictly correct. Other methods, such as tree-ring dating, have been used to calibrate the dates obtained by radiocarbon dating, and all radiocarbon dates reported are now corrected for minor changes in the 14CO2/12CO2 ratio over time.
Example $2$
In 1990, the remains of an apparently prehistoric man were found in a melting glacier in the Italian Alps. Analysis of the 14C content of samples of wood from his tools gave a decay rate of 8.0 dpm/g carbon. How long ago did the man die?
Given: isotope and final activity
Asked for: elapsed time
Strategy:
A Use Equation $\ref{21.4.4}$ to calculate N0/N. Then substitute the value for the half-life of 14C into Equation $\ref{21.4.2}$ to find the rate constant for the reaction.
B Using the values obtained for N0/N and the rate constant, solve Equation $\ref{21.4.6}$ to obtain the elapsed time.
Solution
We know the initial activity from the isotope’s identity (15 dpm/g), the final activity (8.0 dpm/g), and the half-life, so we can use the integrated rate law for a first-order nuclear reaction (Equation $\ref{21.4.6}$) to calculate the elapsed time (the amount of time elapsed since the wood for the tools was cut and began to decay).
\begin{align}\ln\dfrac{N}{N_0}&=-kt \ \dfrac{\ln(N/N_0)}{k}&=t\end{align}
A From Equation $\ref{21.4.4}$, we know that A = kN. We can therefore use the initial and final activities (A0 = 15 dpm and A = 8.0 dpm) to calculate N0/N:
$\dfrac{A_0}{A}=\dfrac{kN_0}{kN}=\dfrac{N_0}{N}=\dfrac{15}{8.0}$
Now we need only calculate the rate constant for the reaction from its half-life (5730 yr) using Equation $\ref{21.4.2}$:
$t_{1/2}=\dfrac{0.693}{k}$
This equation can be rearranged as follows:
$k=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{5730\textrm{ yr}}=1.22\times10^{-4}\textrm{ yr}^{-1}$
B Substituting into the equation for t,
$t=\dfrac{\ln(N_0/N)}{k}=\dfrac{\ln(15/8.0)}{1.22\times10^{-4}\textrm{ yr}^{-1}}=5.2\times10^3\textrm{ yr}$
From our calculations, the man died 5200 yr ago.
Exercise $2$
It is believed that humans first arrived in the Western Hemisphere during the last Ice Age, presumably by traveling over an exposed land bridge between Siberia and Alaska. Archaeologists have estimated that this occurred about 11,000 yr ago, but some argue that recent discoveries in several sites in North and South America suggest a much earlier arrival. Analysis of a sample of charcoal from a fire in one such site gave a 14C decay rate of 0.4 dpm/g of carbon. What is the approximate age of the sample?
Answer
30,000 yr
Summary
• The half-life of a first-order reaction is independent of the concentration of the reactants.
• The half-lives of radioactive isotopes can be used to date objects.
The half-life of a reaction is the time required for the reactant concentration to decrease to one-half its initial value. The half-life of a first-order reaction is a constant that is related to the rate constant for the reaction: t1/2 = 0.693/k. Radioactive decay reactions are first-order reactions. The rate of decay, or activity, of a sample of a radioactive substance is the decrease in the number of radioactive nuclei per unit time. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/21%3A_Nuclear_Chemistry/21.04%3A_Rates_of_Radioactive_Decay.txt |
Learning Objectives
• To calculate a mass-energy balance and a nuclear binding energy.
• To understand the differences between nuclear fission and fusion.
Nuclear reactions, like chemical reactions, are accompanied by changes in energy. The energy changes in nuclear reactions, however, are enormous compared with those of even the most energetic chemical reactions. In fact, the energy changes in a typical nuclear reaction are so large that they result in a measurable change of mass. In this section, we describe the relationship between mass and energy in nuclear reactions and show how the seemingly small changes in mass that accompany nuclear reactions result in the release of enormous amounts of energy.
Mass–Energy Balance
The relationship between mass (m) and energy (E) is expressed in the following equation:
$E = mc^2 \label{Eq1}$
where
• $c$ is the speed of light ($2.998 \times 10^8\; m/s$), and
• $E$ and $m$ are expressed in units of joules and kilograms, respectively.
Albert Einstein first derived this relationship in 1905 as part of his special theory of relativity: the mass of a particle is directly proportional to its energy. Thus according to Equation $\ref{Eq1}$, every mass has an associated energy, and similarly, any reaction that involves a change in energy must be accompanied by a change in mass. This implies that all exothermic reactions should be accompanied by a decrease in mass, and all endothermic reactions should be accompanied by an increase in mass. Given the law of conservation of mass, how can this be true? The solution to this apparent contradiction is that chemical reactions are indeed accompanied by changes in mass, but these changes are simply too small to be detected. As you may recall, all particles exhibit wavelike behavior, but the wavelength is inversely proportional to the mass of the particle (actually, to its momentum, the product of its mass and velocity). Consequently, wavelike behavior is detectable only for particles with very small masses, such as electrons. For example, the chemical equation for the combustion of graphite to produce carbon dioxide is as follows:
$\textrm{C(graphite)} + \frac{1}{2}\textrm O_2(\textrm g)\rightarrow \mathrm{CO_2}(\textrm g)\hspace{5mm}\Delta H^\circ=-393.5\textrm{ kJ/mol} \label{Eq2}$
Combustion reactions are typically carried out at constant pressure, and under these conditions, the heat released or absorbed is equal to ΔH. When a reaction is carried out at constant volume, the heat released or absorbed is equal to ΔE. For most chemical reactions, however, ΔE ≈ ΔH. If we rewrite Einstein’s equation as
$\Delta{E}=(\Delta m)c^2 \label{Eq3}$
we can rearrange the equation to obtain the following relationship between the change in mass and the change in energy:
$\Delta m=\dfrac{\Delta E}{c^2} \label{Eq4}$
Because 1 J = 1 (kg•m2)/s2, the change in mass is as follows:
$\Delta m=\dfrac{-393.5\textrm{ kJ/mol}}{(2.998\times10^8\textrm{ m/s})^2}=\dfrac{-3.935\times10^5(\mathrm{kg\cdot m^2})/(\mathrm{s^2\cdot mol})}{(2.998\times10^8\textrm{ m/s})^2}=-4.38\times10^{-12}\textrm{ kg/mol} \label{Eq5}$
This is a mass change of about 3.6 × 10−10 g/g carbon that is burned, or about 100-millionths of the mass of an electron per atom of carbon. In practice, this mass change is much too small to be measured experimentally and is negligible.
In contrast, for a typical nuclear reaction, such as the radioactive decay of 14C to 14N and an electron (a β particle), there is a much larger change in mass:
$^{14}\textrm C\rightarrow \,^{14}\textrm N+\,^0_{-1}\beta \label{Eq6}$
We can use the experimentally measured masses of subatomic particles and common isotopes given in Table 20.1 to calculate the change in mass directly. The reaction involves the conversion of a neutral 14C atom to a positively charged 14N ion (with six, not seven, electrons) and a negatively charged β particle (an electron), so the mass of the products is identical to the mass of a neutral 14N atom. The total change in mass during the reaction is therefore the difference between the mass of a neutral 14N atom (14.003074 amu) and the mass of a 14C atom (14.003242 amu):
\begin{align} \Delta m &= {\textrm{mass}_{\textrm{products}}- \textrm{mass}_{\textrm{reactants}}} \&=14.003074\textrm{ amu} - 14.003242\textrm{ amu} = - 0.000168\textrm{ amu}\end{align} \label{Eq7}
The difference in mass, which has been released as energy, corresponds to almost one-third of an electron. The change in mass for the decay of 1 mol of 14C is −0.000168 g = −1.68 × 10−4 g = −1.68 × 10−7 kg. Although a mass change of this magnitude may seem small, it is about 1000 times larger than the mass change for the combustion of graphite. The energy change is as follows:
\begin{align}\Delta E &=(\Delta m)c^2=(-1.68\times10^{-7}\textrm{ kg})(2.998\times10^8\textrm{ m/s})^2 \ &=-1.51\times10^{10}(\mathrm{kg\cdot m^2})/\textrm s^2=-1.51\times10^{10}\textrm{ J}=-1.51\times10^7\textrm{ kJ}\end{align} \label{Eq8}
The energy released in this nuclear reaction is more than 100,000 times greater than that of a typical chemical reaction, even though the decay of 14C is a relatively low-energy nuclear reaction.
Because the energy changes in nuclear reactions are so large, they are often expressed in kiloelectronvolts (1 keV = 103 eV), megaelectronvolts (1 MeV = 106 eV), and even gigaelectronvolts (1 GeV = 109 eV) per atom or particle. The change in energy that accompanies a nuclear reaction can be calculated from the change in mass using the relationship 1 amu = 931 MeV. The energy released by the decay of one atom of 14C is thus
$\mathrm{(-1.68\times10^{-4}\, amu) \left(\dfrac{931\, MeV}{amu}\right) = -0.156\, MeV = -156\, keV}\label{Eq9}$
Example $1$
Calculate the changes in mass (in atomic mass units) and energy (in joules per mole and electronvolts per atom) that accompany the radioactive decay of 238U to 234Th and an α particle. The α particle absorbs two electrons from the surrounding matter to form a helium atom.
Given: nuclear decay reaction
Asked for: changes in mass and energy
Strategy:
A Use the mass values in Table 20.1 to calculate the change in mass for the decay reaction in atomic mass units.
B Use Equation $\ref{Eq4}$ to calculate the change in energy in joules per mole.
C Use the relationship between atomic mass units and megaelectronvolts to calculate the change in energy in electronvolts per atom.
Solution
A Using particle and isotope masses from Table 20.1, we can calculate the change in mass as follows:
\begin{align} \Delta m &= {\textrm{mass}_{\textrm{products}}- \textrm{mass}_{\textrm{reactants}}}=(\mathrm{mass \;^{234}Th+mass\;^4_2He})-\mathrm{mass\;^{238}U} \&=(234.043601\textrm{ amu}+4.002603\textrm{ amu}) - 238.050788\textrm{ amu} = - 0.004584\textrm{ amu}\end{align}
B Thus the change in mass for 1 mol of 238U is −0.004584 g or −4.584 × 10−6 kg. The change in energy in joules per mole is as follows:
ΔE = (Δm)c2 = (−4.584 × 10−6 kg)(2.998 × 108 m/s)2 = −4.120 × 1011 J/mol
C The change in energy in electronvolts per atom is as follows:
$\Delta E = -4.584\times10^{-3}\textrm{ amu}\times\dfrac{\textrm{931 MeV}}{\textrm{amu}}\times\dfrac{1\times10^6\textrm{ eV}}{\textrm{1 MeV}}=-4.27\times10^6\textrm{ eV/atom}$
Exercise $1$
Calculate the changes in mass (in atomic mass units) and energy (in kilojoules per mole and kiloelectronvolts per atom) that accompany the radioactive decay of tritium (3H) to 3He and a β particle.
Answer
Δm = −2.0 × 10−5 amu; ΔE = −1.9 × 106 kJ/mol = −19 keV/atom
Nuclear Binding Energies
We have seen that energy changes in both chemical and nuclear reactions are accompanied by changes in mass. Einstein’s equation, which allows us to interconvert mass and energy, has another interesting consequence: The mass of an atom is always less than the sum of the masses of its component particles. The only exception to this rule is hydrogen-1 (1H), whose measured mass of 1.007825 amu is identical to the sum of the masses of a proton and an electron. In contrast, the experimentally measured mass of an atom of deuterium (2H) is 2.014102 amu, although its calculated mass is 2.016490 amu:
\begin{align}m_{^2\textrm H}&=m_{\textrm{neutron}}+m_{\textrm{proton}}+m_{\textrm{electron}} \&=1.008665\textrm{ amu}+1.007276\textrm{ amu}+0.000549\textrm{ amu}=2.016490\textrm{ amu} \end{align}\label{Eq10}
The difference between the sum of the masses of the components and the measured atomic mass is called the mass defect of the nucleus. Just as a molecule is more stable than its isolated atoms, a nucleus is more stable (lower in energy) than its isolated components. Consequently, when isolated nucleons assemble into a stable nucleus, energy is released. According to Equation $\ref{Eq4}$, this release of energy must be accompanied by a decrease in the mass of the nucleus.
The amount of energy released when a nucleus forms from its component nucleons is the nuclear binding energy (Figure $1$). In the case of deuterium, the mass defect is 0.002388 amu, which corresponds to a nuclear binding energy of 2.22 MeV for the deuterium nucleus. Because the magnitude of the mass defect is proportional to the nuclear binding energy, both values indicate the stability of the nucleus.
Just as a molecule is more stable (lower in energy) than its isolated atoms, a nucleus is more stable than its isolated components.
Not all nuclei are equally stable. Chemists describe the relative stability of different nuclei by comparing the binding energy per nucleon, which is obtained by dividing the nuclear binding energy by the mass number (A) of the nucleus. As shown in Figure $2$, the binding energy per nucleon increases rapidly with increasing atomic number until about Z = 26, where it levels off to about 8–9 MeV per nucleon and then decreases slowly. The initial increase in binding energy is not a smooth curve but exhibits sharp peaks corresponding to the light nuclei that have equal numbers of protons and neutrons (e.g., 4He, 12C, and 16O). As mentioned earlier, these are particularly stable combinations.
Because the maximum binding energy per nucleon is reached at 56Fe, all other nuclei are thermodynamically unstable with regard to the formation of 56Fe. Consequently, heavier nuclei (toward the right in Figure $2$) should spontaneously undergo reactions such as alpha decay, which result in a decrease in atomic number. Conversely, lighter elements (on the left in Figure $2$) should spontaneously undergo reactions that result in an increase in atomic number. This is indeed the observed pattern.
Heavier nuclei spontaneously undergo nuclear reactions that decrease their atomic number. Lighter nuclei spontaneously undergo nuclear reactions that increase their atomic number.
Example $2$
Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for 56Fe. The experimental mass of the nuclide is given in Table A4.
Given: nuclide and mass
Asked for: nuclear binding energy and binding energy per nucleon
Strategy:
A Sum the masses of the protons, electrons, and neutrons or, alternatively, use the mass of the appropriate number of 1H atoms (because its mass is the same as the mass of one electron and one proton).
B Calculate the mass defect by subtracting the experimental mass from the calculated mass.
C Determine the nuclear binding energy by multiplying the mass defect by the change in energy in electronvolts per atom. Divide this value by the number of nucleons to obtain the binding energy per nucleon.
Solution
A An iron-56 atom has 26 protons, 26 electrons, and 30 neutrons. We could add the masses of these three sets of particles; however, noting that 26 protons and 26 electrons are equivalent to 26 1H atoms, we can calculate the sum of the masses more quickly as follows:
\begin{align*}\textrm{calculated mass}&=26(\textrm{mass }^1_1\textrm H)+30(\textrm{mass }^1_0 \textrm n)\[4pt] &=26(1.007825)\textrm{amu}+30(1.008665)\textrm{amu}=56.463400\textrm{ amu}\ \textrm{experimental mass} &=55.934938 \end{align*} \nonumber
B We subtract to find the mass defect:
\begin{align*}\textrm{mass defect}&=\textrm{calculated mass}-\textrm{experimental mass} \&=56.463400\textrm{ amu}-55.934938\textrm{ amu}=0.528462\textrm{ amu}\end{align*} \nonumber
C The nuclear binding energy is thus 0.528462 amu × 931 MeV/amu = 492 MeV. The binding energy per nucleon is 492 MeV/56 nucleons = 8.79 MeV/nucleon.
Exercise $2$
Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for 238U.
Answer
1800 MeV/238U; 7.57 MeV/nucleon
Summary
Unlike a chemical reaction, a nuclear reaction results in a significant change in mass and an associated change of energy, as described by Einstein’s equation. Nuclear reactions are accompanied by large changes in energy, which result in detectable changes in mass. The change in mass is related to the change in energy according to Einstein’s equation: ΔE = (Δm)c2. Large changes in energy are usually reported in kiloelectronvolts or megaelectronvolts (thousands or millions of electronvolts). With the exception of 1H, the experimentally determined mass of an atom is always less than the sum of the masses of the component particles (protons, neutrons, and electrons) by an amount called the mass defect of the nucleus. The energy corresponding to the mass defect is the nuclear binding energy, the amount of energy released when a nucleus forms from its component particles. In nuclear fission, nuclei split into lighter nuclei with an accompanying release of multiple neutrons and large amounts of energy. The critical mass is the minimum mass required to support a self-sustaining nuclear chain reaction. Nuclear fusion is a process in which two light nuclei combine to produce a heavier nucleus plus a great deal of energy. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/21%3A_Nuclear_Chemistry/21.05%3A_Energy_Changes_in_Nuclear_Reactions.txt |
Learning Objectives
• Explain nuclear fission
• Relate the concepts of critical mass and nuclear chain reactions
• Summarize basic requirements for nuclear fission
Many heavier elements with smaller binding energies per nucleon can decompose into more stable elements that have intermediate mass numbers and larger binding energies per nucleon—that is, mass numbers and binding energies per nucleon that are closer to the “peak” of the binding energy graph near 56. Sometimes neutrons are also produced. This decomposition is called fission, the breaking of a large nucleus into smaller pieces. The breaking is rather random with the formation of a large number of different products. Fission usually does not occur naturally, but is induced by bombardment with neutrons. The first reported nuclear fission occurred in 1939 when three German scientists, Lise Meitner, Otto Hahn, and Fritz Strassman, bombarded uranium-235 atoms with slow-moving neutrons that split the U-238 nuclei into smaller fragments that consisted of several neutrons and elements near the middle of the periodic table. Since then, fission has been observed in many other isotopes, including most actinide isotopes that have an odd number of neutrons. A typical nuclear fission reaction is shown in Figure \(1\).
Among the products of Meitner, Hahn, and Strassman’s fission reaction were barium, krypton, lanthanum, and cerium, all of which have nuclei that are more stable than uranium-235. Since then, hundreds of different isotopes have been observed among the products of fissionable substances. A few of the many reactions that occur for U-235, and a graph showing the distribution of its fission products and their yields, are shown in Figure \(2\). Similar fission reactions have been observed with other uranium isotopes, as well as with a variety of other isotopes such as those of plutonium.
A tremendous amount of energy is produced by the fission of heavy elements. For instance, when one mole of U-235 undergoes fission, the products weigh about 0.2 grams less than the reactants; this “lost” mass is converted into a very large amount of energy, about 1.8 × 1010 kJ per mole of U-235. Nuclear fission reactions produce incredibly large amounts of energy compared to chemical reactions. The fission of 1 kilogram of uranium-235, for example, produces about 2.5 million times as much energy as is produced by burning 1 kilogram of coal.
As described earlier, when undergoing fission U-235 produces two “medium-sized” nuclei, and two or three neutrons. These neutrons may then cause the fission of other uranium-235 atoms, which in turn provide more neutrons that can cause fission of even more nuclei, and so on. If this occurs, we have a nuclear chain reaction (Figure \(3\)). On the other hand, if too many neutrons escape the bulk material without interacting with a nucleus, then no chain reaction will occur.
Material that can sustain a nuclear fission chain reaction is said to be fissile or fissionable. (Technically, fissile material can undergo fission with neutrons of any energy, whereas fissionable material requires high-energy neutrons.) Nuclear fission becomes self-sustaining when the number of neutrons produced by fission equals or exceeds the number of neutrons absorbed by splitting nuclei plus the number that escape into the surroundings. The amount of a fissionable material that will support a self-sustaining chain reaction is a critical mass. An amount of fissionable material that cannot sustain a chain reaction is a subcritical mass. An amount of material in which there is an increasing rate of fission is known as a supercritical mass. The critical mass depends on the type of material: its purity, the temperature, the shape of the sample, and how the neutron reactions are controlled (Figure \(4\)).
An atomic bomb (Figure \(5\)) contains several pounds of fissionable material, \(\ce{^{235}_{92}U}\) or \(\ce{^{239}_{94}Pu}\), a source of neutrons, and an explosive device for compressing it quickly into a small volume. When fissionable material is in small pieces, the proportion of neutrons that escape through the relatively large surface area is great, and a chain reaction does not take place. When the small pieces of fissionable material are brought together quickly to form a body with a mass larger than the critical mass, the relative number of escaping neutrons decreases, and a chain reaction and explosion result.
Fission Reactors
Chain reactions of fissionable materials can be controlled and sustained without an explosion in a nuclear reactor (Figure \(6\)). Any nuclear reactor that produces power via the fission of uranium or plutonium by bombardment with neutrons must have at least five components: nuclear fuel consisting of fissionable material, a nuclear moderator, reactor coolant, control rods, and a shield and containment system. We will discuss these components in greater detail later in the section. The reactor works by separating the fissionable nuclear material such that a critical mass cannot be formed, controlling both the flux and absorption of neutrons to allow shutting down the fission reactions. In a nuclear reactor used for the production of electricity, the energy released by fission reactions is trapped as thermal energy and used to boil water and produce steam. The steam is used to turn a turbine, which powers a generator for the production of electricity.
Nuclear Fuels
Nuclear fuel consists of a fissionable isotope, such as uranium-235, which must be present in sufficient quantity to provide a self-sustaining chain reaction. In the United States, uranium ores contain from 0.05–0.3% of the uranium oxide U3O8; the uranium in the ore is about 99.3% nonfissionable U-238 with only 0.7% fissionable U-235. Nuclear reactors require a fuel with a higher concentration of U-235 than is found in nature; it is normally enriched to have about 5% of uranium mass as U-235. At this concentration, it is not possible to achieve the supercritical mass necessary for a nuclear explosion. Uranium can be enriched by gaseous diffusion (the only method currently used in the US), using a gas centrifuge, or by laser separation.
In the gaseous diffusion enrichment plant where U-235 fuel is prepared, UF6 (uranium hexafluoride) gas at low pressure moves through barriers that have holes just barely large enough for UF6 to pass through. The slightly lighter 235UF6 molecules diffuse through the barrier slightly faster than the heavier 238UF6 molecules. This process is repeated through hundreds of barriers, gradually increasing the concentration of 235UF6 to the level needed by the nuclear reactor. The basis for this process, Graham’s law, is described in the chapter on gases. The enriched UF6 gas is collected, cooled until it solidifies, and then taken to a fabrication facility where it is made into fuel assemblies. Each fuel assembly consists of fuel rods that contain many thimble-sized, ceramic-encased, enriched uranium (usually UO2) fuel pellets. Modern nuclear reactors may contain as many as 10 million fuel pellets. The amount of energy in each of these pellets is equal to that in almost a ton of coal or 150 gallons of oil.
Nuclear Moderators
Neutrons produced by nuclear reactions move too fast to cause fission (Figure \(4\)). They must first be slowed to be absorbed by the fuel and produce additional nuclear reactions. A nuclear moderator is a substance that slows the neutrons to a speed that is low enough to cause fission. Early reactors used high-purity graphite as a moderator. Modern reactors in the US exclusively use heavy water \(\ce{( ^2_1H2O)}\) or light water (ordinary H2O), whereas some reactors in other countries use other materials, such as carbon dioxide, beryllium, or graphite.
Reactor Coolants
A nuclear reactor coolant is used to carry the heat produced by the fission reaction to an external boiler and turbine, where it is transformed into electricity. Two overlapping coolant loops are often used; this counteracts the transfer of radioactivity from the reactor to the primary coolant loop. All nuclear power plants in the US use water as a coolant. Other coolants include molten sodium, lead, a lead-bismuth mixture, or molten salts.
Control Rods
Nuclear reactors use control rods (Figure \(8\)) to control the fission rate of the nuclear fuel by adjusting the number of slow neutrons present to keep the rate of the chain reaction at a safe level. Control rods are made of boron, cadmium, hafnium, or other elements that are able to absorb neutrons. Boron-10, for example, absorbs neutrons by a reaction that produces lithium-7 and alpha particles:
\[\ce{^{10}_5B + ^1_0n⟶ ^7_3Li + ^4_2He}\]
When control rod assemblies are inserted into the fuel element in the reactor core, they absorb a larger fraction of the slow neutrons, thereby slowing the rate of the fission reaction and decreasing the power produced. Conversely, if the control rods are removed, fewer neutrons are absorbed, and the fission rate and energy production increase. In an emergency, the chain reaction can be shut down by fully inserting all of the control rods into the nuclear core between the fuel rods.
Shield and Containment System
During its operation, a nuclear reactor produces neutrons and other radiation. Even when shut down, the decay products are radioactive. In addition, an operating reactor is thermally very hot, and high pressures result from the circulation of water or another coolant through it. Thus, a reactor must withstand high temperatures and pressures, and must protect operating personnel from the radiation. Reactors are equipped with a containment system (or shield) that consists of three parts:
1. The reactor vessel, a steel shell that is 3–20-centimeters thick and, with the moderator, absorbs much of the radiation produced by the reactor
2. A main shield of 1–3 meters of high-density concrete
3. A personnel shield of lighter materials that protects operators from γ rays and X-rays
In addition, reactors are often covered with a steel or concrete dome that is designed to contain any radioactive materials might be released by a reactor accident.
Video \(1\): Click here to watch a 3-minute video from the Nuclear Energy Institute on how nuclear reactors work.
Nuclear power plants are designed in such a way that they cannot form a supercritical mass of fissionable material and therefore cannot create a nuclear explosion. But as history has shown, failures of systems and safeguards can cause catastrophic accidents, including chemical explosions and nuclear meltdowns (damage to the reactor core from overheating). The following Chemistry in Everyday Life feature explores three infamous meltdown incidents.
Nuclear Accidents
The importance of cooling and containment are amply illustrated by three major accidents that occurred with the nuclear reactors at nuclear power generating stations in the United States (Three Mile Island), the former Soviet Union (Chernobyl), and Japan (Fukushima).
In March 1979, the cooling system of the Unit 2 reactor at Three Mile Island Nuclear Generating Station in Pennsylvania failed, and the cooling water spilled from the reactor onto the floor of the containment building. After the pumps stopped, the reactors overheated due to the high radioactive decay heat produced in the first few days after the nuclear reactor shut down. The temperature of the core climbed to at least 2200 °C, and the upper portion of the core began to melt. In addition, the zirconium alloy cladding of the fuel rods began to react with steam and produced hydrogen:
\[\ce{Zr}(s)+\ce{2H2O}(g)⟶\ce{ZrO2}(s)+\ce{2H2}(g)\]
The hydrogen accumulated in the confinement building, and it was feared that there was danger of an explosion of the mixture of hydrogen and air in the building. Consequently, hydrogen gas and radioactive gases (primarily krypton and xenon) were vented from the building. Within a week, cooling water circulation was restored and the core began to cool. The plant was closed for nearly 10 years during the cleanup process.
Although zero discharge of radioactive material is desirable, the discharge of radioactive krypton and xenon, such as occurred at the Three Mile Island plant, is among the most tolerable. These gases readily disperse in the atmosphere and thus do not produce highly radioactive areas. Moreover, they are noble gases and are not incorporated into plant and animal matter in the food chain. Effectively none of the heavy elements of the core of the reactor were released into the environment, and no cleanup of the area outside of the containment building was necessary (Figure \(8\)).
Another major nuclear accident involving a reactor occurred in April 1986, at the Chernobyl Nuclear Power Plant in Ukraine, which was still a part of the former Soviet Union. While operating at low power during an unauthorized experiment with some of its safety devices shut off, one of the reactors at the plant became unstable. Its chain reaction became uncontrollable and increased to a level far beyond what the reactor was designed for. The steam pressure in the reactor rose to between 100 and 500 times the full power pressure and ruptured the reactor. Because the reactor was not enclosed in a containment building, a large amount of radioactive material spewed out, and additional fission products were released, as the graphite (carbon) moderator of the core ignited and burned. The fire was controlled, but over 200 plant workers and firefighters developed acute radiation sickness and at least 32 soon died from the effects of the radiation. It is predicted that about 4000 more deaths will occur among emergency workers and former Chernobyl residents from radiation-induced cancer and leukemia. The reactor has since been encapsulated in steel and concrete, a now-decaying structure known as the sarcophagus. Almost 30 years later, significant radiation problems still persist in the area, and Chernobyl largely remains a wasteland.
In 2011, the Fukushima Daiichi Nuclear Power Plant in Japan was badly damaged by a 9.0-magnitude earthquake and resulting tsunami. Three reactors up and running at the time were shut down automatically, and emergency generators came online to power electronics and coolant systems. However, the tsunami quickly flooded the emergency generators and cut power to the pumps that circulated coolant water through the reactors. High-temperature steam in the reactors reacted with zirconium alloy to produce hydrogen gas. The gas escaped into the containment building, and the mixture of hydrogen and air exploded. Radioactive material was released from the containment vessels as the result of deliberate venting to reduce the hydrogen pressure, deliberate discharge of coolant water into the sea, and accidental or uncontrolled events.
An evacuation zone around the damaged plant extended over 12.4 miles away, and an estimated 200,000 people were evacuated from the area. All 48 of Japan’s nuclear power plants were subsequently shut down, remaining shuttered as of December 2014. Since the disaster, public opinion has shifted from largely favoring to largely opposing increasing the use of nuclear power plants, and a restart of Japan’s atomic energy program is still stalled (Figure \(1\)0).
The energy produced by a reactor fueled with enriched uranium results from the fission of uranium as well as from the fission of plutonium produced as the reactor operates. As discussed previously, the plutonium forms from the combination of neutrons and the uranium in the fuel. In any nuclear reactor, only about 0.1% of the mass of the fuel is converted into energy. The other 99.9% remains in the fuel rods as fission products and unused fuel. All of the fission products absorb neutrons, and after a period of several months to a few years, depending on the reactor, the fission products must be removed by changing the fuel rods. Otherwise, the concentration of these fission products would increase and absorb more neutrons until the reactor could no longer operate.
Spent fuel rods contain a variety of products, consisting of unstable nuclei ranging in atomic number from 25 to 60, some transuranium elements, including plutonium and americium, and unreacted uranium isotopes. The unstable nuclei and the transuranium isotopes give the spent fuel a dangerously high level of radioactivity. The long-lived isotopes require thousands of years to decay to a safe level. The ultimate fate of the nuclear reactor as a significant source of energy in the United States probably rests on whether or not a politically and scientifically satisfactory technique for processing and storing the components of spent fuel rods can be developed. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/21%3A_Nuclear_Chemistry/21.06%3A_Nuclear_Fission.txt |
Learning Objectives
• Describe the nuclear reactions in a nuclear fusion reaction
• Quantify the energy released or absorbed in a fusion reaction
The process of converting very light nuclei into heavier nuclei is also accompanied by the conversion of mass into large amounts of energy, a process called fusion. The principal source of energy in the sun is a net fusion reaction in which four hydrogen nuclei fuse and produce one helium nucleus and two positrons. This is a net reaction of a more complicated series of events:
$\ce{4^1_1H ⟶ ^4_2He + 2^0_{+1}n}$
A helium nucleus has a mass that is 0.7% less than that of four hydrogen nuclei; this lost mass is converted into energy during the fusion. This reaction produces about 3.6 × 1011 kJ of energy per mole of $\ce{^4_2He}$ produced. This is somewhat larger than the energy produced by the nuclear fission of one mole of U-235 (1.8 × 1010 kJ), and over 3 million times larger than the energy produced by the (chemical) combustion of one mole of octane (5471 kJ).
It has been determined that the nuclei of the heavy isotopes of hydrogen, a deuteron, $^2_1H$ and a triton, $^3_1H$, undergo fusion at extremely high temperatures (thermonuclear fusion). They form a helium nucleus and a neutron:
$\ce{^2_1H + ^3_1H ⟶ ^4_2He + 2^1_0n}$
This change proceeds with a mass loss of 0.0188 amu, corresponding to the release of 1.69 × 109 kilojoules per mole of $\ce{^4_2He}$ formed. The very high temperature is necessary to give the nuclei enough kinetic energy to overcome the very strong repulsive forces resulting from the positive charges on their nuclei so they can collide.
The most important fusion process in nature is the one that powers stars. In the 20th century, it was realized that the energy released from nuclear fusion reactions accounted for the longevity of the Sun and other stars as a source of heat and light. The fusion of nuclei in a star, starting from its initial hydrogen and helium abundance, provides that energy and synthesizes new nuclei as a byproduct of that fusion process. The prime energy producer in the Sun is the fusion of hydrogen to form helium, which occurs at a solar-core temperature of 14 million kelvin. The net result is the fusion of four protons into one alpha particle, with the release of two positrons, two neutrinos (which changes two of the protons into neutrons), and energy (Figure $2$).
Example $1$
Calculate the energy released in each of the following hypothetical processes.
1. $\ce{3 ^4_2He \rightarrow ^{12}_6C}$
2. $\ce{6 ^1_1H + 6 ^1_0n \rightarrow ^{12}_6C}$
3. $\ce{6 ^2_1D \rightarrow ^{12}_6C}$
Solution
1. $Q_a = 3 \times 4.0026 - 12.000) \,amu \times (1.4924\times 10^{-10} \,J/amu) = 1.17 \times 10^{-12} \,J$
2. $Q_b = (6 \times (1.007825 + 1.008665) - 12.00000)\, amu \times (1.4924\times 10^{1-0} J/amu) = 1.476\times 10^{-11} \,J$
3. $Q_c = 6 \times 2.014102 - 12.00000 \, amu \times (1.4924\times 10^{-10} \, J/amu) = 1.263\times 10^{-11}\, J$
Fusion of $\ce{He}$ to give $\ce{C}$ releases the least amount of energy, because the fusion to produce He has released a large amount. The difference between the second and the third is the binding energy of deuterium. The conservation of mass-and-energy is well illustrated in these calculations. On the other hand, the calculation is based on the conservation of mass-and-energy.
Nuclear Reactors
Useful fusion reactions require very high temperatures for their initiation—about 15,000,000 K or more. At these temperatures, all molecules dissociate into atoms, and the atoms ionize, forming plasma. These conditions occur in an extremely large number of locations throughout the universe—stars are powered by fusion. Humans have already figured out how to create temperatures high enough to achieve fusion on a large scale in thermonuclear weapons. A thermonuclear weapon such as a hydrogen bomb contains a nuclear fission bomb that, when exploded, gives off enough energy to produce the extremely high temperatures necessary for fusion to occur.
Another much more beneficial way to create fusion reactions is in a fusion reactor, a nuclear reactor in which fusion reactions of light nuclei are controlled. Because no solid materials are stable at such high temperatures, mechanical devices cannot contain the plasma in which fusion reactions occur. Two techniques to contain plasma at the density and temperature necessary for a fusion reaction are currently the focus of intensive research efforts: containment by a magnetic field and by the use of focused laser beams (Figure $3$). A number of large projects are working to attain one of the biggest goals in science: getting hydrogen fuel to ignite and produce more energy than the amount supplied to achieve the extremely high temperatures and pressures that are required for fusion. At the time of this writing, there are no self-sustaining fusion reactors operating in the world, although small-scale controlled fusion reactions have been run for very brief periods.Contributors | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/21%3A_Nuclear_Chemistry/21.07%3A_Nuclear_Fusion.txt |
Learning Objectives
• To know the differences between ionizing and nonionizing radiation and their effects on matter.
• To identify natural and artificial sources of radiation.
Because nuclear reactions do not typically affect the valence electrons of the atom (although electron capture draws an electron from an orbital of the lowest energy level), they do not directly cause chemical changes. Nonetheless, the particles and the photons emitted during nuclear decay are very energetic, and they can indirectly produce chemical changes in the matter surrounding the nucleus that has decayed. For instance, an α particle is an ionized helium nucleus (He2+) that can act as a powerful oxidant. In this section, we describe how radiation interacts with matter and the some of the chemical and biological effects of radiation.
Ionizing versus Nonionizing Radiation
The effects of radiation on matter are determined primarily by the energy of the radiation, which depends on the nuclear decay reaction that produced it. Nonionizing radiation is relatively low in energy; when it collides with an atom in a molecule or an ion, most or all of its energy can be absorbed without causing a structural or a chemical change. Instead, the kinetic energy of the radiation is transferred to the atom or molecule with which it collides, causing it to rotate, vibrate, or move more rapidly. Because this energy can be transferred to adjacent molecules or ions in the form of heat, many radioactive substances are warm to the touch. Highly radioactive elements such as polonium, for example, have been used as heat sources in the US space program. As long as the intensity of the nonionizing radiation is not great enough to cause overheating, it is relatively harmless, and its effects can be neutralized by cooling.
In contrast, ionizing radiation is higher in energy, and some of its energy can be transferred to one or more atoms with which it collides as it passes through matter. If enough energy is transferred, electrons can be excited to very high energy levels, resulting in the formation of positively charged ions:
$\mathrm{atom + ionizing\: radiation \rightarrow ion^+ + \, {e^-}\label{Eq1}}$
Molecules that have been ionized in this way are often highly reactive, and they can decompose or undergo other chemical changes that create a cascade of reactive molecules that can damage biological tissues and other materials (Figure $1$). Because the energy of ionizing radiation is very high, we often report its energy in units such as megaelectronvolts (MeV) per particle:
$\text{1 MeV/particle} = \text{96 billion J/mol}. \nonumber$
The Effects of Ionizing Radiation on Matter
The effects of ionizing radiation depend on four factors:
1. The type of radiation, which dictates how far it can penetrate into matter
2. The energy of the individual particles or photons
3. The number of particles or photons that strike a given area per unit time
4. The chemical nature of the substance exposed to the radiation
The relative abilities of the various forms of ionizing radiation to penetrate biological tissues are illustrated in Figure $2$. Because of its high charge and mass, α radiation interacts strongly with matter. Consequently, it does not penetrate deeply into an object, and it can be stopped by a piece of paper, clothing, or skin. In contrast, γ rays, with no charge and essentially no mass, do not interact strongly with matter and penetrate deeply into most objects, including the human body. Several inches of lead or more than 12 inches of special concrete are needed to completely stop γ rays. Because β particles are intermediate in mass and charge between α particles and γ rays, their interaction with matter is also intermediate. Beta particles readily penetrate paper or skin, but they can be stopped by a piece of wood or a relatively thin sheet of metal.
Because of their great penetrating ability, γ rays are by far the most dangerous type of radiation when they come from a source outside the body. Alpha particles, however, are the most damaging if their source is inside the body because internal tissues absorb all of their energy. Thus danger from radiation depends strongly on the type of radiation emitted and the extent of exposure, which allows scientists to safely handle many radioactive materials if they take precautions to avoid, for example, inhaling fine particulate dust that contains alpha emitters. Some properties of ionizing radiation are summarized in Table $1$.
Table $1$: Some Properties of Ionizing Radiation
Type Energy Range (MeV) Penetration Distance in Water* Penetration Distance in Air*
*Distance at which half of the radiation has been absorbed.
α particles 3–9 < 0.05 mm < 10 cm
β particles ≤ 3 < 4 mm 1 m
x-rays <10−2 < 1 cm < 3 m
γ rays 10−2–101 < 20 cm > 3 m
There are many different ways to measure radiation exposure, or the dose. The roentgen (R), which measures the amount of energy absorbed by dry air, can be used to describe quantitative exposure.Named after the German physicist Wilhelm Röntgen (1845–1923; Nobel Prize in Physics, 1901), who discovered x-rays. The roentgen is actually defined as the amount of radiation needed to produce an electrical charge of 2.58 × 10−4 C in 1 kg of dry air. Damage to biological tissues, however, is proportional to the amount of energy absorbed by tissues, not air. The most common unit used to measure the effects of radiation on biological tissue is the rad (radiation absorbed dose); the SI equivalent is the gray (Gy). The rad is defined as the amount of radiation that causes 0.01 J of energy to be absorbed by 1 kg of matter, and the gray is defined as the amount of radiation that causes 1 J of energy to be absorbed per kilogram:
$\mathrm{1\: rad = 0.010\: J/kg \hspace{25 pt} 1\: Gy = 1\: J/kg \label{Eq2}}$
Thus a 70 kg human who receives a dose of 1.0 rad over his or her entire body absorbs 0.010 J/70 kg = 1.4 × 10−4 J, or 0.14 mJ. To put this in perspective, 0.14 mJ is the amount of energy transferred to your skin by a 3.8 × 10−5 g droplet of boiling water. Because the energy of the droplet of water is transferred to a relatively large area of tissue, it is harmless. A radioactive particle, however, transfers its energy to a single molecule, which makes it the atomic equivalent of a bullet fired from a high-powered rifle.
Because α particles have a much higher mass and charge than β particles or γ rays, the difference in mass between α and β particles is analogous to being hit by a bowling ball instead of a table tennis ball traveling at the same speed. Thus the amount of tissue damage caused by 1 rad of α particles is much greater than the damage caused by 1 rad of β particles or γ rays. Thus a unit called the rem (roentgen equivalent in man) was devised to describe the actual amount of tissue damage caused by a given amount of radiation. The number of rems of radiation is equal to the number of rads multiplied by the RBE (relative biological effectiveness) factor, which is 1 for β particles, γ rays, and x-rays and about 20 for α particles. Because actual radiation doses tend to be very small, most measurements are reported in millirems (1 mrem = 10−3 rem).
Wilhelm Röntgen
Born in the Lower Rhine Province of Germany, Röntgen was the only child of a cloth manufacturer and merchant. His family moved to the Netherlands where he showed no particular aptitude in school, but where he was fond of roaming the countryside. Röntgen was expelled from technical school in Utrecht after being unjustly accused of drawing a caricature of one of the teachers. He began studying mechanical engineering in Zurich, which he could enter without having the credentials of a regular student, and received a PhD at the University of Zurich in 1869. In 1876 he became professor of physics.
Natural Sources of Radiation
We are continuously exposed to measurable background radiation from a variety of natural sources, which, on average, is equal to about 150–600 mrem/yr (Figure $3$). One component of background radiation is cosmic rays, high-energy particles and $\gamma$ rays emitted by the sun and other stars, which bombard Earth continuously. Because cosmic rays are partially absorbed by the atmosphere before they reach Earth’s surface, the exposure of people living at sea level (about 30 mrem/yr) is significantly less than the exposure of people living at higher altitudes (about 50 mrem/yr in Denver, Colorado). Every 4 hours spent in an airplane at greater than 30,000 ft adds about 1 mrem to a person’s annual radiation exposure.
A second component of background radiation is cosmogenic radiation, produced by the interaction of cosmic rays with gases in the upper atmosphere. When high-energy cosmic rays collide with oxygen and nitrogen atoms, neutrons and protons are released. These, in turn, react with other atoms to produce radioactive isotopes, such as $\ce{^{14}C}$:
$\ce{^{14}_7 N + ^1_0 n \rightarrow ^{14}_6 C + ^1_1p }\label{Eq3}$
The carbon atoms react with oxygen atoms to form CO2, which is eventually washed to Earth’s surface in rain and taken up by plants. About 1 atom in 1 × 1012 of the carbon atoms in our bodies is radioactive 14C, which decays by beta emission. About 5000 14C nuclei disintegrate in your body during the 15 s or so that it takes you to read this paragraph. Tritium (3H) is also produced in the upper atmosphere and falls to Earth in precipitation. The total radiation dose attributable to 14C is estimated to be 1 mrem/yr, while that due to 3H is about 1000 times less.
The third major component of background radiation is terrestrial radiation, which is due to the remnants of radioactive elements that were present on primordial Earth and their decay products. For example, many rocks and minerals in the soil contain small amounts of radioactive isotopes, such as $\ce{^{232}Th}$ and $\ce{^{238}U}$ as well as radioactive daughter isotopes, such as$\ce{^{226}Ra}$. The amount of background radiation from these sources is about the same as that from cosmic rays (approximately 30 mrem/yr). These isotopes are also found in small amounts in building materials derived from rocks and minerals, which significantly increases the radiation exposure for people who live in brick or concrete-block houses (60–160 mrem/yr) instead of houses made of wood (10–20 mrem/yr). Our tissues also absorb radiation (about 40 mrem/yr) from naturally occurring radioactive elements that are present in our bodies. For example, the average adult contains about 140 g of potassium as the $K^+$ ion. Naturally occurring potassium contains 0.0117% $\ce{^{40}K}$, which decays by emitting both a β particle and a (\gamma\) ray. In the last 20 seconds, about the time it took you to read this paragraph, approximately 40,000 $\ce{^{40}K}$ nuclei disintegrated in your body.
By far the most important source of background radiation is radon, the heaviest of the noble gases (group 18). Radon-222 is produced during the decay of238U, and other isotopes of radon are produced by the decay of other heavy elements. Even though radon is chemically inert, all its isotopes are radioactive. For example, 222Rn undergoes two successive alpha-decay events to give 214Pb:
$\ce{^{222}_{86} Rn \rightarrow ^4_2\alpha + ^{218}_{84} Po + ^4_2\alpha + ^{214}_{82} Pb } \label{Eq4}$
Because radon is a dense gas, it tends to accumulate in enclosed spaces such as basements, especially in locations where the soil contains greater-than-average amounts of naturally occurring uranium minerals. Under most conditions, radioactive decay of radon poses no problems because of the very short range of the emitted α particle. If an atom of radon happens to be in your lungs when it decays, however, the chemically reactive daughter isotope polonium-218 can become irreversibly bound to molecules in the lung tissue. Subsequent decay of $\ce{^{218}Po}$ releases an α particle directly into one of the cells lining the lung, and the resulting damage can eventually cause lung cancer. The $\ce{^{218}Po}$ isotope is also readily absorbed by particles in cigarette smoke, which adhere to the surface of the lungs and can hold the radioactive isotope in place. Recent estimates suggest that radon exposure is a contributing factor in about 15% of the deaths due to lung cancer. Because of the potential health problem radon poses, many states require houses to be tested for radon before they can be sold. By current estimates, radon accounts for more than half of the radiation exposure of a typical adult in the United States.
Artificial Sources of Radiation
In addition to naturally occurring background radiation, humans are exposed to small amounts of radiation from a variety of artificial sources. The most important of these are the x-rays used for diagnostic purposes in medicine and dentistry, which are photons with much lower energy than γ rays. A single chest x-ray provides a radiation dose of about 10 mrem, and a dental x-ray about 2–3 mrem. Other minor sources include television screens and computer monitors with cathode-ray tubes, which also produce x-rays. Luminescent paints for watch dials originally used radium, a highly toxic alpha emitter if ingested by those painting the dials. Radium was replaced by tritium (3H) and promethium (147Pr), which emit low-energy β particles that are absorbed by the watch crystal or the glass covering the instrument. Radiation exposure from television screens, monitors, and luminescent dials totals about 2 mrem/yr. Residual fallout from previous atmospheric nuclear-weapons testing is estimated to account for about twice this amount, and the nuclear power industry accounts for less than 1 mrem/yr (about the same as a single 4 h jet flight).
Example $1$
Calculate the annual radiation dose in rads a typical 70 kg chemistry student receives from the naturally occurring 40K in his or her body, which contains about 140 g of potassium (as the K+ ion). The natural abundance of 40K is 0.0117%. Each 1.00 mol of 40K undergoes 1.05 × 107 decays/s, and each decay event is accompanied by the emission of a 1.32 MeV β particle.
Given: mass of student, mass of isotope, natural abundance, rate of decay, and energy of particle
Asked for: annual radiation dose in rads
Strategy:
1. Calculate the number of moles of 40K present using its mass, molar mass, and natural abundance.
2. Determine the number of decays per year for this amount of 40K.
3. Multiply the number of decays per year by the energy associated with each decay event. To obtain the annual radiation dose, use the mass of the student to convert this value to rads.
Solution
A The number of moles of 40K present in the body is the total number of potassium atoms times the natural abundance of potassium atoms present as 40K divided by the atomic mass of 40K:
$\textrm{moles }^{40}\textrm K= 140\textrm{ g K} \times \dfrac{0.0117\textrm{ mol }^{40}\textrm K}{100\textrm{ mol K}}\times\dfrac{1\textrm{ mol K}}{40.0\textrm{ g K}}=4.10\times10^{-4}\mathrm{\,mol\,^{40}K} \nonumber$
B We are given the number of atoms of 40K that decay per second in 1.00 mol of 40K, so the number of decays per year is as follows:
$\dfrac{\textrm{decays}}{\textrm{year}}=4.10\times10^{-4}\mathrm{\,mol^{40}\,K}\times\dfrac{1.05\times10^7\textrm{ decays/s}}{\mathrm{1.00\,mol\,^{40}K}}\times\dfrac{60\textrm{ s}}{1\textrm{ min}}\times\dfrac{60\textrm{ min}}{1\textrm{ h}}\times\dfrac{24\textrm{ h}}{1\textrm{ day}}\times\dfrac{365\textrm{ days}}{1\textrm{ yr}}$
C The total energy the body receives per year from the decay of 40K is equal to the total number of decays per year multiplied by the energy associated with each decay event:
\begin{align*}\textrm{total energy per year}&=\dfrac{1.36\times10^{11}\textrm{ decays}}{\textrm{yr}}\times\dfrac{1.32\textrm{ MeV}}{\textrm{decays}}\times\dfrac{10^6\textrm{ eV}}{\textrm{MeV}}\times\dfrac{1.602\times10^{-19}\textrm{ J}}{\textrm{eV}}\&=2.87\times10^{-2}\textrm{ J/yr}\end{align*} \nonumber
We use the definition of the rad (1 rad = 10−2 J/kg of tissue) to convert this figure to a radiation dose in rads. If we assume the dose is equally distributed throughout the body, then the radiation dose per year is as follows:
\begin{align*}\textrm{radiation dose per year}&=\dfrac{2.87\times10^{-2}\textrm{ J/yr}}{\textrm{70.0 kg}}\times\dfrac{1\textrm{ rad}}{1\times10^{-2}\textrm{ J/kg}}\&=4.10\times10^{-2}\textrm{ rad/yr}=41\textrm{ mrad/yr}\end{align*} \nonumber
This corresponds to almost half of the normal background radiation most people experience.
Exercise $1$
Because strontium is chemically similar to calcium, small amounts of the Sr2+ ion are taken up by the body and deposited in calcium-rich tissues such as bone, using the same mechanism that is responsible for the absorption of Ca2+. Consequently, the radioactive strontium (90Sr) found in fission waste and released by atmospheric nuclear-weapons testing is a major health concern. A normal 70 kg human body has about 280 mg of strontium, and each mole of 90Sr undergoes 4.55 × 1014 decays/s by the emission of a 0.546 MeV β particle. What would be the annual radiation dose in rads for a 70 kg person if 0.10% of the strontium ingested were 90Sr?
Answer
5.7 × 103 rad/yr (which is 10 times the fatal dose)
Assessing the Impact of Radiation Exposure
One of the more controversial public policy issues debated today is whether the radiation exposure from artificial sources, when combined with exposure from natural sources, poses a significant risk to human health. The effects of single radiation doses of different magnitudes on humans are listed in Table $2$. Because of the many factors involved in radiation exposure (length of exposure, intensity of the source, and energy and type of particle), it is difficult to quantify the specific dangers of one radioisotope versus another. Nonetheless, some general conclusions regarding the effects of radiation exposure are generally accepted as valid.
Table $2$: The Effects of a Single Radiation Dose on a 70 kg Human
Dose (rem) Symptoms/Effects
< 5 no observable effect
5–20 possible chromosomal damage
20–100 temporary reduction in white blood cell count
50–100 temporary sterility in men (up to a year)
100–200 mild radiation sickness, vomiting, diarrhea, fatigue; immune system suppressed; bone growth in children retarded
> 300 permanent sterility in women
> 500 fatal to 50% within 30 days; destruction of bone marrow and intestine
> 3000 fatal within hours
Radiation doses of 600 rem and higher are invariably fatal, while a dose of 500 rem kills half the exposed subjects within 30 days. Smaller doses (≤ 50 rem) appear to cause only limited health effects, even though they correspond to tens of years of natural radiation. This does not, however, mean that such doses have no ill effects; they may cause long-term health problems, such as cancer or genetic changes that affect offspring. The possible detrimental effects of the much smaller doses attributable to artificial sources (< 100 mrem/yr) are more difficult to assess.
The tissues most affected by large, whole-body exposures are bone marrow, intestinal tissue, hair follicles, and reproductive organs, all of which contain rapidly dividing cells. The susceptibility of rapidly dividing cells to radiation exposure explains why cancers are often treated by radiation. Because cancer cells divide faster than normal cells, they are destroyed preferentially by radiation. Long-term radiation-exposure studies on fruit flies show a linear relationship between the number of genetic defects and both the magnitude of the dose and the exposure time. In contrast, similar studies on mice show a much lower number of defects when a given dose of radiation is spread out over a long period of time rather than received all at once. Both patterns are plotted in Figure $4$, but which of the two is applicable to humans? According to one hypothesis, mice have very low risk from low doses because their bodies have ways of dealing with the damage caused by natural radiation. At much higher doses, however, their natural repair mechanisms are overwhelmed, leading to irreversible damage. Because mice are biochemically much more similar to humans than are fruit flies, many scientists believe that this model also applies to humans. In contrast, the linear model assumes that all exposure to radiation is intrinsically damaging and suggests that stringent regulation of low-level radiation exposure is necessary. Which view is more accurate? The answer—while yet unknown—has extremely important consequences for regulating radiation exposure.
Summary
Nonionizing radiation is relatively low in energy and can be used as a heat source, whereas ionizing radiation, which is higher in energy, can penetrate biological tissues and is highly reactive. The effects of radiation on matter depend on the energy of the radiation. Nonionizing radiation is relatively low in energy, and the energy is transferred to matter in the form of heat. Ionizing radiation is relatively high in energy, and when it collides with an atom, it can completely remove an electron to form a positively charged ion that can damage biological tissues. Alpha particles do not penetrate very far into matter, whereas γ rays penetrate more deeply. Common units of radiation exposure, or dose, are the roentgen (R), the amount of energy absorbed by dry air, and the rad (radiation absorbed dose), the amount of radiation that produces 0.01 J of energy in 1 kg of matter. The rem (roentgen equivalent in man) measures the actual amount of tissue damage caused by a given amount of radiation. Natural sources of radiation include cosmic radiation, consisting of high-energy particles and γ rays emitted by the sun and other stars; cosmogenic radiation, which is produced by the interaction of cosmic rays with gases in the upper atmosphere; and terrestrial radiation, from radioactive elements present on primordial Earth and their decay products. The risks of ionizing radiation depend on the intensity of the radiation, the mode of exposure, and the duration of the exposure. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/21%3A_Nuclear_Chemistry/21.08%3A_Biological_Effects_of_Radiation.txt |
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
21.1: Radioactivity
Q21.1.1
Why are many radioactive substances warm to the touch? Why do many radioactive substances glow?
Q21.1.2
Describe the differences between nonionizing and ionizing radiation in terms of the intensity of energy emitted and the effect each has on an atom or molecule after collision. Which nuclear decay reactions are more likely to produce ionizing radiation? nonionizing radiation?
Q21.1.3
Would you expect nonionizing or ionizing radiation to be more effective at treating cancer? Why?
S21.1.3
Ionizing radiation is higher in energy and causes greater tissue damage, so it is more likely to destroy cancerous cells.
Q21.1.4
Historically, concrete shelters have been used to protect people from nuclear blasts. Comment on the effectiveness of such shelters.
Q21.1.5
Gamma rays are a very high-energy radiation, yet α particles inflict more damage on biological tissue. Why?
Q21.1.6
List the three primary sources of naturally occurring radiation. Explain the factors that influence the dose that one receives throughout the year. Which is the largest contributor to overall exposure? Which is the most hazardous?
S21.1.6
Three primary naturally occurring radiations are radium, uranium and thorium, each all having long half lives. Inhalation of air, ingestion of food and water,terrestrial radation from the ground and cosmic radiation from space are all factors tat influence the does that a person receives throughout the year. Inhalation of the air is the largest contributor to exposure. Radiation can damage DNA or kill cells. When radiation is exposed to your body, it will collide with atoms and this will change and damage your DNA.
Q21.1.7
Because radon is a noble gas, it is inert and generally unreactive. Despite this, exposure to even low concentrations of radon in air is quite dangerous. Describe the physical consequences of exposure to radon gas. Why are people who smoke more susceptible to these effects?
Q21.1.8
Most medical imaging uses isotopes that have extremely short half-lives. These isotopes usually undergo only one kind of nuclear decay reaction. Which kind of decay reaction is usually used? Why? Why would a short half-life be preferred in these cases?
Beta decay. Alfa decay can be easily stopped by paper, which means it can not be used to see inside people's body. Also, Gamma rays are really dangerous for human, that even a short period of time exploding to it will have negative effect on human body. Thus, Beta decay is the perfect choice. It can be used to see through human's body and stopped by aluminum or some other metals.
Since all these radioactive decays are harmful for human body, if the half time of these reactions are short, the time exploding to these reactions will be short too.
Q21.1.9
Which would you prefer: one exposure of 100 rem, or 10 exposures of 10 rem each? Explain your rationale.
S21.1.9
Ten exposures of 10 rem are less likely to cause major damage.
Q21.1.10
A 2.14 kg sample of rock contains 0.0985 g of uranium. How much energy is emitted over 25 yr if 99.27% of the uranium is 238U, which has a half-life of 4.46 × 109 yr, if each decay event is accompanied by the release of 4.039 MeV? If a 180 lb individual absorbs all of the emitted radiation, how much radiation has been absorbed in rads?
Q21.1.11
There is a story about a “radioactive boy scout” who attempted to convert thorium-232, which he isolated from about 1000 gas lantern mantles, to uranium-233 by bombarding the thorium with neutrons. The neutrons were generated via bombarding an aluminum target with α particles from the decay of americium-241, which was isolated from 100 smoke detectors. Write balanced nuclear reactions for these processes. The “radioactive boy scout” spent approximately 2 h/day with his experiment for 2 yr. Assuming that the alpha emission of americium has an energy of 5.24 MeV/particle and that the americium-241 was undergoing 3.5 × 106 decays/s, what was the exposure of the 60.0 kg scout in rads? The intrepid scientist apparently showed no ill effects from this exposure. Why?
241/95 Am---> 4/2 He + 237/93Np---> 4/2He + 233/91Pa----> 1/0n+ 232/91Th---> 1/1 H + 233/92 U
By adding alpha particles to the products side of the reaction, he was able to reduce the mass number by 4 and the atomic number by 2 to get the products he wanted. Bombardment with neutrons and 1 H was required to lower to the mass number to get Th and then raise both the mass number and the atomic number to yield Uranium.
2 hours*365*2= 1460 hours of exposure.*60min/1hr*60s/1min= 5.26*10^6s of exposure
1MeV= (1.6022*10^-13 Joules) * (5.24 MeV/particle)*2 particles= 1.679*10^-12 Joules.
(1.679*10^-12 Joules) * (1 amu/ 1.4924*10^-10 Joules)= 2.51*10^-13 amu
E=mc^2
E=(2.51*10^-13 amu)(1.66*10^-22kg/amu)(2.9998*10^8m/s)^2= (3.75*10^-18 kgm^2/s)*(3.5*10^6 decays/s)= 1.31*10^-11 joules of exposure per second.
The scientist showed no ill effects from this exposure because if we multiple the energy in joules of exposure per second, 1.31*10^-11, by the total amount of seconds of exposure, 5.26*10^6s, we find that he was only exposed to 6.9*10^-5 joules of radiation throughout the span of two years. This is a very small amount of radiation for such a long span of time.
In order to plug in the values for this equation, we must convert the given MeV to Joules with the known conversion rate. Similarly, we must convert Joules to amu with another known conversion rate. Then we can plug in the values and multiply by c^2 but we must not forget to multiple the amu by the conversion rate to kg in order to yield Joules. After all of this is done, we multiple the amount of Joules of exposure per second by the total amount of exposure in seconds in order to find out the total amount of exposure over the two year span.
21.2: Patterns of Nuclear Stability
Q21.2.1
How do chemical reactions compare with nuclear reactions with respect to mass changes? Does either type of reaction violate the law of conservation of mass? Explain your answers.
Q21.2.2
Why is the amount of energy released by a nuclear reaction so much greater than the amount of energy released by a chemical reaction?
Q21.2.3
Explain why the mass of an atom is less than the sum of the masses of its component particles.
Q21.2.4
The stability of a nucleus can be described using two values. What are they, and how do they differ from each other?
Q21.2.5
In the days before true chemistry, ancient scholars (alchemists) attempted to find the philosopher’s stone, a material that would enable them to turn lead into gold. Is the conversion of Pb → Au energetically favorable? Explain why or why not.
Q21.2.6
Describe the energy barrier to nuclear fusion reactions and explain how it can be overcome.
Q21.2.7
Imagine that the universe is dying, the stars have burned out, and all the elements have undergone fusion or radioactive decay. What would be the most abundant element in this future universe? Why?
Q21.2.8
Numerous elements can undergo fission, but only a few can be used as fuels in a reactor. What aspect of nuclear fission allows a nuclear chain reaction to occur?
Q21.2.9
How are transmutation reactions and fusion reactions related? Describe the main impediment to fusion reactions and suggest one or two ways to surmount this difficulty.
Q21.2.10
Using the information provided in Chapter 33, complete each reaction and calculate the amount of energy released from each in kilojoules.
1. 238Pa → ? + β
2. 216Fr → ? + α
3. 199Bi → ? + β+
Q21.2.22
Using the information provided in Chapter 33, complete each reaction and calculate the amount of energy released from each in kilojoules.
1. 194Tl → ? + β+
2. 171Pt → ? + α
3. 214Pb → ? + β
Q21.2.23
Using the information provided in Chapter 33, complete each reaction and calculate the amount of energy released from each in kilojoules per mole.
1. $_{91}^{234}\textrm{Pa}\rightarrow \,?+\,_{-1}^0\beta$
2. $_{88}^{226}\textrm{Ra}\rightarrow \,?+\,_2^4\alpha$
Q21.2.24
Using the information provided in Chapter 33, complete each reaction and then calculate the amount of energy released from each in kilojoules per mole.
1. $_{27}^{60}\textrm{Co}\rightarrow\,?+\,_{-1}^0\beta$ (The mass of cobalt-60 is 59.933817 amu.)
2. technicium-94 (mass = 93.909657 amu) undergoing fission to produce chromium-52 and potassium-40
Q21.2.25
Using the information provided in Chapter 33, predict whether each reaction is favorable and the amount of energy released or required in megaelectronvolts and kilojoules per mole.
1. the beta decay of bismuth-208 (mass = 207.979742 amu)
2. the formation of lead-206 by alpha decay
Q21.2.26
Using the information provided, predict whether each reaction is favorable and the amount of energy released or required in megaelectronvolts and kilojoules per mole.
1. alpha decay of oxygen-16
2. alpha decay to produce chromium-52
Q21.2.27
Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for 87Sr if the measured mass of 87Sr is 86.908877 amu.
1. the calculated mass
2. the mass defect
3. the nuclear binding energy
4. the nuclear binding energy per nucleon
Q21.2.30
The experimentally determined mass of 29S is 28.996610 amu. Find each of the following.
1. the calculated mass
2. the mass defect
3. the nuclear binding energy
4. the nuclear binding energy per nucleon
Q21.2.31
Calculate the amount of energy that is released by the neutron-induced fission of 235U to give 141Ba, 92Kr (mass = 91.926156 amu), and three neutrons. Report your answer in electronvolts per atom and kilojoules per mole.
Q21.2.31
Calculate the amount of energy that is released by the neutron-induced fission of 235U to give 90Sr, 143Xe, and three neutrons. Report your answer in electronvolts per atom and kilojoules per mole.
Q21.2.33
Calculate the amount of energy released or required by the fusion of helium-4 to produce the unstable beryllium-8 (mass = 8.00530510 amu). Report your answer in kilojoules per mole. Do you expect this to be a spontaneous reaction?
Q21.2.34
Calculate the amount of energy released by the fusion of 6Li and deuterium to give two helium-4 nuclei. Express your answer in electronvolts per atom and kilojoules per mole.
Q21.2.35
How much energy is released by the fusion of two deuterium nuclei to give one tritium nucleus and one proton? How does this amount compare with the energy released by the fusion of a deuterium nucleus and a tritium nucleus, which is accompanied by ejection of a neutron? Express your answer in megaelectronvolts and kilojoules per mole. Pound for pound, which is a better choice for a fusion reactor fuel mixture?
Numerical Answers
1
1. $_{91}^{238}\textrm{Pa}\rightarrow\,_{92}^{238}\textrm{U}+\,_{-1}^0\beta$; −5.540 × 10−16 kJ
2. $_{87}^{216}\textrm{Fr}\rightarrow\,_{85}^{212}\textrm{At}+\,_{2}^4\alpha$; −1.470 × 10−15 kJ
3. $_{83}^{199}\textrm{Bi}\rightarrow\,_{82}^{199}\textrm{Pb}+\,_{+1}^0\beta$; −5.458 × 10−16 kJ
1.
1. $_{91}^{234}\textrm{Pa}\rightarrow\,_{92}^{234}\textrm{U}+\,_{-1}^0\beta$; 2.118 × 108 kJ/mol
2. $_{88}^{226}\textrm{Ra}\rightarrow\,_{86}^{222}\textrm{Rn}+\,_{2}^4\alpha$; 4.700 × 108 kJ/mol
1.
1. The beta decay of bismuth-208 to polonium is endothermic (ΔE = 1.400 MeV/atom, 1.352 × 108 kJ/mol).
2. The formation of lead-206 by alpha decay of polonium-210 is exothermic (ΔE = −5.405 MeV/atom, −5.218 × 108 kJ/mol).
1. 757 MeV/atom, 8.70 MeV/nucleon
1.
1. 53.438245 amu
2. 0.496955 amu
3. 463 MeV/atom
4. 8.74 MeV/nucleon
1. −173 MeV/atom; 1.67 × 1010 kJ/mol
1. ΔE = + 9.0 × 106 kJ/mol beryllium-8; no
1. D–D fusion: ΔE = −4.03 MeV/tritium nucleus formed = −3.89 × 108 kJ/mol tritium; D–T fusion: ΔE = −17.6 MeV/tritium nucleus = −1.70 × 109 kJ/mol; D–T fusion
21.3: Nuclear Transmutations
Q21.3.1
Why do scientists believe that hydrogen is the building block of all other elements? Why do scientists believe that helium-4 is the building block of the heavier elements?
Q21.3.2
How does a star produce such enormous amounts of heat and light? How are elements heavier than Ni formed?
Q21.3.3
Propose an explanation for the observation that elements with even atomic numbers are more abundant than elements with odd atomic numbers.
S21.3.3
The raw material for all elements with Z > 2 is helium (Z = 2), and fusion of helium nuclei will always produce nuclei with an even number of protons.
Q21.3.4
During the lifetime of a star, different reactions that form different elements are used to power the fusion furnace that keeps a star “lit.” Explain the different reactions that dominate in the different stages of a star’s life cycle and their effect on the temperature of the star.
Q21.3.5
A line in a popular song from the 1960s by Joni Mitchell stated, “We are stardust….” Does this statement have any merit or is it just poetic? Justify your answer.
Q21.3.6
If the laws of physics were different and the primary element in the universe were boron-11 (Z = 5), what would be the next four most abundant elements? Propose nuclear reactions for their formation.
Q21.3.7
Write a balanced nuclear reaction for the formation of each isotope.
1. 27Al from two 12C nuclei
2. 9Be from two 4He nuclei
Q21.3.8
At the end of a star’s life cycle, it can collapse, resulting in a supernova explosion that leads to the formation of heavy elements by multiple neutron-capture events. Write a balanced nuclear reaction for the formation of each isotope during such an explosion.
1. 106Pd from nickel-58
2. selenium-79 from iron-56
Q21.3.9
When a star reaches middle age, helium-4 is converted to short-lived beryllium-8 (mass = 8.00530510 amu), which reacts with another helium-4 to produce carbon-12. How much energy is released in each reaction (in megaelectronvolts)? How many atoms of helium must be “burned” in this process to produce the same amount of energy obtained from the fusion of 1 mol of hydrogen atoms to give deuterium?
21.4: Rates of Radioactive Decay
Q21.4.1
What do chemists mean by the half-life of a reaction?
Q21.4.2
If a sample of one isotope undergoes more disintegrations per second than the same number of atoms of another isotope, how do their half-lives compare?
Q21.4.3
Half-lives for the reaction A + B → C were calculated at three values of [A]0, and [B] was the same in all cases. The data are listed in the following table:
[A]0 (M) t½ (s)
0.50 420
0.75 280
1.0 210
Does this reaction follow first-order kinetics? On what do you base your answer?
S21.4.3
1. No; the reaction is second order in A because the half-life decreases with increasing reactant concentration according to t1/2 = 1/k[A0].
Q21.4.4
Ethyl-2-nitrobenzoate (NO2C6H4CO2C2H5) hydrolyzes under basic conditions. A plot of [NO2C6H4CO2C2H5] versus t was used to calculate t½, with the following results:
[NO2C6H4CO2C2H5] (M/cm3) t½ (s)
0.050 240
0.040 300
0.030 400
Is this a first-order reaction? Explain your reasoning.
Q21.4.5
Azomethane (CH3N2CH3) decomposes at 600 K to C2H6 and N2. The decomposition is first order in azomethane. Calculate t½ from the data in the following table:
Time (s)
$P_{\large{\mathrm{CH_3N_2CH_3}}}$ (atm)
0 8.2 × 10−2
2000 3.99 × 10−2
4000 1.94 × 10−2
How long will it take for the decomposition to be 99.9% complete?
S21.4.5
t1/2 = 1.92 × 103 s or 1920 s; 19100 s or 5.32 hrs.
Q21.4.6
The first-order decomposition of hydrogen peroxide has a half-life of 10.7 h at 20°C. What is the rate constant (expressed in s−1) for this reaction? If you started with a solution that was 7.5 × 10−3 M H2O2, what would be the initial rate of decomposition (M/s)? What would be the concentration of H2O2 after 3.3 h?
Application Problems
1. Until the 1940s, uranium glazes were popular on ceramic dishware. One brand, Fiestaware, had bright orange glazes that could contain up to 20% uranium by mass. Although this practice is less common today due to the negative association of radiation, it is still possible to buy Depression-era glassware that is quite radioactive. Aqueous solutions in contact with this “hot” glassware can reach uranium concentrations up to 10 ppm by mass. If 1.0 g of uranium undergoes 1.11 × 107 decays/s, each to an α particle with an energy of 4.03 MeV, what would be your exposure in rem and rad if you drank 1.0 L of water that had been sitting for an extended time in a Fiestaware pitcher? Assume that the water and contaminants are excreted only after 18 h and that you weigh 70.0 kg.
2. Neutrography is a technique used to take the picture of an object using a beam of neutrons. How does the penetrating power of a neutron compare with alpha, beta, and gamma radiation? Do you expect similar penetration for protons? How would the biological damage of each particle compare with the other types of radiation? (Recall that a neutron’s mass is approximately 2000 times the mass of an electron.)
3. Spent fuel elements in a nuclear reactor contain radioactive fission products in addition to heavy metals. The conversion of nuclear fuel in a reactor is shown here:
Neglecting the fission products, write balanced nuclear reactions for the conversion of the original fuel to each product.
1. The first atomic bomb used 235U as a fissile material, but there were immense difficulties in obtaining sufficient quantities of pure 235U. A second fissile element, plutonium, was discovered in 1940, and it rapidly became important as a nuclear fuel. This element was produced by irradiating 238U with neutrons in a nuclear reactor. Complete the series that produced plutonium, all isotopes of which are fissile:
$_{92}^{238}\textrm U+\,_0^1\textrm n\rightarrow\,\textrm U\rightarrow\,\textrm{Np}\rightarrow\,\textrm{Pu}$
1. Boron neutron capture therapy is a potential treatment for many diseases. As the name implies, when boron-10, one of the naturally occurring isotopes of boron, is bombarded with neutrons, it absorbs a neutron and emits an α particle. Write a balanced nuclear reaction for this reaction. One advantage of this process is that neutrons cause little damage on their own, but when they are absorbed by boron-10, they can cause localized emission of alpha radiation. Comment on the utility of this treatment and its potential difficulties.
1. An airline pilot typically flies approximately 80 h per month. If 75% of that time is spent at an altitude of about 30,000 ft, how much radiation is that pilot receiving in one month? over a 30 yr career? Is the pilot receiving toxic doses of radiation?
1. At a breeder reactor plant, a 72 kg employee accidentally inhaled 2.8 mg of 239Pu dust. The isotope decays by alpha decay and has a half-life of 24,100 yr. The energy of the emitted α particles is 5.2 MeV, and the dust stays in the employee’s body for 18 h.
1. How many plutonium atoms are inhaled?
2. What is the energy absorbed by the body?
3. What is the physical dose in rads?
4. What is the dose in rems? Will the dosage be fatal?
1. For many years, the standard source for radiation therapy in the treatment of cancer was radioactive 60Co, which undergoes beta decay to 60Ni and emits two γ rays, each with an energy of 1.2 MeV. Show the sequence of nuclear reactions. If the half-life for beta decay is 5.27 yr, how many 60Co nuclei are present in a typical source undergoing 6000 dps that is used by hospitals? The mass of 60Co is 59.93 amu.
1. It is possible to use radioactive materials as heat sources to produce electricity. These radioisotope thermoelectric generators (RTGs) have been used in spacecraft and many other applications. Certain Cold War–era Russian-made RTGs used a 5.0 kg strontium-90 source. One mole of strontium-90 releases β particles with an energy of 0.545 MeV and undergoes 2.7 × 1013 decays/s. How many watts of power are available from this RTG (1 watt = 1 J/s)?
1. Potassium consists of three isotopes (potassium-39, potassium-40, and potassium-41). Potassium-40 is the least abundant, and it is radioactive, decaying to argon-40, a stable, nonradioactive isotope, by the emission of a β particle with a half-life of precisely 1.25 × 109 yr. Thus the ratio of potassium-40 to argon-40 in any potassium-40–containing material can be used to date the sample. In 1952, fragments of an early hominid, Meganthropus, were discovered near Modjokerto in Java. The bone fragments were lying on volcanic rock that was believed to be the same age as the bones. Potassium–argon dating on samples of the volcanic material showed that the argon-40-to-potassium-40 molar ratio was 0.00281:1. How old were the rock fragments? Could the bones also be the same age? Could radiocarbon dating have been used to date the fragments?
Answers
1. 6.6 × 10−3 rad; 0.13 rem
$_{92}^{235}\textrm U+\,_0^1\textrm n\rightarrow\,_{92}^{236}\textrm U+\,\gamma$
$_{92}^{238}\textrm U+\,_0^1\textrm n\rightarrow\,_{92}^{239}\textrm U+\,\gamma$
$_{92}^{239}\textrm U\rightarrow\,_{93}^{239}\textrm{Np}+\,_{-1}^0\beta$
$_{93}^{239}\textrm{Np}\rightarrow\,_{94}^{239}\textrm{Pu}+\,_{-1}^0\beta$
$_{94}^{239}\textrm{Pu}\rightarrow\,_{95}^{239}\textrm{Am}+\,_{-1}^0\beta$
$_{95}^{239}\textrm{Am}\rightarrow\,_{96}^{239}\textrm{Cm}+\,_{-1}^0\beta$
1. 7.1 × 1018 atoms of Pu
2. 0.35 J
3. 0.49 rad
4. 9.8 rem; this dose is unlikely to be fatal.
1. 130 W | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/21%3A_Nuclear_Chemistry/21.E%3A_Exercises.txt |
21.1: Radioactivity
• nucleons – neutron and proton
• all atoms of a given element have the same number of protons, atomic number
• isotopes – atoms with the same atomic number but different mass numbers
• three isotopes of uranium: uranium-233, uranium-235, uranium-238
• (superscript is mass number, subscript atomic number)
• radionuclides – nuclei that are radioactive
• radioisotopes – atoms containing radionuclides
21.1.1 Nuclear Equations
• alpha particles – helium-4 particles
• alpha radiation – stream of alpha particles
• emission of radiation is one way that an unstable nucleus is transformed into a more stable one
• superscript = mass number
• subscript = atomic number
• radioactive decay – when a nucleus spontaneously decomposes
• sum of the mass numbers is the same on both sides of the equation
• sum of the atomic numbers same on both sides of the equation
• radioactive properties of the nucleus are independent of the state of chemical combination of the atom
• chemical form does not matter when writing nuclear equations
21.1.2 Types of Radioactive Decay
• three most common type of radioactive decay: alpha(α), beta(β), and gamma(γ) radiation
Types of Radiation
Property α β γ
Charge 2+ 1- 0
Mass 6.64x10-24 g 9.11x10-28 g 0
Relative penetrating power 1 100 10,000
Nature of radiation $\ce{^{2}_4 He} \, \text{nuclei}$ electrons High-energy photons
• beta particles – high speed electrons emitted by an unstable nucleus
• beta decay results in increasing the atomic number
• gamma radiation – high-energy protons
• gamma radiation does not change atomic number or mass number or a nucleus
• almost always accompanies other radioactive emission
• represents the energy lost when the remaining nucleons reorganize into more stable arrangements
• positron – particle that has same mass as an electron but opposite charge
• represented by
• emission of a positron has effect of converting a proton to a neutron decreasing atomic number of nucleus by 1
• electron capture – the capture by the nucleus of an inner-shell electron from the electron cloud surrounding the nucleus
• has effect of converting a proton to neutron
Particle Symbol
Neutron $\ce{^{1}_0n}$
Proton $\ce{^{1}_1H}$ or $\ce{^{1}_1p}$
Electron $\ce{^{0}_{-1}e}$
Alpha Particle $\ce{^{4}_2 He}$ or $\ce{^{4}_2 \alpha}$
Beta Particle $\ce{^{0}_{-1} e}$ or $\ce{^{0}_{-1} \beta}$
Positron $\ce{^{0}_1e}$
21.2: Patterns of Nuclear Stability
21.2.1 Neutron-to-Proton Ratio
• strong nuclear force – a strong force of attraction between a large number of protons in the small volume of the nucleus
• stable nuclei with low atomic numbers up to 20 have nearly equal number of neutrons and protons
• for higher atomic numbers, the number of neutrons are greater than the number of protons
• the neutron-to-proton ratio of stable nuclei increase with increasing atomic number
• belt of stability – area where all stable nuclei are found
• ends at bismuth
• all nuclei with 84 or more protons are radioactive
• an even number of protons and neutrons is more stable than an odd number
• determining type of radioactive decay
• 1) nuclei above the belt of stability
• high neutron-to-proton ratios
• move toward belt of stability by emitting a beta particle
• decreases the number of neutrons and increases the number of protons in a nucleus
• 2) nuclei below the belt of stability
• low neutron-to-proton ratios
• move toward belt of stability by positron emission or electron capture
• increase number of neutrons and decrease the number of protons
• positron emission more common with lower nuclear charges
• electron capture becomes more common with increasing nuclear charge
• 3) nuclei with atomic numbers 84
• alpha emission
• decreases both number of neutrons and protons by 2
21.2.2 Radioactive Series
• some nuclei cannot game stability by a single emission
• radioactive series or nuclear disintegration series – series of nuclear reactions that begin with an unstable nucleus to a stable one
• three types of radioactive series found in nature
• uranium-238 to lead-206, uranium-235 to leat-207, and thorium-232 to lead-208
21.2.3 Further Observations
• nuclei with 2, 8, 20, 28, 50, or 82 protons or 2, 8, 20, 28, 50, 82, or 126 neutrons are more stable than with nuclei without these numbers
• numbers called magic numbers
• nuclei with even number of protons and neutrons more stable than with odd number of protons and neutrons
• observations made in terms of the shell model of the nucleus
• nucleons reside in shells
• magic numbers represent closed shells in nuclei
21.3: Nuclear Transmutations
• nuclear transmutations – nuclear reactions caused by the collision of one nucleus with a neutron or by another nucleus
• first conversion of one nucleus into another performed by Ernest Rutherford in 1919
• converted nitrogen-14 to oxygen-17
21.3.1 Using Charged Particles
• particle accelerators – used to accelerate particles at very high speeds
• cyclotron, and synchrotron
21.3.2 Using Neutrons
• neutrons do not need to be accelerated
21.3.4 Transuranium Elements
• transuranium elements – elements with atomic numbers above 92 that are produced by artificial transmutations
21.4: Rates of Radioactive Decay
• radioactive decay is a first-order process
• has characteristic of half life, which is the time required for half of any given quantity of a substance to react
• half-life unaffected by external conditions
21.4.1 Dating
• radiocarbon dating assumes that the ratio of carbon-14 to carbon-12 in the atmosphere has been constant for at least 50,000 years
• age of rocks can be determined by ratio of uranium-238 to lead-206
21.4.2 Calculations Based on Half-life
• rate = kN
• $k$ = decay constant, N = nuclei
• $\ln \dfrac{N_t}{N_o} = -k t \nonumber$
• t = time interval of decay, k = decay constant, N0 = initial number of nuclei at time zero, Nt = number remaining after time interval
• $k= \dfrac{0.693}{t_{1/2}} \nonumber$
21.5 Detection of Radioactivity
• Geiger counter – device used to measure and detect radioactivity
• Based on ionization of matter caused by radiation
• Phosphors – substances that give off light when exposed to radiation
• Scintillation counter – used to detect and measure radiation based on tiny flashes of light produced when radiation strikes a suitable phosphor
21.5.1 Radiotracers
• radioisotopes can be used to follow an element through its chemical reactions
• isotopes of same element have same properties
• radiotracer – radioisotopes used to trace an element
21.6: Energy Changes in Nuclear Reactions
$E=mc^2 \nonumber$
E = energy, m = mass, c = speed of light
If system loses mass, it loses energy (exothermic)
If system gains mass, it gains energy (endothermic)
21.6.1 Nuclear Binding Energies
• masses of nuclei always less than masses of individual nucleons
• mass defect – mass difference between a nucleus and its constituent nucleons
• energy is needed to break nucleus into separated protons and neutrons, addition of energy must also have an increase in mass
• nuclear binding energy – energy required to separate a nucleus into its individual nucleons
• the larger to nuclear binding energy the more stable the nucleus toward decomposition
• fission – energy produced when heavy nuclei split
• fusion – energy produced when light nuclei fuse
21.7: Nuclear Fission
• fission and fusion both exothermic
• chain reaction – reaction in which the neutrons produced in one fission cause further fission reactions
• in order for a fission chain reaction to occur, the sample of fissionable material must have a certain minimum mass
• critical mass – amount of fissionable material large enough to maintain the chain reaction with a constant rate of fission
• supercritical mass – mass in excess of a critical mass
21.7.1 Nuclear Reactors
• nuclear reactors the fission is controlled to generate a constant power
• reactor core consists of fissionable fuel, control rods, a moderator, and cooling fluid
• fission products are extremely radioactive and are thus hard to store
• about 20 half-lives needed for products to react acceptable levels for biological exposure
21.8: Nuclear Fusion
• fusion is appealing because of availability of light isotopes and fusion products are not radioactive
• high energies needed to overcome attraction of nuclei
• thermonuclear reactions – fusion reactions
• lowest temperature required is about 40,000,000 K
21.9: Biological Effects of Radiation
• when matter absorbs radiation, the energy of the radiation can cause either excitation or ionization
• ionization radiation more harmful than nonionization radiation
• most of energy of radiation absorbed by water molecules
• free radical – a substance with one ore more unpaired electrons
• can attack other biomolecules to produce more free radicals
• gamma rays most dangerous
• tissues that take most damage are the ones that reproduce at a rapid rate
• bone marrow, blood forming tissues, lymph nodes
21.9.1 Radiation Doses
• becquerel (Bq) – SI unit for activity of the radiation source; rate at which nuclear disintegrations are occurring
• 1 (Bq) = 1 nuclear disintegration/s
• curie (Ci) = 3.7x1010 disintegrations/s = rate of decay of 1g of radium
• two units used to measure amount of exposure to radiation: gray (Gy) and rad
• gray – SI unit of absorbed dose = absorption of 1 J of energy per kilogram of tissue
• rad (radiation absorbed dose) – absorption of 1x10-2 J of energy per kilogram of tissue
• 1 Gy = 100 rads
• relative biological effectiveness – RBE
• 1 for gamma and beta radiation, 10 for alpha radiation
• exact value varies with dose rate, total dose, and type of tissue affected
• rem (roentgen equivalent for man) – product of the radiation dose in rads and the RBE of the radiation gibes the effective dosage
• rem is unit of radiation damage that is usually used in medicine
• number of rems = (number of rads)(RBE)
• Sievert (Sv) – SI unit for dosage
• 1 Sv = 100 rem
• annual exposure = 360mrem
21.9.2 Radon
• radon exposure estimated to account for more than half annual exposure
• half-life of radon is 3.82 days
• decays into radioisotope polonium
• atoms of polonium can be trapped in lungs giving out alpha radiation causing lung cancer
• recommended levels of radon-222 in homes is to be less than 4 pCi per liter of air | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/21%3A_Nuclear_Chemistry/21.S%3A_Nuclear_Chemistry_%28Summary%29.txt |
The line that divides metals from nonmetals in the periodic table crosses the p block diagonally. As a result, the differences between metallic and nonmetallic properties are evident within each group, even though all members of each group have the same valence electron configuration. The p block is the only portion of the periodic table where we encounter the inert-pair effect. Moreover, as with the s-block elements, the chemistry of the lightest member of each group in the p block differs sharply from that of its heavier congeners but is similar to that of the element immediately below and to the right of it in the next group. Thus diagonal similarities in chemistry are seen across the p block.
A nonmetal is a chemical element that mostly lacks metallic attributes. Physically, nonmetals tend to be highly volatile (easily vaporized), have low elasticity, and are good insulators of heat and electricity; chemically, they tend to have high ionization energy and electronegativity values, and gain or share electrons when they react with other elements or compounds. Seventeen elements are generally classified as nonmetals; most are gases (hydrogen, helium, nitrogen, oxygen, fluorine, neon, chlorine, argon, krypton, xenon and radon); one is a liquid (bromine); and a few are solids (carbon, phosphorus, sulfur, selenium, and iodine).
As you study the periodic trends in properties and the reactivity of the elements in groups 13–18, you will learn how “cobalt blue” glass, rubies, and sapphires are made and why the US military became interested in using boron hydrides as rocket fuels but then abandoned its effort. You will also discover the source of diamonds on Earth, why silicon-based life-forms are likely to exist only in science fiction, and why most compounds with N–N bonds are potentially explosive. You will also learn why phosphorus can cause a painful and lethal condition known as “phossy jaw” and why selenium is used in photocopiers.
• 22.1: General Concepts- Periodic Trends and Reactions
The chemistry of the third-period element in a group is most representative of the chemistry of the group because the chemistry of the second-period elements is dominated by their small radii, energetically unavailable d orbitals, and tendency to form π bonds with other atoms.
• 22.2: Hydrogen
Hydrogen can lose an electron to form a proton, gain an electron to form a hydride ion, or form a covalent bond or polar covalent electron-pair bond. The three isotopes of hydrogen—protium (1H or H), deuterium (2H or D), and tritium (3H or T)—have different physical properties. Deuterium and tritium can be used as tracers, substances that enable biochemists to follow the path of a molecule through an organism or a cell.
• 22.3: Group 18- Noble Gases
The noble gases are characterized by their high ionization energies and low electron affinities. Potent oxidants are needed to oxidize the noble gases to form compounds. The noble gases have a closed-shell valence electron configuration. The ionization energies of the noble gases decrease with increasing atomic number. Only highly electronegative elements can form stable compounds (e.g. F and O) with the noble gases in positive oxidation states without being oxidized themselves.
• 22.4: Group 17- The Halogens
The halogens are highly reactive. All halogens have relatively high ionization energies, and the acid strength and oxidizing power of their oxoacids decreases down the group. The halogens are so reactive that none is found in nature as the free element; instead, all but iodine are found as halide salts with the X− ion. Their chemistry is exclusively that of nonmetals. Consistent with periodic trends, ionization energies decrease down the group.
• 22.5: Oxygen
Oxygen is an element that is widely known by the general public because of the large role it plays in sustaining life. Without oxygen, animals would be unable to breathe and would consequently die. Oxygen is not only important to supporting life, but plays an important role in many other chemical reactions. Oxygen is the most common element in the earth's crust and makes up about 20% of the air we breathe. Historically the discovery of oxygen as an element essential for combustion.
• 22.6: The Other Group 16 Elements- S, Se, Te, and Po
The chalcogens have no stable metallic elements. The tendency to catenate, the strength of single bonds, and the reactivity all decrease moving down the group. Because the electronegativity of the chalcogens decreases down the group, so does their tendency to acquire two electrons to form compounds in the −2 oxidation state. The lightest member, oxygen, has the greatest tendency to form multiple bonds with other elements.
• 22.7: Nitrogen
Nitrogen behaves chemically like nonmetals, Nitrogen forms compounds in nine different oxidation states. Nitrogen does not form stable catenated compounds because of repulsions between lone pairs of electrons on adjacent atoms, but it does form multiple bonds with other second-period atoms. Nitrogen reacts with electropositive elements to produce solids that range from covalent to ionic in character.
• 22.8: The Other Group 15 Elements - P, AS, Sb, and Bi
The reactivity of the heavier group 15 elements decreases down the group, as does the stability of their catenated compounds. In group 15, nitrogen and phosphorus behave chemically like nonmetals, arsenic and antimony behave like semimetals, and bismuth behaves like a metal. The stability of the +5 oxidation state decreases from phosphorus to bismuth because of the inert-pair effect. Due to their higher electronegativity, the lighter pnicogens form compounds in the −3 oxidation state.
• 22.9: Carbon
The stability of the carbon tetrahalides decreases as the halogen increases in size because of poor orbital overlap and steric crowding. Carbon forms three kinds of carbides with less electronegative elements: ionic carbides, which contain metal cations and C4− (methide) or C22− (acetylide) anions; interstitial carbides, which are characterized by covalent metal–carbon interactions and are among the hardest substances known; and covalent carbides, which have three-dimensional covalent network st
• 22.10: The Other Group 14 Elements- Si, Ge, Sn, and Pb
The group 14 elements show the greatest diversity in chemical behavior of any group; covalent bond strengths decease with increasing atomic size, and ionization energies are greater than expected, increasing from C to Pb. Because the covalent bond strength decreases with increasing atomic size and greater-than-expected ionization energies due to an increase in Zeff, the stability of the +2 oxidation state increases from carbon to lead.
• 22.11: Boron
Elemental boron is a semimetal that is remarkably unreactive. Boron forms unique and intricate structures that contain multicenter bonds, in which a pair of electrons holds together three or more atoms. Elemental boron can be induced to react with many nonmetallic elements to give binary compounds that have a variety of applications.
• 22.E: Chemistry of the Nonmetals (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al.
• 22.S: Chemistry of the Nonmetals (Summary)
This the summary for chapter 22 of the Textmap created for "Chemistry: The Central Science" by Brown et al.
22: Chemistry of the Nonmetals
Learning Objectives
• To know important periodic trends in several atomic properties.
As we begin our summary of periodic trends, recall that the single most important unifying principle in understanding the chemistry of the elements is the systematic increase in atomic number, accompanied by the orderly filling of atomic orbitals by electrons, which leads to periodicity in such properties as atomic and ionic size, ionization energy, electronegativity, and electron affinity. The same factors also lead to periodicity in valence electron configurations, which for each group results in similarities in oxidation states and the formation of compounds with common stoichiometries.
The most important periodic trends in atomic properties are summarized in Figure \(1\). Recall that these trends are based on periodic variations in a single fundamental property, the effective nuclear charge (Zeff), which increases from left to right and from top to bottom in the periodic table.
The diagonal line in Figure \(1\) separates the metals (to the left of the line) from the nonmetals (to the right of the line). Because metals have relatively low electronegativities, they tend to lose electrons in chemical reactions to elements that have relatively high electronegativities, forming compounds in which they have positive oxidation states. Conversely, nonmetals have high electronegativities, and they therefore tend to gain electrons in chemical reactions to form compounds in which they have negative oxidation states. The semimetals lie along the diagonal line dividing metals and nonmetals. It is not surprising that they tend to exhibit properties and reactivities intermediate between those of metals and nonmetals. Because the elements of groups 13, 14, and 15 span the diagonal line separating metals and nonmetals, their chemistry is more complex than predicted based solely on their valence electron configurations.
Unique Chemistry of the Lightest Elements
The chemistry of the second-period element of each group (n = 2: Li, Be, B, C, N, O, and F) differs in many important respects from that of the heavier members, or congeners, of the group. Consequently, the elements of the third period (n = 3: Na, Mg, Al, Si, P, S, and Cl) are generally more representative of the group to which they belong. The anomalous chemistry of second-period elements results from three important characteristics: small radii, energetically unavailable d orbitals, and a tendency to form pi (π) bonds with other atoms.
In contrast to the chemistry of the second-period elements, the chemistry of the third-period elements is more representative of the chemistry of the respective group.
Due to their small radii, second-period elements have electron affinities that are less negative than would be predicted from general periodic trends. When an electron is added to such a small atom, increased electron–electron repulsions tend to destabilize the anion. Moreover, the small sizes of these elements prevent them from forming compounds in which they have more than four nearest neighbors. Thus BF3 forms only the four-coordinate, tetrahedral BF4 ion, whereas under the same conditions AlF3 forms the six-coordinate, octahedral AlF63− ion. Because of the smaller atomic size, simple binary ionic compounds of second-period elements also have more covalent character than the corresponding compounds formed from their heavier congeners. The very small cations derived from second-period elements have a high charge-to-radius ratio and can therefore polarize the filled valence shell of an anion. As such, the bonding in such compounds has a significant covalent component, giving the compounds properties that can differ significantly from those expected for simple ionic compounds. As an example, LiCl, which is partially covalent in character, is much more soluble than NaCl in solvents with a relatively low dielectric constant, such as ethanol (ε = 25.3 versus 80.1 for H2O).
Because d orbitals are never occupied for principal quantum numbers less than 3, the valence electrons of second-period elements occupy 2s and 2p orbitals only. The energy of the 3d orbitals far exceeds the energy of the 2s and 2p orbitals, so using them in bonding is energetically prohibitive. Consequently, electron configurations with more than four electron pairs around a central, second-period element are simply not observed. You may recall that the role of d orbitals in bonding in main group compounds with coordination numbers of 5 or higher remains somewhat controversial. In fact, theoretical descriptions of the bonding in molecules such as SF6 have been published without mentioning the participation of d orbitals on sulfur. Arguments based on d-orbital availability and on the small size of the central atom, however, predict that coordination numbers greater than 4 are unusual for the elements of the second period, which is in agreement with experimental results.
One of the most dramatic differences between the lightest main group elements and their heavier congeners is the tendency of the second-period elements to form species that contain multiple bonds. For example, N is just above P in group 15: N2 contains an N≡N bond, but each phosphorus atom in tetrahedral P4 forms three P–P bonds. This difference in behavior reflects the fact that within the same group of the periodic table, the relative energies of the π bond and the sigma (σ) bond differ. A C=C bond, for example, is approximately 80% stronger than a C–C bond. In contrast, an Si=Si bond, with less p-orbital overlap between the valence orbitals of the bonded atoms because of the larger atomic size, is only about 40% stronger than an Si–Si bond. Consequently, compounds that contain both multiple and single C to C bonds are common for carbon, but compounds that contain only sigma Si–Si bonds are more energetically favorable for silicon and the other third-period elements.
Another important trend to note in main group chemistry is the chemical similarity between the lightest element of one group and the element immediately below and to the right of it in the next group, a phenomenon known as the diagonal effect (Figure \(2\)) There are, for example, significant similarities between the chemistry of Li and Mg, Be and Al, and B and Si. Both BeCl2 and AlCl3 have substantial covalent character, so they are somewhat soluble in nonpolar organic solvents. In contrast, although Mg and Be are in the same group, MgCl2 behaves like a typical ionic halide due to the lower electronegativity and larger size of magnesium.
The Inert-Pair Effect
The inert-pair effect refers to the empirical observation that the heavier elements of groups 13–17 often have oxidation states that are lower by 2 than the maximum predicted for their group. For example, although an oxidation state of +3 is common for group 13 elements, the heaviest element in group 13, thallium (Tl), is more likely to form compounds in which it has a +1 oxidation state. There appear to be two major reasons for the inert-pair effect: increasing ionization energies and decreasing bond strengths.
In moving down a group in the p-block, increasing ionization energies and decreasing bond strengths result in an inert-pair effect.
The ionization energies increase because filled (n − 1)d or (n − 2)f subshells are relatively poor at shielding electrons in ns orbitals. Thus the two electrons in the ns subshell experience an unusually high effective nuclear charge, so they are strongly attracted to the nucleus, reducing their participation in bonding. It is therefore substantially more difficult than expected to remove these ns2electrons, as shown in Table \(1\) by the difference between the first ionization energies of thallium and aluminum. Because Tl is less likely than Al to lose its two ns2 electrons, its most common oxidation state is +1 rather than +3.
Table \(1\): Ionization Energies (I) and Average M–Cl Bond Energies for the Group 13 Elements
Element Electron Configuration I1 (kJ/mol) I1 + I2 + I3 (kJ/mol) Average M–Cl Bond Energy (kJ/mol)
B [He] 2s22p1 801 6828 536
Al [Ne] 3s23p1 578 5139 494
Ga [Ar] 3d104s24p1 579 5521 481
In [Kr] 4d105s2p1 558 5083 439
Tl [Xe] 4f145d106s2p1 589 5439 373
Source of data: John A. Dean, Lange’s Handbook of Chemistry, 15th ed. (New York: McGraw-Hill, 1999).
Going down a group, the atoms generally became larger, and the overlap between the valence orbitals of the bonded atoms decreases. Consequently, bond strengths tend to decrease down a column. As shown by the M–Cl bond energies listed in Table \(1\), the strength of the bond between a group 13 atom and a chlorine atom decreases by more than 30% from B to Tl. Similar decreases are observed for the atoms of groups 14 and 15.
The net effect of these two factors—increasing ionization energies and decreasing bond strengths—is that in going down a group in the p-block, the additional energy released by forming two additional bonds eventually is not great enough to compensate for the additional energy required to remove the two ns2 electrons.
Example \(1\)
Based on the positions of the group 13 elements in the periodic table and the general trends outlined in this section,
1. classify these elements as metals, semimetals, or nonmetals.
2. predict which element forms the most stable compounds in the +1 oxidation state.
3. predict which element differs the most from the others in its chemistry.
4. predict which element of another group will exhibit chemistry most similar to that of Al.
Given: positions of elements in the periodic table
Asked for: classification, oxidation-state stability, and chemical reactivity
Strategy:
From the position of the diagonal line in the periodic table separating metals and nonmetals, classify the group 13 elements. Then use the trends discussed in this section to compare their relative stabilities and chemical reactivities.
Solution
1. Group 13 spans the diagonal line separating the metals from the nonmetals. Although Al and B both lie on the diagonal line, only B is a semimetal; the heavier elements are metals.
2. All five elements in group 13 have an ns2np1 valence electron configuration, so they are expected to form ions with a +3 charge from the loss of all valence electrons. The inert-pair effect should be most important for the heaviest element (Tl), so it is most likely to form compounds in an oxidation state that is lower by 2. Thus the +1 oxidation state is predicted to be most important for thallium.
3. Among the main group elements, the lightest member of each group exhibits unique chemistry because of its small size resulting in a high concentration of charge, energetically unavailable d orbitals, and a tendency to form multiple bonds. In group 13, we predict that the chemistry of boron will be quite different from that of its heavier congeners.
4. Within the s and p blocks, similarities between elements in different groups are most marked between the lightest member of one group and the element of the next group immediately below and to the right of it. These elements exhibit similar electronegativities and charge-to-radius ratios. Because Al is the second member of group 13, we predict that its chemistry will be most similar to that of Be, the lightest member of group 2.
Exercise \(1\)
Based on the positions of the group 14 elements C, Si, Ge, Sn, and Pb in the periodic table and the general trends outlined in this section,
1. classify these elements as metals, semimetals, or nonmetals.
2. predict which element forms the most stable compounds in the +2 oxidation state.
3. predict which element differs the most from the others in its chemistry.
4. predict which element of group 14 will be chemically most similar to a group 15 element.
Answer
1. nonmetal: C; semimetals: Si and Ge; metals: Sn and Pb
2. Pb is most stable as M2+.
3. C is most different.
4. C and P are most similar in chemistry.
Summary
The chemistry of the third-period element in a group is most representative of the chemistry of the group because the chemistry of the second-period elements is dominated by their small radii, energetically unavailable d orbitals, and tendency to form π bonds with other atoms. The most important unifying principle in describing the chemistry of the elements is that the systematic increase in atomic number and the orderly filling of atomic orbitals lead to periodic trends in atomic properties. The most fundamental property leading to periodic variations is the effective nuclear charge (Zeff). Because of the position of the diagonal line separating metals and nonmetals in the periodic table, the chemistry of groups 13, 14, and 15 is relatively complex. The second-period elements (n = 2) in each group exhibit unique chemistry compared with their heavier congeners because of their smaller radii, energetically unavailable d orbitals, and greater ability to form π bonds with other atoms. Increasing ionization energies and decreasing bond strengths lead to the inert-pair effect, which causes the heaviest elements of groups 13–17 to have a stable oxidation state that is lower by 2 than the maximum predicted for their respective groups. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/22%3A_Chemistry_of_the_Nonmetals/22.01%3A_General_Concepts-_Periodic_Trends_and_Reactions.txt |
Learning Objectives
• To describe the physical and chemical properties of hydrogen and predict its reactivity.
We now turn from an overview of periodic trends to a discussion of the s-block elements, first by focusing on hydrogen, whose chemistry is sufficiently distinct and important to be discussed in a category of its own. Most versions of the periodic table place hydrogen in the upper left corner immediately above lithium, implying that hydrogen, with a 1s1 electron configuration, is a member of group 1. In fact, the chemistry of hydrogen does not greatly resemble that of the metals of Group 1. Indeed, some versions of the periodic table place hydrogen above fluorine in Group 17 because the addition of a single electron to a hydrogen atom completes its valence shell.
Although hydrogen has an ns1 electron configuration, its chemistry does not resemble that of the Group 1 metals.
Isotopes of Hydrogen
Hydrogen, the most abundant element in the universe, is the ultimate source of all other elements by the process of nuclear fusion. Table $1$ "The Isotopes of Hydrogen" compares the three isotopes of hydrogen, all of which contain one proton and one electron per atom. The most common isotope is protium (1H or H), followed by deuterium (2H or D), which has an additional neutron. The rarest isotope of hydrogen is tritium (3H or T), which is produced in the upper atmosphere by a nuclear reaction when cosmic rays strike nitrogen and other atoms; it is then washed into the oceans by rainfall. Tritium is radioactive, decaying to 3He with a half-life of only 12.32 years. Consequently, the atmosphere and oceans contain only a very low, steady-state level of tritium. The term hydrogen and the symbol H normally refer to the naturally occurring mixture of the three isotopes.
Table $1$: The Isotopes of Hydrogen
Protium Deuterium Tritium
symbol $\mathrm{_1^1H}$ $\mathrm{_1^2H}$ $\mathrm{_1^3H}$
neutrons 0 1 2
mass (amu) 1.00783 2.0140 3.01605
abundance (%) 99.9885 0.0115 ~10−17
half-life (years) 12.32
boiling point of X2 (K) 20.28 23.67 25
melting point/boiling point of X2O (°C) 0.0/100.0 3.8/101.4 4.5/?
The different masses of the three isotopes of hydrogen cause them to have different physical properties. Thus H2, D2, and T2 differ in their melting points, boiling points, densities, and heats of fusion and vaporization. In 1931, Harold Urey and coworkers discovered deuterium by slowly evaporating several liters of liquid hydrogen until a volume of about 1 mL remained. When that remaining liquid was vaporized and its emission spectrum examined, they observed new absorption lines in addition to those previously identified as originating from hydrogen. The natural abundance of tritium, in contrast, is so low that it could not be detected by similar experiments; it was first prepared in 1934 by a nuclear reaction.
Harold Urey (1893–1981)
Urey won the Nobel Prize in Chemistry in 1934 for his discovery of deuterium (2H). Urey was born and educated in rural Indiana. After earning a BS in zoology from the University of Montana in 1917, Urey changed career directions. He earned his PhD in chemistry at Berkeley with G. N. Lewis and subsequently worked with Niels Bohr in Copenhagen. During World War II, Urey was the director of war research for the Atom Bomb Project at Columbia University. In later years, his research focused on the evolution of life. In 1953, he and his graduate student, Stanley Miller, showed that organic compounds, including amino acids, could be formed by passing an electric discharge through a mixture of compounds thought to be present in the atmosphere of primitive Earth.
Because the normal boiling point of D2O is 101.4°C (compared to 100.0°C for H2O), evaporation or fractional distillation can be used to increase the concentration of deuterium in a sample of water by the selective removal of the more volatile H2O. Thus bodies of water that have no outlet, such as the Great Salt Lake and the Dead Sea, which maintain their level solely by evaporation, have significantly higher concentrations of deuterated water than does lake or seawater with at least one outlet. A more efficient way to obtain water highly enriched in deuterium is by prolonged electrolysis of an aqueous solution. Because a deuteron (D+) has twice the mass of a proton (H+), it diffuses more slowly toward the electrode surface. Consequently, the gas evolved at the cathode is enriched in H, the species that diffuses more rapidly, favoring the formation of H2 over D2 or HD. Meanwhile, the solution becomes enriched in deuterium. Deuterium-rich water is called heavy water because the density of D2O (1.1044 g/cm3 at 25°C) is greater than that of H2O (0.99978 g/cm3). Heavy water was an important constituent of early nuclear reactors.
Because deuterons diffuse so much more slowly, D2O will not support life and is actually toxic if administered to mammals in large amounts. The rate-limiting step in many important reactions catalyzed by enzymes involves proton transfer. The transfer of D+ is so slow compared with that of H+ because bonds to D break more slowly than those to H, so the delicate balance of reactions in the cell is disrupted. Nonetheless, deuterium and tritium are important research tools for biochemists. By incorporating these isotopes into specific positions in selected molecules, where they act as labels, or tracers, biochemists can follow the path of a molecule through an organism or a cell. Tracers can also be used to provide information about the mechanism of enzymatic reactions.
Bonding in Hydrogen and Hydrogen-Containing Compounds
The 1s1 electron configuration of hydrogen indicates a single valence electron. Because the 1s orbital has a maximum capacity of two electrons, hydrogen can form compounds with other elements in three ways (Figure $1$):
1. Losing its electron to form a proton (H+) with an empty 1s orbital. The proton is a Lewis acid that can accept a pair of electrons from another atom to form an electron-pair bond. In the acid–base reactions, the proton always binds to a lone pair of electrons on an atom in another molecule to form a polar covalent bond. If the lone pair of electrons belongs to an oxygen atom of a water molecule, the result is the hydronium ion (H3O+).
2. Accepting an electron to form a hydride ion (H), which has a filled 1s2 orbital. Hydrogen reacts with relatively electropositive metals, such as the alkali metals (group 1) and alkaline earth metals (group 2), to form ionic hydrides, which contain metal cations and H ions.
3. Sharing its electron with an electron on another atom to form an electron-pair bond. With a half-filled 1s1 orbital, the hydrogen atom can interact with singly occupied orbitals on other atoms to form either a covalent or a polar covalent electron-pair bond, depending on the electronegativity of the other atom.
Hydrogen can also act as a bridge between two atoms. One familiar example is the hydrogen bond, an electrostatic interaction between a hydrogen bonded to an electronegative atom and an atom that has one or more lone pairs of electrons (Figure $2$). An example of this kind of interaction is the hydrogen bonding network found in water (Figure $2$). Hydrogen can also form a three-center bond (or electron-deficient bond), in which a hydride bridges two electropositive atoms. Compounds that contain hydrogen bonded to boron and similar elements often have this type of bonding. The B–H–B units found in boron hydrides cannot be described in terms of localized electron-pair bonds.
Because the H atom in the middle of such a unit can accommodate a maximum of only two electrons in its 1s orbital, the B–H–B unit can be described as containing a hydride that interacts simultaneously with empty sp3 orbitals on two boron atoms (Figure $3$). In these bonds, only two bonding electrons are used to hold three atoms together, making them electron-deficient bonds. You encountered a similar phenomenon in the discussion of π bonding in ozone and the nitrite ion. Recall that in both these cases, we used the presence of two electrons in a π molecular orbital extending over three atoms to explain the fact that the two O–O bond distances in ozone and the two N–O bond distances in nitrite are the same, which otherwise can be explained only by the use of resonance structures.
Hydrogen can lose its electron to form H+, accept an electron to form H, share its electron, hydrogen bond, or form a three-center bond.
Synthesis, Reactions, and Compounds of Hydrogen
The first known preparation of elemental hydrogen was in 1671, when Robert Boyle dissolved iron in dilute acid and obtained a colorless, odorless, gaseous product. Hydrogen was finally identified as an element in 1766, when Henry Cavendish showed that water was the sole product of the reaction of the gas with oxygen. The explosive properties of mixtures of hydrogen with air were not discovered until early in the 18th century; they partially caused the spectacular explosion of the hydrogen-filled dirigible Hindenburg in 1937 (Figure $4$). Due to its extremely low molecular mass, hydrogen gas is difficult to condense to a liquid (boiling point = 20.3 K), and solid hydrogen has one of the lowest melting points known (13.8 K).
The most common way to produce small amounts of highly pure hydrogen gas in the laboratory was discovered by Boyle: reacting an active metal (M), such as iron, magnesium, or zinc, with dilute acid:
$M_{(s)} + 2H^+_{(aq)} \rightarrow H_{2(g)} + M^{2+}_{(aq)} \label{21.1}$
Hydrogen gas can also be generated by reacting metals such as aluminum or zinc with a strong base:
$\mathrm{Al(s)}+\mathrm{OH^-(aq)}+\mathrm{3H_2O(l)}\rightarrow\frac{3}{2}\mathrm{H_2(g)}+\mathrm{[Al(OH)_4]^-(aq)} \label{21.2}$
Solid commercial drain cleaners such as Drano use this reaction to generate gas bubbles that help break up clogs in a drainpipe. Hydrogen gas is also produced by reacting ionic hydrides with water. Because ionic hydrides are expensive, however, this reaction is generally used for only specialized purposes, such as producing HD gas by reacting a hydride with D2O:
$MH_{(s)} + D_2O(l) \rightarrow HD_{(g)} + M^+(aq) + OD^−_{(aq)} \label{21.3}$
On an industrial scale, H2 is produced from methane by means of catalytic steam reforming, a method used to convert hydrocarbons to a mixture of CO and H2 known as synthesis gas, or syngas. The process is carried out at elevated temperatures (800°C) in the presence of a nickel catalyst:
$\mathrm{CH_4(g)}+\mathrm{H_2O(g)}\xrightarrow{\mathrm{Ni}}\mathrm{CO(g)}+\mathrm{3H_2(g)} \label{21.4}$
Most of the elements in the periodic table form binary compounds with hydrogen, which are collectively referred to as hydrides. Binary hydrides in turn can be classified in one of three ways, each with its own characteristic properties. Covalent hydrides contain hydrogen bonded to another atom via a covalent bond or a polar covalent bond. Covalent hydrides are usually molecular substances that are relatively volatile and have low melting points. Ionic hydrides contain the hydride ion as the anion with cations derived from electropositive metals. Like most ionic compounds, they are typically nonvolatile solids that contain three-dimensional lattices of cations and anions. Unlike most ionic compounds, however, they often decompose to H2(g) and the parent metal after heating. Metallic hydrides are formed by hydrogen and less electropositive metals such as the transition metals. The properties of metallic hydrides are usually similar to those of the parent metal. Consequently, metallic hydrides are best viewed as metals that contain many hydrogen atoms present as interstitial impurities.
Covalent hydrides are relatively volatile and have low melting points; ionic hydrides are generally nonvolatile solids in a lattice framework.
Summary and Key Takeaway
Hydrogen can lose an electron to form a proton, gain an electron to form a hydride ion, or form a covalent bond or polar covalent electron-pair bond. The three isotopes of hydrogen—protium (1H or H), deuterium (2H or D), and tritium (3H or T)—have different physical properties. Deuterium and tritium can be used as tracers, substances that enable biochemists to follow the path of a molecule through an organism or a cell. Hydrogen can form compounds that contain a proton (H+), a hydride ion (H), an electron-pair bond to H, a hydrogen bond, or a three-center bond (or electron-deficient bond), in which two electrons are shared between three atoms. Hydrogen gas can be generated by reacting an active metal with dilute acid, reacting Al or Zn with a strong base, or industrially by catalytic steam reforming, which produces synthesis gas, or syngas. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/22%3A_Chemistry_of_the_Nonmetals/22.02%3A_Hydrogen.txt |
Learning Objectives
• To understand the trends in properties and reactivity of the group 18 elements: the noble gases.
The noble gases were all isolated for the first time within a period of only five years at the end of the 19th century. Their very existence was not suspected until the 18th century, when early work on the composition of air suggested that it contained small amounts of gases in addition to oxygen, nitrogen, carbon dioxide, and water vapor. Helium was the first of the noble gases to be identified, when the existence of this previously unknown element on the sun was demonstrated by new spectral lines seen during a solar eclipse in 1868. Actual samples of helium were not obtained until almost 30 years later, however. In the 1890s, the English physicist J. W. Strutt (Lord Rayleigh) carefully measured the density of the gas that remained after he had removed all O2, CO2, and water vapor from air and showed that this residual gas was slightly denser than pure N2 obtained by the thermal decomposition of ammonium nitrite. In 1894, he and the Scottish chemist William Ramsay announced the isolation of a new “substance” (not necessarily a new element) from the residual nitrogen gas. Because they could not force this substance to decompose or react with anything, they named it argon (Ar), from the Greek argos, meaning “lazy.” Because the measured molar mass of argon was 39.9 g/mol, Ramsay speculated that it was a member of a new group of elements located on the right side of the periodic table between the halogens and the alkali metals. He also suggested that these elements should have a preferred valence of 0, intermediate between the +1 of the alkali metals and the −1 of the halogens.
J. W. Strutt (Lord Rayleigh) (1842–1919)
Lord Rayleigh was one of the few members of British higher nobility to be recognized as an outstanding scientist. Throughout his youth, his education was repeatedly interrupted by his frail health, and he was not expected to reach maturity. In 1861 he entered Trinity College, Cambridge, where he excelled at mathematics. A severe attack of rheumatic fever took him abroad, but in 1873 he succeeded to the barony and was compelled to devote his time to the management of his estates. After leaving the entire management to his younger brother, Lord Rayleigh was able to devote his time to science. He was a recipient of honorary science and law degrees from Cambridge University.
Sir William Ramsay (1852–1916)
Born and educated in Glasgow, Scotland, Ramsay was expected to study for the Calvanist ministry. Instead, he became interested in chemistry while reading about the manufacture of gunpowder. Ramsay earned his PhD in organic chemistry at the University of Tübingen in Germany in 1872. When he returned to England, his interests turned first to physical chemistry and then to inorganic chemistry. He is best known for his work on the oxides of nitrogen and for the discovery of the noble gases with Lord Rayleigh.
In 1895, Ramsey was able to obtain a terrestrial sample of helium for the first time. Then, in a single year (1898), he discovered the next three noble gases: krypton (Kr), from the Greek kryptos, meaning “hidden,” was identified by its orange and green emission lines; neon (Ne), from the Greek neos, meaning “new,” had bright red emission lines; and xenon (Xe), from the Greek xenos, meaning “strange,” had deep blue emission lines. The last noble gas was discovered in 1900 by the German chemist Friedrich Dorn, who was investigating radioactivity in the air around the newly discovered radioactive elements radium and polonium. The element was named radon (Rn), and Ramsay succeeded in obtaining enough radon in 1908 to measure its density (and thus its atomic mass). For their discovery of the noble gases, Rayleigh was awarded the Nobel Prize in Physics and Ramsay the Nobel Prize in Chemistry in 1904. Because helium has the lowest boiling point of any substance known (4.2 K), it is used primarily as a cryogenic liquid. Helium and argon are both much less soluble in water (and therefore in blood) than N2, so scuba divers often use gas mixtures that contain these gases, rather than N2, to minimize the likelihood of the “bends,” the painful and potentially fatal formation of bubbles of N2(g) that can occur when a diver returns to the surface too rapidly.
Preparation and General Properties of the Group 18 Elements
Fractional distillation of liquid air is the only source of all the noble gases except helium. Although helium is the second most abundant element in the universe (after hydrogen), the helium originally present in Earth’s atmosphere was lost into space long ago because of its low molecular mass and resulting high mean velocity. Natural gas often contains relatively high concentrations of helium (up to 7%), however, and it is the only practical terrestrial source.
The elements of group 18 all have closed-shell valence electron configurations, either ns2np6 or 1s2 for He. Consistent with periodic trends in atomic properties, these elements have high ionization energies that decrease smoothly down the group. From their electron affinities, the data in Table $1$ indicate that the noble gases are unlikely to form compounds in negative oxidation states. A potent oxidant is needed to oxidize noble gases and form compounds in positive oxidation states. Like the heavier halogens, xenon and perhaps krypton should form covalent compounds with F, O, and possibly Cl, in which they have even formal oxidation states (+2, +4, +6, and possibly +8). These predictions actually summarize the chemistry observed for these elements.
Table $1$: Selected Properties of the Group 18 Elements
Property Helium Neon Argon Krypton Xenon Radon
*The configuration shown does not include filled d and f subshells. This is the normal boiling point of He. Solid He does not exist at 1 atm pressure, so no melting point can be given.
atomic symbol He Ne Ar Kr Xe Rn
atomic number 2 10 18 36 54 86
atomic mass (amu) 4.00 20.18 39.95 83.80 131.29 222
valence electron configuration* 1s2 2s22p6 3s23p6 4s24p6 5s25p6 6s26p6
triple point/boiling point (°C) —/−269 −249 (at 43 kPa)/−246 −189 (at 69 kPa)/−189 −157/−153 −112 (at 81.6 kPa)/−108 −71/−62
density (g/L) at 25°C 0.16 0.83 1.63 3.43 5.37 9.07
atomic radius (pm) 31 38 71 88 108 120
first ionization energy (kJ/mol) 2372 2081 1521 1351 1170 1037
normal oxidation state(s) 0 0 0 0 (+2) 0 (+2, +4, +6, +8) 0 (+2)
electron affinity (kJ/mol) > 0 > 0 > 0 > 0 > 0 > 0
electronegativity 2.6
product of reaction with O2 none none none none not directly with oxygen, but $\ce{XeO3}$ can be formed by Equation \ref{Eq5}. none
type of oxide acidic
product of reaction with N2 none none none none none none
product of reaction with X2 none none none KrF2 XeF2, XeF4, XeF6 RnF2
product of reaction with H2 none none none none none none
Reactions and Compounds of the Noble Gases
For many years, it was thought that the only compounds the noble gases could form were clathrates. Clathrates are solid compounds in which a gas, the guest, occupies holes in a lattice formed by a less volatile, chemically dissimilar substance, the host (Figure $1$).
Because clathrate formation does not involve the formation of chemical bonds between the guest (Xe) and the host molecules (H2O, in the case of xenon hydrate), the guest molecules are immediately released when the clathrate is melted or dissolved.
Methane Clathrates
In addition to the noble gases, many other species form stable clathrates. One of the most interesting is methane hydrate, large deposits of which occur naturally at the bottom of the oceans. It is estimated that the amount of methane in such deposits could have a major impact on the world’s energy needs later in this century.
The widely held belief in the intrinsic lack of reactivity of the noble gases was challenged when Neil Bartlett, a British professor of chemistry at the University of British Columbia, showed that PtF6, a compound used in the Manhattan Project, could oxidize O2. Because the ionization energy of xenon (1170 kJ/mol) is actually lower than that of O2, Bartlett recognized that PtF6 should also be able to oxidize xenon. When he mixed colorless xenon gas with deep red PtF6 vapor, yellow-orange crystals immediately formed (Figure $3$). Although Bartlett initially postulated that they were $\ce{Xe^{+}PtF6^{−}}$, it is now generally agreed that the reaction also involves the transfer of a fluorine atom to xenon to give the $\ce{XeF^{+}}$ ion:
$\ce{Xe(g) + PtF6(g) -> [XeF^{+}][PtF5^{−}](s)} \label{Eq1}$
Subsequent work showed that xenon reacts directly with fluorine under relatively mild conditions to give XeF2, XeF4, or XeF6, depending on conditions; one such reaction is as follows:
$\ce{Xe(g) + 2F2(g) -> XeF4(s)} \label{Eq2}$
The ionization energies of helium, neon, and argon are so high (Table $1$) that no stable compounds of these elements are known. The ionization energies of krypton and xenon are lower but still very high; consequently only highly electronegative elements (F, O, and Cl) can form stable compounds with xenon and krypton without being oxidized themselves. Xenon reacts directly with only two elements: F2 and Cl2. Although $\ce{XeCl2}$ and $\ce{KrF2}$ can be prepared directly from the elements, they are substantially less stable than the xenon fluorides.
The ionization energies of helium, neon, and argon are so high that no stable compounds of these elements are known.
Because halides of the noble gases are powerful oxidants and fluorinating agents, they decompose rapidly after contact with trace amounts of water, and they react violently with organic compounds or other reductants. The xenon fluorides are also Lewis acids; they react with the fluoride ion, the only Lewis base that is not oxidized immediately on contact, to form anionic complexes. For example, reacting cesium fluoride with XeF6 produces CsXeF7, which gives Cs2XeF8 when heated:
$\ce{XeF6(s) + CsF(s) -> CsXeF7(s)} \label{Eq3}$
$\ce{2CsXeF7(s) ->[\Delta] Cs2XeF8(s) + XeF6(g)} \label{Eq4}$
The $\ce{XeF8^{2-}}$ ion contains eight-coordinate xenon and has the square antiprismatic structure, which is essentially identical to that of the IF8 ion. Cs2XeF8 is surprisingly stable for a polyatomic ion that contains xenon in the +6 oxidation state, decomposing only at temperatures greater than 300°C. Major factors in the stability of Cs2XeF8 are almost certainly the formation of a stable ionic lattice and the high coordination number of xenon, which protects the central atom from attack by other species. (Recall from that this latter effect is responsible for the extreme stability of SF6.)
For a previously “inert” gas, xenon has a surprisingly high affinity for oxygen, presumably because of π bonding between $O$ and $Xe$. Consequently, xenon forms an extensive series of oxides and oxoanion salts. For example, hydrolysis of either $XeF_4$ or $XeF_6$ produces $XeO_3$, an explosive white solid:
$\ce{XeF6(aq) + 3H2O(l) -> XeO3(aq) + 6HF(aq)} \label{Eq5}$
Treating a solution of XeO3 with ozone, a strong oxidant, results in further oxidation of xenon to give either XeO4, a colorless, explosive gas, or the surprisingly stable perxenate ion (XeO64−), both of which contain xenon in its highest possible oxidation state (+8). The chemistry of the xenon halides and oxides is best understood by analogy to the corresponding compounds of iodine. For example, XeO3 is isoelectronic with the iodate ion (IO3), and XeF82− is isoelectronic with the IF8 ion.
Xenon has a high affinity for both fluorine and oxygen.
Because the ionization energy of radon is less than that of xenon, in principle radon should be able to form an even greater variety of chemical compounds than xenon. Unfortunately, however, radon is so radioactive that its chemistry has not been extensively explored.
Example $1$
On a virtual planet similar to Earth, at least one isotope of radon is not radioactive. A scientist explored its chemistry and presented her major conclusions in a trailblazing paper on radon compounds, focusing on the kinds of compounds formed and their stoichiometries. Based on periodic trends, how did she summarize the chemistry of radon?
Given: nonradioactive isotope of radon
Asked for: summary of its chemistry
Strategy:
Based on the position of radon in the periodic table and periodic trends in atomic properties, thermodynamics, and kinetics, predict the most likely reactions and compounds of radon.
Solution
We expect radon to be significantly easier to oxidize than xenon. Based on its position in the periodic table, however, we also expect its bonds to other atoms to be weaker than those formed by xenon. Radon should be more difficult to oxidize to its highest possible oxidation state (+8) than xenon because of the inert-pair effect. Consequently, radon should form an extensive series of fluorides, including RnF2, RnF4, RnF6, and possibly RnF8 (due to its large radius). The ion RnF82− should also exist. We expect radon to form a series of oxides similar to those of xenon, including RnO3 and possibly RnO4. The biggest surprise in radon chemistry is likely to be the existence of stable chlorides, such as RnCl2 and possibly even RnCl4.
Exercise $1$
Predict the stoichiometry of the product formed by reacting XeF6 with a 1:1 stoichiometric amount of KF and propose a reasonable structure for the anion.
Answer
$\ce{KXeF7}$; the xenon atom in XeF7 has 16 valence electrons, which according to the valence-shell electron-pair repulsion model could give either a square antiprismatic structure with one fluorine atom missing or a pentagonal bipyramid if the 5s2 electrons behave like an inert pair that does not participate in bonding.
Summary
The noble gases are characterized by their high ionization energies and low electron affinities. Potent oxidants are needed to oxidize the noble gases to form compounds in positive oxidation states. The noble gases have a closed-shell valence electron configuration. The ionization energies of the noble gases decrease with increasing atomic number. Only highly electronegative elements can form stable compounds with the noble gases in positive oxidation states without being oxidized themselves. Xenon has a high affinity for both fluorine and oxygen, which form stable compounds that contain xenon in even oxidation states up to +8.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/22%3A_Chemistry_of_the_Nonmetals/22.03%3A_Group_18-_Noble_Gases.txt |
Learning Objectives
• To understand the trends in properties and reactivity of the group 18 elements: the noble gases.
The noble gases were all isolated for the first time within a period of only five years at the end of the 19th century. Their very existence was not suspected until the 18th century, when early work on the composition of air suggested that it contained small amounts of gases in addition to oxygen, nitrogen, carbon dioxide, and water vapor. Helium was the first of the noble gases to be identified, when the existence of this previously unknown element on the sun was demonstrated by new spectral lines seen during a solar eclipse in 1868. Actual samples of helium were not obtained until almost 30 years later, however. In the 1890s, the English physicist J. W. Strutt (Lord Rayleigh) carefully measured the density of the gas that remained after he had removed all O2, CO2, and water vapor from air and showed that this residual gas was slightly denser than pure N2 obtained by the thermal decomposition of ammonium nitrite. In 1894, he and the Scottish chemist William Ramsay announced the isolation of a new “substance” (not necessarily a new element) from the residual nitrogen gas. Because they could not force this substance to decompose or react with anything, they named it argon (Ar), from the Greek argos, meaning “lazy.” Because the measured molar mass of argon was 39.9 g/mol, Ramsay speculated that it was a member of a new group of elements located on the right side of the periodic table between the halogens and the alkali metals. He also suggested that these elements should have a preferred valence of 0, intermediate between the +1 of the alkali metals and the −1 of the halogens.
J. W. Strutt (Lord Rayleigh) (1842–1919)
Lord Rayleigh was one of the few members of British higher nobility to be recognized as an outstanding scientist. Throughout his youth, his education was repeatedly interrupted by his frail health, and he was not expected to reach maturity. In 1861 he entered Trinity College, Cambridge, where he excelled at mathematics. A severe attack of rheumatic fever took him abroad, but in 1873 he succeeded to the barony and was compelled to devote his time to the management of his estates. After leaving the entire management to his younger brother, Lord Rayleigh was able to devote his time to science. He was a recipient of honorary science and law degrees from Cambridge University.
Sir William Ramsay (1852–1916)
Born and educated in Glasgow, Scotland, Ramsay was expected to study for the Calvanist ministry. Instead, he became interested in chemistry while reading about the manufacture of gunpowder. Ramsay earned his PhD in organic chemistry at the University of Tübingen in Germany in 1872. When he returned to England, his interests turned first to physical chemistry and then to inorganic chemistry. He is best known for his work on the oxides of nitrogen and for the discovery of the noble gases with Lord Rayleigh.
In 1895, Ramsey was able to obtain a terrestrial sample of helium for the first time. Then, in a single year (1898), he discovered the next three noble gases: krypton (Kr), from the Greek kryptos, meaning “hidden,” was identified by its orange and green emission lines; neon (Ne), from the Greek neos, meaning “new,” had bright red emission lines; and xenon (Xe), from the Greek xenos, meaning “strange,” had deep blue emission lines. The last noble gas was discovered in 1900 by the German chemist Friedrich Dorn, who was investigating radioactivity in the air around the newly discovered radioactive elements radium and polonium. The element was named radon (Rn), and Ramsay succeeded in obtaining enough radon in 1908 to measure its density (and thus its atomic mass). For their discovery of the noble gases, Rayleigh was awarded the Nobel Prize in Physics and Ramsay the Nobel Prize in Chemistry in 1904. Because helium has the lowest boiling point of any substance known (4.2 K), it is used primarily as a cryogenic liquid. Helium and argon are both much less soluble in water (and therefore in blood) than N2, so scuba divers often use gas mixtures that contain these gases, rather than N2, to minimize the likelihood of the “bends,” the painful and potentially fatal formation of bubbles of N2(g) that can occur when a diver returns to the surface too rapidly.
Preparation and General Properties of the Group 18 Elements
Fractional distillation of liquid air is the only source of all the noble gases except helium. Although helium is the second most abundant element in the universe (after hydrogen), the helium originally present in Earth’s atmosphere was lost into space long ago because of its low molecular mass and resulting high mean velocity. Natural gas often contains relatively high concentrations of helium (up to 7%), however, and it is the only practical terrestrial source.
The elements of group 18 all have closed-shell valence electron configurations, either ns2np6 or 1s2 for He. Consistent with periodic trends in atomic properties, these elements have high ionization energies that decrease smoothly down the group. From their electron affinities, the data in Table $1$ indicate that the noble gases are unlikely to form compounds in negative oxidation states. A potent oxidant is needed to oxidize noble gases and form compounds in positive oxidation states. Like the heavier halogens, xenon and perhaps krypton should form covalent compounds with F, O, and possibly Cl, in which they have even formal oxidation states (+2, +4, +6, and possibly +8). These predictions actually summarize the chemistry observed for these elements.
Table $1$: Selected Properties of the Group 18 Elements
Property Helium Neon Argon Krypton Xenon Radon
*The configuration shown does not include filled d and f subshells. This is the normal boiling point of He. Solid He does not exist at 1 atm pressure, so no melting point can be given.
atomic symbol He Ne Ar Kr Xe Rn
atomic number 2 10 18 36 54 86
atomic mass (amu) 4.00 20.18 39.95 83.80 131.29 222
valence electron configuration* 1s2 2s22p6 3s23p6 4s24p6 5s25p6 6s26p6
triple point/boiling point (°C) —/−269 −249 (at 43 kPa)/−246 −189 (at 69 kPa)/−189 −157/−153 −112 (at 81.6 kPa)/−108 −71/−62
density (g/L) at 25°C 0.16 0.83 1.63 3.43 5.37 9.07
atomic radius (pm) 31 38 71 88 108 120
first ionization energy (kJ/mol) 2372 2081 1521 1351 1170 1037
normal oxidation state(s) 0 0 0 0 (+2) 0 (+2, +4, +6, +8) 0 (+2)
electron affinity (kJ/mol) > 0 > 0 > 0 > 0 > 0 > 0
electronegativity 2.6
product of reaction with O2 none none none none not directly with oxygen, but $\ce{XeO3}$ can be formed by Equation \ref{Eq5}. none
type of oxide acidic
product of reaction with N2 none none none none none none
product of reaction with X2 none none none KrF2 XeF2, XeF4, XeF6 RnF2
product of reaction with H2 none none none none none none
Reactions and Compounds of the Noble Gases
For many years, it was thought that the only compounds the noble gases could form were clathrates. Clathrates are solid compounds in which a gas, the guest, occupies holes in a lattice formed by a less volatile, chemically dissimilar substance, the host (Figure $1$).
Because clathrate formation does not involve the formation of chemical bonds between the guest (Xe) and the host molecules (H2O, in the case of xenon hydrate), the guest molecules are immediately released when the clathrate is melted or dissolved.
Methane Clathrates
In addition to the noble gases, many other species form stable clathrates. One of the most interesting is methane hydrate, large deposits of which occur naturally at the bottom of the oceans. It is estimated that the amount of methane in such deposits could have a major impact on the world’s energy needs later in this century.
The widely held belief in the intrinsic lack of reactivity of the noble gases was challenged when Neil Bartlett, a British professor of chemistry at the University of British Columbia, showed that PtF6, a compound used in the Manhattan Project, could oxidize O2. Because the ionization energy of xenon (1170 kJ/mol) is actually lower than that of O2, Bartlett recognized that PtF6 should also be able to oxidize xenon. When he mixed colorless xenon gas with deep red PtF6 vapor, yellow-orange crystals immediately formed (Figure $3$). Although Bartlett initially postulated that they were $\ce{Xe^{+}PtF6^{−}}$, it is now generally agreed that the reaction also involves the transfer of a fluorine atom to xenon to give the $\ce{XeF^{+}}$ ion:
$\ce{Xe(g) + PtF6(g) -> [XeF^{+}][PtF5^{−}](s)} \label{Eq1}$
Subsequent work showed that xenon reacts directly with fluorine under relatively mild conditions to give XeF2, XeF4, or XeF6, depending on conditions; one such reaction is as follows:
$\ce{Xe(g) + 2F2(g) -> XeF4(s)} \label{Eq2}$
The ionization energies of helium, neon, and argon are so high (Table $1$) that no stable compounds of these elements are known. The ionization energies of krypton and xenon are lower but still very high; consequently only highly electronegative elements (F, O, and Cl) can form stable compounds with xenon and krypton without being oxidized themselves. Xenon reacts directly with only two elements: F2 and Cl2. Although $\ce{XeCl2}$ and $\ce{KrF2}$ can be prepared directly from the elements, they are substantially less stable than the xenon fluorides.
The ionization energies of helium, neon, and argon are so high that no stable compounds of these elements are known.
Because halides of the noble gases are powerful oxidants and fluorinating agents, they decompose rapidly after contact with trace amounts of water, and they react violently with organic compounds or other reductants. The xenon fluorides are also Lewis acids; they react with the fluoride ion, the only Lewis base that is not oxidized immediately on contact, to form anionic complexes. For example, reacting cesium fluoride with XeF6 produces CsXeF7, which gives Cs2XeF8 when heated:
$\ce{XeF6(s) + CsF(s) -> CsXeF7(s)} \label{Eq3}$
$\ce{2CsXeF7(s) ->[\Delta] Cs2XeF8(s) + XeF6(g)} \label{Eq4}$
The $\ce{XeF8^{2-}}$ ion contains eight-coordinate xenon and has the square antiprismatic structure, which is essentially identical to that of the IF8 ion. Cs2XeF8 is surprisingly stable for a polyatomic ion that contains xenon in the +6 oxidation state, decomposing only at temperatures greater than 300°C. Major factors in the stability of Cs2XeF8 are almost certainly the formation of a stable ionic lattice and the high coordination number of xenon, which protects the central atom from attack by other species. (Recall from that this latter effect is responsible for the extreme stability of SF6.)
For a previously “inert” gas, xenon has a surprisingly high affinity for oxygen, presumably because of π bonding between $O$ and $Xe$. Consequently, xenon forms an extensive series of oxides and oxoanion salts. For example, hydrolysis of either $XeF_4$ or $XeF_6$ produces $XeO_3$, an explosive white solid:
$\ce{XeF6(aq) + 3H2O(l) -> XeO3(aq) + 6HF(aq)} \label{Eq5}$
Treating a solution of XeO3 with ozone, a strong oxidant, results in further oxidation of xenon to give either XeO4, a colorless, explosive gas, or the surprisingly stable perxenate ion (XeO64−), both of which contain xenon in its highest possible oxidation state (+8). The chemistry of the xenon halides and oxides is best understood by analogy to the corresponding compounds of iodine. For example, XeO3 is isoelectronic with the iodate ion (IO3), and XeF82− is isoelectronic with the IF8 ion.
Xenon has a high affinity for both fluorine and oxygen.
Because the ionization energy of radon is less than that of xenon, in principle radon should be able to form an even greater variety of chemical compounds than xenon. Unfortunately, however, radon is so radioactive that its chemistry has not been extensively explored.
Example $1$
On a virtual planet similar to Earth, at least one isotope of radon is not radioactive. A scientist explored its chemistry and presented her major conclusions in a trailblazing paper on radon compounds, focusing on the kinds of compounds formed and their stoichiometries. Based on periodic trends, how did she summarize the chemistry of radon?
Given: nonradioactive isotope of radon
Asked for: summary of its chemistry
Strategy:
Based on the position of radon in the periodic table and periodic trends in atomic properties, thermodynamics, and kinetics, predict the most likely reactions and compounds of radon.
Solution
We expect radon to be significantly easier to oxidize than xenon. Based on its position in the periodic table, however, we also expect its bonds to other atoms to be weaker than those formed by xenon. Radon should be more difficult to oxidize to its highest possible oxidation state (+8) than xenon because of the inert-pair effect. Consequently, radon should form an extensive series of fluorides, including RnF2, RnF4, RnF6, and possibly RnF8 (due to its large radius). The ion RnF82− should also exist. We expect radon to form a series of oxides similar to those of xenon, including RnO3 and possibly RnO4. The biggest surprise in radon chemistry is likely to be the existence of stable chlorides, such as RnCl2 and possibly even RnCl4.
Exercise $1$
Predict the stoichiometry of the product formed by reacting XeF6 with a 1:1 stoichiometric amount of KF and propose a reasonable structure for the anion.
Answer
$\ce{KXeF7}$; the xenon atom in XeF7 has 16 valence electrons, which according to the valence-shell electron-pair repulsion model could give either a square antiprismatic structure with one fluorine atom missing or a pentagonal bipyramid if the 5s2 electrons behave like an inert pair that does not participate in bonding.
Summary
The noble gases are characterized by their high ionization energies and low electron affinities. Potent oxidants are needed to oxidize the noble gases to form compounds in positive oxidation states. The noble gases have a closed-shell valence electron configuration. The ionization energies of the noble gases decrease with increasing atomic number. Only highly electronegative elements can form stable compounds with the noble gases in positive oxidation states without being oxidized themselves. Xenon has a high affinity for both fluorine and oxygen, which form stable compounds that contain xenon in even oxidation states up to +8.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/22%3A_Chemistry_of_the_Nonmetals/22.04%3A_Group_17-_The_Halogens.txt |
Oxygen is an element that is widely known by the general public because of the large role it plays in sustaining life. Without oxygen, animals would be unable to breathe and would consequently die. Oxygen not only is important to supporting life, but also plays an important role in many other chemical reactions. Oxygen is the most common element in the earth's crust and makes up about 20% of the air we breathe. Historically the discovery of oxygen as an element essential for combustion stands at the heart of the phlogiston controversy (see below).
The Origin and History
Oxygen is found in the group 16 elements and is considered a chalcogen. Named from the Greek oxys + genes, "acid-former", oxygen was discovered in 1772 by Scheele and independently by Priestly in 1774. Oxygen was given its name by the French scientist, Antoine Lavoisier.
Scheele discovered oxygen through an experiment which involved burning manganese oxide. Scheele came to find that the hot manganese oxide produced a gas which he called "fire air". He also came to find that when this gas was able to come into contact with charcoal, it produced beautiful bright sparks. All of the other elements produced the same gas. Although Scheele discovered oxygen, he did not publish his work until three years after another chemist, Joseph Priestly, discovered oxygen. Joseph Priestly, an English chemist, repeated Scheele's experiment in 1774 using a slightly different setup. Priestly used a 12 in burning glass and aimed the sunlight directly towards the compound that he was testing, mercuric oxide. As a result, he was able to "discover better air" that was shown to expand a mouse's lifetime to four times as long and caused a flame to burn with higher intensity. Despite these experiments, both chemists were not able to pinpoint exactly what this element was. It was not until 1775 that Antoine Lavoisier, a French chemist, was able to recognize this unknown gas as an element.
Our atmosphere currently contains about 21% of free oxygen. Oxygen is produced in various ways. The process of photochemical dissociation in which water molecules are broken up by ultraviolet rays produces about 1-2% of our oxygen. Another process that produces oxygen is photosynthesis which is performed by plants and photosynthetic bacteria. Photosynthesis occurs through the following general reaction:
$\ce{CO2 + H2O + h\nu \rightarrow} \text{organic compounds} \ce{+ O2} \nonumber$
The Dangers of Phlogiston
Phlogiston theory is the outdated belief that a fire-like element called phlogiston is contained within combustible bodies and released during combustion. The name comes from the Ancient Greek φλογιστόν phlogistón (burning up), from φλόξ phlóx (flame). It was first stated in 1667 by Johann Joachim Becher, and then put together more formally by Georg Ernst Stahl. The theory attempted to explain burning processes such as combustion and rusting, which are now collectively known as oxidation.
Properties
• Element number: 8
• Atomic weight 15.9994
• Color: gas form- colorless, liquid- pale blue
• Melting point: 54.36K
• Boiling point: 90.2 K
• Density: .001429
• 21% of earth's atmosphere
• Third most abundant element in the universe
• Most abundant element in Earth's crust at 45.4%
• 3 Stable isotopes
• Ionization energy: 13.618 eV
• Oxygen is easily reduced and is a great oxidizing agent making it readily reactive with other elements
Magnetic Properties of Oxygen
Oxygen (O2) is paramagnetic. An oxygen molecule has six valence electrons, so the O2 molecule has 12 valence electrons with the electron configuration shown below:
As shown, there are two unpaired electrons, which causes O2 to be paramagnetic. There are also eight valence electrons in the bonding orbitals and four in antibonding orbitals, which makes the bond order 2. This accounts for the double covalent bond that is present in O2.
Video $1$: A chemical demonstration of the paramagnetism of molecular oxygen, as shown by the attraction of liquid oxygen to magnets.
As shown in Video $1$, since molecular oxygen ($O_2$) has unpaired electrons, it is paramagnetic and is attracted to the magnet. In contrast, molecular nitrogen ($N_2$) has no unpaired electrons and is not attracted to the magnet.
General Chemistry of Oxygen
Oxygen normally has an oxidation state of -2, but is capable of having oxidation states of -2, -1, -1/2, 0, +1, and +2. The oxidation states of oxides, peroxides and superoxides are as follows:
• Oxides: O-2 ,
• peroxides: O2-2 ,
• superoxide: O2-1.
Oxygen does not react with itself, nitrogen, or water under normal conditions. Oxygen does, however, dissolve in water at 20 degrees Celsius and 1 atmosphere. Oxygen also does not normally react with bases or acids. Group 1 metals (alkaline metals) are very reactive with oxygen and must be stored away from oxygen in order to prevent them from becoming oxidized. The metals at the bottom of the group are more reactive than those at the top. The reactions of a few of these metals are explored in more detail below.
Lithium: Reacts with oxygen to form white lithium oxide in the reaction below.
$\ce{4Li + O_2 \rightarrow 2Li_2O} \label{1}$
Sodium: Reacts with oxygen to form a white mixture of sodium oxide and sodium peroxide. The reactions are shown below.
• Sodium oxide: $\ce{4Na + O_2 \rightarrow 2Na_2O} \label{2}$
• Sodium peroxide: $\ce{2Na + O_2 \rightarrow Na_2O_2} \label{3}$
Potassium: Reacts with oxygen to form a mixture of potassium peroxide and potassium superoxide. The reactions are shown below.
• Potassium peroxide: $\ce{2K + O_2 \rightarrow 2K_2O_2} \label{4}$
• Potassium superoxide: $\ce{K + O_2 \rightarrow KO_2} \label{5}$
Rubidium and Cesium: Both metals react to produce superoxides through the same process as that of the potassium superoxide reaction.
The oxides of these metals form metal hydroxides when they react with water. These metal hydroxides make the solution basic or alkaline, hence the name alkaline metals.
Group 2 metals (alkaline earth metals) react with oxygen through the process of burning to form metal oxides but there are a few exceptions.
Beryllium is very difficult to burn because it has a layer of beryllium oxide on its surface which prevents further interaction with oxygen. Strontium and barium react with oxygen to form peroxides. The reaction of barium and oxygen is shown below, and the reaction with strontium would be the same.
$\ce{Ba(s) + O2 (g) \rightarrow BaO2 (s) }\label{6}$
Group 13 reacts with oxygen in order to form oxides and hydroxides that are of the form $X_2O_3$ and $X(OH)_3$. The variable X represents the various group 13 elements. As you go down the group, the oxides and hydroxides get increasingly basic.
Group 14 elements react with oxygen to form oxides. The oxides formed at the top of the group are more acidic than those at the bottom of the group. Oxygen reacts with silicon and carbon to form silicon dioxide and carbon dioxide. Carbon is also able to react with oxygen to form carbon monoxide, which is slightly acidic. Germanium, tin, and lead react with oxygen to form monoxides and dioxides that are amphoteric, which means that they react with both acids and bases.
Group 15 elements react with oxygen to form oxides. The most important are listed below.
• Nitrogen: N2O, NO, N2O3, N2O4, N2O5
• Phosphorus: P4O6, P4O8, P2O5
• Arsenic: As2O3, As2O5
• Antimony: Sb2O3, Sb2O5
• Bismuth: Bi2O3, Bi2O5
Group 16 elements react with oxygen to form various oxides. Some of the oxides are listed below.
• Sulfur: SO, SO2, SO3, S2O7
• Selenium: SeO2, SeO3
• Tellurium: TeO, TeO2, TeO3
• Polonium: PoO, PoO2, PoO3
Group 17 elements (halogens) fluorine, chlorine, bromine, and iodine react with oxygen to form oxides. Fluorine forms two oxides with oxygen: F2O and F2O2. Both fluorine oxides are called oxygen fluorides because fluorine is the more electronegative element. One of the fluorine reactions is shown below.
$\ce{O2 (g) + F2 (g) \rightarrow F2O2 (g)} \label{7}$
Group 18: Some would assume that the Noble Gases would not react with oxygen. However, xenon does react with oxygen to form $\ce{XeO_3}$ and $\ce{XeO_4}$. The ionization energy of xenon is low enough for the electronegative oxygen atom to "steal away" electrons. Unfortunately, $\ce{XeO_3}$ is HIGHLY unstable, and it has been known to spontaneously detonate in a clean, dry environment.
Transition metals react with oxygen to form metal oxides. However, gold, silver, and platinum do not react with oxygen. A few reactions involving transition metals are shown below:
$2Sn_{(s)} + O_{2(g)} \rightarrow 2SnO_{(s)} \label{8}$
$4Fe_{(s)} + 3O_{2(g)} \rightarrow 2Fe_2O_{3(s)} \label{9A}$
$4Al_{(s)} + 3O_{2(g)} \rightarrow 2Al_2O_{3(s)} \label{9B}$
Reaction of Oxides
We will be discussing metal oxides of the form $X_2O$. The variable $X$ represents any metal that is able to bond to oxygen to form an oxide.
• Reaction with water: The oxides react with water to form a metal hydroxide.
$X_2O + H_2O \rightarrow 2XOH \nonumber$
• Reaction with dilute acids: The oxides react with dilute acids to form a salt and water.
$X_2O + 2HCl \rightarrow 2XCl + H_2O \nonumber$
Reactions of Peroxides
The peroxides we will be discussing are of the form $X_2O_2$. The variable $X$ represents any metal that can form a peroxide with oxygen.
Reaction with water: If the temperature of the reaction is kept constant despite the fact that the reaction is exothermic, then the reaction proceeds as follows:
$X_2O_2+ 2H_2O \rightarrow 2XOH + H_2O_2 \nonumber$
If the reaction is not carried out at a constant temperature, then the reaction of the peroxide and water will result in decomposition of the hydrogen peroxide that is produced into water and oxygen.
Reaction with dilute acid: This reaction is more exothermic than that with water. The heat produced causes the hydrogen peroxide to decompose to water and oxygen. The reaction is shown below.
$X_2O_2 + 2HCl \rightarrow 2XCl + H_2O_2 \nonumber$
$2H_2O_2 \rightarrow 2H_2O + O_2 \nonumber$
Reaction of Superoxides
The superoxides we will be talking about are of the form $XO_2$, with $X$ representing any metal that forms a superoxide when reacting with oxygen.
Reaction with water: The superoxide and water react in a very exothermic reaction that is shown below. The heat that is produced in forming the hydrogen peroxide will cause the hydrogen peroxide to decompose to water and oxygen.
$2XO_2 + 2H_2O \rightarrow 2XOH + H_2O_2 + O_2 \nonumber$
Reaction with dilute acids: The superoxide and dilute acid react in a very exothermic reaction that is shown below. The heat produced will cause the hydrogen peroxide to decompose to water and oxygen.
$2XO_2 + 2HCl \rightarrow 2XCl + H_2O_2 + O_2 \nonumber$
Allotropes of Oxygen
There are two allotropes of oxygen; dioxygen (O2) and trioxygen (O3) which is called ozone. The reaction of converting dioxygen into ozone is very endothermic, causing it to occur rarely and only in the lower atmosphere. The reaction is shown below:
$3O_{2 (g)} \rightarrow 2O_{3 (g)} \;\;\; ΔH^o= +285 \;kJ \nonumber$
Ozone is unstable and quickly decomposes back to oxygen but is a great oxidizing agent.
Miscellaneous Reactions
Reaction with Alkanes: The most common reactions that involve alkanes occur with oxygen. Alkanes are able to burn and it is the process of oxidizing the hydrocarbons that makes them important as fuels. An example of an alkane reaction is the reaction of octane with oxygen as shown below.
C8H18(l) + 25/2 O2(g) → 8CO2(g) + 9H2O(l) ∆Ho = -5.48 X 103 kJ
Reaction with ammonia: Oxygen is able to react with ammonia to produce dinitrogen (N2) and water (H2O) through the reaction shown below.
$4NH_3 + 3O_2 \rightarrow 2N_2 + 6H_2O \nonumber$
Reaction with Nitrogen Oxide: Oxygen is able to react with nitrogen oxide in order to produce nitrogen dioxide through the reaction shown below.
$NO + O_2 \rightarrow NO_2 \nonumber$
Problems
1. Is oxygen reactive with noble gases?
2. Which transition metals does oxygen not react with?
3. What is produced when an oxide reacts with water?
4. Is oxygen reactive with alkali metals? Why are the alkali metals named that way?
5. If oxygen is reactive with alkali metals, are oxides, peroxides or superoxides produced?
Solutions
1. No, noble gases are unreactive with oxygen.
2. Oxygen is mostly unreactive with gold and platinum.
3. When an oxide reacts with water, a metal hydroxide is produced.
4. Oxygen is very reactive with alkali metals. Alkali metals are given the name alkali because the oxides of these metals react with water to form a metal hydroxide that is basic or alkaline.
5. Lithium produces an oxide, sodium produces a peroxide, and potassium, cesium, and rubidium produce superoxides.
Contributors and Attributions
• Phillip Ball (UCD), Katharine Williams (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/22%3A_Chemistry_of_the_Nonmetals/22.05%3A_Oxygen.txt |
Learning Objectives
• To understand the trends in properties and reactivity of the group 16 elements: the chalcogens.
The chalcogens are the first group in the p block to have no stable metallic elements. All isotopes of polonium (Po), the only metal in group 16, are radioactive, and only one element in the group, tellurium (Te), can even be described as a semimetal. As in groups 14 and 15, the lightest element of group 16, oxygen, is found in nature as the free element.
Of the group 16 elements, only sulfur was known in ancient times; the others were not discovered until the late 18th and 19th centuries. Sulfur is frequently found as yellow crystalline deposits of essentially pure S8 in areas of intense volcanic activity or around hot springs. As early as the 15th century BC, sulfur was used as a fumigant in Homeric Greece because, when burned, it produces SO2 fumes that are toxic to most organisms, including vermin hiding in the walls and under the floors of houses. Hence references to sulfur are common in ancient literature, frequently in the context of religious purification. In fact, the association of sulfur with the divine was so pervasive that the prefixes thio- (meaning “sulfur”) and theo- (meaning “god”) have the same root in ancient Greek. Though used primarily in the production of sulfuric acid, sulfur is also used to manufacture gunpowder and as a cross-linking agent for rubber, which enables rubber to hold its shape but retain its flexibility.
Group 16 is the first group in the p block with no stable metallic elements.
Oxygen was not discovered until 1771, when the Swedish pharmacist Carl Wilhelm Scheele found that heating compounds such as KNO3, Ag2CO3, and HgO produced a colorless, odorless gas that supported combustion better than air. The results were not published immediately, however, so Scheele’s work remained unknown until 1777. Unfortunately, this was nearly two years after a paper by the English chemist Joseph Priestley had been published, describing the isolation of the same gas by using a magnifying glass to focus the sun’s rays on a sample of HgO. Oxygen is used primarily in the steel industry during the conversion of crude iron to steel using the Bessemer process. Another important industrial use of oxygen is in the production of TiO2, which is commonly used as a white pigment in paints, paper, and plastics.
Tellurium was discovered accidentally in 1782 by the Austrian chemist Franz Joseph Müller von Reichenstein, the chief surveyor of mines in Transylvania who was also responsible for the analysis of ore samples. The silvery-white metal had the same density as antimony but very different properties. Because it was difficult to analyze, Müller called it metallum problematicum (meaning “difficult metal”). The name tellurium (from the Latin tellus, meaning “earth”) was coined by another Austrian chemist, Martin Klaproth, who demonstrated in 1798 that Müller’s “difficult metal” was actually a new element. Tellurium is used to color glass and ceramics, in the manufacture of blasting caps, and in thermoelectric devices.
Jöns Jakob Berzelius (1779–1848)
Berzelius was born into a well-educated Swedish family, but both parents died when he was young. He studied medicine at the University of Uppsala, where his experiments with electroshock therapy caused his interests to turn to electrochemistry. Berzelius devised the system of chemical notation that we use today. In addition, he discovered six elements (cerium, thorium, selenium, silicon, titanium, and zirconium).
The heaviest chalcogen, polonium, was isolated after an extraordinary effort by Marie Curie. Although she was never able to obtain macroscopic quantities of the element, which she named for her native country of Poland, she demonstrated that its chemistry required it to be assigned to group 16. Marie Curie was awarded a second Nobel Prize in Chemistry in 1911 for the discovery of radium and polonium.
Preparation and General Properties of the Group 16 Elements
Oxygen is by far the most abundant element in Earth’s crust and in the hydrosphere (about 44% and 86% by mass, respectively). The same process that is used to obtain nitrogen from the atmosphere produces pure oxygen. Oxygen can also be obtained by the electrolysis of water, the decomposition of alkali metal or alkaline earth peroxides or superoxides, or the thermal decomposition of simple inorganic salts, such as potassium chlorate in the presence of a catalytic amount of MnO2:
$\mathrm{2KClO_3(s)\overset{MnO_2(s)}{\underset{\Delta}\rightleftharpoons}2KCl(s)+3O_2(g)} \label{22.4.1}$
Unlike oxygen, sulfur is not very abundant, but it is found as elemental sulfur in rock formations overlying salt domes, which often accompany petroleum deposits (Figure $1$). Sulfur is also recovered from H2S and organosulfur compounds in crude oil and coal and from metal sulfide ores such as pyrite (FeS2).
Because selenium and tellurium are chemically similar to sulfur, they are usually found as minor contaminants in metal sulfide ores and are typically recovered as by-products. Even so, they are as abundant in Earth’s crust as silver, palladium, and gold. One of the best sources of selenium and tellurium is the “slime” deposited during the electrolytic purification of copper. Both of these elements are notorious for the vile odors of many of their compounds. For example, when the body absorbs even trace amounts of tellurium, dimethyltellurium [(CH3)2Te] is produced and slowly released in the breath and perspiration, resulting in an intense garlic-like smell that is commonly called “tellurium breath.”
With their ns2np4 electron configurations, the chalcogens are two electrons short of a filled valence shell. Thus in reactions with metals, they tend to acquire two additional electrons to form compounds in the −2 oxidation state. This tendency is greatest for oxygen, the chalcogen with the highest electronegativity. The heavier, less electronegative chalcogens can lose either four np electrons or four np and two ns electrons to form compounds in the +4 and +6 oxidation state, respectively, as shown in Table Figure $1$. As with the other groups, the lightest member in the group, in this case oxygen, differs greatly from the others in size, ionization energy, electronegativity, and electron affinity, so its chemistry is unique. Also as in the other groups, the second and third members (sulfur and selenium) have similar properties because of shielding effects. Only polonium is metallic, forming either the hydrated Po2+ or Po4+ ion in aqueous solution, depending on conditions.
Table $1$: Selected Properties of the Group 16 Elements
Property Oxygen Sulfur Selenium Tellurium Polonium
*The configuration shown does not include filled d and f subshells.
The values cited for the hexacations are for six-coordinate ions and are only estimated values.
atomic mass (amu) 16.00 32.07 78.96 127.60 209
atomic number 8 16 34 52 84
atomic radius (pm) 48 88 103 123 135
atomic symbol O S Se Te Po
density (g/cm3) at 25°C 1.31 (g/L) 2.07 4.81 6.24 9.20
electron affinity (kJ/mol) −141 −200 −195 −190 −180
electronegativity 3.4 2.6 2.6 2.1 2.0
first ionization energy (kJ/mol) 1314 1000 941 869 812
ionic radius (pm) 140 (−2) 184 (−2), 29 (+6) 198 (−2), 42 (+6) 221 (−2), 56 (+6) 230 (−2), 97 (+4)
melting point/boiling point (°C) −219/−183 115/445 221/685 450/988 254/962
normal oxidation state(s) −2 +6, +4, −2 +6, +4, −2 +6, +4, −2 +2 (+4)
product of reaction with H2 H2O H2S H2Se none none
product of reaction with N2 NO, NO2 none none none none
product of reaction with O2 SO2 SeO2 TeO2 PoO2
product of reaction with X2 O2F2 SF6, S2Cl2, S2Br2 SeF6, SeX4 TeF6, TeX4 PoF4, PoCl2, PoBr2
standard reduction potential (E°, V) (E0 → H2E in acidic solution) +1.23 +0.14 −0.40 −0.79 −1.00
type of oxide acidic acidic amphoteric basic
valence electron configuration* 2s22p4 3s23p4 4s24p4 5s25p4 6s26p4
Reactions and Compounds of Oxygen
As in groups 14 and 15, the lightest group 16 member has the greatest tendency to form multiple bonds. Thus elemental oxygen is found in nature as a diatomic gas that contains a net double bond: O=O. As with nitrogen, electrostatic repulsion between lone pairs of electrons on adjacent atoms prevents oxygen from forming stable catenated compounds. In fact, except for O2, all compounds that contain O–O bonds are potentially explosive. Ozone, peroxides, and superoxides are all potentially dangerous in pure form. Ozone (O3), one of the most powerful oxidants known, is used to purify drinking water because it does not produce the characteristic taste associated with chlorinated water. Hydrogen peroxide (H2O2) is so thermodynamically unstable that it has a tendency to undergo explosive decomposition when impure:
$2H_2O_{2(l)} \rightarrow 2H_2O_{(l)} + O_{2(g)} \;\;\; ΔG^o = −119\; kJ/mol \label{1}$
As in groups 14 and 15, the lightest element in group 16 has the greatest tendency to form multiple bonds.
Despite the strength of the O=O bond ($D_\mathrm{O_2}$ = 494 kJ/mol), $O_2$ is extremely reactive, reacting directly with nearly all other elements except the noble gases. Some properties of O2 and related species, such as the peroxide and superoxide ions, are in Table $2$. With few exceptions, the chemistry of oxygen is restricted to negative oxidation states because of its high electronegativity (χ = 3.4). Unlike the other chalcogens, oxygen does not form compounds in the +4 or +6 oxidation state. Oxygen is second only to fluorine in its ability to stabilize high oxidation states of metals in both ionic and covalent compounds. For example, AgO is a stable solid that contains silver in the unusual Ag(II) state, whereas OsO4 is a volatile solid that contains Os(VIII). Because oxygen is so electronegative, the O–H bond is highly polar, creating a large bond dipole moment that makes hydrogen bonding much more important for compounds of oxygen than for similar compounds of the other chalcogens.
Table $2$: Some Properties of O2 and Related Diatomic Species
Species Bond Order Number of Unpaired e O–O Distance (pm)*
*Source of data: Lauri Vaska, “Dioxygen-Metal Complexes: Toward a Unified View,” Accounts of Chemical Research 9 (1976): 175.
O2+ 2.5 1 112
O2 2 2 121
O2 1.5 1 133
O22− 1 0 149
Metal oxides are usually basic, and nonmetal oxides are acidic, whereas oxides of elements that lie on or near the diagonal band of semimetals are generally amphoteric. A few oxides, such as CO and PbO2, are neutral and do not react with water, aqueous acid, or aqueous base. Nonmetal oxides are typically covalent compounds in which the bonds between oxygen and the nonmetal are polarized (Eδ+–Oδ−). Consequently, a lone pair of electrons on a water molecule can attack the partially positively charged E atom to eventually form an oxoacid. An example is reacting sulfur trioxide with water to form sulfuric acid:
$H_2O_{(l)} + SO_{3(g)} \rightarrow H_2SO_{4(aq)} \label{2}$
The oxides of the semimetals and of elements such as Al that lie near the metal/nonmetal dividing line are amphoteric, as we expect:
$Al_2O_{3(s)} + 6H^+_{(aq)} \rightarrow 2Al^{3+}_{(aq)} + 3H_2O_{(l)} \label{3}$
$Al_2O_{3(s)} + 2OH^−_{(aq)} + 3H_2O_{(l)} \rightarrow 2Al(OH)^−_{4(aq)} \label{4}$
Oxides of metals tend to be basic, oxides of nonmetals tend to be acidic, and oxides of elements in or near the diagonal band of semimetals are generally amphoteric.
Example $1$
For each reaction, explain why the given products form.
1. Ga2O3(s) + 2OH(aq) + 3H2O(l) → 2Ga(OH)4(aq)
2. 3H2O2(aq) + 2MnO4(aq) + 2H+(aq) → 3O2(g) + 2MnO2(s) + 4H2O(l)
3. KNO3(s) $\xrightarrow{\Delta}$ KNO(s) + O2(g)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. Gallium is a metal. We expect the oxides of metallic elements to be basic and therefore not to react with aqueous base. A close look at the periodic table, however, shows that gallium is close to the diagonal line of semimetals. Moreover, aluminum, the element immediately above gallium in group 13, is amphoteric. Consequently, we predict that gallium will behave like aluminum (Equation $\ref{4}$).
2. Hydrogen peroxide is an oxidant that can accept two electrons per molecule to give two molecules of water. With a strong oxidant, however, H2O2 can also act as a reductant, losing two electrons (and two protons) to produce O2. Because the other reactant is permanganate, which is a potent oxidant, the only possible reaction is a redox reaction in which permanganate is the oxidant and hydrogen peroxide is the reductant. Recall that reducing permanganate often gives MnO2, an insoluble brown solid. Reducing MnO4 to MnO2 is a three-electron reduction, whereas the oxidation of H2O2 to O2 is a two-electron oxidation.
3. This is a thermal decomposition reaction. Because KNO3 contains nitrogen in its highest oxidation state (+5) and oxygen in its lowest oxidation state (−2), a redox reaction is likely. Oxidation of the oxygen in nitrate to atomic oxygen is a two-electron process per oxygen atom. Nitrogen is likely to accept two electrons because oxoanions of nitrogen are known only in the +5 (NO3) and +3 (NO2) oxidation states.
Exercise $2$
Predict the product(s) of each reaction and write a balanced chemical equation for each reaction.
1. SiO2(s) + H+(aq) →
2. NO(g) + O2(g) →
3. SO3(g) + H2O(l) →
4. H2O2(aq) + I(aq) →
Answer
1. SiO2(s) + H+(aq) → no reaction
2. 2NO(g) + O2(g) → 2NO2(g)
3. SO3(g) + H2O(l) → H2SO4(aq)
4. H2O2(aq) + 2I(aq) → I2(aq) + 2OH(aq)
Reactions and Compounds of the Heavier Chalcogens
Because most of the heavier chalcogens (group 16) and pnicogens (group 15) are nonmetals, they often form similar compounds. For example, both third-period elements of these groups (phosphorus and sulfur) form catenated compounds and form multiple allotropes. Consistent with periodic trends, the tendency to catenate decreases as we go down the column.
Sulfur and selenium both form a fairly extensive series of catenated species. For example, elemental sulfur forms S8 rings packed together in a complex “crankshaft” arrangement (Figure $2$), and molten sulfur contains long chains of sulfur atoms connected by S–S bonds. Moreover, both sulfur and selenium form polysulfides (Sn2−) and polyselenides (Sen2−), with n ≤ 6. The only stable allotrope of tellurium is a silvery white substance whose properties and structure are similar to those of one of the selenium allotropes. Polonium, in contrast, shows no tendency to form catenated compounds. The striking decrease in structural complexity from sulfur to polonium is consistent with the decrease in the strength of single bonds and the increase in metallic character as we go down the group.
As in group 15, the reactivity of elements in group 16 decreases from lightest to heaviest. For example, selenium and tellurium react with most elements but not as readily as sulfur does. As expected for nonmetals, sulfur, selenium, and tellurium do not react with water, aqueous acid, or aqueous base, but all dissolve in strongly oxidizing acids such as HNO3 to form oxoacids such as H2SO4. In contrast to the other chalcogens, polonium behaves like a metal, dissolving in dilute HCl to form solutions that contain the Po2+ ion.
Just as with the other groups, the tendency to catenate, the strength of single bonds, and reactivity decrease down the group.
Fluorine reacts directly with all chalcogens except oxygen to produce the hexafluorides (YF6), which are extraordinarily stable and unreactive compounds. Four additional stable fluorides of sulfur are known; thus sulfur oxidation states range from +1 to +6 (Figure $2$). In contrast, only four fluorides of selenium (SeF6, SeF4, FSeSeF, and SeSeF2) and only three of tellurium (TeF4, TeF6, and Te2F10) are known.
Direct reaction of the heavier chalcogens with oxygen at elevated temperatures gives the dioxides (YO2), which exhibit a dramatic range of structures and properties. The dioxides become increasingly metallic in character down the group, as expected, and the coordination number of the chalcogen steadily increases. Thus SO2 is a gas that contains V-shaped molecules (as predicted by the valence-shell electron-pair repulsion model), SeO2 is a white solid with an infinite chain structure (each Se is three coordinate), TeO2 is a light yellow solid with a network structure in which each Te atom is four coordinate, and PoO2 is a yellow ionic solid in which each Po4+ ion is eight coordinate.
The dioxides of sulfur, selenium, and tellurium react with water to produce the weak, diprotic oxoacids (H2YO3—sulfurous, selenous, and tellurous acid, respectively). Both sulfuric acid and selenic acid (H2SeO4) are strong acids, but telluric acid [Te(OH)6] is quite different. Because tellurium is larger than either sulfur or selenium, it forms weaker π bonds to oxygen. As a result, the most stable structure for telluric acid is Te(OH)6, with six Te–OH bonds rather than Te=O bonds. Telluric acid therefore behaves like a weak triprotic acid in aqueous solution, successively losing the hydrogen atoms bound to three of the oxygen atoms. As expected for compounds that contain elements in their highest accessible oxidation state (+6 in this case), sulfuric, selenic, and telluric acids are oxidants. Because the stability of the highest oxidation state decreases with increasing atomic number, telluric acid is a stronger oxidant than sulfuric acid.
The stability of the highest oxidation state of the chalcogens decreases down the column.
Sulfur and, to a lesser extent, selenium react with carbon to form an extensive series of compounds that are structurally similar to their oxygen analogues. For example, CS2 and CSe2 are both volatile liquids that contain C=S or C=Se bonds and have the same linear structure as CO2. Because these double bonds are significantly weaker than the C=O bond, however, CS2, CSe2, and related compounds are less stable and more reactive than their oxygen analogues. The chalcogens also react directly with nearly all metals to form compounds with a wide range of stoichiometries and a variety of structures. Metal chalcogenides can contain either the simple chalcogenide ion (Y2−), as in Na2S and FeS, or polychalcogenide ions (Yn2−), as in FeS2 and Na2S5.
The dioxides of the group 16 elements become increasingly basic, and the coordination number of the chalcogen steadily increases down the group.
Ionic chalcogenides like Na2S react with aqueous acid to produce binary hydrides such as hydrogen sulfide (H2S). Because the strength of the Y–H bond decreases with increasing atomic radius, the stability of the binary hydrides decreases rapidly down the group. It is perhaps surprising that hydrogen sulfide, with its familiar rotten-egg smell, is much more toxic than hydrogen cyanide (HCN), the gas used to execute prisoners in the “gas chamber.” Hydrogen sulfide at relatively low concentrations deadens the olfactory receptors in the nose, which allows it to reach toxic levels without detection and makes it especially dangerous.
Example $2$
For each reaction, explain why the given product forms or no reaction occurs.
1. SO2(g) + Cl2(g) → SO2Cl2(l)
2. SF6(g) + H2O(l) → no reaction
3. 2Se(s) + Cl2(g) → Se2Cl2(l)
Given: balanced chemical equations
Asked for: why the given products (or no products) form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form or why no reaction occurs.
Solution
1. One of the reactants (Cl2) is an oxidant. If the other reactant can be oxidized, then a redox reaction is likely. Sulfur dioxide contains sulfur in the +4 oxidation state, which is 2 less than its maximum oxidation state. Sulfur dioxide is also known to be a mild reducing agent in aqueous solution, producing sulfuric acid as the oxidation product. Hence a redox reaction is probable. The simplest reaction is the formation of SO2Cl2 (sulfuryl chloride), which is a tetrahedral species with two S–Cl and two S=O bonds.
1. Sulfur hexafluoride is a nonmetallic halide. Such compounds normally react vigorously with water to produce an oxoacid of the nonmetal and the corresponding hydrohalic acid. In this case, however, we have a highly stable species, presumably because all of sulfur’s available orbitals are bonding orbitals. Thus SF6 is not likely to react with water.
2. Here we have the reaction of a chalcogen with a halogen. The halogen is a good oxidant, so we can anticipate that a redox reaction will occur. Only fluorine is capable of oxidizing the chalcogens to a +6 oxidation state, so we must decide between SeCl4 and Se2Cl2 as the product. The stoichiometry of the reaction determines which of the two is obtained: SeCl4 or Se2Cl2.
Exercise $2$
Predict the products of each reaction and write a balanced chemical equation for each reaction.
1. Te(s) + Na(s) $\xrightarrow{\Delta}$
2. SF4(g) + H2O(l) →
3. CH3SeSeCH3(soln) + K(s) →
4. Li2Se(s) + H+(aq) →
Answer
1. Te(s) + 2Na(s) → Na2Te(s)
2. SF4(g) + 3H2O(l) → H2SO3(aq) + 4HF(aq)
3. CH3SeSeCH3(soln) + 2K(s) → 2KCH3Se(soln)
4. Li2Se(s) + 2H+(aq) → H2Se(g) + 2Li+(aq)
Summary
The chalcogens have no stable metallic elements. The tendency to catenate, the strength of single bonds, and the reactivity all decrease moving down the group. Because the electronegativity of the chalcogens decreases down the group, so does their tendency to acquire two electrons to form compounds in the −2 oxidation state. The lightest member, oxygen, has the greatest tendency to form multiple bonds with other elements. It does not form stable catenated compounds, however, due to repulsions between lone pairs of electrons on adjacent atoms. Because of its high electronegativity, the chemistry of oxygen is generally restricted to compounds in which it has a negative oxidation state, and its bonds to other elements tend to be highly polar. Metal oxides are usually basic, and nonmetal oxides are acidic, whereas oxides of elements along the dividing line between metals and nonmetals are amphoteric. The reactivity, the strength of multiple bonds to oxygen, and the tendency to form catenated compounds all decrease down the group, whereas the maximum coordination numbers increase. Because Te=O bonds are comparatively weak, the most stable oxoacid of tellurium contains six Te–OH bonds. The stability of the highest oxidation state (+6) decreases down the group. Double bonds between S or Se and second-row atoms are weaker than the analogous C=O bonds because of reduced orbital overlap. The stability of the binary hydrides decreases down the group. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/22%3A_Chemistry_of_the_Nonmetals/22.06%3A_The_Other_Group_16_Elements-_S_Se_Te_and_Po.txt |
Like the group 14 elements, the lightest member of group 15, nitrogen, is found in nature as the free element, and the heaviest elements have been known for centuries because they are easily isolated from their ores. Although nitrogen is the most abundant element in the atmosphere, it was the last of the pnicogens (Group 15 elements) to be obtained in pure form. In 1772, Daniel Rutherford, working with Joseph Black (who discovered CO2), noticed that a gas remained when CO2 was removed from a combustion reaction. Antoine Lavoisier called the gas azote, meaning “no life,” because it did not support life. When it was discovered that the same element was also present in nitric acid and nitrate salts such as KNO3 (nitre), it was named nitrogen. About 90% of the nitrogen produced today is used to provide an inert atmosphere for processes or reactions that are oxygen sensitive, such as the production of steel, petroleum refining, and the packaging of foods and pharmaceuticals.
Preparation and General Properties of Nitrogen
Because the atmosphere contains several trillion tons of elemental nitrogen with a purity of about 80%, it is a huge source of nitrogen gas. Distillation of liquefied air yields nitrogen gas that is more than 99.99% pure, but small amounts of very pure nitrogen gas can be obtained from the thermal decomposition of sodium azide:
$\ce{2NaN3(s) ->[\Delta] 2Na(l) + 3N2(g)} \label{Eq1}$
In contrast, Earth’s crust is relatively poor in nitrogen. The only important nitrogen ores are large deposits of KNO3 and NaNO3 in the deserts of Chile and Russia, which were apparently formed when ancient alkaline lakes evaporated. Consequently, virtually all nitrogen compounds produced on an industrial scale use atmospheric nitrogen as the starting material. Phosphorus, which constitutes only about 0.1% of Earth’s crust, is much more abundant in ores than nitrogen. Like aluminum and silicon, phosphorus is always found in combination with oxygen, and large inputs of energy are required to isolate it.
Reactions and Compounds of Nitrogen
Like carbon, nitrogen has four valence orbitals (one 2s and three 2p), so it can participate in at most four electron-pair bonds by using sp3 hybrid orbitals. Unlike carbon, however, nitrogen does not form long chains because of repulsive interactions between lone pairs of electrons on adjacent atoms. These interactions become important at the shorter internuclear distances encountered with the smaller, second-period elements of groups 15, 16, and 17. Stable compounds with N–N bonds are limited to chains of no more than three N atoms, such as the azide ion (N3).
Nitrogen is the only pnicogen that normally forms multiple bonds with itself and other second-period elements, using π overlap of adjacent np orbitals. Thus the stable form of elemental nitrogen is N2, whose N≡N bond is so strong (DN≡N = 942 kJ/mol) compared with the N–N and N=N bonds (DN–N = 167 kJ/mol; DN=N = 418 kJ/mol) that all compounds containing N–N and N=N bonds are thermodynamically unstable with respect to the formation of N2. In fact, the formation of the N≡N bond is so thermodynamically favored that virtually all compounds containing N–N bonds are potentially explosive.
Again in contrast to carbon, nitrogen undergoes only two important chemical reactions at room temperature: it reacts with metallic lithium to form lithium nitride, and it is reduced to ammonia by certain microorganisms. At higher temperatures, however, N2 reacts with more electropositive elements, such as those in group 13, to give binary nitrides, which range from covalent to ionic in character. Like the corresponding compounds of carbon, binary compounds of nitrogen with oxygen, hydrogen, or other nonmetals are usually covalent molecular substances.
Few binary molecular compounds of nitrogen are formed by direct reaction of the elements. At elevated temperatures, N2 reacts with H2 to form ammonia, with O2 to form a mixture of NO and NO2, and with carbon to form cyanogen (N≡C–C≡N); elemental nitrogen does not react with the halogens or the other chalcogens. Nonetheless, all the binary nitrogen halides (NX3) are known. Except for NF3, all are toxic, thermodynamically unstable, and potentially explosive, and all are prepared by reacting the halogen with NH3 rather than N2. Both nitrogen monoxide (NO) and nitrogen dioxide (NO2) are thermodynamically unstable, with positive free energies of formation. Unlike NO, NO2 reacts readily with excess water, forming a 1:1 mixture of nitrous acid (HNO2) and nitric acid (HNO3):
$\ce{2NO2(g) + H2O(l) -> HNO2(aq) + HNO3(aq)} \label{Eq2}$
Nitrogen also forms $\ce{N2O}$ (dinitrogen monoxide, or nitrous oxide), a linear molecule that is isoelectronic with $\ce{CO2}$ and can be represented as N=N+=O. Like the other two oxides of nitrogen, nitrous oxide is thermodynamically unstable. The structures of the three common oxides of nitrogen are as follows:
Few binary molecular compounds of nitrogen are formed by the direct reaction of the elements.
At elevated temperatures, nitrogen reacts with highly electropositive metals to form ionic nitrides, such as $\ce{Li3N}$ and $\ce{Ca3N2}$. These compounds consist of ionic lattices formed by $\ce{M^{n+}}$ and $\ce{N^{3−}}$ ions. Just as boron forms interstitial borides and carbon forms interstitial carbides, with less electropositive metals nitrogen forms a range of interstitial nitrides, in which nitrogen occupies holes in a close-packed metallic structure. Like the interstitial carbides and borides, these substances are typically very hard, high-melting materials that have metallic luster and conductivity.
Nitrogen also reacts with semimetals at very high temperatures to produce covalent nitrides, such as $\ce{Si3N4}$ and $\ce{BN}$, which are solids with extended covalent network structures similar to those of graphite or diamond. Consequently, they are usually high melting and chemically inert materials.
Ammonia (NH3) is one of the few thermodynamically stable binary compounds of nitrogen with a nonmetal. It is not flammable in air, but it burns in an $\ce{O2}$ atmosphere:
$\ce{4NH3(g) + 3O2(g) -> 2N2(g) + 6H2O(g)} \label{Eq3}$
About 10% of the ammonia produced annually is used to make fibers and plastics that contain amide bonds, such as nylons and polyurethanes, while 5% is used in explosives, such as ammonium nitrate, TNT (trinitrotoluene), and nitroglycerine. Large amounts of anhydrous liquid ammonia are used as fertilizer.
Nitrogen forms two other important binary compounds with hydrogen. Hydrazoic acid ($\ce{HN3}$), also called hydrogen azide, is a colorless, highly toxic, and explosive substance. Hydrazine ($\ce{N2H4}$) is also potentially explosive; it is used as a rocket propellant and to inhibit corrosion in boilers.
B, C, and N all react with transition metals to form interstitial compounds that are hard, high-melting materials.
Example $1$
For each reaction, explain why the given products form when the reactants are heated.
1. $\ce{Sr(s) + N2O(g) ->[\Delta] SrO(s) + N2(g)}$
2. $\ce{NH4NO2(s) ->[\Delta] N2(g) + 2H2O(g)}$
3. $\ce{Pb(NO3)2(s) ->[\Delta] PbO2(s) + 2NO2(g)}$
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. As an alkali metal, strontium is a strong reductant. If the other reactant can act as an oxidant, then a redox reaction will occur. Nitrous oxide contains nitrogen in a low oxidation state (+1), so we would not normally consider it an oxidant. Nitrous oxide is, however, thermodynamically unstable (ΔH°f > 0 and ΔG°f > 0), and it can be reduced to N2, which is a stable species. Consequently, we predict that a redox reaction will occur.
2. When a substance is heated, a decomposition reaction probably will occur, which often involves the release of stable gases. In this case, ammonium nitrite contains nitrogen in two different oxidation states (−3 and +3), so an internal redox reaction is a possibility. Due to its thermodynamic stability, N2 is the probable nitrogen-containing product, whereas we predict that H and O will combine to form H2O.
3. Again, this is probably a thermal decomposition reaction. If one element is in an usually high oxidation state and another in a low oxidation state, a redox reaction will probably occur. Lead nitrate contains the Pb2+ cation and the nitrate anion, which contains nitrogen in its highest possible oxidation state (+5). Hence nitrogen can be reduced, and we know that lead can be oxidized to the +4 oxidation state. Consequently, it is likely that lead(II) nitrate will decompose to lead(IV) oxide and nitrogen dioxide when heated. Even though PbO2 is a powerful oxidant, the release of a gas such as NO2 can often drive an otherwise unfavorable reaction to completion (Le Chatelier’s principle). Note, however, that PbO2 will probably decompose to PbO at high temperatures.
Exercise $1$
Predict the product(s) of each reaction and write a balanced chemical equation for each reaction.
1. $\ce{NO(g) + H2O(l) ->[\Delta]}$
2. $\ce{NH4NO3(s) ->[\Delta]}$
3. $\ce{Sr(s) + N2(g) ->}$
Answer
1. $\ce{NO(g) + H2O(l) ->[\Delta] no reaction}$
2. $\ce{NH4NO3(s) ->[\Delta] N2O(g) + 2H2O(g)}$
3. $\ce{3Sr(s) + N2(g) -> Sr3N2(s)}$
Summary
Nitrogen behaves chemically like nonmetals, Nitrogen forms compounds in nine different oxidation states. Nitrogen does not form stable catenated compounds because of repulsions between lone pairs of electrons on adjacent atoms, but it does form multiple bonds with other second-period atoms. Nitrogen reacts with electropositive elements to produce solids that range from covalent to ionic in character. Reaction with electropositive metals produces ionic nitrides, reaction with less electropositive metals produces interstitial nitrides, and reaction with semimetals produces covalent nitrides. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/22%3A_Chemistry_of_the_Nonmetals/22.07%3A_Nitrogen.txt |
Learning Objectives
• To understand the trends in properties and reactivity of the group 14 elements: the pnicogens.
Antimony (Sb) was probably the first of the pnicogens to be obtained in elemental form and recognized as an element. Its atomic symbol comes from its Roman name: stibium. It is found in stibnite (Sb2S3), a black mineral that has been used as a cosmetic (an early form of mascara) since biblical times, and it is easily reduced to the metal in a charcoal fire (Figure $1$). The Egyptians used antimony to coat copper objects as early as the third millennium BC, and antimony is still used in alloys to improve the tonal quality of bells.
In the form of its yellow sulfide ore, orpiment (As2S3), arsenic (As) has been known to physicians and professional assassins since ancient Greece, although elemental arsenic was not isolated until centuries later. The history of bismuth (Bi), in contrast, is more difficult to follow because early alchemists often confused it with other metals, such as lead, tin, antimony, and even silver (due to its slightly pinkish-white luster). Its name comes from the old German wismut, meaning “white metal.” Bismuth was finally isolated in the 15th century, and it was used to make movable type for printing shortly after the invention of the Gutenberg printing process in 1440. Bismuth is used in printing because it is one of the few substances known whose solid state is less dense than the liquid. Consequently, its alloys expand as they cool, filling a mold completely and producing crisp, clear letters for typesetting.
Phosphorus was discovered in 1669 by the German alchemist Hennig Brandt, who was looking for the “philosophers’ stone,” a mythical substance capable of converting base metals to silver or gold. Believing that human urine was the source of the key ingredient, Brandt obtained several dozen buckets of urine, which he allowed to putrefy. The urine was distilled to dryness at high temperature and then condensed; the last fumes were collected under water, giving a waxy white solid that had unusual properties. For example, it glowed in the dark and burst into flames when removed from the water. (Unfortunately for Brandt, however, it did not turn lead into gold.) The element was given its current name (from the Greek phos, meaning “light,” and phoros, meaning “bringing”) in the 17th century. For more than a century, the only way to obtain phosphorus was the distillation of urine, but in 1769 it was discovered that phosphorus could be obtained more easily from bones. During the 19th century, the demand for phosphorus for matches was so great that battlefields and paupers’ graveyards were systematically scavenged for bones. Early matches were pieces of wood coated with elemental phosphorus that were stored in an evacuated glass tube and ignited when the tube was broken (which could cause unfortunate accidents if the matches were kept in a pocket!).
Unfortunately, elemental phosphorus is volatile and highly toxic. It is absorbed by the teeth and destroys bone in the jaw, leading to a painful and fatal condition called “phossy jaw,” which for many years was accepted as an occupational hazard of working in the match industry.
Preparation and General Properties of the Group 15 Elements
The three non-nitrogen pnicogens are much less abundant than nitrogen: arsenic is found in Earth’s crust at a concentration of about 2 ppm, antimony is an order of magnitude less abundant, and bismuth is almost as rare as gold. All three elements have a high affinity for the chalcogens and are usually found as the sulfide ores (M2S3), often in combination with sulfides of other heavy elements, such as copper, silver, and lead. Hence a major source of antimony and bismuth is flue dust obtained by smelting the sulfide ores of the more abundant metals.
In group 15, as elsewhere in the p block, we see large differences between the lightest element (N) and its congeners in size, ionization energy, electron affinity, and electronegativity (Table $1$). The chemical behavior of the elements can be summarized rather simply: nitrogen and phosphorus behave chemically like nonmetals, arsenic and antimony behave like semimetals, and bismuth behaves like a metal. With their ns2np3 valence electron configurations, all form compounds by losing either the three np valence electrons to form the +3 oxidation state or the three np and the two ns valence electrons to give the +5 oxidation state, whose stability decreases smoothly from phosphorus to bismuth. In addition, the relatively large magnitude of the electron affinity of the lighter pnicogens enables them to form compounds in the −3 oxidation state (such as NH3 and PH3), in which three electrons are formally added to the neutral atom to give a filled np subshell. Nitrogen has the unusual ability to form compounds in nine different oxidation states, including −3, +3, and +5. Because neutral covalent compounds of the trivalent pnicogens contain a lone pair of electrons on the central atom, they tend to behave as Lewis bases.
Table $1$: Selected Properties of the Group 15 Elements
Property Nitrogen Phosphorus Arsenic Antimony Bismuth
*The configuration shown does not include filled d and f subshells.
For white phosphorus.
For gray arsenic.
§The values cited are for six-coordinate ions in the indicated oxidation states. The N5+, P5+, and As5+ ions are not known species.
||The chemical form of the elements in these oxidation states varies considerably. For N, the reaction is NO3 + 3H+ + 2e HNO2 + H2O; for P and As, it is H3EO4 + 2H+ + 2e H3EO3 + H2O; and for Sb it is Sb2O5 + 4e + 10H+ 2Sb3+ + 5H2O.
atomic symbol N P As Sb Bi
atomic number 7 15 33 51 83
atomic mass (amu) 14.01 30.97 74.92 121.76 209.98
valence electron configuration* 2s22p3 3s23p3 4s24p3 5s25p3 6s26p3
melting point/boiling point (°C) −210/−196 44.15/281c 817 (at 3.70 MPa)/603 (sublimes) 631/1587 271/1564
density (g/cm3) at 25°C 1.15 (g/L) 1.82 5.75 6.68 9.79
atomic radius (pm) 56 98 114 133 143
first ionization energy (kJ/mol) 1402 1012 945 831 703
common oxidation state(s) −3 to +5 +5, +3, −3 +5, +3 +5, +3 +3
ionic radius (pm)§ 146 (−3), 16 (+3) 212 (−3), 44 (+3) 58 (+3) 76 (+3), 60 (+5) 103 (+3)
electron affinity (kJ/mol) 0 −72 −78 −101 −91
electronegativity 3.0 2.2 2.2 2.1 1.9
standard reduction potential (E°, V) (EV → EIII in acidic solution)|| +0.93 −0.28 +0.56 +0.65
product of reaction with O2 NO2, NO P4O6, P4O10 As4O6 Sb2O5 Bi2O3
type of oxide acidic (NO2), neutral (NO, N2O) acidic acidic amphoteric basic
product of reaction with N2 none none none none
product of reaction with X2 none PX3, PX5 AsF5, AsX3 SbF5, SbCl5, SbBr3, SbI3 BiF5, BiX3
product of reaction with H2 none none none none none
In group 15, the stability of the +5 oxidation state decreases from P to Bi.
Because neutral covalent compounds of the trivalent group 15 elements have a lone pair of electrons on the central atom, they tend to be Lewis bases.
Reactions and Compounds of the Heavier Pnicogens
Like the heavier elements of group 14, the heavier pnicogens form catenated compounds that contain only single bonds, whose stability decreases rapidly as we go down the group. For example, phosphorus exists as multiple allotropes, the most common of which is white phosphorus, which consists of P4 tetrahedra and behaves like a typical nonmetal. As is typical of a molecular solid, white phosphorus is volatile, has a low melting point (44.1°C), and is soluble in nonpolar solvents. It is highly strained, with bond angles of only 60°, which partially explains why it is so reactive and so easily converted to more stable allotropes. Heating white phosphorus for several days converts it to red phosphorus, a polymer that is air stable, virtually insoluble, denser than white phosphorus, and higher melting, properties that make it much safer to handle. A third allotrope of phosphorus, black phosphorus, is prepared by heating the other allotropes under high pressure; it is even less reactive, denser, and higher melting than red phosphorus. As expected from their structures, white phosphorus is an electrical insulator, and red and black phosphorus are semiconductors. The three heaviest pnicogens—arsenic, antimony, and bismuth—all have a metallic luster, but they are brittle (not ductile) and relatively poor electrical conductors.
As in group 14, the heavier group 15 elements form catenated compounds that contain only single bonds, whose stability decreases as we go down the group.
The reactivity of the heavier pnicogens decreases as we go down the column. Phosphorus is by far the most reactive of the pnicogens, forming binary compounds with every element in the periodic table except antimony, bismuth, and the noble gases. Phosphorus reacts rapidly with O2, whereas arsenic burns in pure O2 if ignited, and antimony and bismuth react with O2 only when heated. None of the pnicogens reacts with nonoxidizing acids such as aqueous HCl, but all dissolve in oxidizing acids such as HNO3. Only bismuth behaves like a metal, dissolving in HNO3 to give the hydrated Bi3+ cation.
The reactivity of the heavier group 15 elements decreases as we go down the column.
The heavier pnicogens can use energetically accessible 3d, 4d, or 5d orbitals to form dsp3 or d2sp3 hybrid orbitals for bonding. Consequently, these elements often have coordination numbers of 5 or higher. Phosphorus and arsenic form halides (e.g., AsCl5) that are generally covalent molecular species and behave like typical nonmetal halides, reacting with water to form the corresponding oxoacids (in this case, H3AsO4). All the pentahalides are potent Lewis acids that can expand their coordination to accommodate the lone pair of a Lewis base:
$AsF_{5(soln)} + F^−_{(soln)} \rightarrow AsF^−_{6(soln)} \label{Eq4}$
In contrast, bismuth halides have extended lattice structures and dissolve in water to produce hydrated ions, consistent with the stronger metallic character of bismuth.
Except for BiF3, which is essentially an ionic compound, the trihalides are volatile covalent molecules with a lone pair of electrons on the central atom. Like the pentahalides, the trihalides react rapidly with water. In the cases of phosphorus and arsenic, the products are the corresponding acids, H3PO3 and H3AsO3, where E is P or As:
$EX_{3(l)} + 3H_2O_{(l)} \rightarrow H_3EO_{3(aq)} + 3HX_{(aq)} \label{Eq5}$
Phosphorus halides are also used to produce insecticides, flame retardants, and plasticizers.
Phosphorus has the greatest ability to form π bonds with elements such as O, N, and C.
With energetically accessible d orbitals, phosphorus and, to a lesser extent, arsenic are able to form π bonds with second-period atoms such as N and O. This effect is even more important for phosphorus than for silicon, resulting in very strong P–O bonds and even stronger P=O bonds. The first four elements in group 15 also react with oxygen to produce the corresponding oxide in the +3 oxidation state. Of these oxides, P4O6 and As4O6 have cage structures formed by inserting an oxygen atom into each edge of the P4 or As4 tetrahedron (part (a) in Figure $2$), and they behave like typical nonmetal oxides. For example, P4O6 reacts with water to form phosphorous acid (H3PO3). Consistent with its position between the nonmetal and metallic oxides, Sb4O6 is amphoteric, dissolving in either acid or base. In contrast, Bi2O3 behaves like a basic metallic oxide, dissolving in acid to give solutions that contain the hydrated Bi3+ ion. The two least metallic elements of the heavier pnicogens, phosphorus and arsenic, form very stable oxides with the formula E4O10 in the +5 oxidation state (part (b) in Figure $2$. In contrast, Bi2O5 is so unstable that there is no absolute proof it exists.
The heavier pnicogens form sulfides that range from molecular species with three-dimensional cage structures, such as P4S3 (part (c) in Figure $2$), to layered or ribbon structures, such as Sb2S3 and Bi2S3, which are semiconductors. Reacting the heavier pnicogens with metals produces substances whose properties vary with the metal content. Metal-rich phosphides (such as M4P) are hard, high-melting, electrically conductive solids with a metallic luster, whereas phosphorus-rich phosphides (such as MP15) are lower melting and less thermally stable because they contain catenated Pn units. Many organic or organometallic compounds of the heavier pnicogens containing one to five alkyl or aryl groups are also known. Because of the decreasing strength of the pnicogen–carbon bond, their thermal stability decreases from phosphorus to bismuth.
The thermal stability of organic or organometallic compounds of group 15 decreases down the group due to the decreasing strength of the pnicogen–carbon bond.
Example $1$
For each reaction, explain why the given products form.
1. $\mathrm{Bi(s) +\frac{3}{2}Br(l)\rightarrow BiBr_3(s)}$
2. 2(CH3)3As(l) + O2(g) → 2(CH3)3As=O(s)
3. PBr3(l) + 3H2O(l) → H3PO3(aq) + 3HBr(aq)
4. As(s) + Ga(s) $\xrightarrow{\Delta}$ GaAs(s)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form.
Solution
1. Bromine is an oxidant, and bismuth is a metal that can be oxidized. Hence a redox reaction is likely to occur. To identify the product, recall that bismuth can form compounds in either the +3 or +5 oxidation state. The heaviest pnicogen, bismuth is rather difficult to oxidize to the +5 oxidation state because of the inert-pair effect. Hence the product will probably be bismuth(III) bromide.
2. Trimethylarsine, with a lone pair of electrons on the arsenic atom, can act as either a Lewis base or a reductant. If arsenic is oxidized by two electrons, then oxygen must be reduced, most probably by two electrons to the −2 oxidation state. Because As(V) forms strong bonds to oxygen due to π bonding, the expected product is (CH3)3As=O.
3. Phosphorus tribromide is a typical nonmetal halide. We expect it to react with water to produce an oxoacid of P(III) and the corresponding hydrohalic acid.Because of the strength of the P=O bond, phosphorous acid (H3PO3) is actually HP(O)(OH)2, which contains a P=O bond and a P–H bond.
4. Gallium is a metal with a strong tendency to act as a reductant and form compounds in the +3 oxidation state. In contrast, arsenic is a semimetal. It can act as a reductant to form compounds in the +3 or +5 oxidation state, or it can act as an oxidant, accepting electrons to form compounds in the −3 oxidation state. If a reaction occurs, then a binary compound will probably form with a 1:1 ratio of the elements. GaAs is an example of a III-V compound, many of which are used in the electronics industry.
Exercise $1$
Predict the products of each reaction and write a balanced chemical equation for each reaction.
1. PCl5(s) + H2O(l) →
2. Bi2O5(s) $\xrightarrow{\Delta}$
3. Ca3P2(s) + H+(aq) →
4. NaNH2(s) + PH3(soln) →
Answer
1. PCl5(s) + 4H2O(l) → H3PO4(aq) + 5HCl(aq)
2. Bi2O5(s) $\xrightarrow{\Delta}$ Bi2O3(s) + O2(g)
3. Ca3P2(s) + 6H+(aq) → 2PH3(g) + 3Ca2+(aq)
4. NaNH2(s) + PH3(soln) → NaPH2(s) + NH3(soln)
Summary
The reactivity of the heavier group 15 elements decreases down the group, as does the stability of their catenated compounds. In group 15, nitrogen and phosphorus behave chemically like nonmetals, arsenic and antimony behave like semimetals, and bismuth behaves like a metal. The stability of the +5 oxidation state decreases from phosphorus to bismuth because of the inert-pair effect. Due to their higher electronegativity, the lighter pnicogens form compounds in the −3 oxidation state. Because of the presence of a lone pair of electrons on the pnicogen, neutral covalent compounds of the trivalent pnicogens are Lewis bases. The reactivity of the pnicogens decreases with increasing atomic number. Compounds of the heavier pnicogens often have coordination numbers of 5 or higher and use dsp3 or d2sp3 hybrid orbitals for bonding. Because phosphorus and arsenic have energetically accessible d orbitals, these elements form π bonds with second-period atoms such as O and N. Phosphorus reacts with metals to produce phosphides. Metal-rich phosphides are hard, high-melting, electrically conductive solids with metallic luster, whereas phosphorus-rich phosphides, which contain catenated phosphorus units, are lower melting and less thermally stable. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/22%3A_Chemistry_of_the_Nonmetals/22.08%3A_The_Other_Group_15_Elements-_P_AS_Sb_and_Bi.txt |
The elements of group 14 show a greater range of chemical behavior than any other family in the periodic table. Three of the five elements—carbon, tin, and lead—have been known since ancient times. For example, some of the oldest known writings are Egyptian hieroglyphics written on papyrus with ink made from lampblack, a finely divided carbon soot produced by the incomplete combustion of hydrocarbons (Figure $1$). Activated carbon is an even more finely divided form of carbon that is produced from the thermal decomposition of organic materials, such as sawdust. Because it adsorbs many organic and sulfur-containing compounds, activated carbon is used to decolorize foods, such as sugar, and to purify gases and wastewater.
Preparation and General Properties of Carbon
Elemental carbon, for example, ranks only 17th on the list of constituents of Earth’s crust. Pure graphite is obtained by reacting coke, an amorphous form of carbon used as a reductant in the production of steel, with silica to give silicon carbide (SiC). This is then thermally decomposed at very high temperatures (2700°C) to give graphite:
$\mathrm{SiO_2(s)}+\mathrm{3C(s)}\xrightarrow{\Delta}\mathrm{SiC(s)}+\mathrm{2CO(g)} \label{$1$}$
$\mathrm{SiC(s)}\xrightarrow{\Delta}\mathrm{Si(s)}+\mathrm{C(graphite)} \label{$2$}$
One allotrope of carbon, diamond, is metastable under normal conditions, with a ΔG°f of 2.9 kJ/mol versus graphite. At pressures greater than 50,000 atm, however, the diamond structure is favored and is the most stable form of carbon. Because the structure of diamond is more compact than that of graphite, its density is significantly higher (3.51 g/cm3 versus 2.2 g/cm3). Because of its high thermal conductivity, diamond powder is used to transfer heat in electronic devices.
The most common sources of diamonds on Earth are ancient volcanic pipes that contain a rock called kimberlite, a lava that solidified rapidly from deep inside the Earth. Most kimberlite formations, however, are much newer than the diamonds they contain. In fact, the relative amounts of different carbon isotopes in diamond show that diamond is a chemical and geological “fossil” older than our solar system, which means that diamonds on Earth predate the existence of our sun. Thus diamonds were most likely created deep inside Earth from primordial grains of graphite present when Earth was formed (Figure $2$). Gem-quality diamonds can now be produced synthetically and have chemical, optical, and physical characteristics identical to those of the highest-grade natural diamonds.
Reactions and Compounds of Carbon
Carbon is the building block of all organic compounds, including biomolecules, fuels, pharmaceuticals, and plastics, whereas inorganic compounds of carbon include metal carbonates, which are found in substances as diverse as fertilizers and antacid tablets, halides, oxides, carbides, and carboranes. Like boron in group 13, the chemistry of carbon differs sufficiently from that of its heavier congeners to merit a separate discussion.
The structures of the allotropes of carbon—diamond, graphite, fullerenes, and nanotubes—are distinct, but they all contain simple electron-pair bonds. Although it was originally believed that fullerenes were a new form of carbon that could be prepared only in the laboratory, fullerenes have been found in certain types of meteorites. Another possible allotrope of carbon has also been detected in impact fragments of a carbon-rich meteorite; it appears to consist of long chains of carbon atoms linked by alternating single and triple bonds, (–C≡C–C≡C–)n. Carbon nanotubes (“buckytubes”) are being studied as potential building blocks for ultramicroscale detectors and molecular computers and as tethers for space stations. They are currently used in electronic devices, such as the electrically conducting tips of miniature electron guns for flat-panel displays in portable computers.
Although all the carbon tetrahalides (CX4) are known, they are generally not obtained by the direct reaction of carbon with the elemental halogens (X2) but by indirect methods such as the following reaction, where X is Cl or Br:
$CH_{4(g)} + 4X_{2(g)} \rightarrow CX_{4(l,s)} + 4HX_{(g)} \label{$7$}$
The carbon tetrahalides all have the tetrahedral geometry predicted by the valence-shell electron-pair repulsion (VSEPR) model, as shown for CCl4 and CI4. Their stability decreases rapidly as the halogen increases in size because of poor orbital overlap and increased crowding. Because the C–F bond is about 25% stronger than a C–H bond, fluorocarbons are thermally and chemically more stable than the corresponding hydrocarbons, while having a similar hydrophobic character. A polymer of tetrafluoroethylene (F2C=CF2), analogous to polyethylene, is the nonstick Teflon lining found on many cooking pans, and similar compounds are used to make fabrics stain resistant (such as Scotch-Gard) or waterproof but breathable (such as Gore-Tex).
The stability of the carbon tetrahalides decreases with increasing size of the halogen due to increasingly poor orbital overlap and crowding.
Carbon reacts with oxygen to form either CO or CO2, depending on the stoichiometry. Carbon monoxide is a colorless, odorless, and poisonous gas that reacts with the iron in hemoglobin to form an Fe–CO unit, which prevents hemoglobin from binding, transporting, and releasing oxygen in the blood (see Figure 23.26). In the laboratory, carbon monoxide can be prepared on a small scale by dehydrating formic acid with concentrated sulfuric acid:
$\mathrm{HCO_2H(l)}\xrightarrow{\mathrm{H_2SO_4(l)}}\mathrm{CO(g)}+\mathrm{H_3O^+(aq)}+\mathrm{HSO^-_4}\label{$8$}$
Carbon monoxide also reacts with the halogens to form the oxohalides (COX2). Probably the best known of these is phosgene (Cl2C=O), which is highly poisonous and was used as a chemical weapon during World War I:
$\mathrm{CO(g)}+\mathrm{Cl_2(g)}\xrightarrow{\Delta}\textrm{Cl}_2\textrm{C=O(g)}\label{$9$}$
Despite its toxicity, phosgene is an important industrial chemical that is prepared on a large scale, primarily in the manufacture of polyurethanes.
Carbon dioxide can be prepared on a small scale by reacting almost any metal carbonate or bicarbonate salt with a strong acid. As is typical of a nonmetal oxide, CO2 reacts with water to form acidic solutions containing carbonic acid (H2CO3). In contrast to its reactions with oxygen, reacting carbon with sulfur at high temperatures produces only carbon disulfide (CS2):
$\mathrm{C(s)}+\mathrm{2S(g)}\xrightarrow{\Delta}\mathrm{CS_2(g)}\label{$1$0}$
The selenium analog CSe2 is also known. Both have the linear structure predicted by the VSEPR model, and both are vile smelling (and in the case of CSe2, highly toxic), volatile liquids. The sulfur and selenium analogues of carbon monoxide, CS and CSe, are unstable because the C≡Y bonds (Y is S or Se) are much weaker than the C≡O bond due to poorer π orbital overlap.
$\pi$ bonds between carbon and the heavier chalcogenides are weak due to poor orbital overlap.
Binary compounds of carbon with less electronegative elements are called carbides. The chemical and physical properties of carbides depend strongly on the identity of the second element, resulting in three general classes: ionic carbides, interstitial carbides, and covalent carbides. The reaction of carbon at high temperatures with electropositive metals such as those of groups 1 and 2 and aluminum produces ionic carbides, which contain discrete metal cations and carbon anions. The identity of the anions depends on the size of the second element. For example, smaller elements such as beryllium and aluminum give methides such as Be2C and Al4C3, which formally contain the C4− ion derived from methane (CH4) by losing all four H atoms as protons. In contrast, larger metals such as sodium and calcium give carbides with stoichiometries of Na2C2 and CaC2. Because these carbides contain the C2− ion, which is derived from acetylene (HC≡CH) by losing both H atoms as protons, they are more properly called acetylides. Reacting ionic carbides with dilute aqueous acid results in protonation of the anions to give the parent hydrocarbons: CH4 or C2H2. For many years, miners’ lamps used the reaction of calcium carbide with water to produce a steady supply of acetylene, which was ignited to provide a portable lantern.
The reaction of carbon with most transition metals at high temperatures produces interstitial carbides. Due to the less electropositive nature of the transition metals, these carbides contain covalent metal–carbon interactions, which result in different properties: most interstitial carbides are good conductors of electricity, have high melting points, and are among the hardest substances known. Interstitial carbides exhibit a variety of nominal compositions, and they are often nonstoichiometric compounds whose carbon content can vary over a wide range. Among the most important are tungsten carbide (WC), which is used industrially in high-speed cutting tools, and cementite (Fe3C), which is a major component of steel.
Elements with an electronegativity similar to that of carbon form covalent carbides, such as silicon carbide (SiC; Equation $1$ ) and boron carbide (B4C). These substances are extremely hard, have high melting points, and are chemically inert. For example, silicon carbide is highly resistant to chemical attack at temperatures as high as 1600°C. Because it also maintains its strength at high temperatures, silicon carbide is used in heating elements for electric furnaces and in variable-temperature resistors.
Carbides formed from group 1 and 2 elements are ionic. Transition metals form interstitial carbides with covalent metal–carbon interactions, and covalent carbides are chemically inert.
Example $1$
For each reaction, explain why the given product forms.
1. CO(g) + Cl2(g) → Cl2C=O(g)
2. CO(g) + BF3(g) → F3B:C≡O(g)
3. Sr(s) + 2C(s) $\xrightarrow{\Delta}$ SrC2(s)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. Because the carbon in CO is in an intermediate oxidation state (+2), CO can be either a reductant or an oxidant; it is also a Lewis base. The other reactant (Cl2) is an oxidant, so we expect a redox reaction to occur in which the carbon of CO is further oxidized. Because Cl2 is a two-electron oxidant and the carbon atom of CO can be oxidized by two electrons to the +4 oxidation state, the product is phosgene (Cl2C=O).
2. Unlike Cl2, BF3 is not a good oxidant, even though it contains boron in its highest oxidation state (+3). Nor can BF3 behave like a reductant. Like any other species with only six valence electrons, however, it is certainly a Lewis acid. Hence an acid–base reaction is the most likely alternative, especially because we know that CO can use the lone pair of electrons on carbon to act as a Lewis base. The most probable reaction is therefore the formation of a Lewis acid–base adduct.
3. Typically, both reactants behave like reductants. Unless one of them can also behave like an oxidant, no reaction will occur. We know that Sr is an active metal because it lies far to the left in the periodic table and that it is more electropositive than carbon. Carbon is a nonmetal with a significantly higher electronegativity; it is therefore more likely to accept electrons in a redox reaction. We conclude, therefore, that Sr will be oxidized, and C will be reduced. Carbon forms ionic carbides with active metals, so the reaction will produce a species formally containing either C4− or C22. Those that contain C4− usually involve small, highly charged metal ions, so Sr2+ will produce the acetylide (SrC2) instead.
Exercise $1$
Predict the products of the reactions and write a balanced chemical equation for each reaction.
1. C(s) + excess O2(g) $\xrightarrow{\Delta}$
2. C(s) + H2O(l) →
3. NaHCO3(s) + H2SO4(aq) →
Answer
1. C(s) + excess O2(g) $\xrightarrow{\Delta}$ CO2(g)
2. C(s) + H2O(l) → no reaction
3. NaHCO3(s) + H2SO4(aq) → CO2(g) + NaHSO4(aq) + H2O(l)
Summary
The stability of the carbon tetrahalides decreases as the halogen increases in size because of poor orbital overlap and steric crowding. Carbon forms three kinds of carbides with less electronegative elements: ionic carbides, which contain metal cations and C4− (methide) or C22 (acetylide) anions; interstitial carbides, which are characterized by covalent metal–carbon interactions and are among the hardest substances known; and covalent carbides, which have three-dimensional covalent network structures that make them extremely hard, high melting, and chemically inert.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/22%3A_Chemistry_of_the_Nonmetals/22.09%3A_Carbon.txt |
Learning Objectives
• To understand the trends in properties and reactivity of the group 14 elements.
Tin and lead oxides and sulfides are easily reduced to the metal by heating with charcoal, a discovery that must have occurred by accident when prehistoric humans used rocks containing their ores for a cooking fire. However, because tin and copper ores are often found together in nature, their alloy—bronze—was probably discovered before either element, a discovery that led to the Bronze Age. The heaviest element in group 14, lead, is such a soft and malleable metal that the ancient Romans used thin lead foils as writing tablets, as well as lead cookware and lead pipes for plumbing. (Recall that the atomic symbols for tin and lead come from their Latin names: Sn for stannum and Pb for plumbum.)
Although the first glasses were prepared from silica (silicon oxide, SiO2) around 1500 BC, elemental silicon was not prepared until 1824 because of its high affinity for oxygen. Jöns Jakob Berzelius was finally able to obtain amorphous silicon by reducing Na2SiF6 with molten potassium. The crystalline element, which has a shiny blue-gray luster, was not isolated until 30 yr later. The last member of the group 14 elements to be discovered was germanium, which was found in 1886 in a newly discovered silver-colored ore by the German chemist Clemens Winkler, who named the element in honor of his native country.
Preparation and General Properties of the Group 14 Elements
The natural abundance of the group 14 elements varies tremendously. Although oxygen is the most abundant element on Earth, the next most abundant is silicon, the next member of group 14. Pure silicon is obtained by reacting impure silicon with Cl2 to give SiCl4, followed by the fractional distillation of the impure SiCl4 and reduction with H2:
$\mathrm{SiCl_4(l)}+\mathrm{2H_2(g)}\xrightarrow{\Delta}\mathrm{Si(s)}+\mathrm{4HCl(g)} \label{3}$
Several million tons of silicon are annually produced with this method. Amorphous silicon containing residual amounts of hydrogen is used in photovoltaic devices that convert light to electricity, and silicon-based solar cells are used to power pocket calculators, boats, and highway signs, where access to electricity by conventional methods is difficult or expensive. Ultrapure silicon and germanium form the basis of the modern electronics industry (Figure $1$).
In contrast to silicon, the concentrations of germanium and tin in Earth’s crust are only 1–2 ppm. The concentration of lead, which is the end product of the nuclear decay of many radionuclides, is 13 ppm, making lead by far the most abundant of the heavy group 14 elements. No concentrated ores of germanium are known; like indium, germanium is generally recovered from flue dusts obtained by processing the ores of metals such as zinc. Because germanium is essentially transparent to infrared radiation, it is used in optical devices.
Tin and lead are soft metals that are too weak for structural applications, but because tin is flexible, corrosion resistant, and nontoxic, it is used as a coating in food packaging. A “tin can,” for example, is actually a steel can whose interior is coated with a thin layer (1–2 µm) of metallic tin. Tin is also used in superconducting magnets and low-melting-point alloys such as solder and pewter. Pure lead is obtained by heating galena (PbS) in air and reducing the oxide (PbO) to the metal with carbon, followed by electrolytic deposition to increase the purity:
$\mathrm{PbS(s)}+\frac{3}{2}\mathrm{O_2(g)}\xrightarrow{\Delta}\mathrm{PbO(s)}+\mathrm{SO_2(g)} \label{$4$}$
$\mathrm{PbO(s)}+\mathrm{C(s)}\xrightarrow{\Delta}\mathrm{Pb(l)}+\mathrm{CO(g)} \label{$5$}$
or
$\mathrm{PbO(s)}+\mathrm{CO(g)}\xrightarrow{\Delta}\mathrm{Pb(l)}+\mathrm{CO_2(g)} \label{$6$}$
By far the single largest use of lead is in lead storage batteries. The group 14 elements all have ns2np2 valence electron configurations. All form compounds in which they formally lose either the two np and the two ns valence electrons or just the two np valence electrons, giving a +4 or +2 oxidation state, respectively. Because covalent bonds decrease in strength with increasing atomic size and the ionization energies for the heavier elements of the group are higher than expected due to a higher Zeff, the relative stability of the +2 oxidation state increases smoothly from carbon to lead.
The relative stability of the +2 oxidation state increases, and the tendency to form catenated compounds decreases, from carbon to lead in group 14.
Recall that many carbon compounds contain multiple bonds formed by π overlap of singly occupied 2p orbitals on adjacent atoms. Compounds of silicon, germanium, tin, and lead with the same stoichiometry as those of carbon, however, tend to have different structures and properties. For example, CO2 is a gas that contains discrete O=C=O molecules, whereas the most common form of SiO2 is the high-melting solid known as quartz, the major component of sand. Instead of discrete SiO2 molecules, quartz contains a three-dimensional network of silicon atoms that is similar to the structure of diamond but with an oxygen atom inserted between each pair of silicon atoms. Thus each silicon atom is linked to four other silicon atoms by bridging oxygen atoms.
The tendency to catenate—to form chains of like atoms—decreases rapidly as we go down group 14 because bond energies for both the E–E and E–H bonds decrease with increasing atomic number (where E is any group 14 element). Consequently, inserting a CH2 group into a linear hydrocarbon such as n-hexane is exergonic (ΔG° = −45 kJ/mol), whereas inserting an SiH2 group into the silicon analogue of n-hexane (Si6H14) actually costs energy (ΔG° ≈ +25 kJ/mol). As a result of this trend, the thermal stability of catenated compounds decreases rapidly from carbon to lead.
In Table $1$ we see, once again, that there is a large difference between the lightest element (C) and the others in size, ionization energy, and electronegativity. As in group 13, the second and third elements (Si and Ge) are similar, and there is a reversal in the trends for some properties, such as ionization energy, between the fourth and fifth elements (Sn and Pb). As for group 13, these effects can be explained by the presence of filled (n − 1)d and (n − 2)f subshells, whose electrons are relatively poor at screening the outermost electrons from the higher nuclear charge.
Table $1$: Selected Properties of the Group 14 Elements
Property Carbon Silicon Germanium Tin Lead
*The configuration shown does not include filled d and f subshells.
The values cited are for six-coordinate +4 ions in the most common oxidation state, except for C4+ and Si4+, for which values for the four-coordinate ion are estimated.
X is Cl, Br, or I. Reaction with F2 gives the tetrafluorides (EF4) for all group 14 elements, where E represents any group 14 element.
atomic symbol C Si Ge Sn Pb
atomic number 6 14 32 50 82
atomic mass (amu) 12.01 28.09 72.64 118.71 207.2
valence electron configuration* 2s22p2 3s23p2 4s24p2 5s25p2 6s26p2
melting point/boiling point (°C) 4489 (at 10.3 MPa)/3825 1414/3265 939/2833 232/2602 327/1749
density (g/cm3) at 25°C 2.2 (graphite), 3.51 (diamond) 2.33 5.32 7.27(white) 11.30
atomic radius (pm) 77 (diamond) 111 125 145 154
first ionization energy (kJ/mol) 1087 787 762 709 716
most common oxidation state +4 +4 +4 +4 +4
ionic radius (pm) ≈29 ≈40 53 69 77.5
electron affinity (kJ/mol) −122 −134 −119 −107 −35
electronegativity 2.6 1.9 2.0 2.0 1.8
standard reduction potential (E°, V) (for EO2 → E in acidic solution) 0.21 −0.86 −0.18 −0.12 0.79
product of reaction with O2 CO2, CO SiO2 GeO2 SnO2 PbO
type of oxide acidic (CO2) acidic neutral (CO) amphoteric amphoteric amphoteric
product of reaction with N2 none Si3N4 none Sn3N4 none
product of reaction with X2 CX4 SiX4 GeX4 SnX4 PbX2
product of reaction with H2 CH4 none none none none
The group 14 elements follow the same pattern as the group 13 elements in their periodic properties.
Reactions and Compounds of the Heavier Group 14 Elements
Although silicon, germanium, tin, and lead in their +4 oxidation states often form binary compounds with the same stoichiometry as carbon, the structures and properties of these compounds are usually significantly different from those of the carbon analogues. Silicon and germanium are both semiconductors with structures analogous to diamond. Tin has two common allotropes: white (β) tin has a metallic lattice and metallic properties, whereas gray (α) tin has a diamond-like structure and is a semiconductor. The metallic β form is stable above 13.2°C, and the nonmetallic α form is stable below 13.2°C. Lead is the only group 14 element that is metallic in both structure and properties under all conditions.
Video $1$: Time lapse tin pest reaction.
Based on its position in the periodic table, we expect silicon to be amphoteric. In fact, it dissolves in strong aqueous base to produce hydrogen gas and solutions of silicates, but the only aqueous acid that it reacts with is hydrofluoric acid, presumably due to the formation of the stable SiF62 ion. Germanium is more metallic in its behavior than silicon. For example, it dissolves in hot oxidizing acids, such as HNO3 and H2SO4, but in the absence of an oxidant, it does not dissolve in aqueous base. Although tin has an even more metallic character than germanium, lead is the only element in the group that behaves purely as a metal. Acids do not readily attack it because the solid acquires a thin protective outer layer of a Pb2+ salt, such as PbSO4.
All group 14 dichlorides are known, and their stability increases dramatically as the atomic number of the central atom increases. Thus CCl2 is dichlorocarbene, a highly reactive, short-lived intermediate that can be made in solution but cannot be isolated in pure form using standard techniques; SiCl2 can be isolated at very low temperatures, but it decomposes rapidly above −150°C, and GeCl2 is relatively stable at temperatures below 20°C. In contrast, SnCl2 is a polymeric solid that is indefinitely stable at room temperature, whereas PbCl2 is an insoluble crystalline solid with a structure similar to that of SnCl2.
The stability of the group 14 dichlorides increases dramatically from carbon to lead.
Although the first four elements of group 14 form tetrahalides (MX4) with all the halogens, only fluorine is able to oxidize lead to the +4 oxidation state, giving PbF4. The tetrahalides of silicon and germanium react rapidly with water to give amphoteric oxides (where M is Si or Ge):
$MX_{4(s,l)} + 2H_2O_{(l)} \rightarrow MO_{2(s)} + 4HX_{(aq)} \label{1}$
In contrast, the tetrahalides of tin and lead react with water to give hydrated metal ions. Because of the stability of its +2 oxidation state, lead reacts with oxygen or sulfur to form PbO or PbS, respectively, whereas heating the other group 14 elements with excess O2 or S8 gives the corresponding dioxides or disulfides, respectively. The dioxides of the group 14 elements become increasingly basic as we go down the group.
The dioxides of the group 14 elements become increasingly basic down the group.
Because the Si–O bond is even stronger than the C–O bond (~452 kJ/mol versus ~358 kJ/mol), silicon has a strong affinity for oxygen. The relative strengths of the C–O and Si–O bonds contradict the generalization that bond strengths decrease as the bonded atoms become larger. This is because we have thus far assumed that a formal single bond between two atoms can always be described in terms of a single pair of shared electrons. In the case of Si–O bonds, however, the presence of relatively low-energy, empty d orbitals on Si and nonbonding electron pairs in the p or spn hybrid orbitals of O results in a partial π bond (Figure $3$). Due to its partial π double bond character, the Si–O bond is significantly stronger and shorter than would otherwise be expected. A similar interaction with oxygen is also an important feature of the chemistry of the elements that follow silicon in the third period (P, S, and Cl). Because the Si–O bond is unusually strong, silicon–oxygen compounds dominate the chemistry of silicon.
Because silicon–oxygen bonds are unusually strong, silicon–oxygen compounds dominate the chemistry of silicon.
Compounds with anions that contain only silicon and oxygen are called silicates, whose basic building block is the SiO44 unit:
The number of oxygen atoms shared between silicon atoms and the way in which the units are linked vary considerably in different silicates. Converting one of the oxygen atoms from terminal to bridging generates chains of silicates, while converting two oxygen atoms from terminal to bridging generates double chains. In contrast, converting three or four oxygens to bridging generates a variety of complex layered and three-dimensional structures, respectively.
Silicon and germanium react with nitrogen at high temperature to form nitrides (M3N4):
$3Si_{(l)} + 2N_{2(g)} \rightarrow Si_3N_{4(s)} \label{2}$
Silicon nitride has properties that make it suitable for high-temperature engineering applications: it is strong, very hard, and chemically inert, and it retains these properties to temperatures of about 1000°C.
Because of the diagonal relationship between boron and silicon, metal silicides and metal borides exhibit many similarities. Although metal silicides have structures that are as complex as those of the metal borides and carbides, few silicides are structurally similar to the corresponding borides due to the significantly larger size of Si (atomic radius 111 pm versus 87 pm for B). Silicides of active metals, such as Mg2Si, are ionic compounds that contain the Si4 ion. They react with aqueous acid to form silicon hydrides such as SiH4:
$Mg_2Si_{(s)} + 4H^+_{(aq)} \rightarrow 2Mg^{2+}_{(aq)} + SiH_{4(g)} \label{3A}$
Unlike carbon, catenated silicon hydrides become thermodynamically less stable as the chain lengthens. Thus straight-chain and branched silanes (analogous to alkanes) are known up to only n = 10; the germanium analogues (germanes) are known up to n = 9. In contrast, the only known hydride of tin is SnH4, and it slowly decomposes to elemental Sn and H2 at room temperature. The simplest lead hydride (PbH4) is so unstable that chemists are not even certain it exists. Because E=E and E≡E bonds become weaker with increasing atomic number (where E is any group 14 element), simple silicon, germanium, and tin analogues of alkenes, alkynes, and aromatic hydrocarbons are either unstable (Si=Si and Ge=Ge) or unknown. Silicon-based life-forms are therefore likely to be found only in science fiction.
The stability of group 14 hydrides decreases down the group, and E=E and E≡E bonds become weaker.
The only important organic derivatives of lead are compounds such as tetraethyllead [(CH3CH2)4Pb]. Because the Pb–C bond is weak, these compounds decompose at relatively low temperatures to produce alkyl radicals (R·), which can be used to control the rate of combustion reactions. For 60 yr, hundreds of thousands of tons of lead were burned annually in automobile engines, producing a mist of lead oxide particles along the highways that constituted a potentially serious public health problem. The use of catalytic converters reduced the amount of carbon monoxide, nitrogen oxides, and hydrocarbons released into the atmosphere through automobile exhausts, but it did nothing to decrease lead emissions. Because lead poisons catalytic converters, however, its use as a gasoline additive has been banned in most of the world.
Compounds that contain Si–C and Si–O bonds are stable and important. High-molecular-mass polymers called silicones contain an (Si–O–)n backbone with organic groups attached to Si (Figure $5$). The properties of silicones are determined by the chain length, the type of organic group, and the extent of cross-linking between the chains. Without cross-linking, silicones are waxes or oils, but cross-linking can produce flexible materials used in sealants, gaskets, car polishes, lubricants, and even elastic materials, such as the plastic substance known as Silly Putty.
Example $2$
For each reaction, explain why the given products form.
1. Pb(s) + Cl2(g) → PbCl2(s)
2. Mg2Si(s) + 4H2O(l) → SiH4(g) + 2Mg(OH)2(s)
3. GeO2(s) + 4OH(aq) → GeO44(aq) + 2H2O(l)
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form.
Solution
1. Lead is a metal, and chlorine is a nonmetal that is a strong oxidant. Thus we can expect a redox reaction to occur in which the metal acts as a reductant. Although lead can form compounds in the +2 and +4 oxidation states, Pb4+ is a potent oxidant (the inert-pair effect). Because lead prefers the +2 oxidation state and chlorine is a weaker oxidant than fluorine, we expect PbCl2 to be the product.
2. This is the reaction of water with a metal silicide, which formally contains the Si4 ion. Water can act as either an acid or a base. Because the other compound is a base, we expect an acid–base reaction to occur in which water acts as an acid. Because Mg2Si contains Si in its lowest possible oxidation state, however, an oxidation–reduction reaction is also a possibility. But water is a relatively weak oxidant, so an acid–base reaction is more likely. The acid (H2O) transfers a proton to the base (Si4), which can accept four protons to form SiH4. Proton transfer from water produces the OH ion, which will combine with Mg2+ to give magnesium hydroxide.
3. We expect germanium dioxide (GeO2) to be amphoteric because of the position of germanium in the periodic table. It should dissolve in strong aqueous base to give an anionic species analogous to silicate.
Exercise $2$
Predict the products of the reactions and write a balanced chemical equation for each reaction.
1. PbO2(s) $\xrightarrow{\Delta}$
2. GeCl4(s) + H2O(l) →
3. Sn(s) + HCl(aq) →
Answer
1. $\mathrm{PbO_2(s)}\xrightarrow{\Delta}\mathrm{PbO(s)}+\frac{1}{2}\mathrm{O_2(g)}$
2. GeCl4(s) + 2H2O(l) → GeO2(s) + 4HCl(aq)
3. Sn(s) + 2HCl(aq) → Sn2+(aq) + H2(g) + 2Cl(aq)
Summary
The group 14 elements show the greatest diversity in chemical behavior of any group; covalent bond strengths decease with increasing atomic size, and ionization energies are greater than expected, increasing from C to Pb. Because the covalent bond strength decreases with increasing atomic size and greater-than-expected ionization energies due to an increase in Zeff, the stability of the +2 oxidation state increases from carbon to lead. The tendency to form multiple bonds and to catenate decreases as the atomic number increases. Consistent with periodic trends, metallic behavior increases down the group. Silicon has a tremendous affinity for oxygen because of partial Si–O π bonding. Dioxides of the group 14 elements become increasingly basic down the group and their metallic character increases. Silicates contain anions that consist of only silicon and oxygen. Aluminosilicates are formed by replacing some of the Si atoms in silicates by Al atoms; aluminosilicates with three-dimensional framework structures are called zeolites. Nitrides formed by reacting silicon or germanium with nitrogen are strong, hard, and chemically inert. The hydrides become thermodynamically less stable down the group. Moreover, as atomic size increases, multiple bonds between or to the group 14 elements become weaker. Silicones, which contain an Si–O backbone and Si–C bonds, are high-molecular-mass polymers whose properties depend on their compositions.
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Learning Objectives
• To understand the trends in properties and the reactivity of the group 13 elements.
Group 13 is the first group to span the dividing line between metals and nonmetals, so its chemistry is more diverse than that of groups 1 and 2, which include only metallic elements. Except for the lightest element (boron), the group 13 elements are all relatively electropositive; that is, they tend to lose electrons in chemical reactions rather than gain them. Although group 13 includes aluminum, the most abundant metal on Earth, none of these elements was known until the early 19th century because they are never found in nature in their free state. Elemental boron and aluminum, which were first prepared by reducing B2O3 and AlCl3, respectively, with potassium, could not be prepared until potassium had been isolated and shown to be a potent reductant. Indium (In) and thallium (Tl) were discovered in the 1860s by using spectroscopic techniques, long before methods were available for isolating them. Indium, named for its indigo (deep blue-violet) emission line, was first observed in the spectrum of zinc ores, while thallium (from the Greek thallos, meaning “a young, green shoot of a plant”) was named for its brilliant green emission line. Gallium (Ga; Mendeleev’s eka-aluminum) was discovered in 1875 by the French chemist Paul Émile Lecoq de Boisbaudran during a systematic search for Mendeleev’s “missing” element in group 13.
Group 13 elements are never found in nature in their free state.
Preparation and General Properties of the Group 13 Elements
As reductants, the group 13 elements are less powerful than the alkali metals and alkaline earth metals. Nevertheless, their compounds with oxygen are thermodynamically stable, and large amounts of energy are needed to isolate even the two most accessible elements—boron and aluminum—from their oxide ores.
Although boron is relatively rare (it is about 10,000 times less abundant than aluminum), concentrated deposits of borax [Na2B4O5(OH)4·8H2O] are found in ancient lake beds (Figure $1$) and were used in ancient times for making glass and glazing pottery. Boron is produced on a large scale by reacting borax with acid to produce boric acid [B(OH)3], which is then dehydrated to the oxide (B2O3). Reduction of the oxide with magnesium or sodium gives amorphous boron that is only about 95% pure:
$\mathrm{Na_2B_4O_5(OH)_4\cdot8H_2O(s)}\xrightarrow{\textrm{acid}}\mathrm{B(OH)_3(s)}\xrightarrow{\Delta}\mathrm{B_2O_3(s)} \label{Eq1}$
$\mathrm{B_2O_3(s)}+\mathrm{3Mg(s)}\xrightarrow{\Delta}\mathrm{2B(s)}+\mathrm{3MgO(s)} \label{Eq2}$
Pure, crystalline boron, however, is extremely difficult to obtain because of its high melting point (2300°C) and the highly corrosive nature of liquid boron. It is usually prepared by reducing pure BCl3 with hydrogen gas at high temperatures or by the thermal decomposition of boron hydrides such as diborane (B2H6):
$\mathrm{BCl_3(g)}+\frac{3}{2}\mathrm{H_2(g)}\rightarrow\mathrm{B(s)}+\mathrm{3HCl(g)} \label{Eq3}$
$B_2H_{6(g)} \rightarrow 2B_{(s)} + 3H_{2(g)} \label{Eq4}$
The reaction shown in Equation $\ref{Eq3}$ is used to prepare boron fibers, which are stiff and light. Hence they are used as structural reinforcing materials in objects as diverse as the US space shuttle and the frames of lightweight bicycles that are used in races such as the Tour de France. Boron is also an important component of many ceramics and heat-resistant borosilicate glasses, such as Pyrex, which is used for ovenware and laboratory glassware.
In contrast to boron, deposits of aluminum ores such as bauxite, a hydrated form of Al2O3, are abundant. With an electrical conductivity about twice that of copper on a weight for weight basis, aluminum is used in more than 90% of the overhead electric power lines in the United States. However, because aluminum–oxygen compounds are stable, obtaining aluminum metal from bauxite is an expensive process. Aluminum is extracted from oxide ores by treatment with a strong base, which produces the soluble hydroxide complex [Al(OH)4]. Neutralization of the resulting solution with gaseous CO2 results in the precipitation of Al(OH)3:
$2[Al(OH)_4]^−_{(aq)} + CO_{2(g)} \rightarrow 2Al(OH)_{3(s)} + CO^{2−}_{3(aq)} + H_2O_{(l)} \label{Eq5}$
Thermal dehydration of Al(OH)3 produces Al2O3, and metallic aluminum is obtained by the electrolytic reduction of Al2O3 using the Hall–Heroult process. Of the group 13 elements, only aluminum is used on a large scale: for example, each Boeing 777 airplane is about 50% aluminum by mass.
The other members of group 13 are rather rare: gallium is approximately 5000 times less abundant than aluminum, and indium and thallium are even scarcer. Consequently, these metals are usually obtained as by-products in the processing of other metals. The extremely low melting point of gallium (29.6°C), however, makes it easy to separate from aluminum. Due to its low melting point and high boiling point, gallium is used as a liquid in thermometers that have a temperature range of almost 2200°C. Indium and thallium, the heavier group 13 elements, are found as trace impurities in sulfide ores of zinc and lead. Indium is used as a crushable seal for high-vacuum cryogenic devices, and its alloys are used as low-melting solders in electronic circuit boards. Thallium, on the other hand, is so toxic that the metal and its compounds have few uses. Both indium and thallium oxides are released in flue dust when sulfide ores are converted to metal oxides and SO2. Until relatively recently, these and other toxic elements were allowed to disperse in the air, creating large “dead zones” downwind of a smelter. The flue dusts are now trapped and serve as a relatively rich source of elements such as In and Tl (as well as Ge, Cd, Te, and As).
Table $1$ summarizes some important properties of the group 13 elements. Notice the large differences between boron and aluminum in size, ionization energy, electronegativity, and standard reduction potential, which is consistent with the observation that boron behaves chemically like a nonmetal and aluminum like a metal. All group 13 elements have ns2np1 valence electron configurations, and all tend to lose their three valence electrons to form compounds in the +3 oxidation state. The heavier elements in the group can also form compounds in the +1 oxidation state formed by the formal loss of the single np valence electron. Because the group 13 elements generally contain only six valence electrons in their neutral compounds, these compounds are all moderately strong Lewis acids.
Table $1$: Selected Properties of the Group 13 Elements
Property Boron Aluminum* Gallium Indium Thallium
*This is the name used in the United States; the rest of the world inserts an extra i and calls it aluminium.
The configuration shown does not include filled d and f subshells.
The values cited are for six-coordinate ions in the most common oxidation state, except for Al3+, for which the value for the four-coordinate ion is given. The B3+ ion is not a known species; the radius cited is an estimated four-coordinate value.
§X is Cl, Br, or I. Reaction with F2 gives the trifluorides (MF3) for all group 13 elements.
atomic symbol B Al Ga In Tl
atomic number 5 13 31 49 81
atomic mass (amu) 10.81 26.98 69.72 114.82 204.38
valence electron configuration 2s22p1 3s23p1 4s24p1 5s25p1 6s26p1
melting point/boiling point (°C) 2075/4000 660/2519 29.7/2204 156.6/2072 304/1473
density (g/cm3) at 25°C 2.34 2.70 5.91 7.31 11.8
atomic radius (pm) 87 118 136 156 156
first ionization energy (kJ/mol) 801 578 579 558 589
most common oxidation state +3 +3 +3 +3 +1
ionic radius (pm) −25 54 62 80 162
electron affinity (kJ/mol) −27 −42 −40 −39 −37
electronegativity 2.0 1.6 1.8 1.8 1.8
standard reduction potential (E°, V) −0.87 −1.66 −0.55 −0.34 +0.741 of M3+(aq)
product of reaction with O2 B2O3 Al2O3 Ga2O3 In2O3 Tl2O
type of oxide acidic amphoteric amphoteric amphoteric basic
product of reaction with N2 BN AlN GaN InN none
product of reaction with X2§ BX3 Al2X6 Ga2X6 In2X6 TlX
Neutral compounds of the group 13 elements are electron deficient, so they are generally moderately strong Lewis acids.
In contrast to groups 1 and 2, the group 13 elements show no consistent trends in ionization energies, electron affinities, and reduction potentials, whereas electronegativities actually increase from aluminum to thallium. Some of these anomalies, especially for the series Ga, In, Tl, can be explained by the increase in the effective nuclear charge (Zeff) that results from poor shielding of the nuclear charge by the filled (n − 1)d10 and (n − 2)f14 subshells. Consequently, although the actual nuclear charge increases by 32 as we go from indium to thallium, screening by the filled 5d and 4f subshells is so poor that Zeff increases significantly from indium to thallium. Thus the first ionization energy of thallium is actually greater than that of indium.
Anomalies in periodic trends among Ga, In, and Tl can be explained by the increase in the effective nuclear charge due to poor shielding.
Reactions and Compounds of Boron
Elemental boron is a semimetal that is remarkably unreactive; in contrast, the other group 13 elements all exhibit metallic properties and reactivity. We therefore consider the reactions and compounds of boron separately from those of other elements in the group. All group 13 elements have fewer valence electrons than valence orbitals, which generally results in delocalized, metallic bonding. With its high ionization energy, low electron affinity, low electronegativity, and small size, however, boron does not form a metallic lattice with delocalized valence electrons. Instead, boron forms unique and intricate structures that contain multicenter bonds, in which a pair of electrons holds together three or more atoms.
Elemental boron forms multicenter bonds, whereas the other group 13 elements exhibit metallic bonding.
The basic building block of elemental boron is not the individual boron atom, as would be the case in a metal, but rather the B12 icosahedron. Because these icosahedra do not pack together very well, the structure of solid boron contains voids, resulting in its low density (Figure $3$). Elemental boron can be induced to react with many nonmetallic elements to give binary compounds that have a variety of applications. For example, plates of boron carbide (B4C) can stop a 30-caliber, armor-piercing bullet, yet they weigh 10%–30% less than conventional armor. Other important compounds of boron with nonmetals include boron nitride (BN), which is produced by heating boron with excess nitrogen (Equation $\ref{Eq22.6}$); boron oxide (B2O3), which is formed when boron is heated with excess oxygen (Equation $\ref{Eq22.7}$); and the boron trihalides (BX3), which are formed by heating boron with excess halogen (Equation $\ref{Eq22.8}$).
$\mathrm{2B(s)}+\mathrm{N_2(g)}\xrightarrow{\Delta}\mathrm{2BN(s)} \label{Eq22.6}$
$\mathrm{4B(s)} + \mathrm{3O_2(g)}\xrightarrow{\Delta}\mathrm{2B_2O_3(s)}\label{Eq22.7}$
$\mathrm{2B(s)} +\mathrm{3X_2(g)}\xrightarrow{\Delta}\mathrm{2BX_3(g)}\label{Eq22.8}$
As is typical of elements lying near the dividing line between metals and nonmetals, many compounds of boron are amphoteric, dissolving in either acid or base.
Boron nitride is similar in many ways to elemental carbon. With eight electrons, the B–N unit is isoelectronic with the C–C unit, and B and N have the same average size and electronegativity as C. The most stable form of BN is similar to graphite, containing six-membered B3N3 rings arranged in layers. At high temperature and pressure, hexagonal BN converts to a cubic structure similar to diamond, which is one of the hardest substances known. Boron oxide (B2O3) contains layers of trigonal planar BO3 groups (analogous to BX3) in which the oxygen atoms bridge two boron atoms. It dissolves many metal and nonmetal oxides, including SiO2, to give a wide range of commercially important borosilicate glasses. A small amount of CoO gives the deep blue color characteristic of “cobalt blue” glass.
At high temperatures, boron also reacts with virtually all metals to give metal borides that contain regular three-dimensional networks, or clusters, of boron atoms. The structures of two metal borides—ScB12 and CaB6—are shown in Figure $4$. Because metal-rich borides such as ZrB2 and TiB2 are hard and corrosion resistant even at high temperatures, they are used in applications such as turbine blades and rocket nozzles.
Boron hydrides were not discovered until the early 20th century, when the German chemist Alfred Stock undertook a systematic investigation of the binary compounds of boron and hydrogen, although binary hydrides of carbon, nitrogen, oxygen, and fluorine have been known since the 18th century. Between 1912 and 1936, Stock oversaw the preparation of a series of boron–hydrogen compounds with unprecedented structures that could not be explained with simple bonding theories. All these compounds contain multicenter bonds. The simplest example is diborane (B2H6), which contains two bridging hydrogen atoms (part (a) in Figure $5$. An extraordinary variety of polyhedral boron–hydrogen clusters is now known; one example is the B12H122− ion, which has a polyhedral structure similar to the icosahedral B12 unit of elemental boron, with a single hydrogen atom bonded to each boron atom.
A related class of polyhedral clusters, the carboranes, contain both CH and BH units; an example is shown here. Replacing the hydrogen atoms bonded to carbon with organic groups produces substances with novel properties, some of which are currently being investigated for their use as liquid crystals and in cancer chemotherapy.
The enthalpy of combustion of diborane (B2H6) is −2165 kJ/mol, one of the highest values known:
$B_2H_{6(g)} + 3O_{2(g)} \rightarrow B_2O_{3(s)} + 3H_2O(l)\;\;\; ΔH_{comb} = −2165\; kJ/mol \label{Eq 22.9}$
Consequently, the US military explored using boron hydrides as rocket fuels in the 1950s and 1960s. This effort was eventually abandoned because boron hydrides are unstable, costly, and toxic, and, most important, B2O3 proved to be highly abrasive to rocket nozzles. Reactions carried out during this investigation, however, showed that boron hydrides exhibit unusual reactivity.
Because boron and hydrogen have almost identical electronegativities, the reactions of boron hydrides are dictated by minor differences in the distribution of electron density in a given compound. In general, two distinct types of reaction are observed: electron-rich species such as the BH4 ion are reductants, whereas electron-deficient species such as B2H6 act as oxidants.
Example $1$
For each reaction, explain why the given products form.
1. B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(l)
2. BCl3(l) + 3H2O(l) → B(OH)3(aq) + 3HCl(aq)
3. $\mathrm{2BI_3(s)}+\mathrm{3H_2(g)}\xrightarrow{\Delta}\frac{1}{6}\mathrm{B_{12}(s)}+\mathrm{6HI(g)}$
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form.
Solution
1. Molecular oxygen is an oxidant. If the other reactant is a potential reductant, we expect that a redox reaction will occur. Although B2H6 contains boron in its highest oxidation state (+3), it also contains hydrogen in the −1 oxidation state (the hydride ion). Because hydride is a strong reductant, a redox reaction will probably occur. We expect that H will be oxidized to H+ and O2 will be reduced to O2−, but what are the actual products? A reasonable guess is B2O3 and H2O, both stable compounds.
2. Neither BCl3 nor water is a powerful oxidant or reductant, so a redox reaction is unlikely; a hydrolysis reaction is more probable. Nonmetal halides are acidic and react with water to form a solution of the hydrohalic acid and a nonmetal oxide or hydroxide. In this case, the most probable boron-containing product is boric acid [B(OH)3].
3. We normally expect a boron trihalide to behave like a Lewis acid. In this case, however, the other reactant is elemental hydrogen, which usually acts as a reductant. The iodine atoms in BI3 are in the lowest accessible oxidation state (−1), and boron is in the +3 oxidation state. Consequently, we can write a redox reaction in which hydrogen is oxidized and boron is reduced. Because compounds of boron in lower oxidation states are rare, we expect that boron will be reduced to elemental boron. The other product of the reaction must therefore be HI.
Exercise $1$
Predict the products of the reactions and write a balanced chemical equation for each reaction.
1. $\mathrm{B_2H_6(g)}+\mathrm{H_2O(l)}\xrightarrow{\Delta}$
2. $\mathrm{BBr_3(l)}+\mathrm{O_2(g)}\rightarrow$
3. $\mathrm{B_2O_3(s)}+\mathrm{Ca(s)}\xrightarrow{\Delta}$
Answer
1. $\mathrm{B_2H_6(g)}+\mathrm{H_2O(l)}\xrightarrow{\Delta}\mathrm{2B(OH)_3(s)}+\mathrm{6H_2(g)}$
2. $\mathrm{BBr_3(l)}+\mathrm{O_2(g)}\rightarrow\textrm{no reaction}$
3. $\mathrm{6B_2O_3(s)}+18\mathrm{Ca(s)}\xrightarrow{\Delta}\mathrm{B_{12}(s)}+\mathrm{18CaO(s)}$
Reactions and Compounds of the Heavier Group 13 Elements
All four of the heavier group 13 elements (Al, Ga, In, and Tl) react readily with the halogens to form compounds with a 1:3 stoichiometry:
$2M_{(s)} + 3X_{2(s,l,g)} \rightarrow 2MX_{3(s)} \text{ or } M_2X_6 \label{Eq10}$
The reaction of Tl with iodine is an exception: although the product has the stoichiometry TlI3, it is not thallium(III) iodide, but rather a thallium(I) compound, the Tl+ salt of the triiodide ion (I3). This compound forms because iodine is not a powerful enough oxidant to oxidize thallium to the +3 oxidation state.
Of the halides, only the fluorides exhibit behavior typical of an ionic compound: they have high melting points (>950°C) and low solubility in nonpolar solvents. In contrast, the trichorides, tribromides, and triiodides of aluminum, gallium, and indium, as well as TlCl3 and TlBr3, are more covalent in character and form halogen-bridged dimers (part (b) in Figure $4$). Although the structure of these dimers is similar to that of diborane (B2H6), the bonding can be described in terms of electron-pair bonds rather than the delocalized electron-deficient bonding found in diborane. Bridging halides are poor electron-pair donors, so the group 13 trihalides are potent Lewis acids that react readily with Lewis bases, such as amines, to form a Lewis acid–base adduct:
$Al_2Cl_{6(soln)} + 2(CH_3)_3N_{(soln)} \rightarrow 2(CH_3)_3N:AlCl_{3(soln)} \label{Eq11}$
In water, the halides of the group 13 metals hydrolyze to produce the metal hydroxide ($M(OH)_3\)): \[MX_{3(s)} + 3H_2O_{(l)} \rightarrow M(OH)_{3(s)} + 3HX_{(aq)} \label{Eq12}$
In a related reaction, Al2(SO4)3 is used to clarify drinking water by the precipitation of hydrated Al(OH)3, which traps particulates. The halides of the heavier metals (In and Tl) are less reactive with water because of their lower charge-to-radius ratio. Instead of forming hydroxides, they dissolve to form the hydrated metal complex ions: [M(H2O)6]3+.
Of the group 13 halides, only the fluorides behave as typical ionic compounds.
Like boron (Equation $\ref{Eq22.7}$), all the heavier group 13 elements react with excess oxygen at elevated temperatures to give the trivalent oxide (M2O3), although Tl2O3 is unstable:
$\mathrm{4M(s)}+\mathrm{3O_2(g)}\xrightarrow{\Delta}\mathrm{2M_2O_3(s)} \label{Eq13}$
Aluminum oxide (Al2O3), also known as alumina, is a hard, high-melting-point, chemically inert insulator used as a ceramic and as an abrasive in sandpaper and toothpaste. Replacing a small number of Al3+ ions in crystalline alumina with Cr3+ ions forms the gemstone ruby, whereas replacing Al3+ with a mixture of Fe2+, Fe3+, and Ti4+ produces blue sapphires. The gallium oxide compound MgGa2O4 gives the brilliant green light familiar to anyone who has ever operated a xerographic copy machine. All the oxides dissolve in dilute acid, but Al2O3 and Ga2O3 are amphoteric, which is consistent with their location along the diagonal line of the periodic table, also dissolving in concentrated aqueous base to form solutions that contain M(OH)4 ions.
Group 13 trihalides are potent Lewis acids that react with Lewis bases to form a Lewis acid–base adduct.
Aluminum, gallium, and indium also react with the other group 16 elements (chalcogens) to form chalcogenides with the stoichiometry M2Y3. However, because Tl(III) is too strong an oxidant to form a stable compound with electron-rich anions such as S2−, Se2−, and Te2−, thallium forms only the thallium(I) chalcogenides with the stoichiometry Tl2Y. Only aluminum, like boron, reacts directly with N2 (at very high temperatures) to give AlN, which is used in transistors and microwave devices as a nontoxic heat sink because of its thermal stability; GaN and InN can be prepared using other methods. All the metals, again except Tl, also react with the heavier group 15 elements (pnicogens) to form the so-called III–V compounds, such as GaAs. These are semiconductors, whose electronic properties, such as their band gaps, differ from those that can be achieved using either pure or doped group 14 elements. For example, nitrogen- and phosphorus-doped gallium arsenide (GaAs1−x−yPxNy) is used in the displays of calculators and digital watches.
All group 13 oxides dissolve in dilute acid, but Al2O3 and Ga2O3 are amphoteric.
Unlike boron, the heavier group 13 elements do not react directly with hydrogen. Only the aluminum and gallium hydrides are known, but they must be prepared indirectly; AlH3 is an insoluble, polymeric solid that is rapidly decomposed by water, whereas GaH3 is unstable at room temperature.
Complexes of Group 13 Elements
Boron has a relatively limited tendency to form complexes, but aluminum, gallium, indium, and, to some extent, thallium form many complexes. Some of the simplest are the hydrated metal ions [M(H2O)63+], which are relatively strong Brønsted–Lowry acids that can lose a proton to form the M(H2O)5(OH)2+ ion:
$[M(H_2O)_6]^{3+}_{(aq)} \rightarrow M(H_2O)_5(OH)^{2+}_{(aq)} + H^+_{(aq)} \label{Eq14}$
Group 13 metal ions also form stable complexes with species that contain two or more negatively charged groups, such as the oxalate ion. The stability of such complexes increases as the number of coordinating groups provided by the ligand increases.
Example $2$
For each reaction, explain why the given products form.
1. $\mathrm{2Al(s)} + \mathrm{Fe_2O_3(s)}\xrightarrow{\Delta}\mathrm{2Fe(l)} + \mathrm{Al_2O_3(s)}$
2. $\mathrm{2Ga(s)} + \mathrm{6H_2O(l)}+ \mathrm{2OH^-(aq)}\xrightarrow{\Delta}\mathrm{3H_2(g)} + \mathrm{2Ga(OH)^-_4(aq)}$
3. $\mathrm{In_2Cl_6(s)}\xrightarrow{\mathrm{H_2O(l)}}\mathrm{2In^{3+}(aq)}+\mathrm{6Cl^-(aq)}$
Given: balanced chemical equations
Asked for: why the given products form
Strategy:
Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form.
Solution
1. Aluminum is an active metal and a powerful reductant, and Fe2O3 contains Fe(III), a potential oxidant. Hence a redox reaction is probable, producing metallic Fe and Al2O3. Because Al is a main group element that lies above Fe, which is a transition element, it should be a more active metal than Fe. Thus the reaction should proceed to the right. In fact, this is the thermite reaction, which is so vigorous that it produces molten Fe and can be used for welding.
2. Gallium lies immediately below aluminum in the periodic table and is amphoteric, so it will dissolve in either acid or base to produce hydrogen gas. Because gallium is similar to aluminum in many of its properties, we predict that gallium will dissolve in the strong base.
3. The metallic character of the group 13 elements increases with increasing atomic number. Indium trichloride should therefore behave like a typical metal halide, dissolving in water to form the hydrated cation.
Exercise $2$
Predict the products of the reactions and write a balanced chemical equation for each reaction.
1. LiH(s) + Al2Cl6(soln)→
2. Al2O3(s) + OH(aq)→
3. Al(s) + N2(g) $\xrightarrow{\Delta}$
4. Ga2Cl6(soln) + Cl(soln)→
Answer
1. 8LiH(s) + Al2Cl6(soln)→2LiAlH4(soln) + 6LiCl(s)
2. Al2O3(s) + 2OH(aq) + 3H2O(l) → 2Al(OH)4(aq)
3. 2Al(s) + N2(g) $\xrightarrow{\Delta}$ 2AlN(s)
4. Ga2Cl6(soln) + 2Cl(soln) → 2GaCl4(soln)
Summary
Compounds of the group 13 elements with oxygen are thermodynamically stable. Many of the anomalous properties of the group 13 elements can be explained by the increase in Zeff moving down the group. Isolation of the group 13 elements requires a large amount of energy because compounds of the group 13 elements with oxygen are thermodynamically stable. Boron behaves chemically like a nonmetal, whereas its heavier congeners exhibit metallic behavior. Many of the inconsistencies observed in the properties of the group 13 elements can be explained by the increase in Zeff that arises from poor shielding of the nuclear charge by the filled (n − 1)d10 and (n − 2)f14 subshells. Instead of forming a metallic lattice with delocalized valence electrons, boron forms unique aggregates that contain multicenter bonds, including metal borides, in which boron is bonded to other boron atoms to form three-dimensional networks or clusters with regular geometric structures. All neutral compounds of the group 13 elements are electron deficient and behave like Lewis acids. The trivalent halides of the heavier elements form halogen-bridged dimers that contain electron-pair bonds, rather than the delocalized electron-deficient bonds characteristic of diborane. Their oxides dissolve in dilute acid, although the oxides of aluminum and gallium are amphoteric. None of the group 13 elements reacts directly with hydrogen, and the stability of the hydrides prepared by other routes decreases as we go down the group. In contrast to boron, the heavier group 13 elements form a large number of complexes in the +3 oxidation state. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/22%3A_Chemistry_of_the_Nonmetals/22.11%3A_Boron.txt |
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
22.3: Group 18: Nobel Gases
Questions
1. The chemistry of the noble gases is largely dictated by a balance between two competing properties. What are these properties? How do they affect the reactivity of these elements?
2. Of the group 18 elements, only krypton, xenon, and radon form stable compounds with other atoms and then only with very electronegative elements. Why?
3. Give the type of hybrid orbitals used by xenon in each species.
1. XeF2
2. XeF4
3. XeO3
4. XeOF4
5. XeO4
6. XeO64−
1. Which element is the least metallic—B, Ga, Tl, Pb, Ne, or Ge?
2. Of Br, N, Ar, Bi, Se, He, and S, which would you expect to form positive ions most easily? negative ions most easily?
3. Of BCl3, BCl4, CH4, H3N·BF3, PCl3, PCl5, XeO3, H2O, and F, which species do you expect to be
1. electron donors?
2. electron acceptors?
3. neither electron donors nor acceptors?
4. both electron donors and acceptors?
1. Of HCl, HClO4, HBr, H2S, HF, KrF2, and PH3, which is the strongest acid?
2. Of CF4, NH3, NF3, H2O, OF2, SiF4, H2S, XeF4, and SiH4, which is the strongest base?
Structure and Reactivity
1. Write a balanced chemical equation showing how you would prepare each compound from its elements and other common compounds.
1. XeF2
2. XeF4
3. XeF6
4. XeOF4
5. XeO3
1. Write a balanced chemical equation showing how you would make each compound.
1. XeF2 from Xe gas
2. NaXeF7 from its elements
3. RnO3 from Rn
1. In an effort to synthesize XeF6, a chemist passed fluorine gas through a glass tube containing xenon gas. However, the product was not the one expected. What was the actual product?
2. Write a balanced chemical equation to describe the reaction of each species with water.
1. B2H6
2. F2
3. C4+
1. Using heavy water (D2O) as the source of deuterium, how could you prepare each compound?
1. LiAlD4
2. D2SO4
3. SiD4
4. DF
2. Predict the product(s) of each reaction and write a balanced chemical equation for each reaction.
1. Al2O3(s) in OH(aq)
2. Ar(g) + F2(g)
3. PI3(s) + H2O(l)
4. H3PO3(l) + OH(aq)
5. Bi(s) + excess Br2(l)
Answers
1. Xe(g) + F2(g) → XeF2(s)
2. Xe(g) + 2F2(g) → XeF4(s)
3. Xe(g) + 3F2(g) → XeF6(s)
4. 2XeF6(s) + SiO2(s) → 2XeOF4(l) + SiF4(l)
5. XeF6(s) + 3H2O(l) → XeO3(s) + 6HF(aq)
1. SiF4; SiO2(s) + 2F2(g) → SiF4(l) + 2O2(g)
1. 2Na(s) + 2D2O(l) → D2(g) + 2NaOD(aq)
2Li(s) + D2(g) → 2LiD(s)
4LiD(s) + AlCl3(soln) → LiAlD4(s) + 3LiCl(soln)
1. D2O(l) + SO3(g) → D2SO4(l)
2. SiCl4(l) + LiAlD4(s) [from part (a)] → SiD4(g) + LiCl(s) + AlCl3(s)
3. CaF2(s) + D2SO4(l) [from part (b)] → 2DF(g) + CaSO4(s)
22.4: Group 7A: The Halogens
Questions
1. The lightest elements of groups 15, 16, and 17 form unusually weak single bonds. Why are their bonds so weak?
2. Fluorine has an anomalously low F–F bond energy. Why? Why does fluorine form compounds only in the −1 oxidation state, whereas the other halogens exist in multiple oxidation states?
3. Compare AlI3, InCl3, GaF3, and LaBr3 with respect to the type of M–X bond formed, melting point, and solubility in nonpolar solvents.
4. What are the formulas of the interhalogen compounds that will most likely contain the following species in the indicated oxidation states: I (+3), Cl (+3), I (−1), Br (+5)?
5. Consider this series of bromides: AlBr3, SiBr4, and PBr5. Does the ionic character of the bond between the Br atoms and the central atom decrease or increase in this series?
6. Chromium forms compounds in the +6, +3, and +2 oxidation states. Which halogen would you use to produce each oxidation state? Justify your selections.
7. Of ClF7, BrF5, IF7, BrF3, ICl3, IF3, and IF5, which one is least likely to exist? Justify your selection.
Answers
1. Electrostatic repulsions between lone pairs on adjacent atoms decrease bond strength.
1. Ionic character decreases as Δχ decreases from Al to P.
1. ClF7
Structure and Reactivity
1. SiF4 reacts easily with NaF to form SiF62−. In contrast, CF4 is totally inert and shows no tendency to form CF62− under even extreme conditions. Explain this difference.
2. Predict the products of each reaction and then balance each chemical equation.
1. Xe(g) + excess F2(g) →
2. Se(s) + Cl2(g) →
3. SO2(g) + Br2(g) →
4. NaBH4(s) + BF3(soln) →
1. Write a balanced chemical equation for the reaction of aqueous HF with
1. SiO2.
2. Na2CO3.
3. CaO.
1. Oxyhalides of sulfur, such as the thionyl halides (SOX2, where X is F, Cl, or Br), are well known. Because the thionyl halides react vigorously with trace amounts of water, they are used for dehydrating hydrated metal salts. Write a balanced chemical equation to show the products of reaction of SOCl2 with water.
1. Write a balanced chemical equation describing each reaction.
1. the burning of sulfur in a chlorine atmosphere
2. the dissolution of iodine in a potassium iodide solution
3. the hydrolysis of PCl3
4. the preparation of HF from calcium fluoride and sulfuric acid
5. the thermal decomposition of KClO3
6. the oxidation of sulfide ion by elemental iodine
1. Write the complete Lewis electron structure, the type of hybrid used by the central atom, and the number of lone pair electrons present on the central atom for each compound.
1. CF4
2. PCl3
3. XeF4
Answers
1. Carbon has no low energy d orbitals that can be used to form a set of d2sp3 hybrid orbitals. It is also so small that it is impossible for six fluorine atoms to fit around it at a distance that would allow for formation of strong C–F bonds.
1. SiO2(s) + 6HF(aq) → SiF62−(aq) + 2H+(aq) + 2H2O(l)
2. Na2CO3(s) + 2HF(aq) → CO2(g) + 2NaF(aq) + H2O(l)
3. CaO(s) + 2HF(aq) → CaF2(s) + H2O(l)
1. S8(s) + 4Cl2(g) → 4S2Cl2(l)
2. I2(s) + KI(aq) → I3(aq) + K+(aq)
3. PCl3(l) + 3H2O(l) → H3PO3(aq) + 3HCl(aq)
4. CaF2(s) + H2SO4(aq) → 2HF(aq) + CaSO4(s)
5. 2KClO3(s) \(\xrightarrow{\Delta}\) 2KCl(s) + 3O2(g)
6. 8S2−(aq) + 8I2(aq) → S8(s) + 16I(aq)
22.7: Nitrogen
Problems
• Complete and balance the following equations
N2 + ___H2→ ___NH_
H2N2O2 → ?
2NH3 + CO2 → ?
__Mg + N2 → Mg_N_
N2H5 + H2O → ?
• What are the different isotopes of Nitrogen?
• List the oxiadation states of various nitrogen oxides: N2O, NO, N2O3, N2O4, N2O5
• List the different elements that Nitrogen will react with to make it basic or acidic....
• Uses of nitrogen
Answers
• Complete and balance the following equations
N2 + 3H2→ 2NH3(Haber process)
H2N2O2 → HNO
2NH3 + CO2 → (NH2)2CO + H2O
2Mg + 3N2 → Mg3N2
N2H5 + H2O → N2+ H+ + H2O
• What are the different isotopes of Nitrogen?
Stable forms include nitrogen-14 and nitrogen-15
• List the oxidation states of various nitrogen oxides: +1, +2, +3, +4, +5 respectively
• List the different elements that Nitrogen will react with to make it basic or acidic :Nitride ion is a strong base when reacted with water, Ammonia is generally a weak acid
• Uses of nitrogen include anesthetic, Refrigerant, metal protector | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/22%3A_Chemistry_of_the_Nonmetals/22.E%3A_Chemistry_of_the_Nonmetals_%28Exercises%29.txt |
Descriptive chemistry of the elements is consistent with the various principles discussed earlier. We will focus on trends in and explanations for the observed behavior of the elements.
22.1: General Concepts: Periodic Trends and Reactions
Main group elements -- the valence electrons are filling s and p orbitals
• Three types of main-group elements: metal, metalloid, nonmetal
• Increasing metallic character going down a group and from right to left across a period
• Group Work: List the relative properties of the metals:
• volatility
• melting and boiling points
• density
• thermal conductance
• electrical conductance
• appearance as solids
• brittleness
• typical structure
• bonding
• tendency to lose or gain electrons
• acidity/basicity of oxides
Metals
Metals are nonvolatile, high melting and boiling points, high density and thermal and electrical conductance, characteristic shine or luster as crystalline solids, malleable and ductile, consist of large arrays of atoms with long-range bonding forces (the metallic bond) between free electrons and cations, tend to lose electrons, oxides are alkaline
• Group Work: List the relative properties of the non-metals:
• volatility
• melting and boiling points
• density
• thermal conductance
• electrical conductance
• appearance as solids
• brittleness
• typical structure
• bonding
• tendency to lose or gain electrons
• acidity/basicity of oxides
NonMetals
Nonmetals are volatile, low melting and boiling points, low density, brittle or soft as solids, low thermal and electrical conductance, solids dull in appearance, many are discrete small molecules with atoms joined by strong covalent bonds, intermolecular forces between molecules are weak, chemical properties characterized by their tendency to gain electrons, oxides are usually acidic
• Group Work: Which group of metals is the most reactive?
• Alkali metals - why?
• Which group of nonmetals is the most reactive?
• Halogens - why?
• Compare the metalloids to the metals and the non-metals.
Metalloids
• Metalloids: physical properties more like those of metals, but chemical reactivity is more like that of nonmetals; many atomic properties are intermediate between those of metals and of nonmetals Period 2 Elements are Unique
• Compounds formed between nonmetals tend to be molecular.
• As we move down the periodic table bonding changes.
• The third period onwards has accessible d-orbitals that can participate in bonding.
• Therefore, the octet rule can be broken for the third period onwards.
• The first member of a group can form π bonds more readily than subsequent members.
• Si is much larger than C and the 3p orbital is much larger than the 2p orbital, so the overlap between 3p orbitals to form a π3p bond is significantly poorer than for a π2p bond.
• Since the Si-Si π bond is much weaker than the C-C π bond, Si tends to form σ bonds.
• Example: CO2 is a gas with O=C=O bonds. SiO2 is a network solid with Si-O bonds.
Lightest Elements are Unique
• Properties of the first element in each group are usually more distinctive, while the rest of the elements in a group have similar properties
• The unusual properties of the first element in a group can be explained on the basis of its unusually small size, which arises because the valence electrons are not shielded from the nucleus and the electrons are held relatively tightly in the atom Metals with Covalent Bonds
• Small metal ions such as Be2+ are polarizing and pull electrons from nonmetal anions to form partly covalent bonds.
• Small nonmetals (such as N, O, F) are highly electronegative because of the increased attraction of the nucleus for the electrons at short distances and form multiple bonds because p orbitals can overlap. Chapter 22 22-3
Diagonal Relationships
In addition to horizontal and vertical trends, there is a diagonal relationship between elements such as Li and Mg, Be and Al, B and Si, that have an adjacent upper left/lower right relative location in the periodic table.
• These pairs of elements have similar size and electronegativity, resulting in similar properties. 22.2 Hydrogen Isotopes of Hydrogen
• There are three isotopes of hydrogen: Protium 1 1H, deuterium 2 1H, and tritium 3 1H.
• Deuterium (D) is about 0.0156 % of naturally occurring H.
• Tritium (T) is radioactive with a half-life of 12.3 yr.
22.2: Hydrogen
• Hydrogen is unique.
• Hydrogen has a 1s1 electron configuration so it is placed above Li in the periodic table.
• However, H is significantly less reactive than the alkali metals.
• Hydrogen can gain an electron to form H- , which has a He electron configuration. Therefore, H could be placed above the halogens.
• However, the electron affinity of H is lower than any halogen.
• Elemental hydrogen is a colorless, odorless gas at room temperature.
• Since H2 is nonpolar and only has two electrons, the intermolecular forces are weak (boiling point -253°C, melting point -259°C).
• The H-H bond enthalpy is high (436 kJ/mol). Therefore, reactions with hydrogen are slow and a catalyst needs to be used.
• When hydrogen reacts with air explosions result (Hindenburg exploded in 1937):
$2H2(g) + O2(g) → 2H2O(l) \nonumber$
with $∆H = -571.7 \,kJ/mol$
Preparation of Hydrogen
• In the laboratory hydrogen is usually prepared by reduction of an acid.
• Zn is added to an acidic solution and hydrogen bubbles are formed.
• The hydrogen bubbles out of solution and is collected in a flask.
• The collection flask is usually filled with water so the volume of hydrogen collected is the volume of water displaced.
In larger quantities, hydrogen can be prepared by the reduction of methane in the presence of steam at 1100°C:
$CH4(g) + H2O(g) → CO(g) + 3H2(g) \nonumber$
$CO(g) + H2O(g) → CO2(g) + H2(g) \nonumber$
Uses of Hydrogen
• Hydrogen is used for ammonia production and to hydrogenate vegetable oils to make margarine and shortening.
• Hydrogen is used to manufacture methanol: [CO(g) + 2H2(g) → CH3OH(g) \nonumber \]
• Chapter 22
Binary Hydrogen Compounds
• There types of binary hydrogen compounds are formed:
• ionic hydrides (e.g. LiH, made between metals and H);
• metallic hydrides (e.g. TiH2, made between transition metals and H); and
• molecular hydrides (e.g. CH4, made between nonmetals and metalloids and H).
• Thermal stability (measured by ∆G°f) decreases as we go down a group and increases across a period.
• Most stable is HF.
• Metal hydrides, such as CaH2, react with water to give H2 and metal hydroxide.
22.3: Group 18: Nobel Gases
• Noble gases are very unreactive.
• All elements have high ionization energies.
• He is the most important noble gas as liquid helium is used as a coolant.
• The heavier noble gases react more readily than the lighter ones.
• The most common compounds of noble gases are xenon fluorides.
• Xenon fluorides have Xe in the +2 to +8 oxidation states.
• Noble gas compounds violate the octet rule.
• In the presence of water, xenon fluorides form oxyfluorides:
• XeF6(s) + H2O(l) → XeOF4(l) + 2HF
• XeF6(s) + 3H2O(l) → XeO3(aq) + 6HF
• The only other noble gas compound known is KrF2, which decomposes at -10°C.
• Xenon fluorides are more stable than the oxides and oxyfluorides.
22.4: Group 17: The Halogens
• F, Cl, Br, I, At
• Most common are chlorine, bromine, and iodine
• Fluorine has properties atypical of the group
• Astatine is radioactive and exists naturally only in very small amounts
• The halogens exist as diatomic molecules
• At room temperature, fluorine is a yellow gas, chlorine is a pale green gas, bromine is a red liquid, and iodine is a purple solid
• The elements have very high ionization energies, typical of nonmetals
Properties of the Halogens
Outer electron configurations: ns2np5.
All halogens have large electron affinities.
Most common oxidation state is -1, but oxidation states of +1, +3, +5 and +7 are possible.
Halogens are good oxidizing agents.
Each halogen is the most electronegative element in its row.
The properties of the halogens vary regularly with their atomic number. Chapter 22 22-5 Fluorine
The bond enthalpy of F2 is low. Hence fluorine is very reactive.
Water is oxidized more readily than fluorine, so F2 cannot be prepared by electrolysis of a salt solution. F2 is an extremely reactive gas, which reacts with all the elements, except oxygen and the lighter noble gases, to form stable fluorides, often explosively.
F2 is such a strong oxidizing agent that it can convert oxides, including water, to molecular oxygen
Write a balanced equation for the reaction of F2 with H2O (to form O2 and ??).
Chlorine
• Chlorine exists as chlorides in sea water, salt lakes, and brine deposits
• Cl2 gas prepared industrially by the electrolysis of sodium chloride solutions
• Also a by-product of the preparation of metals by electrolysis of molten salts such as NaCl, MgCl2, and CaCl2
• Most Cl2 is used as a raw material in the production of other chemicals, in the synthesis of herbicides and insecticides, in the bleaching of textiles and paper, in purifying drinking water, and in the production of plastics such as polyvinyl chloride (PVC)
Bromine
• Bromine exists in small quantities in the form of bromides co-existing with chlorides
• Prepared by reacting a solution containing bromide ion with chlorine
• Uses of bromine:
• as a bleach
• in the manufacture of bromide compounds
• ethylene bromide, $C_2H_4Br_2$, used as an antiknock agent in gasoline
Iodine
• Iodine exists as iodides in brines and seaweed, and as iodates in deposits of sodium nitrate (NaNO3, or Chile saltpeter)
• Recovered by oxidation of I- with Cl2 or by reduction of IO3 - with HSO3 -
• Used as an antiseptic and disinfectant and as a reagent for chemical analysis
Astatine
• Not much is known about its chemistry
• It is highly radioactive; all chemical studies have used small quantities added to iodine solutions and measure behavior by determining where the radioactivity ends up.
• The total amount in the Earth's crust is estimated to be < 30 g at any one time.
Halogens
• In addition to the common -1 and 0 oxidation numbers, the halogens (except for fluorine) exist with each positive oxidation number through +7
• Halogen oxides are known with oxidation numbers as high as +7; most are very strong oxidizing agents
22-6 Oxyacids and Oxyanions
• Fluorine only forms one oxyacid: HOF. Oxygen is in the zero oxidation state.
• All are strong oxidizing agents.
• All are unstable and decompose readily.
• Oxyanions are more stable than oxyacids.
• Acid strength increases as the oxidation state of the halogen increases.
Periodic Acid
• Periodic (HIO4) and paraperiodic (H5IO6) acid have iodine in the +7 oxidation state.
• Periodic acid is a strong acid, paraperiodic acid is a weak acid (Ka1 = 2.8 × 10-2, Ka2 = 4.9 × 10-9).
• The large iodine atom allows 6 oxygen atoms around it.
• Smaller halogens cannot form this type of compound.
Halogens
Many aqueous halogen species are susceptible to disproportionation:
• ClO2
• HClO2
• HOCl
• Cl2 (in base)
• HOBr
• Br2 (in base)
• HOI
• IO-
• I2 (in base)
How do we decide which species will disproportionate?
22.5: Oxygen
Properties of Oxygen
• Oxygen has two allotropes: O2 and O3.
• O2 is a colorless, odorless gas at room temperature.
• The electron configuration is [He]2s2 2p4 , which means the dominant oxidation state is 2-.
• The O=O bond is strong (bond enthalpy 495 kJ/mol). Preparation of Oxygen
• Commercially: obtained by fractional distillation of air. (Normal boiling point of O2 is -183°C and N2 -196°C.)
• Laboratory preparation of oxygen is the catalytic decomposition of KClO3 in the presence of $MnO2: 2KClO3(s) → 2KCl(s) + 3O2(g). \nonumber$
• Atmospheric oxygen is replenished by photosynthesis (process in plants where CO2 is converted to O2 in the presence of sunlight).
Uses of Oxygen
• Most widely used as an oxidizing agent. (E.g. in the steel industry to remove impurities.)
• Oxygen is used in medicine.
• It is used with acetylene, C2H2 for oxyacetylene welding: $2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g) \nonumber$
Ozone
• Pale blue, poisonous gas.
• Ozone dissociates to form oxygen: O3(g) → O2(g) + O(g) ∆H° = 107 kJ
• Ozone is a stronger oxidizing agent than oxygen: O3(g) + 2H+ (aq) + 2e- → O2(g) + H2O(l) E° = 2.07 V O2(g) + 4H+ (aq) + 4e- → 2H2O(l) E° = 1.23 V
• Ozone can be made by passing an electric current through dry O2: 3O2(g) → 2O3(g)
Peroxides and Superoxides
• Peroxides: have an O-O bond and O in the -1 oxidation state.
• Hydrogen peroxide is unstable and decomposes into water and oxygen: 2H2O2(l) → 2H2O(l) + O2(g) ∆H = -196.0 kJ
• The oxygen produced will kill bacteria.
• Peroxides are important in biochemistry: it is produced when O2 is metabolized.
Peroxides
• Disproportionation occurs when an element is simultaneous oxidized and reduced: 2H+ (aq) + H2O2(aq) + 2e- → 2H2O(l) E° = 1.78 V O2(g) + 2H+ (aq) + 2e- → H2O2(aq) E° = 0.68 V
• Disproportionation: 2H2O2(aq) → 2H2O(l) + O2(g) E° = 1.10 V
Superoxides
• Superoxides: have an O-O bond and O in an oxidation state of -½ (superoxide ion is O2- ).
• Usually form with active metals (KO2, RbO2 and CsO2).
• Potassium superoxide reacts with water vapor from the breath to form oxygen gas. 4KO2 + 2H2O → 3O2 + 4KOH 22.6 The Other Group 6A Elements: S, Se, Te, and Po (Chalcogens)
• Trend that affects other properties is the increase in metallic character down the group, indicated by the decreases in ionization energy and electronegativity
22.6: The Other Group 16 Elements: S, Se, Te, and Po
• Nonmetallic character dominates in this group
• Nonmetallic O exists as diatomic molecules
• Nonmetallic S exists as various covalently bonded polyatomic forms
• Metalloids Se and Te are more metallic than S, but bear some resemblance to S
• Po is even more metallic, but its behavior is not well known since it is a rare, radioactive element
• Oxygen dominated by oxidation number -2
• The other elements have oxidation numbers from -2 through +6, the higher ones being quite common, especially in combination with oxygen
Sulfur
• S found in earth's crust as sulfide and sulfate minerals and as the free element
• Also a small but critical constituent of plant and animal tissue
• Occurs as sulfur dioxide and sulfur trioxide in the atmosphere
• Elemental sulfur is a tasteless, odorless, combustible yellow solid existing in a variety of allotropes with different molecular structures
• Rhombic and monoclinic forms consist of S8 rings
• Major use of sulfur is the preparation of sulfuric acid, which is used primarily to make phosphate fertilizers and impure phosphoric acid from phosphate rock
• Sulfur forms binary compounds with all the elements except iodine and the noble gases
• Hydrogen sulfide (H2S):
• usually prepared by reaction of a metal sulfide with an acid
• gas well known for its "rotten-egg" odor
• extremely poisonous
• largest source of sulfur in the atmosphere
• Sulfur reacts with oxygen to form two oxides, sulfur dioxide and sulfur trioxide, which form oxoanions (SO3 2- and SO4 2-) and oxoacids (H2SO3 and H2SO4) by reaction with metal oxides or water
• Sulfuric acid is a powerful dehydrating agent, strong acid and moderate oxidizer.
• Sulfuric acid removes H2O from the sugar leaving a black mass of C. Steam is produced because the reaction is very exothermic.
Exercise $1$
What are the structures of:
• SO2
• SO3
• SO3 2-
• SO4 2-
• H2SO3
• H2SO4
22.7: Nitrogen
• Electronic configuration of the Group VA (15) elements is ns2 np3
• Little resemblance between the chemistry of nitrogen and the other elements in this group Chapter
Nitrogen
• Nitrogen as an element is the colorless, odorless, diatomic molecule N2, the major constituent of air
• An essential component of all living matter in protein and amino acids
• Nitrogen compounds are important components of chemical fertilizers
• Most uses of nitrogen involve its compounds, such as ammonia and nitrogen oxides.
Exercise $2$
• List the formulas for all the oxides that you expect nitrogen to form.
Preparation of Nitrogen
• N2 is produced by fractional distillation of air.
• Nitrogen is fixed by forming NH3 (Haber Process).
• NH3 is converted into other useful chemicals (NO, NO2, nitrites and nitrates).
Nitrogen
• Positive oxidation numbers of nitrogen occur in the oxides, including N2O, NO, N2O3, NO2, N2O4, and N2O5.
• Aqueous N2O3 is converted to nitrous acid (HNO2), and N2O5 to nitric acid (HNO3).
• Nitrous oxide, N2O, occurs naturally in the atmosphere, as a result of the natural degradation of proteins.
• N2O can be formed by thermal decomposition of NH4NO3
• N2O is used as an anesthetic (in laughing gas)
• Nitric oxide, NO, is formed by reaction of Cu metal with dilute aqueous nitric acid or in high temperature combustion processes and in the oxidation of ammonia gas commercially
• NO reacts rapidly with O2 to form reddish-brown NO2.
• Dinitrogen trioxide results from reaction between NO and NO2.
• N2O3 is blue as a liquid.
• Gaseous N2O3 reacts with water to form the weak acid, nitrous acid, HNO2.
Exercise $3$
• Write an equation for the preparation of nitrogen dioxide from common laboratory chemicals.
Nitrogen
• Nitrogen dioxide, NO2, is a poisonous, reddish-brown gas with an irritating odor, which exists in equilibrium with colorless N2O4
• Dinitrogen pentoxide, N2O5, is a volatile low-melting white solid; dissolved in water, it forms HNO3, nitric acid
22.8: The Other Group 15 Elements: P, AS, Sb, and Bi
• Although nitrogen is found in nature primarily as unreactive N2, the other elements are found only in compounds
• Metallic nature increases down the group
Exercise $43$
Predict the following properties for phosphorus:
• metallic character
• type of bonding
• acid/base character of oxides
• formulas of oxides Phosphorus
• Phosphorus is essentially nonmetallic, forms covalent bonds and has acidic oxides Group Work
• Predict the acid/base character of the oxides of As, Sb, and Bi, relative to those of P. How can we explain the predicted trend? Group 5A
• Arsenic has properties between those of a nonmetal and a metalloid, with amphoteric (though more acidic than basic) oxides
• Antimony is mostly metallic, but with some properties of a metalloid, and amphoteric (more basic than acidic) oxides
• Bismuth, is metallic, with basic oxides
• These trends are consistent with dramatic decreases in ionization energy down the group
• Trends in oxidation number:
• N and P range from -3 to +5
• As primarily +3 and +5
• Sb mostly +3, occasionally +5
• Bi almost exclusively +3, rarely as +5
• The lower, more metallic elements in the group have fewer stable oxidation numbers than N
Phosphorus
• Phosphorus found as phosphates in 190 different minerals, most importantly apatite, Ca5(PO4)3OH
• Phosphates are an important constituent of all bone tissue
• Mainly occurs in phosphorus minerals (e.g. phosphate rock, Ca3(PO4)2).
• Elemental P4 produced by reduction $2 Ca3(PO4)2(s) + 6SiO2(s) + 10C(s) → P4(g) + 6CaSiO3(l) + 10CO(g) \nonumber$
• Phosphoric acid, made from phosphate rock, is one of the ingredients in Coca-Cola.
• Phosphorus occurs as 19 allotropes, the principal ones being white (tetrahedral P4), red, and black phosphorus
• White phosphorus is a soft solid, with molecules held together by weak intermolecular forces.
• White phosphorus is poisonous and causes painful skin burns.
• White phosphorus is quite reactive and ignites spontaneously in air
• Red phosphorus is an amorphous solid formed by heating white phosphorus
• Red phosphorus involves P4 tetrahedral molecules bonded to one another in long chains
• Black phosphorus is still less reactive
• Black phosphorus may be amorphous or have a graphite-like structure; it is metallic in appearance and an electrical conductor
Exercise $1$
What acids are formed when the oxides, P4O6 and P4O10, are reacted with excess water?
Phosphorus
• Two oxides of phosphorus, P4O6 and P4O10, formed by burning phosphorus
• React with water to form phosphorous acid, H3PO3 and phosphoric acid, H3PO4
• P4O10 is used as a drying agent because of its affinity for water. 22.9 Carbon
• Carbon constitutes about 0.027 % of the earthís crust.
• Carbon is the main constituent of living matter.
• Study of carbon compounds is called organic chemistry.
• Carbon forms more compounds than all other elements except hydrogen; typical compounds are the hydrocarbons and their derivatives
• Exists as diamond, graphite, and an amorphous form, as well as the recently discovered allotrope C60, called buckminsterfullerene
• Diamond: clear crystalline form of carbon, one of the hardest substances known, can be synthesized from graphite with high temperature and pressure and a metal catalyst
Exercise $1$
Why is diamond so hard, while graphite is soft and slippery, even though both are pure carbon?
22.9: Carbon
• Graphite:
• slippery gray-black solid
• strong covalent bonds hold atoms together in each layer, but the layers are bonded only by weak van der Waals forces, so the layers slide across one another readily
• found widely distributed in the earth's crust and synthesized from amorphous carbon
• Uses of graphite:
• crucibles
• lubricants
• pencils
• nuclear reactors (to slow down fast neutrons)
• electrodes for electrolysis reactions
• Amorphous carbon:
• carbon blacks
• charcoal
• activated carbon
• soot
• coke
• Essentially microcrystalline forms of graphite with no layering
• Formed by thermal decomposition or partial decomposition of coal, petroleum, natural gas, and wood with an insufficient supply of oxygen
• burning oil gives lampblack
• heating coal in the absence of air produces coke
• heating wood in the absence of air gives charcoal
• Carbon black used as a filler in rubber tires to increase toughness and prevent brittleness
• Lampblack used in inks, paints, coating on carbon paper
• Charcoal used:
• in filters to adsorb odors
• in gas masks to adsorb poisonous gases
• in the decolorizing of sugar
• in water treatment
• in the reclamation of dry-cleaning solvents
• Coke used in the extraction of metals from their oxide ores
• Buckminsterfullerene consists of molecules of C60 formed by laser or high- temperature carbon arc vaporization of graphite
• One member of a class of new forms of carbon called fullerenes, which consist of clusters of carbon atoms containing even numbers from 44 to 84
• C60 exists as a truncated icosahedron, which contains 12 pentagonal faces and 20 hexagonal faces
• Remarkable physical stability, but chemically reactive
• Now being prepared as tubes, into which metals can be inserted. These are the thinnest capillary tubes known.
• Elemental carbon is relatively unreactive at room temperature
• Insoluble in water, dilute acids and bases, and organic solvents
• At high temperatures, carbon becomes highly reactive and combines directly with many elements, including oxygen
• Carbon at high temperatures also reduces water, metal oxides, oxoanions (e.g., phosphate in phosphate rock), hydrogen
Oxides of Carbon
• Carbon forms CO and CO2.
• CO is very toxic (binds irreversibly to Fe in hemoglobin, causing respiratory arrest).
• CO also has a lone pair on C, which is unusual.
• CO is a good Lewis base Ni(CO)4 forms when Ni is warmed in CO
• CO can be used as a fuel 2CO(g) + O2(g) → 2CO2(g) ∆H = -566 kJ Chapter 22 22-13
• CO is a good reducing agent Fe3O4(s) + 4CO(g) → 3Fe(s) + 4CO2(g)
• CO2 is produced when organic compounds are burned in oxygen: C(s) + O2(g) → CO2(g) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
• CO2 is produced by treating carbonates with acid.
• Fermentation of sugar to produce alcohol also produces CO2: C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g)
• At atmospheric pressure, CO2 condenses to form CO2(s) or dry ice.
• CO2 is used as dry ice (refrigeration), carbonation of beverages, washing soda (Na2CO3.10H2O) and baking soda (NaHCO3.10H2O).
Carbonic Acid and Carbonates
• When CO2 dissolves in water (moderately soluble) carbonic acid forms: $CO2(aq) + H2O(l) ! H2CO3(aq) \nonumber$
• Carbonic acid is responsible for giving carbonated beverages a sharp acidic taste.
• Partial neutralization of H2CO3 gives hydrogen carbonates (bicarbonates) and full neutralization gives carbonates.
• Many minerals contain CO3 2-.
• At elevated temperatures CaCO3 decomposes: $CaCO3(s) → CaO(s) + 2CO2(g) \nonumber$
• This reaction is the commercial source of lime, CaO.
• CaO reacts with water and CO2 to form CaCO3 which binds the sand in mortar: CaO(s) + H2O(l) ! Ca2+(aq) + 2OH- (aq) Ca2+(aq) + 2OH- (aq) + CO2(aq) → CaCO3(s) + H2O(l)
Carbides
• Carbon combines with elements with a lower electronegativity to form carbides, which exist in three classes
• Salt-like carbides form with the most electropositive metals and are ionic, so they are hydrolyzed by water or dilute acid to give hydrocarbons
• Interstitial carbides are formed with transition metals, are very hard and have very high melting points, high metallic conductivity, and metallic luster; they consist of a metal with carbon atoms located in some of the interstitial sites (or holes) in the metal structure
• Covalent carbides include those of silicon and boron, which are close in size and electronegativity to carbon; they are completely covalent and form infinite network structures, are exceptionally hard materials widely used as abrasives 22.10 The Other Group 4A Elements: Si, Ge, Sn, and Pb General Characteristics of Group 4A Elements
• The electronegativities are low.
• The dominant oxidation state for Ge, Sn and Pb is +2.
• Carbon has a coordination number of 4, the other members have higher coordination numbers.
• C-C bonds are very strong, so C tends to form long chains.
• Because the Si-O bond is stronger than the Si-Si bond, Si tends to form oxides (silicates).
Exercise $1$
Compare the following properties on going down Group 4A:
• Metallic character
• Ionization energies
• Melting points
• Acid/base nature of oxides
22.10: The Other Group 14 Elements: Si, Ge, Sn, and Pb
• Si is nonmetallic/metalloid with only some of its chemistry similar to carbon
• Ge is a metalloid
• Sn and Pb are metallic
• Ionization energies and melting points decrease down the group, reflecting the change from nonmetal to metal
• Carbon bonds readily to itself
• This tendency diminishes on going down the group because the bond strength decreases considerably as the elements get larger
• Oxidation number +4 dominates near the top of the group, +2 becomes more stable down the group
Silicon
• Silicon exists in the earth's crust as silicon dioxide and over 800 silicate minerals
• Elemental silicon obtained by reduction of SiO2 with C or CaC2 at high temperature, purified by zone refining $SiO2(l) + 2C(s) → Si(l) + 2CO(g) \nonumber$
• Silicon:
• brittle gray-black metallic-looking solid
• quite hard
• high melting point
• diamond-like tetrahedral network structure
• inert at room temperature but reactive at high temperatures
• Used in semiconductor devices
• Silicon hydrides or silanes arise from reaction of Mg2Si with acids, giving a mixture of SiH4, Si2H6, Si3H8, Si4H10, Si5H12, and Si6H14
Exercise $1$
What is the structure and bonding for $Si_2H_6$?
Silicon
• Silanes are quite reactive:
• The first two are stable. Chapter 22 22-15
• The others decompose to give SiH4, Si2H6, and H2.
• Much more reactive than the corresponding alkanes because of the availability of empty 3d orbitals that can be used to form bonds with another reactant
• They are spontaneously flammable in air. Group Work
Exercise $1$
• SiO2, found as quartz, is quite hard. What feature of its structure gives rise to this hardness?
Silicon
• SiO2 exists as polymeric (SiO2)n, which has silicon covalently bonded to four bridging oxygen atoms.
• Extended covalent bonding network
• Hard, high-melting solid
• Glass is formed by heating together silicon dioxide, alkali metal and alkaline earth metal oxides, and sometimes other oxides, the particular mixture controlling the properties of the glass.
• Silicate structure is somewhat random in contrast to crystalline silicates
• Some bonds are under more strain than others, so the glass melts over a range of temperatures and can be softened without melting
• 90 % of the earthís crust is composed of compounds of Si and O.
• Silicates are compounds where Si has four O atoms surrounding it in a tetrahedral arrangement.
• The oxidation state of Si is +4.
• The silicate tetrahedra are building blocks for more complicated structures.
• Many silicates naturally with the basic structural unit being the SiO4 tetrahedron occurring in several varieties:
• singly
• small groups sharing oxygen atoms
• small cyclic groups
• infinite chains
• double-stranded chains (or bands)
• infinite sheets
• These few structures form many hundreds of minerals by combination of silicate anions with metal cations.
Silicon
• The silicate structure is reflected in the physical structure of minerals such as mica and asbestos.
22.11: Boron
• Black crystalline element, extremely hard and brittle, low density, high melting point and boiling point, low electrical conductivity (so classified as a semiconductor), used in semiconductor electronics and added to steel to increase strength and to copper to increase electrical conductivity
• Similarities of some properties to those of carbon, of others to those of silicon (due to Chapter 22 22-16 the diagonal relationship leading to similar size and electronegativity)
• Carbon, silicon, and boron all form covalently-bonded extended network solids and covalent halides
• Occurs in deposits of borax, Na2B4O7 . 10H2O
• Borax used as an additive to laundry detergents to soften the water.
• Which U.S. president was associated with Twenty Mule Team Borax?
• Resembles metals in its physical properties but is more like nonmetals chemically
• Chemical behavior is complex and unusual: ionization energy is unusually high, so formation of a cation is difficult, high electronegativity (comparable to nonmetals)
• All its compounds are covalent
Exercise $1$
• What is the valence electron configuration of boron? How many covalent bonds will it normally form?
Boron
• Valence electron configuration 2s2 2p1 , so forms only three normal covalent bonds, but electron deficiency makes it a good Lewis acid
• Oxidation number +3 common, but others found in boranes
• B3+ does not exist in aqueous solution. Why not?
• Reacts with F2 and Cl2 to give trihalides
• Great affinity for O2, which is used to remove oxygen from metal oxides to purify molten metals
• B reacts with N2 at high temperature to give solid BN.
• BN is very stable due to its graphite-like structure, which arises from the presence of only 3 valence electrons and the tendency to use sp2 hybrid orbitals.
• BN is also known in a diamond-like structure, which is formed by application of high temperature and pressure, and is extremely hard and is used as an abrasive.
• Large number of hydrides have been prepared
• BH3 is known but very unstable (reactive)
• Simplest stable one is diborane, B2H6, which decomposes to other boranes, e.g. B5H9, when heated
• Unusual structure and bonding in the boranes, in which hydrogens act as bridges between boron and the B-H-B arrangement shares two electrons between three atoms (called three center bonding)
• Because of their electron deficiencies, boranes are highly reactive
• Diborane is very reactive:
$B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(g) \nonumber$
with ∆H = -2030 kJ
• Some boranes are reactive (B5H9) while some are stable in air at room temperature (B10H14).
• B2O3 is the only important boron oxide.
• Boric acid, H3BO3 or B(OH)3 is a weak acid (Ka = 5.8 × 10-10).
• Boric acid is used as an eye wash. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/22%3A_Chemistry_of_the_Nonmetals/22.S%3A_Chemistry_of_the_Nonmetals_%28Summary%29.txt |
Transition metals are defined as those elements that have (or readily form) partially filled d orbitals. These include the d-block (groups 3–11) and f-block element elements. The variety of properties exhibited by transition metals is due to their complex valence shells. Unlike most main group metals where one oxidation state is normally observed, the valence shell structure of transition metals means that they usually occur in several different stable oxidation states. In addition, electron transitions in these elements can correspond with absorption of photons in the visible electromagnetic spectrum, leading to colored compounds. Because of these behaviors, transition metals exhibit a rich and fascinating chemistry.
• 23.1: Metal Complexes
A complex ion has a metal ion at its center with a number of other molecules or ions surrounding it. These can be considered to be attached to the central ion by coordinate (dative covalent) bonds (in some cases, the bonding is actually more complicated than that. The molecules or ions surrounding the central metal ion are called ligands. Simple ligands include water, ammonia and chloride ions.
• 23.2: Ligands with more than one Donor Atom
Ligands can be further characterized as monodentate, bidentate, tridentate etc. where the concept of teeth (dent) is introduced. Monodentate ligands bind through only one donor atom. Monodentate means "one-toothed." The halides, phosphines, ammonia and amines seen previously are monodentate ligands. Bidentate ligands bind through two donor sites. Bidentate means "two-toothed." It can bind to a metal via two donor atoms at once.
• 23.3: Nomenclature of Coordination Chemistry
Coordination complexes have their own classes of isomers, different magnetic properties and colors, and various applications (photography, cancer treatment, etc), so it makes sense that they would have a naming system as well.
• 23.4: Isomerization
Two compounds that have the same formula and the same connectivity do not always have the same shape. There are two reasons why this may happen. In one case, the molecule may be flexible, so that it can twist into different shapes via rotation around individual sigma bonds. This phenomenon is called conformation, and it is covered in a different chapter. The second case occurs when two molecules appear to be connected the same way on paper, but are connected in two different ways in three dimens
• 23.5: Color and Magnetism
Crystal field theory, which assumes that metal–ligand interactions are only electrostatic in nature, explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity.
• 23.6: Crystal Field Theory
Crystal field theory treats interactions between the electrons on the metal and the ligands as a simple electrostatic effect. The presence of the ligands near the metal ion changes the energies of the metal d orbitals relative to their energies in the free ion. Both the color and the magnetic properties of a complex can be attributed to this crystal field splitting. The magnitude of the splitting depends on the nature of the ligands bonded to the metal.
• 23.E: Chemistry of Coordination Chemistry (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al.
Thumbnail: Structure of the \(trans-[CoCl_2(NH_3)_4]^+\) complex ion. (Public domain; Benjah-bmm27).
23: Chemistry of Coordination Chemistry
Learning Objectives
• To introduce complex ions and the basic principles of metal-ligand bonding
A complex ion has a metal ion at its center with a number of other molecules or ions surrounding it. These can be considered to be attached to the central ion by coordinate (dative covalent) bonds (in some cases, the bonding is actually more complicated than that. The molecules or ions surrounding the central metal ion are called ligands. Simple ligands include water, ammonia and chloride ions.
What all these have got in common is active lone pairs of electrons in the outer energy level. These are used to form co-ordinate bonds with the metal ion. All ligands are lone pair donors. In other words, all ligands function as Lewis bases.
Bonding in Simple Complex Ions
We are going to look in detail at the bonding in the complex ion formed when water molecules attach themselves to an aluminum ion to give Al(H2O)63+. Start by thinking about the structure of a naked aluminum ion before the water molecules bond to it.
Example \(1\): \(Al(H_2O)_6^{3+}\)
Aluminum has the electronic structure
1s22s22p63s23px1
When it forms an Al3+ ion it loses the 3-level electrons to leave
1s22s22p6
That means that all the 3-level orbitals are now empty. The aluminium uses all six of these empty 3-level orbitals to accept lone pairs from six water molecules. It re-organizes (hybridizes) the 3s, the three 3p, and two of the 3d orbitals to produce six new orbitals all with the same energy.
You might wonder why it chooses to use six orbitals rather than four or eight or whatever. Six is the maximum number of water molecules that will fit around an aluminum ion (and most other metal ions) due to steric constraints. By making the maximum number of bonds, it releases most energy and so becomes most energetically stable.
Only one lone pair is shown on each water molecule. The other lone pair is pointing away from the aluminum and so isn't involved in the bonding. The resulting ion looks like this:
Because of the movement of electrons towards the center of the ion, the 3+ charge is no longer located entirely on the aluminum, but is now spread over the whole of the ion. Because the aluminum is forming 6 bonds, the co-ordination number of the aluminum is said to be 6. The co-ordination number of a complex ion counts the number of co-ordinate bonds being formed by the metal ion at its center.
In a simple case like this, that obviously also counts the number of ligands - but that is not necessarily so, as you will see later. Some ligands can form more than one co-ordinate bond with the metal ion.
Example \(2\): \(Fe(H_2O)_6^{3+}\)
Iron has the electronic structure
1s22s22p63s23p63d64s2
When it forms an Fe3+ ion it loses the 4s electrons and one of the 3d electrons to leave
1s22s22p63s23p63d5
Looking at this as electrons-in-boxes, at the bonding level:
The single electrons in the 3d level are NOT involved in the bonding in any way. Instead, the ion uses 6 orbitals from the 4s, 4p and 4d levels to accept lone pairs from the water molecules. Before they are used, the orbitals are re-organized (hybridized) to produce 6 orbitals of equal energy.
Once the co-ordinate bonds have been formed, the ion looks exactly the same as the equivalent aluminium ion.
Because the iron is forming 6 bonds, the co-ordination number of the iron is 6.
Example \(3\): \(CuCl_4^{2-}\)
This is a simple example of the formation of a complex ion with a negative charge.
Copper has the electronic structure
1s22s22p63s23p63d104s1
When it forms a Cu2+ ion it loses the 4s electron and one of the 3d electrons to leave
1s22s22p63s23p63d9
To bond the four chloride ions as ligands, the empty 4s and 4p orbitals are used (in a hybridized form) to accept a lone pair of electrons from each chloride ion. Because chloride ions are bigger than water molecules, you can't fit 6 of them around the central ion - that's why you only use 4.
Only one of the 4 lone pairs on each chloride ion is shown. The other three are pointing away from the copper ion, and aren't involved in the bonding. That gives you the complex ion:
The ion carries 2 negative charges overall. That comes from a combination of the 2 positive charges on the copper ion and the 4 negative charges from the 4 chloride ions. In this case, the co-ordination number of the copper is 4. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/23%3A_Chemistry_of_Coordination_Chemistry/23.01%3A_Metal_Complexes.txt |
Ligands can be further characterized as monodentate, bidentate, tridentate etc. where the concept of teeth (dent) is introduced. Monodentate ligands bind through only one donor atom. Monodentate means "one-toothed." The halides, phosphines, ammonia and amines seen previously are monodentate ligands. Bidentate ligands bind through two donor sites. Bidentate means "two-toothed." An example of a bidentate ligand is ethylenediamine. It can bind to a metal via two donor atoms at once.
Bidentate binding allows a ligand to bind more tightly. Tridentate ligands, which bind through three donors, can bind even more tightly, and so on. This phenomenon is generally called the "chelate effect." This term comes from the Greek chelos, meaning "crab." A crab does not have any teeth at all, but it does have two claws for tightly holding onto something for a couple of reasons. A very simple analogy is that, if you are holding something with two hands rather than one, you are not as likely to drop it.
Complex metal ions containing more complicated ligands
In the examples previously disccussed, each ligand only forms one bond with the central metal ion to give the complex ion. Such a ligand is said to be unidentate. That means literally that it only has one tooth! It only has one pair of electrons that it can use to bond to the metal - any other lone pairs are pointing in the wrong direction. Some ligands, however, have rather more teeth! These are known generally as multidentate or polydentate ligands, but can be broken down into a number of different types.
Bidentate ligands
Bidentate ligands have two lone pairs, both of which can bond to the central metal ion. The two commonly used examples are 1,2-diaminoethane (old name: ethylenediamine - often given the abbreviation "en"), and the ethanedioate ion (old name: oxalate).
In the ethanedioate ion, there are lots more lone pairs than the two shown, but these are the only ones we are interested in. You can think of these bidentate ligands rather as if they were a pair of headphones, carrying lone pairs on each of the "ear pieces". These will then fit snuggly around a metal ion.
Example $1$: $Ni (NH_2CH_2CH_2NH_2)_3^{2+}$
You might find this abbreviated to $[Ni(en)_3]^{2+}$. The structure of the ion looks like this:
In this case, the "ear pieces" are the nitrogen atoms of the NH2 groups - and the "bit that goes over your head " is the $-CH_2CH_2-$ group. If you were going to draw this in an exam, you would obviously want to draw it properly - but for learning purposes, drawing all the atoms makes the diagram look unduly complicated!
Notice that the arrangement of the bonds around the central metal ion is exactly the same as it was with the ions with 6 water molecules attached. The only difference is that this time each ligand uses up two of the positions - at right angles to each other.
Because the nickel is forming 6 co-ordinate bonds, the coordination number of this ion is 6, despite the fact that it is only joined to 3 ligands. Coordination number counts the number of bonds, not the number of ligands.
Example 5: $Cr(C_2O_4)_3^{3-}$
This is the complex ion formed by attaching 3 ethanedioate (oxalate) ions to a chromium(III) ion. The shape is exactly the same as the previous nickel complex. The only real difference is the number of charges. The original chromium ion carried 3+ charges, and each ethanedioate ion carried -2, i.e.,
$(+3) + (3 \times -2) = -3. \nonumber$
The structure of the ion looks like this:
Again, if you drew this in an exam, you would want to show all the atoms properly. If you need to be able to do this, practice drawing it so that it looks clear and tidy! Refer back to the diagram of the ethanedioate ion further up the page to help you.
A Quadridentate Ligand
A quadridentate ligand has four lone pairs, all of which can bond to the central metal ion. An example of this occurs in haemoglobin (American: hemoglobin). The functional part of this is an iron(II) ion surrounded by a complicated molecule called heme. This is a sort of hollow ring of carbon and hydrogen atoms, at the center of which are 4 nitrogen atoms with lone pairs on them. Heme is one of a group of similar compounds called porphyrins. They all have the same sort of ring system, but with different groups attached to the outside of the ring. You aren't going to need to know the exact structure of the haem at this level.
We could simplify the heme with the trapped iron ion as:
Each of the lone pairs on the nitrogen can form a co-ordinate bond with the iron(II) ion - holding it at the center of the complicated ring of atoms. The iron forms 4 co-ordinate bonds with the heme, but still has space to form two more - one above and one below the plane of the ring. The protein globin attaches to one of these positions using a lone pair on one of the nitrogens in one of its amino acids. The interesting bit is the other position.
The water molecule which is bonded to the bottom position in the diagram is easily replaced by an oxygen molecule (again via a lone pair on one of the oxygens in $O_2$) - and this is how oxygen gets carried around the blood by the haemoglobin. When the oxygen gets to where it is needed, it breaks away from the haemoglobin which returns to the lungs to get some more.
Carbon Monoxide Poisoning
You probably know that carbon monoxide is poisonous because it reacts with hemeoglobin. It bonds to the same site that would otherwise be used by the oxygen - but it forms a very stable complex. The carbon monoxide doesn't break away again, and that makes that hemeoglobin molecule useless for any further oxygen transfer.
A Hexadentate Ligand
A hexadentate ligand has 6 lone pairs of electrons - all of which can form co-ordinate bonds with the same metal ion. The best example is EDTA. The diagram shows the structure of the ion with the important atoms and lone pairs picked out.
The EDTA ion entirely wraps up a metal ion using all 6 of the positions that we have seen before. The co-ordination number is again 6 because of the 6 co-ordinate bonds being formed by the central metal ion. The diagram below shows this happening with a copper(II) ion. Here is a simplified version. Make sure that you can see how this relates to the full structure above.
The overall charge, of course, comes from the 2+ on the original copper(II) ion and the 4- on the $EDTA^{4-}$ ion. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/23%3A_Chemistry_of_Coordination_Chemistry/23.02%3A_Ligands_with_more_than_one_Donor_Atom.txt |
Learning Objectives
• To learn the basis for complex ion and compound nomenclature
Coordination complexes have their own classes of isomers, different magnetic properties and colors, and various applications (photography, cancer treatment, etc), so it makes sense that they would have a naming system as well. Consisting of a metal and ligands, their formulas follow the pattern [Metal ligands]±Charge, while names are written Prefix Ligands Metal (Oxidation State).
Introduction
According to the Lewis base theory, ligands are Lewis bases since they can donate electrons to the central metal atom. The metals, in turn, are Lewis acids since they accept electrons. Coordination complexes consist of a ligand and a metal center cation. The overall charge can be positive, negative, or neutral. Coordination compounds are complex or contain complex ions, for example:
• Complex Cation: \(\ce{[Co(NH3)6]^{3+}}\)
• Complex Anion: \(\ce{[CoCl4(NH3)2]^{-}}\)
• Neutral Complex: \(\ce{[CoCl3(NH3)3]}\)
• Coordination Compound: \(\ce{K4[Fe(CN)6]}\)
A ligand can be an anion or a neutral molecule that donates an electron pair to the complex (NH3, H2O, Cl-). The number of ligands that attach to a metal depends on whether the ligand is monodentate or polydentate. To begin naming coordination complexes, here are some things to keep in mind.
1. Ligands are named first in alphabetical order.
2. The name of the metal comes next.
3. The oxidation state of the metal follows, noted by a Roman numeral in parentheses (II, IV).
Rule 1: Anionic Ligands
Ligands that act as anions which end in "-ide" are replaced with an ending "-o" (e.g., Chloride → Chloro). Anions ending with "-ite" and "-ate" are replaced with endings "-ito" and "-ato" respectively (e.g., Nitrite → Nitrito, Nitrate → Nitrato).
Table \(1\): Anionic Monodentate Ligands
Molecular Formula Ligand Name Molecular Formula Ligand Name
F- Fluoro OH- Hydroxo
Cl- Chloro SO42- Sulfato
Br- Bromo S2O32- Thiosulfato
I- Iodo NO2- Nitrito-N-; Nitro
O2- Oxo ONO- Nitrito-O-; Nitrito
CN- Cyano SCN- Thiocyanato-S-; Thiocyanato
NC- Isocyano NCS- Thiocyanato-N-; Isothiocyanato
Rule 2: Neutral Ligands
Most neutral molecules that are ligands carry their normal name. The few exceptions are the first four on the chart: ammine, aqua, carbonyl, and nitrosyl.
Table \(2\): Select Neutral Monodentate Ligands. Note: Ammine is spelled with two m's when referring to a ligand. Amines are a class of organic nitrogen-containing compounds.
Molecular Formula of Ligand Ligand Name
NH3 Ammine
H2O Aqua
CO Carbonyl
NO Nitrosyl
CH3NH2 Methylamine
C5H5N Pyridine
Polydentate ligands follow the same rules for anions and neutral molecules.
Table \(3\): Select Polydentate ligands
Short name Extended name
en Ethylenediamine
ox2- Oxalato
EDTA4- Ethylenediaminetetraacetato
Rule 3: Ligand Multiplicity
The number of ligands present in the complex is indicated with the prefixes di, tri, etc. The exceptions are polydentates that have a prefix already in their name (en and EDTA4- are the most common). When indicating how many of these are present in a coordination complex, put the ligand's name in parentheses and use bis (for two ligands), tris (for three ligands), and tetrakis (for four ligands).
Table \(4\): Prefixes for indicating number of ligands in a complex.
Number of Ligands Monodentate Ligands Polydentate Ligands
1 mono -
2 di bis
3 tri tris
4 tetra tetrakis
5 penta pentakis
6 hexa hexakis
Prefixes always go before the ligand name; they are not taken into account when putting ligands in alphabetical order. Note that "mono" often is not used. For example, \(\ce{[FeCl(CO)2(NH3)3]^{2+}}\) would be called triamminedicarbonylchloroiron(III) ion. Remember that ligands are always named first, before the metal is.
Example \(1\)
What is the name of this complex ion: \(\ce{[CrCl2(H2O)4]^{+}}\)?
Solution
Let's start by identifying the ligands. The ligands here are Cl and H2O. Therefore, we will use the monodentate ligand names of "chloro" and "aqua". Alphabetically, aqua comes before chloro, so this will be their order in the complex's name. There are 4 aqua's and 2 chloro's, so we will add the number prefixes before the names. Since both are monodentate ligands, we will say "tetra[aqua]di[chloro]".
Now that the ligands are named, we will name the metal itself. The metal is Cr, which is chromium. Therefore, this coordination complex is called tetraaquadichlorochromium(III) ion. See the next section for an explanation of the (III).
Exercise \(1\)
What is the name of this complex ion: \(\ce{[CoCl_2(en)_2]^{+}}\)?
Answer
We take the same approach. There are two chloro and ethylenediamine ligands. The metal is Co, cobalt. We follow the same steps, except that \(en\) is a polydentate ligand with a prefix in its name (ethylenediamine), so "bis" is used instead of "di", and parentheses are added. Therefore, this coordination complex is called dichlorobis(ethylenediamine)cobalt(III) ion.
Rule 4: The Metals
When naming the metal center, you must know the formal metal name and the oxidation state. To show the oxidation state, we use Roman numerals inside parenthesis. For example, in the problems above, chromium and cobalt have the oxidation state of +3, so that is why they have (III) after them. Copper, with an oxidation state of +2, is denoted as copper(II). If the overall coordination complex is an anion, the ending "-ate" is attached to the metal center. Some metals also change to their Latin names in this situation. Copper +2 will change into cuprate(II). The following change to their Latin names when part of an anion complex:
Table \(5\): Latin terms for Select Metal Ion
Transition Metal Latin
Iron Ferrate
Copper Cuprate
Tin Stannate
Silver Argentate
Lead Plumbate
Gold Aurate
The rest of the metals simply have -ate added to the end (cobaltate, nickelate, zincate, osmate, cadmate, platinate, mercurate, etc. Note that the -ate tends to replace -um or -ium, if present).
Finally, when a complex has an overall charge, "ion" is written after it. This is not necessary if it is neutral or part of a coordination compound (Example \(3\)). Here are some examples with determining oxidation states, naming a metal in an anion complex, and naming coordination compounds.
Example \(2\)
What is the name of [Cr(OH)4]- ?
Solution
Immediately we know that this complex is an anion. There is only one monodentate ligand, hydroxide. There are four of them, so we will use the name "tetrahydroxo". The metal is chromium, but since the complex is an anion, we will have to use the "-ate" ending, yielding "chromate". The oxidation state of the metal is 3 (x+(-1)4=-1). Write this with Roman numerals and parentheses (III) and place it after the metal to get tetrahydroxochromate(III) ion.
Exercise \(2\)
What is the name of \(\ce{[CuCl4]^{2-}}\)?
Answer
tetrachlorocuprate(II) ion
A last little side note: when naming a coordination compound, it is important that you name the cation first, then the anion. You base this on the charge of the ligand. Think of NaCl. Na, the positive cation, comes first and Cl, the negative anion, follows.
Example \(3\)
What is the name of \([\ce{Pt(NH3)4}][\ce{Pt(Cl)4}]\)?
Solution
NH3 is neutral, making the first complex positively charged overall. Cl has a -1 charge, making the second complex the anion. Therefore, you will write the complex with NH3 first, followed by the one with Cl (the same order as the formula). This coordination compound is called tetraammineplatinum(II) tetrachloroplatinate(II).
Exercise \(3\): The Nitro/Nitrito Ambidentate Ligand
What is the name of \(\ce{[CoCl(NO2)(NH3)4]^{+}}\) ?
Answer
This coordination complex is called tetraamminechloronitrito-N-cobalt(III). N comes before the O in the symbol for the nitrite ligand, so it is called nitrito-N. If an O came first, as in [CoCl(ONO)(NH3)4]+, the ligand would be called nitrito-O, yielding the name tetraamminechloronitrito-O-cobalt(III).
Nitro (for NO2) and nitrito (for ONO) can also be used to describe the nitrite ligand, yielding the names tetraamminechloronitrocobalt(III) and tetraamminechloronitritocobalt(III).
Writing Formulas of Coordination Complexes
While chemistry typically follow the nomenclature rules for naming complexes and compounds, there is disagreement with the rules for constructing formulas of inorganic complex. The order of ligand names in their formula has been ambiguous with different conventions being used (charged vs neutral, number of each ligand, etc.). In 2005, IUPAC adopted the recommendation that all ligand names in formulas be listed alphabetically (in the same way as in the naming convention) irrespective of the charge or number of each ligand type.
However, this rule is not adhered to in many chemistry laboratories. For practice, the order of the ligands in chemical formulas does not matter as long as you write the transition metal first, which is the stance taken here.
Examples \(4\)
Write the chemical formulas for:
1. Amminetetraaquachromium(II) ion
2. Amminesulfatochromium(II)
Solution
1. Amminetetraaquachromium(II) ion could be written as \(\ce{[Cr(H2O)4(NH3)]^{+2}}\) or \(\ce{[Cr(NH3)(H2O)4]^{+2}}\).
2. Amminesulfatochromium (II) could be written as \(\ce{[Cr(SO4)(NH3)]}\) or \(\ce{[Cr(NH3)(SO4)]}\).
Exercise \(4\)
Write the chemical formulas for
1. Amminetetraaquachromium (II) sulfate
2. Potassium hexacyanoferrate (III)
Answer
1. Amminetetraaquachromium (II) sulfate can be written as \(\ce{[Cr(H2O)4(NH3)]SO4}\). Although \(\ce{[Cr(NH3)(H2O)4]SO4}\) is also acceptable.
2. Potassium hexacyanoferrate (III) is be written as \(\ce{K3[Fe(CN)6]}\)
Contributors and Attributions
• Justin Hosung Lee (UCD), Sophia Muller (UCD) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/23%3A_Chemistry_of_Coordination_Chemistry/23.03%3A_Nomenclature_of_Coordination_Chemistry.txt |
Learning Objectives
• To understand that there may be more than one way to arrange the same groups around the same atom with the same geometry (stereochemistry).
Two compounds that have the same formula and the same connectivity do not always have the same shape. There are two reasons why this may happen. In one case, the molecule may be flexible, so that it can twist into different shapes via rotation around individual sigma bonds. This phenomenon is called conformation, and it is covered in a different chapter. The second case occurs when two molecules appear to be connected the same way on paper, but are connected in two different ways in three dimensional space. These two, different molecules are called stereoisomers.
One simple example of stereoisomers from inorganic chemistry is diammine platinum dichloride, (NH3)2PtCl2. This important compound is sometimes called "platin" for short. As the formula implies, it contains a platinum ion that is coordinated to two ammonia ligands and two chloride ligands (remember, a ligand in inorganic chemistry is an electron donor that is attached to a metal atom, donating a pair of electrons to form a bond).
Platin is an example of a coordination compound. The way the different pieces of coordination compounds bond together is discussed in the chapter of Lewis acids and bases. For reasons arising from molecular orbital interactions, platin has a square planar geometry at the platinum atom. That arrangement results in two possible ways the ligands could be connected. The two sets of like ligands could be connected on the same side of the square or on opposite corners.
These two arrangements result in two different compounds; they are isomers that differ only in three-dimensional space.
• The one with the two amines beside each other is called cis-platin.
• These two ligands are 90 degrees from each other.
• The one with the amines across from each other is trans-platin.
• These two ligands are 180 degrees from each other.
CIS/TRANS isomers have different physical properties
Although these two compounds are very similar, they have slightly different physical properties. Both are yellow compounds that decompose when heated to 270 degrees C, but trans-platin forms pale yellow crystals and is more soluble than cis-platin in water.
CIS/TRANS isomers have different biological properties
Cis-platin has clinical importance in the treatment of ovarian and testicular cancers. The biological mechanism of the drug's action was long suspected to involve binding of the platinum by DNA. Further details were worked out by MIT chemist Steve Lippard and graduate student Amy Rosenzweig in the 1990's. Inside the cell nucleus, the two ammines in cis-platin can be replaced by nitrogen donors from a DNA strand. To donate to the Lewis acidic platinum, the DNA molecule must bend slightly. Normally that bend is detected and repaired by proteins in the cell. However, ovarian and testicular cells happen to contain a protein that is just the right shape to fit around this slightly bent DNA strand. The DNA strand becomes lodged in the protein and can't be displaced, and so it is unable to bind with other proteins used in DNA replication. The cell becomes unable to replicate, and so cancerous growth is stopped.
Exercise \(1\)
Draw the cis and trans isomers of the following compounds:
1. \(\ce{(NH3)2IrCl(CO)}\)
2. \(\ce{(H3P)2PtHBr}\)
3. \(\ce{(AsH3)2PtH(CO)}\)
Exercise \(2\)
Only one isomer of (tmeda)PtCl2 is possible [tmeda = (CH3)2NCH2CH2N(CH3)2; both nitrogens connect to the platinum]. Draw this isomer and explain why the other isomer is not possible.
Geometric Isomers
The existence of coordination compounds with the same formula but different arrangements of the ligands was crucial in the development of coordination chemistry. Two or more compounds with the same formula but different arrangements of the atoms are called isomers. Because isomers usually have different physical and chemical properties, it is important to know which isomer we are dealing with if more than one isomer is possible. Recall that in many cases more than one structure is possible for organic compounds with the same molecular formula; examples discussed previously include n-butane versus isobutane and cis-2-butene versus trans-2-butene. As we will see, coordination compounds exhibit the same types of isomers as organic compounds, as well as several kinds of isomers that are unique.
Planar Isomers
Metal complexes that differ only in which ligands are adjacent to one another (cis) or directly across from one another (trans) in the coordination sphere of the metal are called geometrical isomers. They are most important for square planar and octahedral complexes.
Because all vertices of a square are equivalent, it does not matter which vertex is occupied by the ligand B in a square planar MA3B complex; hence only a single geometrical isomer is possible in this case (and in the analogous MAB3 case). All four structures shown here are chemically identical because they can be superimposed simply by rotating the complex in space:
For an MA2B2 complex, there are two possible isomers: either the A ligands can be adjacent to one another (cis), in which case the B ligands must also be cis, or the A ligands can be across from one another (trans), in which case the B ligands must also be trans. Even though it is possible to draw the cis isomer in four different ways and the trans isomer in two different ways, all members of each set are chemically equivalent:
The anticancer drug cisplatin and its inactive trans isomer. Cisplatin is especially effective against tumors of the reproductive organs, which primarily affect individuals in their 20s and were notoriously difficult to cure. For example, after being diagnosed with metastasized testicular cancer in 1991 and given only a 50% chance of survival, Lance Armstrong was cured by treatment with cisplatin.
Square planar complexes that contain symmetrical bidentate ligands, such as [Pt(en)2]2+, have only one possible structure, in which curved lines linking the two N atoms indicate the ethylenediamine ligands:
Octahedral Isomers
Octahedral complexes also exhibit cis and trans isomers. Like square planar complexes, only one structure is possible for octahedral complexes in which only one ligand is different from the other five (MA5B). Even though we usually draw an octahedron in a way that suggests that the four “in-plane” ligands are different from the two “axial” ligands, in fact all six vertices of an octahedron are equivalent. Consequently, no matter how we draw an MA5B structure, it can be superimposed on any other representation simply by rotating the molecule in space. Two of the many possible orientations of an MA5B structure are as follows:
If two ligands in an octahedral complex are different from the other four, giving an MA4B2 complex, two isomers are possible. The two B ligands can be cis or trans. Cis- and trans-[Co(NH3)4Cl2]Cl are examples of this type of system:
Replacing another A ligand by B gives an MA3B3 complex for which there are also two possible isomers. In one, the three ligands of each kind occupy opposite triangular faces of the octahedron; this is called the fac isomer (for facial). In the other, the three ligands of each kind lie on what would be the meridian if the complex were viewed as a sphere; this is called the mer isomer (for meridional):
Example \(1\)
Draw all the possible geometrical isomers for the complex [Co(H2O)2(ox)BrCl], where ox is O2CCO2, which stands for oxalate.
Given: formula of complex
Asked for: structures of geometrical isomers
Solution
This complex contains one bidentate ligand (oxalate), which can occupy only adjacent (cis) positions, and four monodentate ligands, two of which are identical (H2O). The easiest way to attack the problem is to go through the various combinations of ligands systematically to determine which ligands can be trans. Thus either the water ligands can be trans to one another or the two halide ligands can be trans to one another, giving the two geometrical isomers shown here:
In addition, two structures are possible in which one of the halides is trans to a water ligand. In the first, the chloride ligand is in the same plane as the oxalate ligand and trans to one of the oxalate oxygens. Exchanging the chloride and bromide ligands gives the other, in which the bromide ligand is in the same plane as the oxalate ligand and trans to one of the oxalate oxygens:
This complex can therefore exist as four different geometrical isomers.
Exercise \(1\)
Draw all the possible geometrical isomers for the complex [Cr(en)2(CN)2]+.
Answer
Two geometrical isomers are possible: trans and cis.
Summary
Many metal complexes form isomers, which are two or more compounds with the same formula but different arrangements of atoms. Structural isomers differ in which atoms are bonded to one another, while geometrical isomers differ only in the arrangement of ligands around the metal ion. Ligands adjacent to one another are cis, while ligands across from one another are trans. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/23%3A_Chemistry_of_Coordination_Chemistry/23.04%3A_Isomerization.txt |
Learning Objectives
• To get a simple overview of the origin of color and magnetism in complex ions.
Electromagnetic radiation is a form of energy that is produced by oscillating electric and magnetic disturbance, or by the movement of electrically charged particles traveling through a vacuum or matter. Electron radiation is released as photons, which are bundles of light energy that travel at the speed of light as quantized harmonic waves. This energy is then grouped into categories based on its wavelength into the electromagnetic spectrum and have certain characteristics, including amplitude, wavelength, and frequency (Figure $1$).
General properties of all electromagnetic radiation include:
1. Electromagnetic radiation can travel through empty space, while most other types of waves must travel through some sort of substance. For example, sound waves need either a gas, solid, or liquid to pass through to be heard.
2. The speed of light ($c$) is always a constant (2.99792458 x 108 m s-1).
3. Wavelengths ($\lambda$) are measured between the distances of either crests or troughs.
The energy of a photon is expressed by Planck's law in terms of the frequency ($u$) of the photon
$E=h u \label{24.5.1}$
since $\lambda u =c$ for all light Plancks law can be also expressed in terms of the wavelength of the photon
$E = h u = \dfrac{hc}{\lambda} \label{24.5.2}$
If white light is passed through a prism, it splits into all the colors of the rainbow (Figure $2$). Visible light is simply a small part of an electromagnetic spectrum most of which we cannot see - gamma rays, X-rays, infra-red, radio waves and so on. Each of these has a particular wavelength, ranging from 10-16 meters for gamma rays to several hundred meters for radio waves. Visible light has wavelengths from about 400 to 750 nm (1 nanometer = 10-9 meters).
Example $1$: Blue Color of Copper (II) Sulfate in Solution
If white light (ordinary sunlight, for example) passes through copper(II) sulfate solution, some wavelengths in the light are absorbed by the solution. Copper(II) ions in solution absorb light in the red region of the spectrum. The light which passes through the solution and out the other side will have all the colors in it except for the red. We see this mixture of wavelengths as pale blue (cyan). The diagram gives an impression of what happens if you pass white light through copper(II) sulfate solution.
Working out what color you will see is not easy if you try to do it by imagining "mixing up" the remaining colors. You would not have thought that all the other colors apart from some red would look cyan, for example. Sometimes what you actually see is quite unexpected. Mixing different wavelengths of light doesn't give you the same result as mixing paints or other pigments. You can, however, sometimes get some estimate of the color you would see using the idea of complementary colors.
Origin of Colors
The process of absorption involves the excitation of the valence electrons in the molecule typically from the low lying level called the Highest Occupied Molecular Orbital (HOMO) into a higher lying state called the the Lowest Unoccupied Molecular Orbital (LUMO). When this HOMO and LUMO transition (Figure $3$) involves the absorption of visible light, the sample is colored.
The HOMO-LUMO energy difference
$\Delta E = E_{HOMO} - E_{LUMO} \label{24.5.3A}$
depends on the nature of the molecule and can be connected to the wavelength of the light absorbed
$\Delta E = h u = \dfrac{hc}{\lambda} \label{24.5.3B}$
Equation $\ref{24.5.3B}$ is the most important equation in the field of light-matter interactions (spectroscopy).
As Example $1$ demonstrated, when white light passes through or is reflected by a colored substance, a characteristic portion of the mixed wavelengths is absorbed. The remaining light will then assume the complementary color to the wavelength(s) absorbed. This relationship is demonstrated by the color wheel shown below. Here, complementary colors are diametrically opposite each other (Figure $5$). Thus, absorption of 420-430 nm light renders a substance yellow, and absorption of 500-520 nm light makes it red. Green is unique in that it can be created by absorption close to 400 nm as well as absorption near 800 nm.
Colors directly opposite each other on the color wheel are said to be complementary colors. Blue and yellow are complementary colors; red and cyan are complementary; and so are green and magenta. Mixing together two complementary colors of light will give you white light. What this all means is that if a particular color is absorbed from white light, what your eye detects by mixing up all the other wavelengths of light is its complementary color. Copper(II) sulfate solution is pale blue (cyan) because it absorbs light in the red region of the spectrum and cyan is the complementary color of red (Table $1$).
Table $1$: The Visible Spectrum
Color Wavelength (nm) ΔE HOMO - LUMO gap (eV)
UV 100 - 400 12.4 - 3.10
Violet 400 - 425 3.10 - 2.92
Blue 425 - 492 2.92 - 2.52
Green 492 - 575 2.52 - 2.15
Yellow 575 - 585 2.15 - 2.12
Orange 585 - 647 2.12 - 1.92
Red 647 - 700 1.92 - 1.77
Near IR 700 - 10,000 1.77 - 0.12
If the compound absorbs in one region of the spectra, it appears with the opposite (complementary) color, since all of the absorbed color has been removed. For example:
• the material absorbs violet light ⇒ color is yellow
• the material absorbs blue light ⇒ color is orange
• the material absorbs yellow-green light ⇒ color is red-violet
The Origin of Color in Complex Ions
We often casually talk about the transition metals as being those in the middle of the Periodic Table where d orbitals are being filled, but these should really be called d block elements rather than transition elements (or metals). The definition of a transition metal is one which forms one or more stable ions which have incompletely filled d orbitals. Zinc with the electronic structure [Ar] 3d104s2 does not count as a transition metal whichever definition you use. In the metal, it has a full 3d level. When it forms an ion, the 4s electrons are lost - again leaving a completely full 3d level. At the other end of the row, scandium ([Ar] 3d14s2) does not really counts as a transition metal either. Although there is a partially filled d level in the metal, when it forms its ion, it loses all three outer electrons. The Sc3+ ion does not count as a transition metal ion because its 3d level is empty.
Example $3$: Hexaaqua Metal Ions
The diagrams show the approximate colors of some typical hexaaqua metal ions, with the formula [ M(H2O)6 ]n+. The charge on these ions is typically 2+ or 3+.
• Non-transition metal ions
• Transition metal ions
The corresponding transition metal ions are colored. Some, like the hexaaquamanganese(II) ion (not shown) and the hexaaquairon(II) ion, are quite faintly colored - but they are colored.
So, what causes transition metal ions to absorb wavelengths from visible light (causing color) whereas non-transition metal ions do not? And why does the color vary so much from ion to ion? This is discussed in the next sections.
Magnetism
The magnetic moment of a system measures the strength and the direction of its magnetism. The term itself usually refers to the magnetic dipole moment. Anything that is magnetic, like a bar magnet or a loop of electric current, has a magnetic moment. A magnetic moment is a vector quantity, with a magnitude and a direction. An electron has an electron magnetic dipole moment, generated by the electron's intrinsic spin property, making it an electric charge in motion. There are many different magnetic forms: including paramagnetism, and diamagnetism, ferromagnetism, and anti-ferromagnetism. Only the first two are introduced below.
Paramagnetism
Paramagnetism refers to the magnetic state of an atom with one or more unpaired electrons. The unpaired electrons are attracted by a magnetic field due to the electrons' magnetic dipole moments. Hund's Rule states that electrons must occupy every orbital singly before any orbital is doubly occupied. This may leave the atom with many unpaired electrons. Because unpaired electrons can spin in either direction, they display magnetic moments in any direction. This capability allows paramagnetic atoms to be attracted to magnetic fields. Diatomic oxygen, $O_2$ is a good example of paramagnetism (described via molecular orbital theory). The following video shows liquid oxygen attracted into a magnetic field created by a strong magnet:
A chemical demonstration of the paramagnetism of oxygen, as shown by the attraction of liquid oxygen to a magnet. Carleton University, Ottawa, Canada.
As shown in the video, molecular oxygen ($O_2$ is paramagnetic and is attracted to the magnet. Incontrast, Molecular nitrogen, $N_2$, however, has no unpaired electrons and it is diamagnetic (this concept is discussed below); it is therefore unaffected by the magnet.
There are some exceptions to the paramagnetism rule; these concern some transition metals, in which the unpaired electron is not in a d-orbital. Examples of these metals include $Sc^{3+}$, $Ti^{4+}$, $Zn^{2+}$, and $Cu^+$. These metals are the not defined as paramagnetic: they are considered diamagnetic because all d-electrons are paired. Paramagnetic compounds sometimes display bulk magnetic properties due to the clustering of the metal atoms. This phenomenon is known as ferromagnetism, but this property is not discussed here.
Diamagnetism
Diamagnetic substances are characterized by paired electrons—except in the previously-discussed case of transition metals, there are no unpaired electrons. According to the Pauli Exclusion Principle which states that no two identical electrons may take up the same quantum state at the same time, the electron spins are oriented in opposite directions. This causes the magnetic fields of the electrons to cancel out; thus there is no net magnetic moment, and the atom cannot be attracted into a magnetic field. In fact, diamagnetic substances are weakly repelled by a magnetic field. In fact, diamagnetic substances are weakly repelled by a magnetic field as demonstrated with the pyrolytic carbon sheet in Figure $6$.
Figure $6$: Levitating pyrolytic carbon: A small (~6 mm) piece of pyrolytic graphite levitating over a permanent neodymium magnet array (5 mm cubes on a piece of steel). Note that the poles of the magnets are aligned vertically and alternate (two with north facing up, and two with south facing up, diagonally). from Wikipedia.
How to Tell if a Substance is Paramagnetic or Diamagnetic
The magnetic form of a substance can be determined by examining its electron configuration: if it shows unpaired electrons, then the substance is paramagnetic; if all electrons are paired, the substance is diamagnetic. This process can be broken into four steps:
1. Find the electron configuration
2. Draw the valence orbitals
3. Look for unpaired electrons
4. Determine whether the substance is paramagnetic (one or more unpaired electrons) or diamagnetic (all electrons paired)
Example $4$: Chlorine atoms
Are chlorine atoms paramagnetic or diamagnetic?
Step 1: Find the electron configuration
For Cl atoms, the electron configuration is 3s23p5
Step 2: Draw the valence orbitals
Ignore the core electrons and focus on the valence electrons only.
Step 3: Look for unpaired electrons
There is one unpaired electron.
Step 4: Determine whether the substance is paramagnetic or diamagnetic
Since there is an unpaired electron, Cl atoms are paramagnetic (but is quite weak).
Example 2: Zinc Atoms
Step 1: Find the electron configuration
For Zn atoms, the electron configuration is 4s23d10
Step 2: Draw the valence orbitals
Step 3: Look for unpaired electrons
There are no unpaired electrons.
Step 4: Determine whether the substance is paramagnetic or diamagnetic
Because there are no unpaired electrons, Zn atoms are diamagnetic. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/23%3A_Chemistry_of_Coordination_Chemistry/23.05%3A_Color_and_Magnetism.txt |
Learning Objectives
• To understand how crystal field theory explains the electronic structures and colors of metal complexes.
One of the most striking characteristics of transition-metal complexes is the wide range of colors they exhibit. In this section, we describe crystal field theory (CFT), a bonding model that explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. The central assumption of CFT is that metal–ligand interactions are purely electrostatic in nature. Even though this assumption is clearly not valid for many complexes, such as those that contain neutral ligands like CO, CFT enables chemists to explain many of the properties of transition-metal complexes with a reasonable degree of accuracy. The Learning Objective of this Module is to understand how crystal field theory explains the electronic structures and colors of metal complexes.
d-Orbital Splittings
CFT focuses on the interaction of the five (n − 1)d orbitals with ligands arranged in a regular array around a transition-metal ion. We will focus on the application of CFT to octahedral complexes, which are by far the most common and the easiest to visualize. Other common structures, such as square planar complexes, can be treated as a distortion of the octahedral model. According to CFT, an octahedral metal complex forms because of the electrostatic interaction of a positively charged metal ion with six negatively charged ligands or with the negative ends of dipoles associated with the six ligands. In addition, the ligands interact with one other electrostatically. As you learned in our discussion of the valence-shell electron-pair repulsion (VSEPR) model, the lowest-energy arrangement of six identical negative charges is an octahedron, which minimizes repulsive interactions between the ligands.
We begin by considering how the energies of the d orbitals of a transition-metal ion are affected by an octahedral arrangement of six negative charges. Recall that the five d orbitals are initially degenerate (have the same energy). If we distribute six negative charges uniformly over the surface of a sphere, the d orbitals remain degenerate, but their energy will be higher due to repulsive electrostatic interactions between the spherical shell of negative charge and electrons in the d orbitals (Figure $\PageIndex{1a}$). Placing the six negative charges at the vertices of an octahedron does not change the average energy of the d orbitals, but it does remove their degeneracy: the five d orbitals split into two groups whose energies depend on their orientations. As shown in Figure $\PageIndex{1b}$, the dz2 and dx2−y2 orbitals point directly at the six negative charges located on the x, y, and z axes. Consequently, the energy of an electron in these two orbitals (collectively labeled the eg orbitals) will be greater than it will be for a spherical distribution of negative charge because of increased electrostatic repulsions. In contrast, the other three d orbitals (dxy, dxz, and dyz, collectively called the t2g orbitals) are all oriented at a 45° angle to the coordinate axes, so they point between the six negative charges. The energy of an electron in any of these three orbitals is lower than the energy for a spherical distribution of negative charge.
The difference in energy between the two sets of d orbitals is called the crystal field splitting energy (Δo), where the subscript o stands for octahedral. As we shall see, the magnitude of the splitting depends on the charge on the metal ion, the position of the metal in the periodic table, and the nature of the ligands. (Crystal field splitting energy also applies to tetrahedral complexes: Δt.) It is important to note that the splitting of the d orbitals in a crystal field does not change the total energy of the five d orbitals: the two eg orbitals increase in energy by 0.6Δo, whereas the three t2g orbitals decrease in energy by 0.4Δo. Thus the total change in energy is
$2(0.6Δ_o) + 3(−0.4Δ_o) = 0. \nonumber$
Crystal field splitting does not change the total energy of the d orbitals.
Thus far, we have considered only the effect of repulsive electrostatic interactions between electrons in the d orbitals and the six negatively charged ligands, which increases the total energy of the system and splits the d orbitals. Interactions between the positively charged metal ion and the ligands results in a net stabilization of the system, which decreases the energy of all five d orbitals without affecting their splitting (as shown at the far right in Figure $\PageIndex{1a}$).
Electronic Structures of Metal Complexes
We can use the d-orbital energy-level diagram in Figure $1$ to predict electronic structures and some of the properties of transition-metal complexes. We start with the Ti3+ ion, which contains a single d electron, and proceed across the first row of the transition metals by adding a single electron at a time. We place additional electrons in the lowest-energy orbital available, while keeping their spins parallel as required by Hund’s rule. As shown in Figure 24.6.2, for d1–d3 systems—such as [Ti(H2O)6]3+, [V(H2O)6]3+, and [Cr(H2O)6]3+, respectively—the electrons successively occupy the three degenerate t2g orbitals with their spins parallel, giving one, two, and three unpaired electrons, respectively. We can summarize this for the complex [Cr(H2O)6]3+, for example, by saying that the chromium ion has a d3 electron configuration or, more succinctly, Cr3+ is a d3 ion.
When we reach the d4 configuration, there are two possible choices for the fourth electron: it can occupy either one of the empty eg orbitals or one of the singly occupied t2g orbitals. Recall that placing an electron in an already occupied orbital results in electrostatic repulsions that increase the energy of the system; this increase in energy is called the spin-pairing energy (P). If Δo is less than P, then the lowest-energy arrangement has the fourth electron in one of the empty eg orbitals. Because this arrangement results in four unpaired electrons, it is called a high-spin configuration, and a complex with this electron configuration, such as the [Cr(H2O)6]2+ ion, is called a high-spin complex. Conversely, if Δo is greater than P, then the lowest-energy arrangement has the fourth electron in one of the occupied t2g orbitals. Because this arrangement results in only two unpaired electrons, it is called a low-spin configuration, and a complex with this electron configuration, such as the [Mn(CN)6]3− ion, is called a low-spin complex. Similarly, metal ions with the d5, d6, or d7 electron configurations can be either high spin or low spin, depending on the magnitude of Δo.
In contrast, only one arrangement of d electrons is possible for metal ions with d8–d10 electron configurations. For example, the [Ni(H2O)6]2+ ion is d8 with two unpaired electrons, the [Cu(H2O)6]2+ ion is d9 with one unpaired electron, and the [Zn(H2O)6]2+ ion is d10 with no unpaired electrons.
If Δo is less than the spin-pairing energy, a high-spin configuration results. Conversely, if Δo is greater, a low-spin configuration forms.
Factors That Affect the Magnitude of Δo
The magnitude of Δo dictates whether a complex with four, five, six, or seven d electrons is high spin or low spin, which affects its magnetic properties, structure, and reactivity. Large values of Δo (i.e., Δo > P) yield a low-spin complex, whereas small values of Δo (i.e., Δo < P) produce a high-spin complex. As we noted, the magnitude of Δo depends on three factors: the charge on the metal ion, the principal quantum number of the metal (and thus its location in the periodic table), and the nature of the ligand. Values of Δo for some representative transition-metal complexes are given in Table $1$.
Table $1$: Crystal Field Splitting Energies for Some Octahedral (Δo)* and Tetrahedral (Δt) Transition-Metal Complexes
Octahedral Complexes Δo (cm−1) Octahedral Complexes Δo (cm−1) Tetrahedral Complexes Δt (cm−1)
*Energies obtained by spectroscopic measurements are often given in units of wave numbers (cm−1); the wave number is the reciprocal of the wavelength of the corresponding electromagnetic radiation expressed in centimeters: 1 cm−1 = 11.96 J/mol.
[Ti(H2O)6]3+ 20,300 [Fe(CN)6]4− 32,800 VCl4 9010
[V(H2O)6]2+ 12,600 [Fe(CN)6]3− 35,000 [CoCl4]2− 3300
[V(H2O)6]3+ 18,900 [CoF6]3− 13,000 [CoBr4]2− 2900
[CrCl6]3− 13,000 [Co(H2O)6]2+ 9300 [CoI4]2− 2700
[Cr(H2O)6]2+ 13,900 [Co(H2O)6]3+ 27,000
[Cr(H2O)6]3+ 17,400 [Co(NH3)6]3+ 22,900
[Cr(NH3)6]3+ 21,500 [Co(CN)6]3− 34,800
[Cr(CN)6]3− 26,600 [Ni(H2O)6]2+ 8500
Cr(CO)6 34,150 [Ni(NH3)6]2+ 10,800
[MnCl6]4− 7500 [RhCl6]3− 20,400
[Mn(H2O)6]2+ 8500 [Rh(H2O)6]3+ 27,000
[MnCl6]3− 20,000 [Rh(NH3)6]3+ 34,000
[Mn(H2O)6]3+ 21,000 [Rh(CN)6]3− 45,500
[Fe(H2O)6]2+ 10,400 [IrCl6]3− 25,000
[Fe(H2O)6]3+ 14,300 [Ir(NH3)6]3+ 41,000
Source of data: Duward F. Shriver, Peter W. Atkins, and Cooper H. Langford, Inorganic Chemistry, 2nd ed. (New York: W. H. Freeman and Company, 1994).
Charge on the Metal Ion
Increasing the charge on a metal ion has two effects: the radius of the metal ion decreases, and negatively charged ligands are more strongly attracted to it. Both factors decrease the metal–ligand distance, which in turn causes the negatively charged ligands to interact more strongly with the d orbitals. Consequently, the magnitude of Δo increases as the charge on the metal ion increases. Typically, Δo for a tripositive ion is about 50% greater than for the dipositive ion of the same metal; for example, for [V(H2O)6]2+, Δo = 11,800 cm−1; for [V(H2O)6]3+, Δo = 17,850 cm−1.
Principal Quantum Number of the Metal
For a series of complexes of metals from the same group in the periodic table with the same charge and the same ligands, the magnitude of Δo increases with increasing principal quantum number: Δo (3d) < Δo (4d) < Δo (5d). The data for hexaammine complexes of the trivalent group 9 metals illustrate this point:
[Co(NH3)6]3+: Δo = 22,900 cm−1
[Rh(NH3)6]3+: Δo = 34,100 cm−1
[Ir(NH3)6]3+: Δo = 40,000 cm−1
The increase in Δo with increasing principal quantum number is due to the larger radius of valence orbitals down a column. In addition, repulsive ligand–ligand interactions are most important for smaller metal ions. Relatively speaking, this results in shorter M–L distances and stronger d orbital–ligand interactions.
The Nature of the Ligands
Experimentally, it is found that the Δo observed for a series of complexes of the same metal ion depends strongly on the nature of the ligands. For a series of chemically similar ligands, the magnitude of Δo decreases as the size of the donor atom increases. For example, Δo values for halide complexes generally decrease in the order F > Cl > Br > I− because smaller, more localized charges, such as we see for F, interact more strongly with the d orbitals of the metal ion. In addition, a small neutral ligand with a highly localized lone pair, such as NH3, results in significantly larger Δo values than might be expected. Because the lone pair points directly at the metal ion, the electron density along the M–L axis is greater than for a spherical anion such as F. The experimentally observed order of the crystal field splitting energies produced by different ligands is called the spectrochemical series, shown here in order of decreasing Δo:
$\mathrm{\underset{\textrm{strong-field ligands}}{CO\approx CN^->}NO_2^->en>NH_3>\underset{\textrm{intermediate-field ligands}}{SCN^->H_2O>oxalate^{2-}}>OH^->F>acetate^->\underset{\textrm{weak-field ligands}}{Cl^->Br^->I^-}}$
The values of Δo listed in Table $1$ illustrate the effects of the charge on the metal ion, the principal quantum number of the metal, and the nature of the ligand.
The largest Δo splittings are found in complexes of metal ions from the third row of the transition metals with charges of at least +3 and ligands with localized lone pairs of electrons.
Colors of Transition-Metal Complexes
The striking colors exhibited by transition-metal complexes are caused by excitation of an electron from a lower-energy d orbital to a higher-energy d orbital, which is called a d–d transition (Figure 24.6.3). For a photon to effect such a transition, its energy must be equal to the difference in energy between the two d orbitals, which depends on the magnitude of Δo.
Recall that the color we observe when we look at an object or a compound is due to light that is transmitted or reflected, not light that is absorbed, and that reflected or transmitted light is complementary in color to the light that is absorbed. Thus a green compound absorbs light in the red portion of the visible spectrum and vice versa, as indicated by the color wheel. Because the energy of a photon of light is inversely proportional to its wavelength, the color of a complex depends on the magnitude of Δo, which depends on the structure of the complex. For example, the complex [Cr(NH3)6]3+ has strong-field ligands and a relatively large Δo. Consequently, it absorbs relatively high-energy photons, corresponding to blue-violet light, which gives it a yellow color. A related complex with weak-field ligands, the [Cr(H2O)6]3+ ion, absorbs lower-energy photons corresponding to the yellow-green portion of the visible spectrum, giving it a deep violet color.
We can now understand why emeralds and rubies have such different colors, even though both contain Cr3+ in an octahedral environment provided by six oxide ions. Although the chemical identity of the six ligands is the same in both cases, the Cr–O distances are different because the compositions of the host lattices are different (Al2O3 in rubies and Be3Al2Si6O18 in emeralds). In ruby, the Cr–O distances are relatively short because of the constraints of the host lattice, which increases the d orbital–ligand interactions and makes Δo relatively large. Consequently, rubies absorb green light and the transmitted or reflected light is red, which gives the gem its characteristic color. In emerald, the Cr–O distances are longer due to relatively large [Si6O18]12− silicate rings; this results in decreased d orbital–ligand interactions and a smaller Δo. Consequently, emeralds absorb light of a longer wavelength (red), which gives the gem its characteristic green color. It is clear that the environment of the transition-metal ion, which is determined by the host lattice, dramatically affects the spectroscopic properties of a metal ion.
Crystal Field Stabilization Energies
Recall that stable molecules contain more electrons in the lower-energy (bonding) molecular orbitals in a molecular orbital diagram than in the higher-energy (antibonding) molecular orbitals. If the lower-energy set of d orbitals (the t2g orbitals) is selectively populated by electrons, then the stability of the complex increases. For example, the single d electron in a d1 complex such as [Ti(H2O)6]3+ is located in one of the t2g orbitals. Consequently, this complex will be more stable than expected on purely electrostatic grounds by 0.4Δo. The additional stabilization of a metal complex by selective population of the lower-energy d orbitals is called its crystal field stabilization energy (CFSE). The CFSE of a complex can be calculated by multiplying the number of electrons in t2g orbitals by the energy of those orbitals (−0.4Δo), multiplying the number of electrons in eg orbitals by the energy of those orbitals (+0.6Δo), and summing the two. Table $2$ gives CFSE values for octahedral complexes with different d electron configurations. The CFSE is highest for low-spin d6 complexes, which accounts in part for the extraordinarily large number of Co(III) complexes known. The other low-spin configurations also have high CFSEs, as does the d3 configuration.
Table $2$: CFSEs for Octahedral Complexes with Different Electron Configurations (in Units of Δo)
High Spin CFSE (Δo) Low Spin CFSE (Δo)
d 0 0
d 1 0.4
d 2 ↿ ↿ 0.8
d 3 ↿ ↿ ↿ 1.2
d 4 ↿ ↿ ↿ 0.6 ↿⇂ ↿ ↿ 1.6
d 5 ↿ ↿ ↿ ↿ ↿ 0.0 ↿⇂ ↿⇂ ↿ 2.0
d 6 ↿⇂ ↿ ↿ ↿ ↿ 0.4 ↿⇂ ↿⇂ ↿⇂ 2.4
d 7 ↿⇂ ↿⇂ ↿ ↿ ↿ 0.8 ↿⇂ ↿⇂ ↿⇂ 1.8
d 8 ↿⇂ ↿⇂ ↿⇂ ↿ ↿ 1.2
d 9 ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿ 0.6
d 10 ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ 0.0
CFSEs are important for two reasons. First, the existence of CFSE nicely accounts for the difference between experimentally measured values for bond energies in metal complexes and values calculated based solely on electrostatic interactions. Second, CFSEs represent relatively large amounts of energy (up to several hundred kilojoules per mole), which has important chemical consequences.
Octahedral d3 and d8 complexes and low-spin d6, d5, d7, and d4 complexes exhibit large CFSEs.
Example $1$
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
1. [CoF6]3−
2. [Rh(CO)2Cl2]
Given: complexes
Asked for: structure, high spin versus low spin, and the number of unpaired electrons
Strategy:
1. From the number of ligands, determine the coordination number of the compound.
2. Classify the ligands as either strong field or weak field and determine the electron configuration of the metal ion.
3. Predict the relative magnitude of Δo and decide whether the compound is high spin or low spin.
4. Place the appropriate number of electrons in the d orbitals and determine the number of unpaired electrons.
Solution
1. A With six ligands, we expect this complex to be octahedral.
B The fluoride ion is a small anion with a concentrated negative charge, but compared with ligands with localized lone pairs of electrons, it is weak field. The charge on the metal ion is +3, giving a d6 electron configuration.
C Because of the weak-field ligands, we expect a relatively small Δo, making the compound high spin.
D In a high-spin octahedral d6 complex, the first five electrons are placed individually in each of the d orbitals with their spins parallel, and the sixth electron is paired in one of the t2g orbitals, giving four unpaired electrons.
1. A This complex has four ligands, so it is either square planar or tetrahedral.
B C Because rhodium is a second-row transition metal ion with a d8 electron configuration and CO is a strong-field ligand, the complex is likely to be square planar with a large Δo, making it low spin. Because the strongest d-orbital interactions are along the x and y axes, the orbital energies increase in the order dz2dyz, and dxz (these are degenerate); dxy; and dx2−y2.
D The eight electrons occupy the first four of these orbitals, leaving the dx2−y2. orbital empty. Thus there are no unpaired electrons.
Exercise $1$
For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.
1. [Mn(H2O)6]2+
2. [PtCl4]2−
Answers
1. octahedral; high spin; five
2. square planar; low spin; no unpaired electrons
Summary
Crystal field theory, which assumes that metal–ligand interactions are only electrostatic in nature, explains many important properties of transition-metal complexes, including their colors, magnetism, structures, stability, and reactivity. Crystal field theory (CFT) is a bonding model that explains many properties of transition metals that cannot be explained using valence bond theory. In CFT, complex formation is assumed to be due to electrostatic interactions between a central metal ion and a set of negatively charged ligands or ligand dipoles arranged around the metal ion. Depending on the arrangement of the ligands, the d orbitals split into sets of orbitals with different energies. The difference between the energy levels in an octahedral complex is called the crystal field splitting energy (Δo), whose magnitude depends on the charge on the metal ion, the position of the metal in the periodic table, and the nature of the ligands. The spin-pairing energy (P) is the increase in energy that occurs when an electron is added to an already occupied orbital. A high-spin configuration occurs when the Δo is less than P, which produces complexes with the maximum number of unpaired electrons possible. Conversely, a low-spin configuration occurs when the Δo is greater than P, which produces complexes with the minimum number of unpaired electrons possible. Strong-field ligands interact strongly with the d orbitals of the metal ions and give a large Δo, whereas weak-field ligands interact more weakly and give a smaller Δo. The colors of transition-metal complexes depend on the environment of the metal ion and can be explained by CFT. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/23%3A_Chemistry_of_Coordination_Chemistry/23.06%3A_Crystal_Field_Theory.txt |
Carbon is unique among the elements in its ability to catenate, to form a wide variety of compounds that contain long chains and/or rings of carbon atoms. Some of the most complex chemical structures known are those of the organic molecules found in living organisms. In spite of their size and complexity, these biological molecules obey the same chemical principles as simpler organic molecules. Thus we can use Lewis electron structures to understand the preferred mode of reactivity of a variety of organic compounds, relative electronegativities and bond polarities to predict how certain groups of atoms will react, and molecular orbital theory to explain why certain organic species that contain multiple bonds are especially stable or undergo particular reactions when they interact with light.
In this chapter, we continue our description of organic compounds by focusing on their molecular structures and reactivity; we will also introduce some of the fundamental types of reactions and reaction mechanisms you will encounter in organic and biological chemistry. We discuss why butter is a solid and oils are liquids despite the apparent similarities in their structures, why the widely used anti-inflammatory drug ibuprofen takes longer than half an hour to relieve pain, and the identity of the major carcinogen in grilled meats and cigarette smoke. The chapter concludes with a brief introduction to the molecules of life, which will explain how the consumption of lactose can result in mental retardation and cirrhosis of the liver in some individuals, how hibernating animals survive during the winter, and how certain groups of antibiotics kill bacteria that are harmful to humans.
• 24.1: General Characteristics of Organic Molecules
Organic chemistry is the study of carbon compounds, nearly all of which also contain hydrogen atoms.
• 24.2: Introduction to Hydrocarbons
Hydrocarbons are organic compounds that contain only carbon and hydrogen. The four general classes of hydrocarbons are: alkanes, alkenes, alkynes and arenes. Aromatic compounds derive their names from the fact that many of these compounds in the early days of discovery were grouped because they were oils with fragrant odors.
• 24.3: Alkanes
Simple alkanes exist as a homologous series, in which adjacent members differ by a CH2 unit.
• 24.4: Unsaturated Hydrocarbons
As noted before, alkenes are hydrocarbons with carbon-to-carbon double bonds (R2C=CR2) and alkynes are hydrocarbons with carbon-to-carbon triple bonds (R–C≡C–R). Collectively, they are called unsaturated hydrocarbons because they have fewer hydrogen atoms than does an alkane with the same number of carbon atoms, as is indicated in the following general formulas:
• 24.5: Functional Groups
Functional groups are atoms or small groups of atoms (two to four) that exhibit a characteristic reactivity. A particular functional group will almost always display its characteristic chemical behavior when it is present in a compound. Because of their importance in understanding organic chemistry, functional groups have characteristic names that often carry over in the naming of individual compounds incorporating specific groups
• 24.6: Compounds with a Carbonyl Group
Aldehydes and ketones are characterized by the presence of a carbonyl group (C=O), and their reactivity can generally be understood by recognizing that the carbonyl carbon contains a partial positive charge ( δ+ ) and the carbonyl oxygen contains a partial negative charge ( δ− ). Aldehydes are typically more reactive than ketones.
• 24.7: Chirality in Organic Chemistry
Stereoisomers are isomers that differ in spatial arrangement of atoms, rather than order of atomic connectivity. One of their most interesting type of isomer is the mirror-image stereoisomers, a non-superimposable set of two molecules that are mirror image of one another. The existence of these molecules are determined by concept known as chirality.
• 24.8: Introduction to Biochemistry
Biochemistry is the study of chemical processes in living organisms, including, but not limited to, living matter. Biochemistry governs all living organisms and living processes. By controlling information flow through biochemical signaling and the flow of chemical energy through metabolism, biochemical processes give rise to the incredible complexity of life.
• 24.9: Proteins
The proteins in all living species are constructed from the same set of 20 amino acids, so called because each contains an amino group attached to a carboxylic acid. The amino acids in proteins are α-amino acids, which means the amino group is attached to the α-carbon of the carboxylic acid. Humans can synthesize only about half of the needed amino acids; the remainder must be obtained from the diet and are known as essential amino acids.
• 24.10: Carbohydrates
All carbohydrates consist of carbon, hydrogen, and oxygen atoms and are polyhydroxy aldehydes or ketones or are compounds that can be broken down to form such compounds. Examples of carbohydrates include starch, fiber, the sweet-tasting compounds called sugars, and structural materials such as cellulose. The term carbohydrate had its origin in a misinterpretation of the molecular formulas of many of these substances.
• 24.11: Nucleic Acids
Nucleotides are composed of phosphoric acid, a pentose sugar (ribose or deoxyribose), and a nitrogen-containing base (adenine, cytosine, guanine, thymine, or uracil). Ribonucleotides contain ribose, while deoxyribonucleotides contain deoxyribose.
• 24.E: Organic and Biological Chemistry (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al.
• 24.S: Organic and Biological Chemistry (Summary)
Thumbnail: DNA Double Helix. (Public Domain; National Human Genome Research Institute)
24: Chemistry of Life- Organic and Biological Chemistry
Learning Objectives
• To recognize the composition and properties typical of organic and inorganic compounds.
Organic substances have been used throughout this text to illustrate the differences between ionic and covalent bonding and to demonstrate the intimate connection between the structures of compounds and their chemical reactivity. You learned, for example, that even though NaOH and alcohols (ROH) both have OH in their formula, NaOH is an ionic compound that dissociates completely in water to produce a basic solution containing Na+ and OH ions, whereas alcohols are covalent compounds that do not dissociate in water and instead form neutral aqueous solutions. You also learned that an amine (RNH2), with its lone pairs of electrons, is a base, whereas a carboxylic acid (RCO2H), with its dissociable proton, is an acid.
Scientists of the 18th and early 19th centuries studied compounds obtained from plants and animals and labeled them organic because they were isolated from “organized” (living) systems. Compounds isolated from nonliving systems, such as rocks and ores, the atmosphere, and the oceans, were labeled inorganic. For many years, scientists thought organic compounds could be made by only living organisms because they possessed a vital force found only in living systems. The vital force theory began to decline in 1828, when the German chemist Friedrich Wöhler synthesized urea from inorganic starting materials. He reacted silver cyanate (AgOCN) and ammonium chloride (NH4Cl), expecting to get ammonium cyanate (NH4OCN). What he expected is described by the following equation.
$AgOCN + NH_4Cl \rightarrow AgCl + NH_4OCN \label{25.1.1}$
Instead, he found the product to be urea (NH2CONH2), a well-known organic material readily isolated from urine. This result led to a series of experiments in which a wide variety of organic compounds were made from inorganic starting materials. The vital force theory gradually went away as chemists learned that they could make many organic compounds in the laboratory.
Today organic chemistry is the study of the chemistry of the carbon compounds, and inorganic chemistry is the study of the chemistry of all other elements. It may seem strange that we divide chemistry into two branches—one that considers compounds of only one element and one that covers the 100-plus remaining elements. However, this division seems more reasonable when we consider that of tens of millions of compounds that have been characterized, the overwhelming majority are carbon compounds.
The word organic has different meanings. Organic fertilizer, such as cow manure, is organic in the original sense; it is derived from living organisms. Organic foods generally are foods grown without synthetic pesticides or fertilizers. Organic chemistry is the chemistry of compounds of carbon.
Carbon is unique among the other elements in that its atoms can form stable covalent bonds with each other and with atoms of other elements in a multitude of variations. The resulting molecules can contain from one to millions of carbon atoms. We previously surveyed organic chemistry by dividing its compounds into families based on functional groups. We begin with the simplest members of a family and then move on to molecules that are organic in the original sense—that is, they are made by and found in living organisms. These complex molecules (all containing carbon) determine the forms and functions of living systems and are the subject of biochemistry.
Organic compounds, like inorganic compounds, obey all the natural laws. Often there is no clear distinction in the chemical or physical properties among organic and inorganic molecules. Nevertheless, it is useful to compare typical members of each class, as in Table $1$.
Table $1$: General Contrasting Properties and Examples of Organic and Inorganic Compounds
Organic Hexane Inorganic NaCl
low melting points −95°C high melting points 801°C
low boiling points 69°C high boiling points 1,413°C
low solubility in water; high solubility in nonpolar solvents insoluble in water; soluble in gasoline greater solubility in water; low solubility in nonpolar solvents soluble in water; insoluble in gasoline
flammable highly flammable nonflammable nonflammable
aqueous solutions do not conduct electricity nonconductive aqueous solutions conduct electricity conductive in aqueous solution
exhibit covalent bonding covalent bonds exhibit ionic bonding ionic bonds
Keep in mind, however, that there are exceptions to every category in this table. To further illustrate typical differences among organic and inorganic compounds, Table $1$ also lists properties of the inorganic compound sodium chloride (common table salt, NaCl) and the organic compound hexane (C6H14), a solvent that is used to extract soybean oil from soybeans (among other uses). Many compounds can be classified as organic or inorganic by the presence or absence of certain typical properties, as illustrated in Table $1$.
Key Takeaway
• Organic chemistry is the study of carbon compounds, nearly all of which also contain hydrogen atoms. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/24%3A_Chemistry_of_Life-_Organic_and_Biological_Chemistry/24.01%3A_General_Characteristics_of_Organic_Molecules.txt |
Learning Objectives
• To get an overview of hydrocarbons molecules and their four primary classifications
Hydrocarbons are organic compounds that contain only carbon and hydrogen. The four general classes of hydrocarbons are: alkanes, alkenes, alkynes and arenes. Aromatic compounds derive their names from the fact that many of these compounds in the early days of discovery were grouped because they were oils with fragrant odors.
The classifications for hydrocarbons, defined by IUPAC nomenclature of organic chemistry are as follows:
1. Saturated hydrocarbons (alkanes) are the simplest of the hydrocarbon species. They are composed entirely of single bonds and are saturated with hydrogen. The general formula for saturated hydrocarbons is \(C_nH_{2n+2}\) (assuming non-cyclic structures). Saturated hydrocarbons are the basis of petroleum fuels and are found as either linear or branched species.The simplest alkanes have their C atoms bonded in a straight chain; these are called normal alkanes. They are named according to the number of C atoms in the chain. The smallest alkane is methane:
2. Unsaturated hydrocarbons have one or more double or triple bonds between carbon atoms. Those with double bond are called alkenes and those with one double bond have the formula \(C_nH_{2n}\) (assuming non-cyclic structures). Those containing triple bonds are called alkynes, with general formula \(C_nH_{2n-2}\). The smallest alkene—ethene—has two C atoms and is also known by its common name ethylene:
The smallest alkyne is ethyne, which is also known as acetylene:
1. Cycloalkanes are hydrocarbons containing one or more carbon rings to which hydrogen atoms are attached. The general formula for a saturated hydrocarbon containing one ring is \(C_nH_{2n}\).
2. Aromatic hydrocarbons, also known as arenes, are hydrocarbons that have at least one aromatic ring. Aromatic compounds contain the benzene unit. Benzene itself is composed of six C atoms in a ring, with alternating single and double C–C bonds:
Because of differences in molecular structure, the empirical formula remains different between hydrocarbons; in linear, or "straight-run" alkanes, alkenes and alkynes, the amount of bonded hydrogen lessens in alkenes and alkynes due to the "self-bonding" or catenation of carbon preventing entire saturation of the hydrocarbon by the formation of double or triple bonds.
The inherent ability of hydrocarbons to bond to themselves is known as catenation, and allows hydrocarbon to form more complex molecules, such as cyclohexane, and in rarer cases, arenes such as benzene. This ability comes from the fact that the bond character between carbon atoms is entirely non-polar, in that the distribution of electrons between the two elements is somewhat even due to the same electronegativity values of the elements (~0.30).
• Wikipedia
24.03: Alkanes
Learning Objectives
• To identify and name simple (straight-chain) alkanes given formulas and write formulas for straight-chain alkanes given their names.
We begin our study of organic chemistry with the hydrocarbons, the simplest organic compounds, which are composed of carbon and hydrogen atoms only. As we noted, there are several different kinds of hydrocarbons. They are distinguished by the types of bonding between carbon atoms and the properties that result from that bonding. Hydrocarbons with only carbon-to-carbon single bonds (C–C) and existing as a continuous chain of carbon atoms also bonded to hydrogen atoms are called alkanes (or saturated hydrocarbons). Saturated, in this case, means that each carbon atom is bonded to four other atoms (hydrogen or carbon)—the most possible; there are no double or triple bonds in the molecules.
The word saturated has the same meaning for hydrocarbons as it does for the dietary fats and oils: the molecule has no carbon-to-carbon double bonds (C=C).
We previously introduced the three simplest alkanes—methane (CH4), ethane (C2H6), and propane (C3H8) and they are shown again in Figure $1$.
The flat representations shown do not accurately portray bond angles or molecular geometry. Methane has a tetrahedral shape that chemists often portray with wedges indicating bonds coming out toward you and dashed lines indicating bonds that go back away from you. An ordinary solid line indicates a bond in the plane of the page. Recall that the VSEPR theory correctly predicts a tetrahedral shape for the methane molecule (Figure $2$).
Methane (CH4), ethane (C2H6), and propane (C3H8) are the beginning of a series of compounds in which any two members in a sequence differ by one carbon atom and two hydrogen atoms—namely, a CH2 unit. The first 10 members of this series are given in Table $1$.
Table $1$: The First 10 Straight-Chain Alkanes
Name Molecular Formula (CnH2n + 2) Condensed Structural Formula Number of Possible Isomers
methane CH4 CH4
ethane C2H6 CH3CH3
propane C3H8 CH3CH2CH3
butane C4H10 CH3CH2CH2CH3 2
pentane C5H12 CH3CH2CH2CH2CH3 3
hexane C6H14 CH3CH2CH2CH2CH2CH3 5
heptane C7H16 CH3CH2CH2CH2CH2CH2CH3 9
octane C8H18 CH3CH2CH2CH2CH2CH2CH2CH3 18
nonane C9H20 CH3CH2CH2CH2CH2CH2CH2CH2CH3 35
decane C10H22 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3 75
Consider the series in Figure $3$. The sequence starts with C3H8, and a CH2 unit is added in each step moving up the series. Any family of compounds in which adjacent members differ from each other by a definite factor (here a CH2 group) is called a homologous series. The members of such a series, called homologs, have properties that vary in a regular and predictable manner. The principle of homology gives organization to organic chemistry in much the same way that the periodic table gives organization to inorganic chemistry. Instead of a bewildering array of individual carbon compounds, we can study a few members of a homologous series and from them deduce some of the properties of other compounds in the series.
The principle of homology allows us to write a general formula for alkanes: CnH2n + 2. Using this formula, we can write a molecular formula for any alkane with a given number of carbon atoms. For example, an alkane with eight carbon atoms has the molecular formula C8H(2 × 8) + 2 = C8H18.
Key Takeaway
• Simple alkanes exist as a homologous series, in which adjacent members differ by a CH2 unit. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/24%3A_Chemistry_of_Life-_Organic_and_Biological_Chemistry/24.02%3A_Introduction_to_Hydrocarbons.txt |
Learning Objectives
• To name alkenes given formulas and write formulas for alkenes given names.
As noted before, alkenes are hydrocarbons with carbon-to-carbon double bonds (R2C=CR2) and alkynes are hydrocarbons with carbon-to-carbon triple bonds (R–C≡C–R). Collectively, they are called unsaturated hydrocarbons because they have fewer hydrogen atoms than does an alkane with the same number of carbon atoms, as is indicated in the following general formulas:
Some representative alkenes—their names, structures, and physical properties—are given in Table \(1\).
Table \(1\): Physical Properties of Some Selected Alkenes
IUPAC Name Molecular Formula Condensed Structural Formula Melting Point (°C) Boiling Point (°C)
ethene C2H4 CH2=CH2 –169 –104
propene C3H6 CH2=CHCH3 –185 –47
1-butene C4H8 CH2=CHCH2CH3 –185 –6
1-pentene C5H10 CH2=CH(CH2)2CH3 –138 30
1-hexene C6H12 CH2=CH(CH2)3CH3 –140 63
1-heptene C7H14 CH2=CH(CH2)4CH3 –119 94
1-octene C8H16 CH2=CH(CH2)5CH3 –102 121
We used only condensed structural formulas in Table \(1\). Thus, CH2=CH2 stands for
The double bond is shared by the two carbons and does not involve the hydrogen atoms, although the condensed formula does not make this point obvious. Note that the molecular formula for ethene is C2H4, whereas that for ethane is C2H6.
The first two alkenes in Table \(1\), ethene and propene, are most often called by their common names—ethylene and propylene, respectively (Figure \(1\)). Ethylene is a major commercial chemical. The US chemical industry produces about 25 billion kilograms of ethylene annually, more than any other synthetic organic chemical. More than half of this ethylene goes into the manufacture of polyethylene, one of the most familiar plastics. Propylene is also an important industrial chemical. It is converted to plastics, isopropyl alcohol, and a variety of other products.
Although there is only one alkene with the formula C2H4 (ethene) and only one with the formula C3H6 (propene), there are several alkenes with the formula C4H8.
Here are some basic rules for naming alkenes from the International Union of Pure and Applied Chemistry (IUPAC):
1. The longest chain of carbon atoms containing the double bond is considered the parent chain. It is named using the same stem as the alkane having the same number of carbon atoms but ends in -ene to identify it as an alkene. Thus the compound CH2=CHCH3 is propene.
2. If there are four or more carbon atoms in a chain, we must indicate the position of the double bond. The carbons atoms are numbered so that the first of the two that are doubly bonded is given the lower of the two possible numbers.The compound CH3CH=CHCH2CH3, for example, has the double bond between the second and third carbon atoms. Its name is 2-pentene (not 3-pentene).
3. Substituent groups are named as with alkanes, and their position is indicated by a number. Thus, the structure below is 5-methyl-2-hexene. Note that the numbering of the parent chain is always done in such a way as to give the double bond the lowest number, even if that causes a substituent to have a higher number. The double bond always has priority in numbering.
Example \(1\)
Name each compound.
Solution
1. The longest chain containing the double bond has five carbon atoms, so the compound is a pentene (rule 1). To give the first carbon atom of the double bond the lowest number (rule 2), we number from the left, so the compound is a 2-pentene. There is a methyl group on the fourth carbon atom (rule 3), so the compound’s name is 4-methyl-2-pentene.
2. The longest chain containing the double bond has five carbon atoms, so the parent compound is a pentene (rule 1). To give the first carbon atom of the double bond the lowest number (rule 2), we number from the left, so the compound is a 2-pentene. There is a methyl group on the third carbon atom (rule 3), so the compound’s name is 3-methyl-2-pentene.
Exercise \(1\)
Name each compound.
1. CH3CH2CH2CH2CH2CH=CHCH3
Just as there are cycloalkanes, there are cycloalkenes. These compounds are named like alkenes, but with the prefix cyclo- attached to the beginning of the parent alkene name.
Example \(2\)
Draw the structure for each compound.
1. 3-methyl-2-pentene
2. cyclohexene
Solution
1. First write the parent chain of five carbon atoms: C–C–C–C–C. Then add the double bond between the second and third carbon atoms:
Now place the methyl group on the third carbon atom and add enough hydrogen atoms to give each carbon atom a total of four bonds.
• First, consider what each of the three parts of the name means. Cyclo means a ring compound, hex means 6 carbon atoms, and -ene means a double bond.
Exercise \(2\)
Draw the structure for each compound.
1. 2-ethyl-1-hexene
2. cyclopentene
Key Takeaway
• Alkenes are hydrocarbons with a carbon-to-carbon double bond. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/24%3A_Chemistry_of_Life-_Organic_and_Biological_Chemistry/24.04%3A_Unsaturated_Hydrocarbons.txt |
Learning Objectives
• To know the major classes of organic compounds and identify important functional groups.
You were previously introduced to several structural units that chemists use to classify organic compounds and predict their reactivities. These functional groups, which determine the chemical reactivity of a molecule under a given set of conditions, can consist of a single atom (such as Cl) or a group of atoms (such as CO2H). The major families of organic compounds are characterized by their functional groups. Figure \(1\) summarizes five families introduced in earlier chapters, gives examples of compounds that contain each functional group, and lists the suffix or prefix used in the systematic nomenclature of compounds that contain each functional group.
The first family listed in Figure \(1\) is the hydrocarbons. These include alkanes, with the general molecular formula CnH2n+2 where n is an integer; alkenes, represented by CnH2n; alkynes, represented by CnH2n−2; and arenes. Halogen-substituted alkanes, alkenes, and arenes form a second major family of organic compounds, which include the alkyl halides and the aryl halides. Oxygen-containing organic compounds, a third family, may be divided into two main types: those that contain at least one C–O bond, which include alcohols, phenols (derivatives of benzene), and ethers, and those that contain a carbonyl group (C=O), which include aldehydes, ketones, and carboxylic acids. Carboxylic acid derivatives, the fourth family listed, are compounds in which the OH of the –CO2H functional group is replaced by either an alkoxy (–OR) group, producing an ester, or by an amido (–NRR′, where R and R′ can be H and/or alkyl groups), forming an amide. Nitrogen-containing organic compounds, the fifth family, include amines; nitriles, which have a C≡N bond; and nitro compounds, which contain the –NO2 group.
The systematic nomenclature of organic compounds indicates the positions of substituents using the lowest numbers possible to identify their locations in the carbon chain of the parent compound. If two compounds have the same systematic name, then they are the same compound. Although systematic names are preferred because they are unambiguous, many organic compounds are known by their common names rather than their systematic names. Common nomenclature uses the prefix form—for a compound that contains no carbons other than those in the functional group, and acet—for those that have one carbon atom in addition [two in the case of acetone, (CH3)2C=O]. Thus methanal and ethanal, respectively, are the systematic names for formaldehyde and acetaldehyde.
Recall that in the systematic nomenclature of aromatic compounds, the positions of groups attached to the aromatic ring are indicated by numbers, starting with 1 and proceeding around the ring in the direction that produces the lowest possible numbers. For example, the position of the first CH3 group in dimethyl benzene is indicated with a 1, but the second CH3 group, which can be placed in any one of three positions, produces 1,2-dimethylbenzene, 1,3-dimethylbenzene, or 1,4-dimethylbenzene (Figure \(2\)). In common nomenclature, in contrast, the prefixes ortho-, meta-, and para- are used to describe the relative positions of groups attached to an aromatic ring. If the CH3 groups in dimethylbenzene, whose common name is xylene, are adjacent to each other, the compound is commonly called ortho-xylene, abbreviated o-xylene. If they are across from each other on the ring, the compound is commonly called para-xylene or p-xylene. When the arrangement is intermediate between those of ortho- and para- compounds, the name is meta-xylene or m-xylene.
We begin our discussion of the structure and reactivity of organic compounds by exploring structural variations in the simple saturated hydrocarbons known as alkanes. These compounds serve as the scaffolding to which the various functional groups are most often attached.
Summary
Functional groups determine the chemical reactivity of an organic molecule. Functional groups are structural units that determine the chemical reactivity of a molecule under a given set of conditions. Organic compounds are classified into several major categories based on the functional groups they contain. In the systematic names of organic compounds, numbers indicate the positions of functional groups in the basic hydrocarbon framework. Many organic compounds also have common names, which use the prefix form—for a compound that contains no carbons other than those in the functional group and acet—for those that have one additional carbon atom.
Conceptual Problems
1. Can two substances have the same systematic name and be different compounds?
2. Is a carbon–carbon multiple bond considered a functional group?
24.06: Compounds with a Carbonyl Group
Aldehydes and Ketones
There are a number of functional groups that contain a carbon-oxygen double bond, which is commonly referred to as a carbonyl. Ketones and aldehydes are two closely related carbonyl-based functional groups that react in very similar ways. In a ketone, the carbon atom of a carbonyl is bonded to two other carbons. In an aldehyde, the carbonyl carbon is bonded on one side to a hydrogen, and on the other side to a carbon. The exception to this definition is formaldehyde, in which the carbonyl carbon has bonds to two hydrogens.
Molecules with carbon-nitrogen double bonds are called imines, or Schiff bases.
Carboxylic acids and acid derivatives
If a carbonyl carbon is bonded on one side to a carbon (or hydrogen) and on the other side to a heteroatom (in organic chemistry, this term generally refers to oxygen, nitrogen, sulfur, or one of the halogens), the functional group is considered to be one of the ‘carboxylic acid derivatives’, a designation that describes a grouping of several functional groups. The eponymous member of this grouping is the carboxylic acid functional group, in which the carbonyl is bonded to a hydroxyl (OH) group.
As the name implies, carboxylic acids are acidic, meaning that they are readily deprotonated to form the conjugate base form, called a carboxylate (much more about carboxylic acids in the acid-base chapter!).
In amides, the carbonyl carbon is bonded to a nitrogen. The nitrogen in an amide can be bonded either to hydrogens, to carbons, or to both. Another way of thinking of an amide is that it is a carbonyl bonded to an amine.
In esters, the carbonyl carbon is bonded to an oxygen which is itself bonded to another carbon. Another way of thinking of an ester is that it is a carbonyl bonded to an alcohol. Thioesters are similar to esters, except a sulfur is in place of the oxygen.
In an acyl phosphate, the carbonyl carbon is bonded to the oxygen of a phosphate, and in an acid chloride, the carbonyl carbon is bonded to a chlorine.
Finally, in a nitrile group, a carbon is triple-bonded to a nitrogen. Nitriles are also often referred to as cyano groups.
A single compound often contains several functional groups. The six-carbon sugar molecules glucose and fructose, for example, contain aldehyde and ketone groups, respectively, and both contain five alcohol groups (a compound with several alcohol groups is often referred to as a ‘polyol’).
Capsaicin, the compound responsible for the heat in hot peppers, contains phenol, ether, amide, and alkene functional groups.
The male sex hormone testosterone contains ketone, alkene, and secondary alcohol groups, while acetylsalicylic acid (aspirin) contains aromatic, carboxylic acid, and ester groups.
While not in any way a complete list, this section has covered most of the important functional groups that we will encounter in biological and laboratory organic chemistry. The table on the inside back cover provides a summary of all of the groups listed in this section, plus a few more that will be introduced later in the text. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/24%3A_Chemistry_of_Life-_Organic_and_Biological_Chemistry/24.05%3A_Functional_Groups.txt |
Organic compounds, molecules created around a chain of carbon atom (more commonly known as carbon backbone), play an essential role in the chemistry of life. These molecules derive their importance from the energy they carry, mainly in a form of potential energy between atomic molecules. Since such potential force can be widely affected due to changes in atomic placement, it is important to understand the concept of an isomer, a molecule sharing same atomic make up as another but differing in structural arrangement.
Stereoisomers are isomers that differ in spatial arrangement of atoms, rather than order of atomic connectivity. One of their most interesting type of isomer is the mirror-image stereoisomers, a non-superimposable set of two molecules that are mirror image of one another. The existence of these molecules are determined by concept known as chirality.
Stereoisomers
Molecules with the same connectivity but different arrangements of the atoms in space are called stereoisomers. There are two types of stereoisomers: geometric and optical. Geometric isomers differ in the relative position(s) of substituents in a rigid molecule. Simple rotation about a C–C σ bond in an alkene, for example, cannot occur because of the presence of the π bond. The substituents are therefore rigidly locked into a particular spatial arrangement. Thus a carbon–carbon multiple bond, or in some cases a ring, prevents one geometric isomer from being readily converted to the other. The members of an isomeric pair are identified as either cis or trans, and interconversion between the two forms requires breaking and reforming one or more bonds. Because their structural difference causes them to have different physical and chemical properties, cis and trans isomers are actually two distinct chemical compounds.
Stereoisomers have the same connectivity, but different arrangements of atoms in space.
Optical isomers are molecules whose structures are mirror images but cannot be superimposed on one another in any orientation. Optical isomers have identical physical properties, although their chemical properties may differ in asymmetric environments. Molecules that are nonsuperimposable mirror images of each other are said to be chiral (pronounced “ky-ral,” from the Greek cheir, meaning “hand”). Examples of some familiar chiral objects are your hands, feet, and ears. As shown in Figure \(\PageIndex{1a}\), your left and right hands are nonsuperimposable mirror images. (Try putting your right shoe on your left foot—it just doesn’t work.) An achiral object is one that can be superimposed on its mirror image, as shown by the superimposed flasks in Figure \(\PageIndex{1b}\).
Most chiral organic molecules have at least one carbon atom that is bonded to four different groups, as occurs in the bromochlorofluoromethane molecule shown in part (a) in Figure \(2\). This carbon, often designated by an asterisk in structural drawings, is called a chiral center or asymmetric carbon atom. If the bromine atom is replaced by another chlorine (Figure \(\PageIndex{2b}\)), the molecule and its mirror image can now be superimposed by simple rotation. Thus the carbon is no longer a chiral center. Asymmetric carbon atoms are found in many naturally occurring molecules, such as lactic acid, which is present in milk and muscles, and nicotine, a component of tobacco. A molecule and its nonsuperimposable mirror image are called enantiomers (from the Greek enantiou, meaning “opposite”).
Thalidomide
In the 1960’s, a drug called thalidomide was widely prescribed in the Western Europe to alleviate morning sickness in pregnant women.
Thalidomide had previously been used in other countries as an antidepressant, and was believed to be safe and effective for both purposes. The drug was not approved for use in the U.S.A. It was not long, however, before doctors realized that something had gone horribly wrong: many babies born to women who had taken thalidomide during pregnancy suffered from severe birth defects.
Researchers later realized the that problem lay in the fact that thalidomide was being provided as a mixture of two different isomeric forms.
One of the isomers is an effective medication, the other caused the side effects. Both isomeric forms have the same molecular formula and the same atom-to-atom connectivity, so they are not constitutional isomers. Where they differ is in the arrangement in three-dimensional space about one tetrahedral, sp3-hybridized carbon. These two forms of thalidomide are stereoisomers.
Looking for planes of symmetry in a molecule is useful, but often difficult in practice. In most cases, the easiest way to decide whether a molecule is chiral or achiral is to look for one or more stereocenters - with a few rare exceptions, the general rule is that molecules with at least one stereocenter are chiral, and molecules with no stereocenters are achiral. Carbon stereocenters are also referred to quite frequently as chiral carbons.
When evaluating a molecule for chirality, it is important to recognize that the question of whether or not the dashed/solid wedge drawing convention is used is irrelevant. Chiral molecules are sometimes drawn without using wedges (although obviously this means that stereochemical information is being omitted). Conversely, wedges may be used on carbons that are not stereocenters – look, for example, at the drawings of glycine and citrate in the figure above. Just because you see dashed and solid wedges in a structure, do not automatically assume that you are looking at a stereocenter.
24.08: Introduction to Biochemistry
Biochemistry is the study of chemical processes in living organisms, including, but not limited to, living matter. Biochemistry governs all living organisms and living processes. By controlling information flow through biochemical signaling and the flow of chemical energy through metabolism, biochemical processes give rise to the incredible complexity of life.
Over the last decades of the 20th century, biochemistry become so successful at explaining living processes that now almost all areas of the life sciences from botany to medicine to genetics are engaged in biochemical research. Today, the main focus of pure biochemistry is in understanding how biological molecules give rise to the processes that occur within living cells, which in turn relates greatly to the study and understanding of whole organisms.
Biochemistry is closely related to molecular biology, the study of the molecular mechanisms by which genetic information encoded in DNA is able to result in the processes of life. Depending on the exact definition of the terms used, molecular biology can be thought of as a branch of biochemistry, or biochemistry as a tool with which to investigate and study molecular biology.
Much of biochemistry deals with the structures, functions and interactions of biological macromolecules, such as proteins, nucleic acids, carbohydrates and lipids, which provide the structure of cells and perform many of the functions associated with life. The chemistry of the cell also depends on the reactions of smaller molecules and ions. These can be inorganic, for example water and metal ions, or organic, for example the amino acids which are used to synthesize proteins. The mechanisms by which cells harness energy from their environment via chemical reactions are known as metabolism. The findings of biochemistry are applied primarily in medicine, nutrition, and agriculture. In medicine, biochemists investigate the causes and cures of disease. In nutrition, they study how to maintain health and study the effects of nutritional deficiencies. In agriculture, biochemists investigate soil and fertilizers, and try to discover ways to improve crop cultivation, crop storage and pest control. Much of biochemistry deals with the structures and functions of cellular components such as proteins, carbohydrates, lipids, nucleic acids and other biomolecules—although increasingly processes rather than individual molecules are the main focus.
• Wikipedia | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/24%3A_Chemistry_of_Life-_Organic_and_Biological_Chemistry/24.07%3A_Chirality_in_Organic_Chemistry.txt |
Learning Objectives
• To recognize amino acids and classify them based on the characteristics of their side chains.
The proteins in all living species, from bacteria to humans, are constructed from the same set of 20 amino acids, so called because each contains an amino group attached to a carboxylic acid. The amino acids in proteins are α-amino acids, which means the amino group is attached to the α-carbon of the carboxylic acid. Humans can synthesize only about half of the needed amino acids; the remainder must be obtained from the diet and are known as essential amino acids. However, two additional amino acids have been found in limited quantities in proteins: Selenocysteine was discovered in 1986, while pyrrolysine was discovered in 2002.
The amino acids are colorless, nonvolatile, crystalline solids, melting and decomposing at temperatures above 200°C. These melting temperatures are more like those of inorganic salts than those of amines or organic acids and indicate that the structures of the amino acids in the solid state and in neutral solution are best represented as having both a negatively charged group and a positively charged group. Such a species is known as a zwitterion.
Classification
In addition to the amino and carboxyl groups, amino acids have a side chain or R group attached to the α-carbon. Each amino acid has unique characteristics arising from the size, shape, solubility, and ionization properties of its R group. As a result, the side chains of amino acids exert a profound effect on the structure and biological activity of proteins. Although amino acids can be classified in various ways, one common approach is to classify them according to whether the functional group on the side chain at neutral pH is nonpolar, polar but uncharged, negatively charged, or positively charged. The structures and names of the 20 amino acids, their one- and three-letter abbreviations, and some of their distinctive features are given in Table \(1\).
Table \(1\): Common Amino Acids Found in Proteins
Common Name Abbreviation Structural Formula (at pH 6) Molar Mass Distinctive Feature
Amino acids with a nonpolar R group
glycine gly (G) 75 the only amino acid lacking a chiral carbon
alanine ala (A) 89
valine val (V) 117 a branched-chain amino acid
leucine leu (L) 131 a branched-chain amino acid
isoleucine ile (I) 131 an essential amino acid because most animals cannot synthesize branched-chain amino acids
phenylalanine phe (F) 165 also classified as an aromatic amino acid
tryptophan trp (W) 204 also classified as an aromatic amino acid
methionine met (M) 149 side chain functions as a methyl group donor
proline pro (P) 115 contains a secondary amine group; referred to as an α-imino acid
Amino acids with a polar but neutral R group
serine ser (S) 105 found at the active site of many enzymes
threonine thr (T) 119 named for its similarity to the sugar threose
cysteine cys (C) 121 oxidation of two cysteine molecules yields cystine
tyrosine tyr (Y) 181 also classified as an aromatic amino acid
asparagine asn (N) 132 the amide of aspartic acid
glutamine gln (Q) 146 the amide of glutamic acid
Amino acids with a negatively charged R group
aspartic acid asp (D) 132 carboxyl groups are ionized at physiological pH; also known as aspartate
glutamic acid glu (E) 146 carboxyl groups are ionized at physiological pH; also known as glutamate
Amino acids with a positively charged R group
histidine his (H) 155 the only amino acid whose R group has a pKa (6.0) near physiological pH
lysine lys (K) 147
arginine arg (R) 175 almost as strong a base as sodium hydroxide
The first amino acid to be isolated was asparagine in 1806. It was obtained from protein found in asparagus juice (hence the name). Glycine, the major amino acid found in gelatin, was named for its sweet taste (Greek glykys, meaning “sweet”). In some cases an amino acid found in a protein is actually a derivative of one of the common 20 amino acids (one such derivative is hydroxyproline). The modification occurs after the amino acid has been assembled into a protein.
Configuration
Notice in Table \(1\) that glycine is the only amino acid whose α-carbon is not chiral. Therefore, with the exception of glycine, the amino acids could theoretically exist in either the D- or the L-enantiomeric form and rotate plane-polarized light. As with sugars, chemists used L-glyceraldehyde as the reference compound for the assignment of absolute configuration to amino acids. Its structure closely resembles an amino acid structure except that in the latter, an amino group takes the place of the OH group on the chiral carbon of the L-glyceraldehyde and a carboxylic acid replaces the aldehyde. Modern stereochemistry assignments using the Cahn-Ingold-Prelog priority rules used ubiquitously in chemistry show that all of the naturally occurring chiral amino acids are S except Cys which is R.
We learned that all naturally occurring sugars belong to the D series. It is interesting, therefore, that nearly all known plant and animal proteins are composed entirely of L-amino acids. However, certain bacteria contain D-amino acids in their cell walls, and several antibiotics (e.g., actinomycin D and the gramicidins) contain varying amounts of D-leucine, D-phenylalanine, and D-valine.
Summary
Amino acids can be classified based on the characteristics of their distinctive side chains as nonpolar, polar but uncharged, negatively charged, or positively charged. The amino acids found in proteins are L-amino acids.
Learning Objectives
• Describe the four levels of protein structure.
• Identify the types of attractive interactions that hold proteins in their most stable three-dimensional structure.
• Explain what happens when proteins are denatured.
• Identify how a protein can be denatured.
Each of the thousands of naturally occurring proteins has its own characteristic amino acid composition and sequence that result in a unique three-dimensional shape. Since the 1950s, scientists have determined the amino acid sequences and three-dimensional conformation of numerous proteins and thus obtained important clues on how each protein performs its specific function in the body.
Proteins are compounds of high molar mass consisting largely or entirely of chains of amino acids. Because of their great complexity, protein molecules cannot be classified on the basis of specific structural similarities, as carbohydrates and lipids are categorized. The two major structural classifications of proteins are based on far more general qualities: whether the protein is (1) fiberlike and insoluble or (2) globular and soluble. Some proteins, such as those that compose hair, skin, muscles, and connective tissue, are fiberlike. These fibrous proteins are insoluble in water and usually serve structural, connective, and protective functions. Examples of fibrous proteins are keratins, collagens, myosins, and elastins. Hair and the outer layer of skin are composed of keratin. Connective tissues contain collagen. Myosins are muscle proteins and are capable of contraction and extension. Elastins are found in ligaments and the elastic tissue of artery walls.
Globular proteins, the other major class, are soluble in aqueous media. In these proteins, the chains are folded so that the molecule as a whole is roughly spherical. Familiar examples include egg albumin from egg whites and serum albumin in blood. Serum albumin plays a major role in transporting fatty acids and maintaining a proper balance of osmotic pressures in the body. Hemoglobin and myoglobin, which are important for binding oxygen, are also globular proteins.
Levels of Protein Structure
The structure of proteins is generally described as having four organizational levels. The first of these is the primary structure, which is the number and sequence of amino acids in a protein’s polypeptide chain or chains, beginning with the free amino group and maintained by the peptide bonds connecting each amino acid to the next. The primary structure of insulin, composed of 51 amino acids, is shown in Figure \(1\).
A protein molecule is not a random tangle of polypeptide chains. Instead, the chains are arranged in unique but specific conformations. The term secondary structure refers to the fixed arrangement of the polypeptide backbone. On the basis of X ray studies, Linus Pauling and Robert Corey postulated that certain proteins or portions of proteins twist into a spiral or a helix. This helix is stabilized by intrachain hydrogen bonding between the carbonyl oxygen atom of one amino acid and the amide hydrogen atom four amino acids up the chain (located on the next turn of the helix) and is known as a right-handed α-helix. X ray data indicate that this helix makes one turn for every 3.6 amino acids, and the side chains of these amino acids project outward from the coiled backbone (Figure \(2\)). The α-keratins, found in hair and wool, are exclusively α-helical in conformation. Some proteins, such as gamma globulin, chymotrypsin, and cytochrome c, have little or no helical structure. Others, such as hemoglobin and myoglobin, are helical in certain regions but not in others.
Another common type of secondary structure, called the β-pleated sheet conformation, is a sheetlike arrangement in which two or more extended polypeptide chains (or separate regions on the same chain) are aligned side by side. The aligned segments can run either parallel or antiparallel—that is, the N-terminals can face in the same direction on adjacent chains or in different directions—and are connected by interchain hydrogen bonding (Figure \(3\)). The β-pleated sheet is particularly important in structural proteins, such as silk fibroin. It is also seen in portions of many enzymes, such as carboxypeptidase A and lysozyme.
Tertiary structure refers to the unique three-dimensional shape of the protein as a whole, which results from the folding and bending of the protein backbone. The tertiary structure is intimately tied to the proper biochemical functioning of the protein. Figure \(4\) shows a depiction of the three-dimensional structure of insulin.
Four major types of attractive interactions determine the shape and stability of the tertiary structure of proteins. You studied several of them previously.
1. Ionic bonding. Ionic bonds result from electrostatic attractions between positively and negatively charged side chains of amino acids. For example, the mutual attraction between an aspartic acid carboxylate ion and a lysine ammonium ion helps to maintain a particular folded area of a protein (part (a) of Figure \(5\)).
2. Hydrogen bonding. Hydrogen bonding forms between a highly electronegative oxygen atom or a nitrogen atom and a hydrogen atom attached to another oxygen atom or a nitrogen atom, such as those found in polar amino acid side chains. Hydrogen bonding (as well as ionic attractions) is extremely important in both the intra- and intermolecular interactions of proteins (part (b) of Figure \(5\)).
3. Disulfide linkages. Two cysteine amino acid units may be brought close together as the protein molecule folds. Subsequent oxidation and linkage of the sulfur atoms in the highly reactive sulfhydryl (SH) groups leads to the formation of cystine (part (c) of Figure \(5\)). Intrachain disulfide linkages are found in many proteins, including insulin (yellow bars in Figure \(1\)) and have a strong stabilizing effect on the tertiary structure.
1. Dispersion forces. Dispersion forces arise when a normally nonpolar atom becomes momentarily polar due to an uneven distribution of electrons, leading to an instantaneous dipole that induces a shift of electrons in a neighboring nonpolar atom. Dispersion forces are weak but can be important when other types of interactions are either missing or minimal (part (d) of Figure \(5\)). This is the case with fibroin, the major protein in silk, in which a high proportion of amino acids in the protein have nonpolar side chains. The term hydrophobic interaction is often misused as a synonym for dispersion forces. Hydrophobic interactions arise because water molecules engage in hydrogen bonding with other water molecules (or groups in proteins capable of hydrogen bonding). Because nonpolar groups cannot engage in hydrogen bonding, the protein folds in such a way that these groups are buried in the interior part of the protein structure, minimizing their contact with water.
When a protein contains more than one polypeptide chain, each chain is called a subunit. The arrangement of multiple subunits represents a fourth level of structure, the quaternary structure of a protein. Hemoglobin, with four polypeptide chains or subunits, is the most frequently cited example of a protein having quaternary structure (Figure \(6\)). The quaternary structure of a protein is produced and stabilized by the same kinds of interactions that produce and maintain the tertiary structure. A schematic representation of the four levels of protein structure is in Figure \(7\).
The primary structure consists of the specific amino acid sequence. The resulting peptide chain can twist into an α-helix, which is one type of secondary structure. This helical segment is incorporated into the tertiary structure of the folded polypeptide chain. The single polypeptide chain is a subunit that constitutes the quaternary structure of a protein, such as hemoglobin that has four polypeptide chains.
Denaturation of Proteins
The highly organized structures of proteins are truly masterworks of chemical architecture. But highly organized structures tend to have a certain delicacy, and this is true of proteins. Denaturation is the term used for any change in the three-dimensional structure of a protein that renders it incapable of performing its assigned function. A denatured protein cannot do its job. (Sometimes denaturation is equated with the precipitation or coagulation of a protein; our definition is a bit broader.) A wide variety of reagents and conditions, such as heat, organic compounds, pH changes, and heavy metal ions can cause protein denaturation (Figure \(1\)).
Figure \(1\): Protein Denaturation Methods
Method Effect on Protein Structure
Heat above 50°C or ultraviolet (UV) radiation Heat or UV radiation supplies kinetic energy to protein molecules, causing their atoms to vibrate more rapidly and disrupting relatively weak hydrogen bonding and dispersion forces.
Use of organic compounds, such as ethyl alcohol These compounds are capable of engaging in intermolecular hydrogen bonding with protein molecules, disrupting intramolecular hydrogen bonding within the protein.
Salts of heavy metal ions, such as mercury, silver, and lead These ions form strong bonds with the carboxylate anions of the acidic amino acids or SH groups of cysteine, disrupting ionic bonds and disulfide linkages.
Alkaloid reagents, such as tannic acid (used in tanning leather) These reagents combine with positively charged amino groups in proteins to disrupt ionic bonds.
Anyone who has fried an egg has observed denaturation. The clear egg white turns opaque as the albumin denatures and coagulates. No one has yet reversed that process. However, given the proper circumstances and enough time, a protein that has unfolded under sufficiently gentle conditions can refold and may again exhibit biological activity (Figure \(8\)). Such evidence suggests that, at least for these proteins, the primary structure determines the secondary and tertiary structure. A given sequence of amino acids seems to adopt its particular three-dimensional arrangement naturally if conditions are right.
The primary structures of proteins are quite sturdy. In general, fairly vigorous conditions are needed to hydrolyze peptide bonds. At the secondary through quaternary levels, however, proteins are quite vulnerable to attack, though they vary in their vulnerability to denaturation. The delicately folded globular proteins are much easier to denature than are the tough, fibrous proteins of hair and skin.
Summary
Proteins can be divided into two categories: fibrous, which tend to be insoluble in water, and globular, which are more soluble in water. A protein may have up to four levels of structure. The primary structure consists of the specific amino acid sequence. The resulting peptide chain can form an α-helix or β-pleated sheet (or local structures not as easily categorized), which is known as secondary structure. These segments of secondary structure are incorporated into the tertiary structure of the folded polypeptide chain. The quaternary structure describes the arrangements of subunits in a protein that contains more than one subunit. Four major types of attractive interactions determine the shape and stability of the folded protein: ionic bonding, hydrogen bonding, disulfide linkages, and dispersion forces. A wide variety of reagents and conditions can cause a protein to unfold or denature. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/24%3A_Chemistry_of_Life-_Organic_and_Biological_Chemistry/24.09%3A_Proteins.txt |
Learning Objectives
• To recognize carbohydrates and classify them as mono-, di-, or polysaccharides.
All carbohydrates consist of carbon, hydrogen, and oxygen atoms and are polyhydroxy aldehydes or ketones or are compounds that can be broken down to form such compounds. Examples of carbohydrates include starch, fiber, the sweet-tasting compounds called sugars, and structural materials such as cellulose. The term carbohydrate had its origin in a misinterpretation of the molecular formulas of many of these substances. For example, because its formula is C6H12O6, glucose was once thought to be a “carbon hydrate” with the structure C6·6H2O.
Example $1$
Which compounds would be classified as carbohydrates?
Solution
1. This is a carbohydrate because the molecule contains an aldehyde functional group with OH groups on the other two carbon atoms.
2. This is not a carbohydrate because the molecule does not contain an aldehyde or a ketone functional group.
3. This is a carbohydrate because the molecule contains a ketone functional group with OH groups on the other two carbon atoms.
4. This is not a carbohydrate; although it has a ketone functional group, one of the other carbons atoms does not have an OH group attached.
Exercise $1$
Which compounds would be classified as carbohydrates?
Green plants are capable of synthesizing glucose (C6H12O6) from carbon dioxide (CO2) and water (H2O) by using solar energy in the process known as photosynthesis:
$\ce{6CO_2 + 6H_2O} + \text{686 kcal} \rightarrow \ce{C_6H_{12}O_6 + 6O_2} \label{$1$}$
(The 686 kcal come from solar energy.) Plants can use the glucose for energy or convert it to larger carbohydrates, such as starch or cellulose. Starch provides energy for later use, perhaps as nourishment for a plant’s seeds, while cellulose is the structural material of plants. We can gather and eat the parts of a plant that store energy—seeds, roots, tubers, and fruits—and use some of that energy ourselves. Carbohydrates are also needed for the synthesis of nucleic acids and many proteins and lipids.
Animals, including humans, cannot synthesize carbohydrates from carbon dioxide and water and are therefore dependent on the plant kingdom to provide these vital compounds. We use carbohydrates not only for food (about 60%–65% by mass of the average diet) but also for clothing (cotton, linen, rayon), shelter (wood), fuel (wood), and paper (wood).
The simplest carbohydrates—those that cannot be hydrolyzed to produce even smaller carbohydrates—are called monosaccharides. Two or more monosaccharides can link together to form chains that contain from two to several hundred or thousand monosaccharide units. Prefixes are used to indicate the number of such units in the chains. Disaccharide molecules have two monosaccharide units, trisaccharide molecules have three units, and so on. Chains with many monosaccharide units joined together are called polysaccharides. All these so-called higher saccharides can be hydrolyzed back to their constituent monosaccharides.
Compounds that cannot be hydrolyzed will not react with water to form two or more smaller compounds.
Summary
Carbohydrates are an important group of biological molecules that includes sugars and starches. Photosynthesis is the process by which plants use energy from sunlight to synthesize carbohydrates. A monosaccharide is the simplest carbohydrate and cannot be hydrolyzed to produce a smaller carbohydrate molecule. Disaccharides contain two monosaccharide units, and polysaccharides contain many monosaccharide units. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/24%3A_Chemistry_of_Life-_Organic_and_Biological_Chemistry/24.10%3A_Carbohydrates.txt |
Learning Objectives
• To identify the different molecules that combine to form nucleotides.
The repeating, or monomer, units that are linked together to form nucleic acids are known as nucleotides. The deoxyribonucleic acid (DNA) of a typical mammalian cell contains about 3 × 109 nucleotides. Nucleotides can be further broken down to phosphoric acid (H3PO4), a pentose sugar (a sugar with five carbon atoms), and a nitrogenous base (a base containing nitrogen atoms).
$\mathrm{nucleic\: acids \underset{down\: into}{\xrightarrow{can\: be\: broken}} nucleotides \underset{down\: into}{\xrightarrow{can\: be\: broken}} H_3PO_4 + nitrogen\: base + pentose\: sugar} \nonumber$
If the pentose sugar is ribose, the nucleotide is more specifically referred to as a ribonucleotide, and the resulting nucleic acid is ribonucleic acid (RNA). If the sugar is 2-deoxyribose, the nucleotide is a deoxyribonucleotide, and the nucleic acid is DNA.
The nitrogenous bases found in nucleotides are classified as pyrimidines or purines. Pyrimidines are heterocyclic amines with two nitrogen atoms in a six-member ring and include uracil, thymine, and cytosine. Purines are heterocyclic amines consisting of a pyrimidine ring fused to a five-member ring with two nitrogen atoms. Adenine and guanine are the major purines found in nucleic acids (Figure $1$).
The formation of a bond between C1′ of the pentose sugar and N1 of the pyrimidine base or N9 of the purine base joins the pentose sugar to the nitrogenous base. In the formation of this bond, a molecule of water is removed. Table $1$ summarizes the similarities and differences in the composition of nucleotides in DNA and RNA.
The numbering convention is that primed numbers designate the atoms of the pentose ring, and unprimed numbers designate the atoms of the purine or pyrimidine ring.
Table $1$: Composition of Nucleotides in DNA and RNA
Composition DNA RNA
purine bases adenine and guanine adenine and guanine
pyrimidine bases cytosine and thymine cytosine and uracil
pentose sugar 2-deoxyribose ribose
inorganic acid phosphoric acid (H3PO4) H3PO4
The names and structures of the major ribonucleotides and one of the deoxyribonucleotides are given in Figure $2$.
Apart from being the monomer units of DNA and RNA, the nucleotides and some of their derivatives have other functions as well. Adenosine diphosphate (ADP) and adenosine triphosphate (ATP), shown in Figure $3$, have a role in cell metabolism. Moreover, a number of coenzymes, including flavin adenine dinucleotide (FAD), nicotinamide adenine dinucleotide (NAD+), and coenzyme A, contain adenine nucleotides as structural components.
Summary
Nucleotides are composed of phosphoric acid, a pentose sugar (ribose or deoxyribose), and a nitrogen-containing base (adenine, cytosine, guanine, thymine, or uracil). Ribonucleotides contain ribose, while deoxyribonucleotides contain deoxyribose.
24.E: Organic and Biological Chemistry (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
25.1: General Characteristics of Organic Molecules
Concept Review Exercises
1. Classify each compound as organic or inorganic.
1. C3H8O
2. CaCl2
3. Cr(NH3)3Cl3
4. C30H48O3N
2. Which compound is likely organic and which is likely inorganic?
1. a flammable compound that boils at 80°C and is insoluble in water
2. a compound that does not burn, melts at 630°C, and is soluble in water
1. organic
2. inorganic
3. inorganic
4. organic
1. organic
2. inorganic
Exercises
1. Classify each compound as organic or inorganic.
1. C6H10
2. CoCl2
3. C12H22O11
2. Classify each compound as organic or inorganic.
1. CH3NH2
2. NaNH2
3. Cu(NH3)6Cl2
3. Which member of each pair has a higher melting point?
1. CH3OH and NaOH
2. CH3Cl and KCl
4. Which member of each pair has a higher melting point?
1. C2H6 and CoCl2
2. CH4 and LiH
1. organic
2. inorganic
3. organic
1. NaOH
2. KCl
25.5: Functional Groups
Concept Review Exercises
1. What is the functional group of an alkene? An alkyne?
2. Does CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH3 have a functional group? Explain.
Answers
1. carbon-to-carbon double bond; carbon-to-carbon triple bond
2. No; it has nothing but carbon and hydrogen atoms and all single bonds.
Exercises
1. What is the functional group of 1-butanol (CH3CH2CH2CH2OH)?
2. What is the functional group of butyl bromide, CH3CH2CH2CH2Br?
1. OH | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/24%3A_Chemistry_of_Life-_Organic_and_Biological_Chemistry/24.11%3A_Nucleic_Acids.txt |
• organic chemistry – the study of carbon compounds
• biochemistry – the study of the chemistry of living species
25.2: Introduction to Hydrocarbons
• made of only hydrogen and carbon
• 4 types: alkanes, alkenes, alkynes, and aromatic hydrocarbons
• alkanes – only single bonds
• also called saturated hydrocarbons
• have the largest amount of hydrogen atoms bonded to a single carbon atom
• alkenes (olefins) – have double carbon bonds
• alkynes – triple carbon bonds
• aromatic hydrocarbons – carbon atoms connected in a planar ring structure, joined by s and p bonds between carbon atoms
• alkenes, alkynes and aromatic hydrocarbons – unsaturated hydrocarbons
• less volatile with increasing molar mass
• very low molecular weight = gas at room temperature
• moderate molecular weight = liquid
• high molecular weight = solid
25.3: Alkanes
• methane major part of natural gas
• used in home heating, gas stoves, hot-water heaters
• propane major part of bottled gas
• used for home heating, cooking
• butane – used in disposable lighters, fuel canisters
• alkanes with 5-12 carbon atoms are found in gasoline
• formla for alkanes is called condensed structural formulas
Condensed Structural Formulas
Molecular Formula Condensed Structural Formula Name Boiling point (°C)
CH4 CH4 Methane -161
C2H6 CH3 CH3 Ethane -89
C3H8 CH3 CH2 CH3 Propane -44
C4H10 CH3 CH2 CH2 CH3 Butane -0.5
C5H12 CH3 CH2 CH2 CH2 CH3 Pentane 36
C6H14 CH3 CH2 CH2 CH2 CH2 CH3 Hexane 68
C7H16 CH3 CH2 CH2 CH2 CH2 CH2 CH3 Heptane 98
C8H18 CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH3 Octane 125
C9H20 CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH3 Nonane 151
C10H22 CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH3 Decane 174
Lewis Structure
1. Structures of Alkanes
• tetrahedral geometry
• carbon-carbon single bond rotates easily at room temperature
• long chains tend to change shape
1. Structural Isomers
• straight-chained hydrocarbons – carbons atoms that are joined in a continuous chain
• branched-chain hydrocarbons – hydrocarbons with branched chains, 4 or more carbon atoms
• structural isomers – compounds with the same molecular formula but different bonding arrangements
1. Nomenclature of Alkanes
• 1) Find the longest continuous chain of carbon atoms, and use the name of this chain as the base name of the compound.
• 2) Number the carbon atoms in the longest chain, beginning with the end of the chain that is nearest to a substituent
• 3) Name and give the location of each substituent group
• 4) When two or more substituents are present list them in alphabetical order
1. Cycloalkanes
• alkanes that form rings or cycles
• carbon rings with fewer than five carbon atoms are strained
• more reactive
1. Reactions of Alkanes
• at room temperature alkanes do not react with acids, bases, or strong oxidizing agents
• important for it’s combustion in air; bases for fuels
25.4: Unsaturated Hydrocarbons
• Alkenes
• double carbon bonds
• ethene or ethylene simplest alkene\
• nomenclature of alkenes come from the name of the corresponding alkane
• the –ane ending is changed to –ene
• geometrical isomers – compounds that have the same molecular formula and the same groups bonded to one another but differ in the spatial arrangement of these groups
• geometrical isomers have distinct physical properties and differ in chemical behavior
• the double carbon bond is resistant to twisting
• rotation about a double bond is a key process in the chemistry of vision
1. Alkynes
• one or more triple bonds
• simplest alkyne is acetylene
• highly reactive
• named by changing the alkane ending (-ane) to –yne
1. Addition Reactions of Alkenes and Alkynes
• addition reactions – a reactant is added to the two atoms that form the multiple bond
• hyrogenation – reaction between al alkene and H2
1. Aromatic Hydrocarbons
• simplest is benzene
• stability comes from the stabilization of the p electrons through delocalization in the p orbitals
• represented by a hexagon with an inscribed circle
• substitution reactions – one atom of a molecule is removed and replaced by another atom or group of atoms
25.5: Functional Groups
functional group – site of reactivity in an organic molecule; controls how the molecule behaves or functions
• chemistry of an organic molecule is determined by the functional groups it contains
1. Alcohols (R-OH)
• alcohols – hydrocarbon derivatives in which one ore more hydrogens of a parent hydrocarbon have been replaced by a hydroxyle or alcohol functional group
1. Ethers (R-O-R’)
• ethers – two hyrocarbon groups are bonded to one oxygen
• formed from two molecules of alcohol by splitting out a molecule of water
• condensation reaction – reaction where water is split out from two substances
• used in solvents
25.6: Compounds with a Carbonyl Group
• carbon oxygen double bonds
1. Aldehydes and Ketones
• carbonyl group in aldehydes has at least one attached hydrogen atom
formaldehyde
• carbonyl group in ketones occurs at the interior of a carbon chain
Acetone
• prepared by oxidizing alcohols
1. Carboxylic Acids
• carboxylic acids contain carboxyl functional group (COOH)
• important for manufacturing of polymers
• produced by oxidation in which the OH group is attached to the CH2
1. Esters
• carboxylic acids that undergo condensation with alcohols
• have pleasant odors
• saponification – hydrolysis of an ester in a base
1. Amines and Amides
• amines – organic bases; general formula R3N
• amides – anines containing a hydrogen bonded to nitrogen that undergoes condensation
25.8: Introduction to Biochemistry
• biosphere – part of the earth where living organisms are formed and living
• includes influences on life of the atmosphere, natural waters, solid earth
• living organisms require a large amount of energy
• biopolymers – three categories: proteins, polysaccharides (carbohydrates), and nucleic acids
25.9: Proteins
• macromolecular substances
• 50% body’s dry weight is proteins
• composed of amino acids
1. Amino Acids
• differ in the R group
• building block of all proteins - a -amino acid
• general form:
• chiral - any molecule containing a carbon with four different attached groups
• enantiomers – mirror-image of chiral
• enantiomers and chiral have the same physical properties
• differ in chemical reactivity toward other chiral molecules
1. Polypeptides and Proteins
• peptide bond – condesation reaction between the carboxyl group of one amino acid and the amino group of another amino acid
• polypeptides – large number of amino acids that are linked together by peptide bonds
• proteins are polypeptide molecules
• weighs from 6000 to over 50 million amu
1. Protein Structure
• primary structure – arrangement of amino acids along a protein chain
• makes the protein unique
• secondary structure – the way segments of the protein chain are oriented in a regular pattern
• a -helix – most important and common secondary structure arrangement
• first propoesed by Linus Pauling and R. B. Corey
• tertiary structure – overal shape of a protein
• globular protein – a protein that folds into a compact spherical shape
• soluble in water, mobile within cells
• enzymes – large protein molecules that serve as catalysts
25.10: Carbohydrates
• written as Cx(H2O)y
• glucose the most abundant carbohydrate C6H12O6
• not really hydrates of carbon but polyhydroxly aldehydes and ketones
1. Disaccharides
• monosaccharides – simple sugars that can’t be broken into smaller molecules by hydrolysis with aqueous acids
• disaccharide – two linked monosaccharides
• two common disaccharides: sucrose, lactose
1. Polysaccharides
• made of several monosaccharide units
• starch – group of polysaccharides
• food storage in plant seeds and tuers
• glycogen – starchlike substance synthesized in the body
• 5000 to more than 5 million amu
• energy bank in the body; muscles and liver
• cellulose – major strructural unit of plants
• straight chains of glucose units
• more than 500,000 amu
25.11: Nucleic Acids
• nucleic acids – biopolymers that are chemical carriers of an organisn’t genetic information
• deoxyribonucleic acids (DNA) – huge molecules with molar weights of 6-16million amu
• ribonucleic acids (RNA) – smaller molecules with molecular weights of 20,000 to 40,000 amu
• DNA found inside the nucleus of a cell, RNA found outside nucleus in the cytoplasm
• DNA stores the genetic information of the cell and controls the production of proteins
• RNA carries information from the DNA out of the nucleus
• Monomers of nucleic acids (nucleotides) have three parts:
• A phosphoric acid molecule, H3PO4
• A five-carbon sugar
ribose
• A nitrogen-containing organic base
deoxyribose
• deoxyribose has one less oxygen atom at carbon 2 than ribose
• nucleic acids are polynucleotides
• DNA molecules made of two DNA chains that form a double helix | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/24%3A_Chemistry_of_Life-_Organic_and_Biological_Chemistry/24.S%3A_Organic_and_Biological_Chemistry_%28Summary%29.txt |
Learning Objectives
• To classify matter.
Chemists study the structures, physical properties, and chemical properties of material substances. These consist of matter, which is anything that occupies space and has mass. Gold and iridium are matter, as are peanuts, people, and postage stamps. Smoke, smog, and laughing gas are matter. Energy, light, and sound, however, are not matter; ideas and emotions are also not matter.
The mass of an object is the quantity of matter it contains. Do not confuse an object’s mass with its weight, which is a force caused by the gravitational attraction that operates on the object. Mass is a fundamental property of an object that does not depend on its location.In physical terms, the mass of an object is directly proportional to the force required to change its speed or direction. A more detailed discussion of the differences between weight and mass and the units used to measure them is included in Essential Skills 1 (Section 1.9). Weight, on the other hand, depends on the location of an object. An astronaut whose mass is 95 kg weighs about 210 lb on Earth but only about 35 lb on the moon because the gravitational force he or she experiences on the moon is approximately one-sixth the force experienced on Earth. For practical purposes, weight and mass are often used interchangeably in laboratories. Because the force of gravity is considered to be the same everywhere on Earth’s surface, 2.2 lb (a weight) equals 1.0 kg (a mass), regardless of the location of the laboratory on Earth.
Under normal conditions, there are three distinct states of matter: solids, liquids, and gases. Solids are relatively rigid and have fixed shapes and volumes. A rock, for example, is a solid. In contrast, liquids have fixed volumes but flow to assume the shape of their containers, such as a beverage in a can. Gases, such as air in an automobile tire, have neither fixed shapes nor fixed volumes and expand to completely fill their containers. Whereas the volume of gases strongly depends on their temperature and pressure (the amount of force exerted on a given area), the volumes of liquids and solids are virtually independent of temperature and pressure. Matter can often change from one physical state to another in a process called a physical change. For example, liquid water can be heated to form a gas called steam, or steam can be cooled to form liquid water. However, such changes of state do not affect the chemical composition of the substance.
Pure Substances and Mixtures
A pure chemical substance is any matter that has a fixed chemical composition and characteristic properties. Oxygen, for example, is a pure chemical substance that is a colorless, odorless gas at 25°C. Very few samples of matter consist of pure substances; instead, most are mixtures, which are combinations of two or more pure substances in variable proportions in which the individual substances retain their identity. Air, tap water, milk, blue cheese, bread, and dirt are all mixtures. If all portions of a material are in the same state, have no visible boundaries, and are uniform throughout, then the material is homogeneous. Examples of homogeneous mixtures are the air we breathe and the tap water we drink. Homogeneous mixtures are also called solutions. Thus air is a solution of nitrogen, oxygen, water vapor, carbon dioxide, and several other gases; tap water is a solution of small amounts of several substances in water. The specific compositions of both of these solutions are not fixed, however, but depend on both source and location; for example, the composition of tap water in Boise, Idaho, is not the same as the composition of tap water in Buffalo, New York. Although most solutions we encounter are liquid, solutions can also be solid. The gray substance still used by some dentists to fill tooth cavities is a complex solid solution that contains 50% mercury and 50% of a powder that contains mostly silver, tin, and copper, with small amounts of zinc and mercury. Solid solutions of two or more metals are commonly called alloys.
If the composition of a material is not completely uniform, then it is heterogeneous (e.g., chocolate chip cookie dough, blue cheese, and dirt). Mixtures that appear to be homogeneous are often found to be heterogeneous after microscopic examination. Milk, for example, appears to be homogeneous, but when examined under a microscope, it clearly consists of tiny globules of fat and protein dispersed in water. The components of heterogeneous mixtures can usually be separated by simple means. Solid-liquid mixtures such as sand in water or tea leaves in tea are readily separated by filtration, which consists of passing the mixture through a barrier, such as a strainer, with holes or pores that are smaller than the solid particles. In principle, mixtures of two or more solids, such as sugar and salt, can be separated by microscopic inspection and sorting. More complex operations are usually necessary, though, such as when separating gold nuggets from river gravel by panning. First solid material is filtered from river water; then the solids are separated by inspection. If gold is embedded in rock, it may have to be isolated using chemical methods.
Homogeneous mixtures (solutions) can be separated into their component substances by physical processes that rely on differences in some physical property, such as differences in their boiling points. Two of these separation methods are distillation and crystallization. Distillation makes use of differences in volatility, a measure of how easily a substance is converted to a gas at a given temperature. A simple distillation apparatus for separating a mixture of substances, at least one of which is a liquid. The most volatile component boils first and is condensed back to a liquid in the water-cooled condenser, from which it flows into the receiving flask. If a solution of salt and water is distilled, for example, the more volatile component, pure water, collects in the receiving flask, while the salt remains in the distillation flask.
Mixtures of two or more liquids with different boiling points can be separated with a more complex distillation apparatus. One example is the refining of crude petroleum into a range of useful products: aviation fuel, gasoline, kerosene, diesel fuel, and lubricating oil (in the approximate order of decreasing volatility). Another example is the distillation of alcoholic spirits such as brandy or whiskey. (This relatively simple procedure caused more than a few headaches for federal authorities in the 1920s during the era of Prohibition, when illegal stills proliferated in remote regions of the United States!)
Crystallization separates mixtures based on differences in solubility, a measure of how much solid substance remains dissolved in a given amount of a specified liquid. Most substances are more soluble at higher temperatures, so a mixture of two or more substances can be dissolved at an elevated temperature and then allowed to cool slowly. Alternatively, the liquid, called the solvent, may be allowed to evaporate. In either case, the least soluble of the dissolved substances, the one that is least likely to remain in solution, usually forms crystals first, and these crystals can be removed from the remaining solution by filtration.
Most mixtures can be separated into pure substances, which may be either elements or compounds. An element, such as gray, metallic sodium, is a substance that cannot be broken down into simpler ones by chemical changes; a compound, such as white, crystalline sodium chloride, contains two or more elements and has chemical and physical properties that are usually different from those of the elements of which it is composed. With only a few exceptions, a particular compound has the same elemental composition (the same elements in the same proportions) regardless of its source or history. The chemical composition of a substance is altered in a process called a chemical change. The conversion of two or more elements, such as sodium and chlorine, to a chemical compound, sodium chloride, is an example of a chemical change, often called a chemical reaction. Currently, about 118 elements are known, but millions of chemical compounds have been prepared from these 118 elements. The known elements are listed in the periodic table.
Different Definitions of Matter: Different Definitions of Matter, YouTube (opens in new window) [youtu.be]
In general, a reverse chemical process breaks down compounds into their elements. For example, water (a compound) can be decomposed into hydrogen and oxygen (both elements) by a process called electrolysis. In electrolysis, electricity provides the energy needed to separate a compound into its constituent elements (Figure $5$). A similar technique is used on a vast scale to obtain pure aluminum, an element, from its ores, which are mixtures of compounds. Because a great deal of energy is required for electrolysis, the cost of electricity is by far the greatest expense incurred in manufacturing pure aluminum. Thus recycling aluminum is both cost-effective and ecologically sound.
The overall organization of matter and the methods used to separate mixtures are summarized in Figure $6$.
Example $1$
Identify each substance as a compound, an element, a heterogeneous mixture, or a homogeneous mixture (solution).
1. filtered tea
2. freshly squeezed orange juice
3. a compact disc
4. aluminum oxide, a white powder that contains a 2:3 ratio of aluminum and oxygen atoms
5. selenium
Given: a chemical substance
Asked for: its classification
Strategy:
1. Decide whether a substance is chemically pure. If it is pure, the substance is either an element or a compound. If a substance can be separated into its elements, it is a compound.
2. If a substance is not chemically pure, it is either a heterogeneous mixture or a homogeneous mixture. If its composition is uniform throughout, it is a homogeneous mixture.
Solution
1. A Tea is a solution of compounds in water, so it is not chemically pure. It is usually separated from tea leaves by filtration. B Because the composition of the solution is uniform throughout, it is a homogeneous mixture.
2. A Orange juice contains particles of solid (pulp) as well as liquid; it is not chemically pure. B Because its composition is not uniform throughout, orange juice is a heterogeneous mixture.
3. A A compact disc is a solid material that contains more than one element, with regions of different compositions visible along its edge. Hence a compact disc is not chemically pure. B The regions of different composition indicate that a compact disc is a heterogeneous mixture.
4. A Aluminum oxide is a single, chemically pure compound.
5. A Selenium is one of the known elements.
Exercise $1$
Identify each substance as a compound, an element, a heterogeneous mixture, or a homogeneous mixture (solution).
1. white wine
2. mercury
3. ranch-style salad dressing
4. table sugar (sucrose)
Answer A
solution
Answer B
element
Answer C
heterogeneous mixture
Answer D
compound
Different Definitions of Changes: Different Definitions of Changes, YouTube(opens in new window) [youtu.be] (opens in new window)
Summary
Matter can be classified according to physical and chemical properties. Matter is anything that occupies space and has mass. The three states of matter are solid, liquid, and gas. A physical change involves the conversion of a substance from one state of matter to another, without changing its chemical composition. Most matter consists of mixtures of pure substances, which can be homogeneous (uniform in composition) or heterogeneous (different regions possess different compositions and properties). Pure substances can be either chemical compounds or elements. Compounds can be broken down into elements by chemical reactions, but elements cannot be separated into simpler substances by chemical means. The properties of substances can be classified as either physical or chemical. Scientists can observe physical properties without changing the composition of the substance, whereas chemical properties describe the tendency of a substance to undergo chemical changes (chemical reactions) that change its chemical composition. Physical properties can be intensive or extensive. Intensive properties are the same for all samples; do not depend on sample size; and include, for example, color, physical state, and melting and boiling points. Extensive properties depend on the amount of material and include mass and volume. The ratio of two extensive properties, mass and volume, is an important intensive property called density.
Contributors and Attributions
Learning Objectives
• To separate physical from chemical properties and changes
All matter has physical and chemical properties. Physical properties are characteristics that scientists can measure without changing the composition of the sample under study, such as mass, color, and volume (the amount of space occupied by a sample). Chemical properties describe the characteristic ability of a substance to react to form new substances; they include its flammability and susceptibility to corrosion. All samples of a pure substance have the same chemical and physical properties. For example, pure copper is always a reddish-brown solid (a physical property) and always dissolves in dilute nitric acid to produce a blue solution and a brown gas (a chemical property).
Physical properties can be extensive or intensive. Extensive properties vary with the amount of the substance and include mass, weight, and volume. Intensive properties, in contrast, do not depend on the amount of the substance; they include color, melting point, boiling point, electrical conductivity, and physical state at a given temperature. For example, elemental sulfur is a yellow crystalline solid that does not conduct electricity and has a melting point of 115.2 °C, no matter what amount is examined (Figure $1$). Scientists commonly measure intensive properties to determine a substance’s identity, whereas extensive properties convey information about the amount of the substance in a sample.
Although mass and volume are both extensive properties, their ratio is an important intensive property called density ($\rho$). Density is defined as mass per unit volume and is usually expressed in grams per cubic centimeter (g/cm3). As mass increases in a given volume, density also increases. For example, lead, with its greater mass, has a far greater density than the same volume of air, just as a brick has a greater density than the same volume of Styrofoam. At a given temperature and pressure, the density of a pure substance is a constant:
\begin{align*} \text{density} &={\text{mass} \over \text{volume}} \[4pt] \rho &={m \over V} \label{Eq1} \end{align*}
Pure water, for example, has a density of 0.998 g/cm3 at 25 °C. The average densities of some common substances are in Table $1$. Notice that corn oil has a lower mass to volume ratio than water. This means that when added to water, corn oil will “float” (Figure $2$).
Table $1$: Densities of Common Substances
Substance Density at 25 °C (g/cm3) Substance Density at 25 °C (g/cm3)
blood 1.035 corn oil 0.922
body fat 0.918 mayonnaise 0.910
whole milk 1.030 honey 1.420
Physical Property and Change
Physical changes are changes in which no chemical bonds are broken or formed. This means that the same types of compounds or elements that were there at the beginning of the change are there at the end of the change. Because the ending materials are the same as the beginning materials, the properties (such as color, boiling point, etc) will also be the same. Physical changes involve moving molecules around, but not changing them. Some types of physical changes include:
• Changes of state (changes from a solid to a liquid or a gas and vice versa)
• Separation of a mixture
• Physical deformation (cutting, denting, stretching)
• Making solutions (special kinds of mixtures) .
As an ice cube melts, its shape changes as it acquires the ability to flow. However, its composition does not change. Melting is an example of a physical change (Figure $3$), since some properties of the material change, but the identity of the matter does not. Physical changes can further be classified as reversible or irreversible. The melted ice cube may be refrozen, so melting is a reversible physical change. Physical changes that involve a change of state are all reversible. Other changes of state include vaporization (liquid to gas), freezing (liquid to solid), and condensation (gas to liquid). Dissolving is also a reversible physical change. When salt is dissolved into water, the salt is said to have entered the aqueous state. The salt may be regained by boiling off the water, leaving the salt behind.
Figure $3$: Ice Melting is a physical change. When solid water ($\ce{H_2O}$) as ice melts into a liquid (water), it appears changed. However, this change is only physical as the the composition of the constituent molecules is the same: 11.19% hydrogen and 88.81% oxygen by mass.
Chemical Properties and Change
Chemical changes occur when bonds are broken and/or formed between molecules or atoms. This means that one substance with a certain set of properties (such as melting point, color, taste, etc) is turned into a different substance with different properties. Chemical changes are frequently harder to reverse than physical changes.
One good example of a chemical change is burning paper. In contrast to the act of ripping paper, the act of burning paper actually results in the formation of new chemicals (carbon dioxide and water, to be exact). Another example of chemical change occurs when water is formed. Each molecule contains two atoms of hydrogen and one atom of oxygen chemically bonded.
Another example of a chemical change is what occurs when natural gas is burned in your furnace. This time, before the reaction we have a molecule of methane, $\ce{CH_4}$, and two molecules of oxygen, $\ce{O_2}$, while after the reaction we have two molecules of water, $\ce{H_2O}$, and one molecule of carbon dioxide, $\ce{CO_2}$. In this case, not only has the appearance changed, but the structure of the molecules has also changed. The new substances do not have the same chemical properties as the original ones. Therefore, this is a chemical change.
The combustion of magnesium metal is also chemical change (Magnesium + Oxygen → Magnesium Oxide):
$\ce{2 Mg + O_2 \rightarrow 2 MgO } \nonumber$
as is the rusting of iron (Iron + Oxygen → Iron Oxide/ Rust):
$\ce{4 Fe + 3O_2 \rightarrow 2 Fe_2O_3} \nonumber$
Using the components of composition and properties, we have the ability to distinguish one sample of matter from the others.
Different Definitions of Changes: Different Definitions of Changes, YouTube(opens in new window) [youtu.be]
Different Definitions of Properties: Different Definitions of Properties, YouTube(opens in new window) [youtu.be]
Contributors and Attributions
• Samantha Ma (UC Davis) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/01%3A_Keys_to_the_Study_of_Chemistry/1.01%3A_Some_Fundamental_Definitions.txt |
Learning Objectives
• To identify the components of the scientific method
Scientists search for answers to questions and solutions to problems by using a procedure called the scientific method. This procedure consists of making observations, formulating hypotheses, and designing experiments, which in turn lead to additional observations, hypotheses, and experiments in repeated cycles (Figure \(1\)).
Observations can be qualitative or quantitative. Qualitative observations describe properties or occurrences in ways that do not rely on numbers. Examples of qualitative observations include the following: the outside air temperature is cooler during the winter season, table salt is a crystalline solid, sulfur crystals are yellow, and dissolving a penny in dilute nitric acid forms a blue solution and a brown gas. Quantitative observations are measurements, which by definition consist of both a number and a unit. Examples of quantitative observations include the following: the melting point of crystalline sulfur is 115.21 °C, and 35.9 grams of table salt—whose chemical name is sodium chloride—dissolve in 100 grams of water at 20 °C. An example of a quantitative observation was the initial observation leading to the modern theory of the dinosaurs’ extinction: iridium concentrations in sediments dating to 66 million years ago were found to be 20–160 times higher than normal. The development of this theory is a good exemplar of the scientific method in action (see Figure \(2\) below).
After deciding to learn more about an observation or a set of observations, scientists generally begin an investigation by forming a hypothesis, a tentative explanation for the observation(s). The hypothesis may not be correct, but it puts the scientist’s understanding of the system being studied into a form that can be tested. For example, the observation that we experience alternating periods of light and darkness corresponding to observed movements of the sun, moon, clouds, and shadows is consistent with either of two hypotheses:
1. Earth rotates on its axis every 24 hours, alternately exposing one side to the sun, or
2. The sun revolves around Earth every 24 hours.
Suitable experiments can be designed to choose between these two alternatives. For the disappearance of the dinosaurs, the hypothesis was that the impact of a large extraterrestrial object caused their extinction. Unfortunately (or perhaps fortunately), this hypothesis does not lend itself to direct testing by any obvious experiment, but scientists collected additional data that either support or refute it.
After a hypothesis has been formed, scientists conduct experiments to test its validity. Experiments are systematic observations or measurements, preferably made under controlled conditions—that is, under conditions in which a single variable changes. For example, in the dinosaur extinction scenario, iridium concentrations were measured worldwide and compared. A properly designed and executed experiment enables a scientist to determine whether the original hypothesis is valid. Experiments often demonstrate that the hypothesis is incorrect or that it must be modified. More experimental data are then collected and analyzed, at which point a scientist may begin to think that the results are sufficiently reproducible (i.e., dependable) to merit being summarized in a law, a verbal or mathematical description of a phenomenon that allows for general predictions. A law simply says what happens; it does not address the question of why.
One example of a law, the Law of Definite Proportions, which was discovered by the French scientist Joseph Proust (1754–1826), states that a chemical substance always contains the same proportions of elements by mass. Thus sodium chloride (table salt) always contains the same proportion by mass of sodium to chlorine, in this case 39.34% sodium and 60.66% chlorine by mass, and sucrose (table sugar) is always 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. Some solid compounds do not strictly obey the law of definite proportions. The law of definite proportions should seem obvious—we would expect the composition of sodium chloride to be consistent—but the head of the US Patent Office did not accept it as a fact until the early 20th century.
Whereas a law states only what happens, a theory attempts to explain why nature behaves as it does. Laws are unlikely to change greatly over time unless a major experimental error is discovered. In contrast, a theory, by definition, is incomplete and imperfect, evolving with time to explain new facts as they are discovered. The theory developed to explain the extinction of the dinosaurs, for example, is that Earth occasionally encounters small- to medium-sized asteroids, and these encounters may have unfortunate implications for the continued existence of most species. This theory is by no means proven, but it is consistent with the bulk of evidence amassed to date. Figure \(2\) summarizes the application of the scientific method in this case.
Example \(1\)
Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation.
1. Ice always floats on liquid water.
2. Birds evolved from dinosaurs.
3. Hot air is less dense than cold air, probably because the components of hot air are moving more rapidly.
4. When 10 g of ice were added to 100 mL of water at 25 °C, the temperature of the water decreased to 15.5 °C after the ice melted.
5. The ingredients of Ivory soap were analyzed to see whether it really is 99.44% pure, as advertised.
Given: components of the scientific method
Asked for: statement classification
Strategy: Refer to the definitions in this section to determine which category best describes each statement.
Solution
1. This is a general statement of a relationship between the properties of liquid and solid water, so it is a law.
2. This is a possible explanation for the origin of birds, so it is a hypothesis.
3. This is a statement that tries to explain the relationship between the temperature and the density of air based on fundamental principles, so it is a theory.
4. The temperature is measured before and after a change is made in a system, so these are quantitative observations.
5. This is an analysis designed to test a hypothesis (in this case, the manufacturer’s claim of purity), so it is an experiment.
Exercise \(1\)
Classify each statement as a law, a theory, an experiment, a hypothesis, a qualitative observation, or a quantitative observation.
1. Measured amounts of acid were added to a Rolaids tablet to see whether it really “consumes 47 times its weight in excess stomach acid.”
2. Heat always flows from hot objects to cooler ones, not in the opposite direction.
3. The universe was formed by a massive explosion that propelled matter into a vacuum.
4. Michael Jordan is the greatest pure shooter ever to play professional basketball.
5. Limestone is relatively insoluble in water but dissolves readily in dilute acid with the evolution of a gas.
6. Gas mixtures that contain more than 4% hydrogen in air are potentially explosive.
Answer a
experiment
Answer b
law
Answer c
theory
Answer d
hypothesis
Answer e
qualitative observation
Answer f
quantitative observation
Because scientists can enter the cycle shown in Figure \(1\) at any point, the actual application of the scientific method to different topics can take many different forms. For example, a scientist may start with a hypothesis formed by reading about work done by others in the field, rather than by making direct observations.
It is important to remember that scientists have a tendency to formulate hypotheses in familiar terms simply because it is difficult to propose something that has never been encountered or imagined before. As a result, scientists sometimes discount or overlook unexpected findings that disagree with the basic assumptions behind the hypothesis or theory being tested. Fortunately, truly important findings are immediately subject to independent verification by scientists in other laboratories, so science is a self-correcting discipline. When the Alvarezes originally suggested that an extraterrestrial impact caused the extinction of the dinosaurs, the response was almost universal skepticism and scorn. In only 20 years, however, the persuasive nature of the evidence overcame the skepticism of many scientists, and their initial hypothesis has now evolved into a theory that has revolutionized paleontology and geology.
Summary
Chemists expand their knowledge by making observations, carrying out experiments, and testing hypotheses to develop laws to summarize their results and theories to explain them. In doing so, they are using the scientific method.
Fundamental Definitions in Chemistry: https://youtu.be/SBwjbkFNkdw | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/01%3A_Keys_to_the_Study_of_Chemistry/1.03%3A_The_Scientific_Approach-_Developing_a_Model.txt |
Learning Objectives
• To identify the basic units of measurement of the seven fundamental properties
• Describe the names and abbreviations of the SI base units and the SI decimal prefixes.
• Define the liter and the metric ton in these units.
• Explain the meaning and use of unit dimensions; state the dimensions of volume.
• State the quantities that are needed to define a temperature scale, and show how these apply to the Celsius, Kelvin, and Fahrenheit temperature scales.
• Explain how a Torricellian barometer works.
Have you ever estimated a distance by “stepping it off”— that is, by counting the number of steps required to take you a certain distance? Or perhaps you have used the width of your hand, or the distance from your elbow to a fingertip to compare two dimensions. If so, you have engaged in what is probably the first kind of measurement ever undertaken by primitive mankind. The results of a measurement are always expressed on some kind of a scale that is defined in terms of a particular kind of unit. The first scales of distance were likely related to the human body, either directly (the length of a limb) or indirectly (the distance a man could walk in a day).
As civilization developed, a wide variety of measuring scales came into existence, many for the same quantity (such as length), but adapted to particular activities or trades. Eventually, it became apparent that in order for trade and commerce to be possible, these scales had to be defined in terms of standards that would allow measures to be verified, and, when expressed in different units (bushels and pecks, for example), to be correlated or converted.
History of Units
Over the centuries, hundreds of measurement units and scales have developed in the many civilizations that achieved some literate means of recording them. Some, such as those used by the Aztecs, fell out of use and were largely forgotten as these civilizations died out. Other units, such as the various systems of measurement that developed in England, achieved prominence through extension of the Empire and widespread trade; many of these were confined to specific trades or industries. The examples shown here are only some of those that have been used to measure length or distance. The history of measuring units provides a fascinating reflection on the history of industrial development.
The most influential event in the history of measurement was undoubtedly the French Revolution and the Age of Rationality that followed. This led directly to the metric system that attempted to do away with the confusing multiplicity of measurement scales by reducing them to a few fundamental ones that could be combined in order to express any kind of quantity. The metric system spread rapidly over much of the world, and eventually even to England and the rest of the U.K. when that country established closer economic ties with Europe in the latter part of the 20th Century. The United States is presently the only major country in which “metrication” has made little progress within its own society, probably because of its relative geographical isolation and its vibrant internal economy.
Science, being a truly international endeavor, adopted metric measurement very early on; engineering and related technologies have been slower to make this change, but are gradually doing so. Even the within the metric system, however, a variety of units were employed to measure the same fundamental quantity; for example, energy could be expressed within the metric system in units of ergs, electron-volts, joules, and two kinds of calories. This led, in the mid-1960s, to the adoption of a more basic set of units, the Systeme Internationale (SI) units that are now recognized as the standard for science and, increasingly, for technology of all kinds.
The Seven SI Base Units and Decimal Prefixes
In principle, any physical quantity can be expressed in terms of only seven base units (Table $1$), with each base unit defined by a standard described in the NIST Web site.
Table $1$: The Seven Base Units
Property Unit Symbol
length meter m
mass kilogram kg
time second s
temperature (absolute) kelvin K
amount of substance mole mol
electric current ampere A
luminous intensity candela cd
A few special points about some of these units are worth noting:
• The base unit of mass is unique in that a decimal prefix (Table $2$) is built into it; i.e., the base SI unit is not the gram.
• The base unit of time is the only one that is not metric. Numerous attempts to make it so have never garnered any success; we are still stuck with the 24:60:60 system that we inherited from ancient times. The ancient Egyptians of around 1500 BC invented the 12-hour day, and the 60:60 part is a remnant of the base-60 system that the Sumerians used for their astronomical calculations around 100 BC.
• Of special interest to Chemistry is the mole, the base unit for expressing the quantity of matter. Although the number is not explicitly mentioned in the official definition, chemists define the mole as Avogadro’s number (approximately 6.02x1023) of anything.
Owing to the wide range of values that quantities can have, it has long been the practice to employ prefixes such as milli and mega to indicate decimal fractions and multiples of metric units. As part of the SI standard, this system has been extended and formalized (Table $2$).
Table $2$: Prefixes used to scale up or down base units
Prefix Abbreviation Multiplier Prefix Abbreviation Multiplier
peta P 1015 deci d 10–1
tera T 1012 centi c 10–2
giga G 109 milli m 10–3
mega M 106 micro μ 10–6
kilo k 103 nano n 10–9
hecto h 102 pico p 10–12
deca da 10 femto f 10–15
Pseudo-Si Units
There is a category of units that are “honorary” members of the SI in the sense that it is acceptable to use them along with the base units defined above. These include such mundane units as the hour, minute, and degree (of angle), etc., but the three shown here are of particular interest to chemistry, and you will need to know them.
Pseudo-Si Units
liter (litre) L 1 L = 1 dm3 = 10–3 m3
metric ton t 1 t = 103 kg
united atomic mass unit (amu) u 1 u = 1.66054×10–27 kg
Derived Units and Dimensions
Most of the physical quantities we actually deal with in science and also in our daily lives, have units of their own: volume, pressure, energy and electrical resistance are only a few of hundreds of possible examples. It is important to understand, however, that all of these can be expressed in terms of the SI base units; they are consequently known as derived units. In fact, most physical quantities can be expressed in terms of one or more of the following five fundamental units:
• mass (M)
• length (L)
• time (T)
• electric charge (Q)
• temperature (Θ theta)
Consider, for example, the unit of volume, which we denote as V. To measure the volume of a rectangular box, we need to multiply the lengths as measured along the three coordinates:
$V = x · y · z \nonumber$
We say, therefore, that volume has the dimensions of length-cubed:
$dim\{V\} = L^3 \nonumber$
Thus the units of volume will be m3 (in the SI) or cm3, ft3 (English), etc. Moreover, any formula that calculates a volume must contain within it the L3 dimension; thus the volume of a sphere is $4/3 πr^3$. The dimensions of a unit are the powers which M, L, t, Q and Q must be given in order to express the unit. Thus,
$dim\{V\} = M^0L^3T^0Q^0 Θ^0 \nonumber$
as given above.
There are several reasons why it is worthwhile to consider the dimensions of a unit.
1. Perhaps the most important use of dimensions is to help us understand the relations between various units of measure and thereby get a better understanding of their physical meaning. For example, a look at the dimensions of the frequently confused electrical terms resistance and resistivity should enable you to explain, in plain words, the difference between them.
2. By the same token, the dimensions essentially tell you how to calculate any of these quantities, using whatever specific units you wish. (Note here the distinction between dimensions and units.)
3. Just as you cannot add apples to oranges, an expression such as $a = b + cx^2$ is meaningless unless the dimensions of each side are identical. (Of course, the two sides should work out to the same units as well.)
4. Many quantities must be dimensionless— for example, the variable x in expressions such as $\log x$, $e^x$, and $\sin x$. Checking through the dimensions of such a quantity can help avoid errors.
The formal, detailed study of dimensions is known as dimensional analysis and is a topic in any basic physics course.
Example $1$
Find the dimensions of energy.
Solution
When mechanical work is performed on a body, its energy increases by the amount of work done, so the two quantities are equivalent and we can concentrate on work. The latter is the product of the force applied to the object and the distance it is displaced. From Newton’s law, force is the product of mass and acceleration, and the latter is the rate of change of velocity, typically expressed in meters per second per second. Combining these quantities and their dimensions yields the result shown in Table $1$.
Table $3$: Dimensions of units commonly used in Chemistry
Q
M
L
t quantity SI unit, other typical units
1 - - - electric charge coulomb
- 1 - - mass kilogram, gram, metric ton, pound
- - 1 - length meter, foot, mile
- - - 1 time second, day, year
- - 3 - volume liter, cm3, quart, fluidounce
- 1 –3 - density kg m–3, g cm–3
- 1 1 –2 force newton, dyne
- 1 –1 –2 pressure pascal, atmosphere, torr
- 1 2 –2 energy joule, erg, calorie, electron-volt
- 1 2 –3 power watt
1 1 2 –2 electric potential volt
1 - - –1 electric current ampere
1 1 1 –2 electric field intensity volt m–1
–2 1 2 –1 electric resistance ohm
2 1 3 –1 electric resistivity -
2 –1 –2 1 electric conductance siemens, mho
Dimensional analysis is widely employed when it is necessary to convert one kind of unit into another, and chemistry students often use it in "chemical arithmetic" calculations, in which context it is also known as the "Factor-Label" method. In this section, we will look at some of the quantities that are widely encountered in Chemistry, and at the units in which they are commonly expressed. In doing so, we will also consider the actual range of values these quantities can assume, both in nature in general, and also within the subset of nature that chemistry normally addresses. In looking over the various units of measure, it is interesting to note that their unit values are set close to those encountered in everyday human experience
Mass is not weight
These two quantities are widely confused. Although they are often used synonymously in informal speech and writing, they have different dimensions: weight is the force exerted on a mass by the local gravational field:
$f = m a = m g \label{Eq1}$
where g is the acceleration of gravity. While the nominal value of the latter quantity is 9.80 m s–2 at the Earth’s surface, its exact value varies locally. Because it is a force, the SI unit of weight is properly the newton, but it is common practice (except in physics classes!) to use the terms "weight" and "mass" interchangeably, so the units kilograms and grams are acceptable in almost all ordinary laboratory contexts.
Please note that in this diagram and in those that follow, the numeric scale represents the logarithm of the number shown. For example, the mass of the electron is 10–30 kg.
The range of masses spans 90 orders of magnitude, more than any other unit. The range that chemistry ordinarily deals with has greatly expanded since the days when a microgram was an almost inconceivably small amount of material to handle in the laboratory; this lower limit has now fallen to the atomic level with the development of tools for directly manipulating these particles. The upper level reflects the largest masses that are handled in industrial operations, but in the recently developed fields of geochemistry and enivonmental chemistry, the range can be extended indefinitely. Flows of elements between the various regions of the environment (atmosphere to oceans, for example) are often quoted in teragrams.
Length
Chemists tend to work mostly in the moderately-small part of the distance range. Those who live in the lilliputian world of crystal- and molecular structures and atomic radii find the picometer a convenient currency, but one still sees the older non-SI unit called the Ångstrom used in this context; 1Å = 10–10 m = 100pm. Nanotechnology, the rage of the present era, also resides in this realm. The largest polymeric molecules and colloids define the top end of the particulate range; beyond that, in the normal world of doing things in the lab, the centimeter and occasionally the millimeter commonly rule.
Time
For humans, time moves by the heartbeat; beyond that, it is the motions of our planet that count out the hours, days, and years that eventually define our lifetimes. Beyond the few thousands of years of history behind us, those years-to-the-powers-of-tens that are the fare for such fields as evolutionary biology, geology, and cosmology, cease to convey any real meaning for us. Perhaps this is why so many people are not very inclined to accept their validity.
Most of what actually takes place in the chemist’s test tube operates on a far shorter time scale, although there is no limit to how slow a reaction can be; the upper limits of those we can directly study in the lab are in part determined by how long a graduate student can wait around before moving on to gainful employment. Looking at the microscopic world of atoms and molecules themselves, the time scale again shifts us into an unreal world where numbers tend to lose their meaning. You can gain some appreciation of the duration of a nanosecond by noting that this is about how long it takes a beam of light to travel between your two outstretched hands. In a sense, the material foundations of chemistry itself are defined by time: neither a new element nor a molecule can be recognized as such unless it lasts long enough to have its “picture” taken through measurement of its distinguishing properties.
Temperature
Temperature, the measure of thermal intensity, spans the narrowest range of any of the base units of the chemist’s measurement toolbox. The reason for this is tied into temperature’s meaning as a measure of the intensity of thermal kinetic energy. Chemical change occurs when atoms are jostled into new arrangements, and the weakness of these motions brings most chemistry to a halt as absolute zero is approached. At the upper end of the scale, thermal motions become sufficiently vigorous to shake molecules into atoms, and eventually, as in stars, strip off the electrons, leaving an essentially reaction-less gaseous fluid, or plasma, of bare nuclei (ions) and electrons.
The degree is really an increment of temperature, a fixed fraction of the distance between two defined reference points on a temperature scale.
Although rough means of estimating and comparing temperatures have been around since AD 170, the first mercury thermometer and temperature scale were introduced in Holland in 1714 by Gabriel Daniel Fahrenheit. Fahrenheit established three fixed points on his thermometer. Zero degrees was the temperature of an ice, water, and salt mixture, which was about the coldest temperature that could be reproduced in a laboratory of the time. When he omitted salt from the slurry, he reached his second fixed point when the water-ice combination stabilized at "the thirty-second degree." His third fixed point was "found at the ninety-sixth degree, and the spirit expands to this degree when the thermometer is held in the mouth or under the armpit of a living man in good health."
After Fahrenheit died in 1736, his thermometer was recalibrated using 212 degrees, the temperature at which water boils, as the upper fixed point. Normal human body temperature registered 98.6 rather than 96. In 1743, the Swedish astronomer Anders Celsius devised the aptly-named centigrade scale that places exactly 100 degrees between the two reference points defined by the freezing and boiling points of water.
When we say that the temperature is so many degrees, we must specify the particular scale on which we are expressing that temperature. A temperature scale has two defining characteristics, both of which can be chosen arbitrarily:
• The temperature that corresponds to 0° on the scale;
• The magnitude of the unit increment of temperature– that is, the size of the degree.
To express a temperature given on one scale in terms of another, it is necessary to take both of these factors into account. The key to temperature conversions is easy if you bear in mind that between the so-called ice- and steam-points of water there are 180 Fahrenheit degrees, but only 100 Celsius degrees, making the F° 100/180 = 5/9 the magnitude of the C°. Note the distinction between “°C” (a temperature) and “C°” (a temperature increment). Because the ice point is at 32°F, the two scales are offset by this amount. If you remember this, there is no need to memorize a conversion formula; you can work it out whenever you need it.
Near the end of the 19th Century when the physical significance of temperature began to be understood, the need was felt for a temperature scale whose zero really means zero— that is, the complete absence of thermal motion. This gave rise to the absolute temperature scale whose zero point is –273.15 °C, but which retains the same degree magnitude as the Celsius scale. This eventually got renamed after Lord Kelvin (William Thompson); thus the Celsius degree became the kelvin. Thus we can now express an increment such as five C° as “five kelvins”
The "other" Absolute Scale
In 1859 the Scottish engineer and physicist William J. M. Rankine proposed an absolute temperature scale based on the Fahrenheit degree. Absolute zero (0° Ra) corresponds to –459.67°F. The Rankine scale has been used extensively by those same American and English engineers who delight in expressing heat capacities in units of BTUs per pound per F°.
The importance of absolute temperature scales is that absolute temperatures can be entered directly in all the fundamental formulas of physics and chemistry in which temperature is a variable.
Units of Temperature: Units of Temperature, YouTube(opens in new window) [youtu.be] (opens in new window)
Pressure
Pressure is the measure of the force exerted on a unit area of surface. Its SI units are therefore newtons per square meter, but we make such frequent use of pressure that a derived SI unit, the pascal, is commonly used:
$1\; Pa = 1\; N \;m^{–2} \nonumber$
The concept of pressure first developed in connection with studies relating to the atmosphere and vacuum that were carried out in the 17th century.
Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car below. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant.
The molecules of a gas are in a state of constant thermal motion, moving in straight lines until experiencing a collision that exchanges momentum between pairs of molecules and sends them bouncing off in other directions. This leads to a completely random distribution of the molecular velocities both in speed and direction— or it would in the absence of the Earth’s gravitational field which exerts a tiny downward force on each molecule, giving motions in that direction a very slight advantage. In an ordinary container this effect is too small to be noticeable, but in a very tall column of air the effect adds up: the molecules in each vertical layer experience more downward-directed hits from those above it. The resulting force is quickly randomized, resulting in an increased pressure in that layer which is then propagated downward into the layers below.
At sea level, the total mass of the sea of air pressing down on each 1-cm2 of surface is about 1034 g, or 10340 kg m–2. The force (weight) that the Earth’s gravitional acceleration g exerts on this mass is
$f = ma = mg = (10340 \;kg)(9.81\; m\; s^{–2}) = 1.013 \times 10^5 \;kg \;m \;s^{–2} = 1.013 \times 10^5\; N \nonumber$
resulting in a pressure of 1.013 × 105 n m–2 = 1.013 × 105 Pa. The actual pressure at sea level varies with atmospheric conditions, so it is customary to define standard atmospheric pressure as 1 atm = 1.01325 x 105 Pa or 101.325 kPa. Although the standard atmosphere is not an SI unit, it is still widely employed. In meteorology, the bar, exactly 1.000 × 105 = 0.967 atm, is often used.
The Barometer
In the early 17th century, the Italian physicist and mathematician Evangalisto Torricelli invented a device to measure atmospheric pressure. The Torricellian barometer consists of a vertical glass tube closed at the top and open at the bottom. It is filled with a liquid, traditionally mercury, and is then inverted, with its open end immersed in the container of the same liquid. The liquid level in the tube will fall under its own weight until the downward force is balanced by the vertical force transmitted hydrostatically to the column by the downward force of the atmosphere acting on the liquid surface in the open container. Torricelli was also the first to recognize that the space above the mercury constituted a vacuum, and is credited with being the first to create a vacuum.
One standard atmosphere will support a column of mercury that is 760 mm high, so the “millimeter of mercury”, now more commonly known as the torr, has long been a common pressure unit in the sciences: 1 atm = 760 torr.
International System of Units (SI Units): International System of Units (SI Units), YouTube(opens in new window) [youtu.be]
Summary
The natural sciences begin with observation, and this usually involves numerical measurements of quantities such as length, volume, density, and temperature. Most of these quantities have units of some kind associated with them, and these units must be retained when you use them in calculations. Measuring units can be defined in terms of a very small number of fundamental ones that, through "dimensional analysis", provide insight into their derivation and meaning, and must be understood when converting between different unit systems.
Contributions
Learning Objectives
• To be introduced to the dimensional analysis and how it can be used to aid basic chemistry problem solving.
• To use dimensional analysis to identify whether an equation is set up correctly in a numerical calculation
• To use dimensional analysis to facilitate the conversion of units.
Dimensional analysis is amongst the most valuable tools physical scientists use. Simply put, it is the conversion between an amount in one unit to the corresponding amount in a desired unit using various conversion factors. This is valuable because certain measurements are more accurate or easier to find than others.
A Macroscopic Example: Party Planning
If you have every planned a party, you have used dimensional analysis. The amount of beer and munchies you will need depends on the number of people you expect. For example, if you are planning a Friday night party and expect 30 people you might estimate you need to go out and buy 120 bottles of sodas and 10 large pizza's. How did you arrive at these numbers? The following indicates the type of dimensional analysis solution to party problem:
$(30 \; \cancel{humans}) \times \left( \dfrac{\text{4 sodas}}{1 \; \cancel{human}} \right) = 120 \; \text{sodas} \label{Eq1}$
$(30 \; \cancel{humans}) \times \left( \dfrac{\text{0.333 pizzas}}{1 \; \cancel{human}} \right) = 10 \; \text{pizzas} \label{Eq2}$
Notice that the units that canceled out are lined out and only the desired units are left (discussed more below). Finally, in going to buy the soda, you perform another dimensional analysis: should you buy the sodas in six-packs or in cases?
$(120\; { sodas}) \times \left( \dfrac{\text{1 six pack}}{6\; {sodas}} \right) = 20 \; \text{six packs} \label{Eq3}$
$(120\; {sodas}) \times \left( \dfrac{\text{1 case }}{24\; {sodas}} \right) = 5 \; \text{cases} \label{Eq4}$
Realizing that carrying around 20 six packs is a real headache, you get 5 cases of soda instead.
In this party problem, we have used dimensional analysis in two different ways:
• In the first application (Equations $\ref{Eq1}$ and Equation $\ref{Eq2}$), dimensional analysis was used to calculate how much soda is needed need. This is based on knowing: (1) how much soda we need for one person and (2) how many people we expect; likewise for the pizza.
• In the second application (Equations $\ref{Eq3}$ and $\ref{Eq4}$), dimensional analysis was used to convert units (i.e. from individual sodas to the equivalent amount of six packs or cases)
Using Dimensional Analysis to Convert Units
Consider the conversion in Equation $\ref{Eq3}$:
$(120\; {sodas}) \times \left( \dfrac{\text{1 six pack}}{6\; {sodas}} \right) = 20 \; \text{six packs} \label{Eq3a}$
If we ignore the numbers for a moment, and just look at the units (i.e. dimensions), we have:
$\text{soda} \times \left(\dfrac{\text{six pack}}{\text{sodas}}\right) \nonumber$
We can treat the dimensions in a similar fashion as other numerical analyses (i.e. any number divided by itself is 1). Therefore:
$\text{soda} \times \left(\dfrac{\text{six pack}}{\text{sodas}}\right) = \cancel{\text{soda}} \times \left(\dfrac{\text{six pack}}{\cancel{\text{sodas}}}\right) \nonumber$
So, the dimensions of the numerical answer will be "six packs".
How can we use dimensional analysis to be sure we have set up our equation correctly? Consider the following alternative way to set up the above unit conversion analysis:
$120 \cancel{\text{soda}} \times \left(\dfrac{\text{6 sodas}}{\cancel{\text{six pack}}}\right) = 720 \; \dfrac{\text{sodas}^2}{\text{1 six pack}} \nonumber$
• While it is correct that there are 6 sodas in one six pack, the above equation yields a value of 720 with units of sodas2/six pack.
• These rather bizarre units indicate that the equation has been setup incorrectly (and as a consequence you will have a ton of extra soda at the party).
Using Dimensional Analysis in Calculations
In the above case it was relatively straightforward keeping track of units during the calculation. What if the calculation involves powers, etc? For example, the equation relating kinetic energy to mass and velocity is:
$E_{kinetics} = \dfrac{1}{2} \text{mass} \times \text{velocity}^2 \label{KE}$
An example of units of mass is kilograms (kg) and velocity might be in meters/second (m/s). What are the dimensions of $E_{kinetic}$?
$(kg) \times \left( \dfrac{m}{s} \right)^2 = \dfrac{kg \; m^2}{s^2} \nonumber$
The $\frac{1}{2}$ factor in Equation \ref{KE} is neglected since pure numbers have no units. Since the velocity is squared in Equation \ref{KE}, the dimensions associated with the numerical value of the velocity are also squared. We can double check this by knowing the the Joule ($J$) is a measure of energy, and as a composite unit can be decomposed thusly:
$1\; J = kg \dfrac{m^2}{s^2} \nonumber$
Units of Pressure
Pressure (P) is a measure of the Force (F) per unit area (A):
$P =\dfrac{F}{A} \nonumber$
Force, in turn, is a measure of the acceleration ($a$) on a mass ($m$):
$F= m \times a \nonumber$
Thus, pressure ($P$) can be written as:
$P= \dfrac{m \times a}{A} \nonumber$
What are the units of pressure from this relationship? (Note: acceleration is the change in velocity per unit time)
$P =\dfrac{kg \times \frac{\cancel{m}}{s^2}}{m^{\cancel{2}}} \nonumber$
We can simplify this description of the units of Pressure by dividing numerator and denominator by $m$:
$P =\dfrac{\frac{kg}{s^2}}{m}=\dfrac{kg}{m\; s^2} \nonumber$
In fact, these are the units of a the composite Pascal (Pa) unit and is the SI measure of pressure.
Performing Dimensional Analysis
The use of units in a calculation to ensure that we obtain the final proper units is called dimensional analysis. For example, if we observe experimentally that an object’s potential energy is related to its mass, its height from the ground, and to a gravitational force, then when multiplied, the units of mass, height, and the force of gravity must give us units corresponding to those of energy.
Energy is typically measured in joules, calories, or electron volts (eV), defined by the following expressions:
• 1 J = 1 (kg·m2)/s2 = 1 coulomb·volt
• 1 cal = 4.184 J
• 1 eV = 1.602 × 10−19 J
Performing dimensional analysis begins with finding the appropriate conversion factors. Then, you simply multiply the values together such that the units cancel by having equal units in the numerator and the denominator. To understand this process, let us walk through a few examples.
Example $1$
Imagine that a chemist wants to measure out 0.214 mL of benzene, but lacks the equipment to accurately measure such a small volume. The chemist, however, is equipped with an analytical balance capable of measuring to $\pm 0.0001 \;g$. Looking in a reference table, the chemist learns the density of benzene ($\rho=0.8765 \;g/mL$). How many grams of benzene should the chemist use?
Solution
$0.214 \; \cancel{mL} \left( \dfrac{0.8765\; g}{1\;\cancel{mL}}\right)= 0.187571\; g \nonumber$
Notice that the mL are being divided by mL, an equivalent unit. We can cancel these our, which results with the 0.187571 g. However, this is not our final answer, since this result has too many significant figures and must be rounded down to three significant digits. This is because 0.214 mL has three significant digits and the conversion factor had four significant digits. Since 5 is greater than or equal to 5, we must round the preceding 7 up to 8.
Hence, the chemist should weigh out 0.188 g of benzene to have 0.214 mL of benzene.
Example $2$
To illustrate the use of dimensional analysis to solve energy problems, let us calculate the kinetic energy in joules of a 320 g object traveling at 123 cm/s.
Solution
To obtain an answer in joules, we must convert grams to kilograms and centimeters to meters. Using Equation \ref{KE}, the calculation may be set up as follows:
\begin{align*} KE&=\dfrac{1}{2}mv^2=\dfrac{1}{2}(g) \left(\dfrac{kg}{g}\right) \left[\left(\dfrac{cm}{s}\right)\left(\dfrac{m}{cm}\right) \right]^2 \[4pt] &= (\cancel{g})\left(\dfrac{kg}{\cancel{g}}\right) \left(\dfrac{\cancel{m^2}}{s^2}\right) \left(\dfrac{m^2}{\cancel{cm^2}}\right) = \dfrac{kg⋅m^2}{s^2} \[4pt] &=\dfrac{1}{2}320\; \cancel{g} \left( \dfrac{1\; kg}{1000\;\cancel{g}}\right) \left[\left(\dfrac{123\;\cancel{cm}}{1 \;s}\right) \left(\dfrac{1 \;m}{100\; \cancel{cm}}\right) \right]^2=\dfrac{0.320\; kg}{2}\left[\dfrac{123 m}{s(100)}\right]^2 \[4pt] &=\dfrac{1}{2} 0.320\; kg \left[ \dfrac{(123)^2 m^2}{s^2(100)^2} \right]= 0.242 \dfrac{kg⋅m^2}{s^2} = 0.242\; J \end{align*} \nonumber
Alternatively, the conversions may be carried out in a stepwise manner:
Step 1: convert $g$ to $kg$
$320\; \cancel{g} \left( \dfrac{1\; kg}{1000\;\cancel{g}}\right) = 0.320 \; kg \nonumber$
Step 2: convert $cm$ to $m$
$123\;\cancel{cm} \left(\dfrac{1 \;m}{100\; \cancel{cm}}\right) = 1.23\ m \nonumber$
Now the natural units for calculating joules is used to get final results
\begin{align*} KE &=\dfrac{1}{2} 0.320\; kg \left(1.23 \;ms\right)^2 \[4pt] &=\dfrac{1}{2} 0.320\; kg \left(1.513 \dfrac{m^2}{s^2}\right)= 0.242\; \dfrac{kg⋅m^2}{s^2}= 0.242\; J \end{align*} \nonumber
Of course, steps 1 and 2 can be done in the opposite order with no effect on the final results. However, this second method involves an additional step.
Example $3$
Now suppose you wish to report the number of kilocalories of energy contained in a 7.00 oz piece of chocolate in units of kilojoules per gram.
Solution
To obtain an answer in kilojoules, we must convert 7.00 oz to grams and kilocalories to kilojoules. Food reported to contain a value in Calories actually contains that same value in kilocalories. If the chocolate wrapper lists the caloric content as 120 Calories, the chocolate contains 120 kcal of energy. If we choose to use multiple steps to obtain our answer, we can begin with the conversion of kilocalories to kilojoules:
$120 \cancel{kcal} \left(\dfrac{1000 \;\cancel{cal}}{\cancel{kcal}}\right)\left(\dfrac{4.184 \;\cancel{J}}{1 \cancel{cal}}\right)\left(\dfrac{1 \;kJ}{1000 \cancel{J}}\right)= 502\; kJ \nonumber$
We next convert the 7.00 oz of chocolate to grams:
$7.00\;\cancel{oz} \left(\dfrac{28.35\; g}{1\; \cancel{oz}}\right)= 199\; g \nonumber$
The number of kilojoules per gram is therefore
$\dfrac{ 502 \;kJ}{199\; g}= 2.52\; kJ/g \nonumber$
Alternatively, we could solve the problem in one step with all the conversions included:
$\left(\dfrac{120\; \cancel{kcal}}{7.00\; \cancel{oz}}\right)\left(\dfrac{1000 \;\cancel{cal}}{1 \;\cancel{kcal}}\right)\left(\dfrac{4.184 \;\cancel{J}}{1 \; \cancel{cal}}\right)\left(\dfrac{1 \;kJ}{1000 \;\cancel{J}}\right)\left(\dfrac{1 \;\cancel{oz}}{28.35\; g}\right)= 2.53 \; kJ/g \nonumber$
The discrepancy between the two answers is attributable to rounding to the correct number of significant figures for each step when carrying out the calculation in a stepwise manner. Recall that all digits in the calculator should be carried forward when carrying out a calculation using multiple steps. In this problem, we first converted kilocalories to kilojoules and then converted ounces to grams.
Converting Between Units: Converting Between Units, YouTube(opens in new window) [youtu.be]
Summary
Dimensional analysis is used in numerical calculations, and in converting units. It can help us identify whether an equation is set up correctly (i.e. the resulting units should be as expected). Units are treated similarly to the associated numerical values, i.e., if a variable in an equation is supposed to be squared, then the associated dimensions are squared, etc.
Contributors and Attributions
• Mark Tye (Diablo Valley College) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/01%3A_Keys_to_the_Study_of_Chemistry/1.04%3A_Measurement_and_Chemical_Problem_Solving.txt |
Learning Objectives
• To introduce the fundamental mathematical skills you will need to complete basic chemistry questions and problems
Measurements may be accurate, meaning that the measured value is the same as the true value; they may be precise, meaning that multiple measurements give nearly identical values (i.e., reproducible results); they may be both accurate and precise; or they may be neither accurate nor precise. The goal of scientists is to obtain measured values that are both accurate and precise.
Suppose, for example, that the mass of a sample of gold was measured on one balance and found to be 1.896 g. On a different balance, the same sample was found to have a mass of 1.125 g. Which was correct? Careful and repeated measurements, including measurements on a calibrated third balance, showed the sample to have a mass of 1.895 g. The masses obtained from the three balances are in the following table:
Masses Obtained from the three balances
Balance 1 Balance 2 Balance 3
1.896 g 1.125 g 1.893 g
1.895 g 1.158 g 1.895 g
1.894 g 1.067 g 1.895 g
Whereas the measurements obtained from balances 1 and 3 are reproducible (precise) and are close to the accepted value (accurate), those obtained from balance 2 are neither. Even if the measurements obtained from balance 2 had been precise (if, for example, they had been 1.125, 1.124, and 1.125), they still would not have been accurate. We can assess the precision of a set of measurements by calculating the average deviation of the measurements as follows:
1. Calculate the average value of all the measurements:
$\text{average} = \dfrac{\text{sum of measurements} }{\text{number of measurements}} \label{Eq1}$
2. Calculate the deviation of each measurement, which is the absolute value of the difference between each measurement and the average value:
$\text{deviation} = |\text{measurement − average}| \label{Eq2}$
where $|\, |$ means absolute value (i.e., convert any negative number to a positive number).
3. Add all the deviations and divide by the number of measurements to obtain the average deviation:
$\text{average} = \dfrac{\text{sum of deviations}}{\text{number of measurements}} \label{Eq3}$
Then we can express the precision as a percentage by dividing the average deviation by the average value of the measurements and multiplying the result by 100. In the case of balance 2, the average value is
${1.125 \;g + 1.158 \;g + 1.067\; g \over 3} = 1.117 \;g \nonumber$
The deviations are
• $|1.125\; g − 1.117 \;g| = 0.008\; g$
• $|1.158\; g − 1.117\; g| = 0.041 \:g$, and
• $|1.067\; g − 1.117\; g| = 0.050 \;g$.
So the average deviation is
${0.008 \:g + 0.041 \;g + 0.050 \;g \over 3} = 0.033\; g \nonumber$
The precision of this set of measurements is therefore
${0.033\;g \over 1.117\;g} \times 100 = 3.0 \% \nonumber$
When a series of measurements is precise but not accurate, the error is usually systematic. Systematic errors can be caused by faulty instrumentation or faulty technique.
Example $1$
The following archery targets show marks that represent the results of four sets of measurements. Which target shows
1. a precise but inaccurate set of measurements?
2. an accurate but imprecise set of measurements?
3. a set of measurements that is both precise and accurate?
4. a set of measurements that is neither precise nor accurate?
Example $2$
1. A 1-carat diamond has a mass of 200.0 mg. When a jeweler repeatedly weighed a 2-carat diamond, he obtained measurements of 450.0 mg, 459.0 mg, and 463.0 mg. Were the jeweler’s measurements accurate? Were they precise?
2. A single copper penny was tested three times to determine its composition. The first analysis gave a composition of 93.2% zinc and 2.8% copper, the second gave 92.9% zinc and 3.1% copper, and the third gave 93.5% zinc and 2.5% copper. The actual composition of the penny was 97.6% zinc and 2.4% copper. Were the results accurate? Were they precise?
Solution
a. The expected mass of a 2-carat diamond is 2 × 200.0 mg = 400.0 mg. The average of the three measurements is 457.3 mg, about 13% greater than the true mass. These measurements are not particularly accurate.
The deviations of the measurements are 7.3 mg, 1.7 mg, and 5.7 mg, respectively, which give an average deviation of 4.9 mg and a precision of
${4.9 mg \over 457.3 mg } \times 100 = 1.1 \% \nonumber$
These measurements are rather precise.
b. The average values of the measurements are 93.2% zinc and 2.8% copper versus the true values of 97.6% zinc and 2.4% copper. Thus these measurements are not very accurate, with errors of −4.5% and + 17% for zinc and copper, respectively. (The sum of the measured zinc and copper contents is only 96.0% rather than 100%, which tells us that either there is a significant error in one or both measurements or some other element is present.)
The deviations of the measurements are 0.0%, 0.3%, and 0.3% for both zinc and copper, which give an average deviation of 0.2% for both metals. We might therefore conclude that the measurements are equally precise, but that is not the case. Recall that precision is the average deviation divided by the average value times 100. Because the average value of the zinc measurements is much greater than the average value of the copper measurements (93.2% versus 2.8%), the copper measurements are much less precise.
\begin{align*} \text {precision (Zn)} &= \dfrac {0.2 \%}{93.2 \% } \times 100 = 0.2 \% \[4pt] \text {precision (Cu)} &= \dfrac {0.2 \%}{2.8 \% } \times 100 = 7 \% \end{align*} \nonumber
Significant Figures
No measurement is free from error. Error is introduced by the limitations of instruments and measuring devices (such as the size of the divisions on a graduated cylinder) and the imperfection of human senses (i.e., detection). Although errors in calculations can be enormous, they do not contribute to uncertainty in measurements. Chemists describe the estimated degree of error in a measurement as the uncertainty of the measurement, and they are careful to report all measured values using only significant figures, numbers that describe the value without exaggerating the degree to which it is known to be accurate. Chemists report as significant all numbers known with absolute certainty, plus one more digit that is understood to contain some uncertainty. The uncertainty in the final digit is usually assumed to be ±1, unless otherwise stated.
Significant Figure Rules
The following rules have been developed for counting the number of significant figures in a measurement or calculation:
1. Any nonzero digit is significant.
2. Any zeros between nonzero digits are significant. The number 2005, for example, has four significant figures.
3. Any zeros used as a placeholder preceding the first nonzero digit are not significant. So 0.05 has one significant figure because the zeros are used to indicate the placement of the digit 5. In contrast, 0.050 has two significant figures because the last two digits correspond to the number 50; the last zero is not a placeholder. As an additional example, 5.0 has two significant figures because the zero is used not to place the 5 but to indicate 5.0.
4. When a number does not contain a decimal point, zeros added after a nonzero number may or may not be significant. An example is the number 100, which may be interpreted as having one, two, or three significant figures. (Note: treat all trailing zeros in exercises and problems in this text as significant unless you are specifically told otherwise.)
5. Integers obtained either by counting objects or from definitions are exact numbers, which are considered to have infinitely many significant figures. If we have counted four objects, for example, then the number 4 has an infinite number of significant figures (i.e., it represents 4.000…). Similarly, 1 foot (ft) is defined to contain 12 inches (in), so the number 12 in the following equation has infinitely many significant figures: $1 \, \text{ft} = 1, \text{in} \nonumber$
An effective method for determining the number of significant figures is to convert the measured or calculated value to scientific notation because any zero used as a placeholder is eliminated in the conversion. When 0.0800 is expressed in scientific notation as 8.00 × 10−2, it is more readily apparent that the number has three significant figures rather than five; in scientific notation, the number preceding the exponential (i.e., N) determines the number of significant figures.
Example $3$
Give the number of significant figures in each. Identify the rule for each.
1. 5.87
2. 0.031
3. 52.90
4. 00.2001
5. 500
6. 6 atoms
Solution
1. three (rule 1)
2. two (rule 3); in scientific notation, this number is represented as 3.1 × 10−2, showing that it has two significant figures.
3. four (rule 3)
4. four (rule 2); this number is 2.001 × 10−1 in scientific notation, showing that it has four significant figures.
5. one, two, or three (rule 4)
6. infinite (rule 5)
Example $4$
Which measuring apparatus would you use to deliver 9.7 mL of water as accurately as possible? To how many significant figures can you measure that volume of water with the apparatus you selected?
Answer
Use the 10 mL graduated cylinder, which will be accurate to two significant figures.
Mathematical operations are carried out using all the digits given and then rounding the final result to the correct number of significant figures to obtain a reasonable answer. This method avoids compounding inaccuracies by successively rounding intermediate calculations. After you complete a calculation, you may have to round the last significant figure up or down depending on the value of the digit that follows it. If the digit is 5 or greater, then the number is rounded up. For example, when rounded to three significant figures, 5.215 is 5.22, whereas 5.213 is 5.21. Similarly, to three significant figures, 5.005 kg becomes 5.01 kg, whereas 5.004 kg becomes 5.00 kg. The procedures for dealing with significant figures are different for addition and subtraction versus multiplication and division.
When we add or subtract measured values, the value with the fewest significant figures to the right of the decimal point determines the number of significant figures to the right of the decimal point in the answer. Drawing a vertical line to the right of the column corresponding to the smallest number of significant figures is a simple method of determining the proper number of significant figures for the answer:
$3240.7 + 21.236 = 3261.9|36 \nonumber$
The line indicates that the digits 3 and 6 are not significant in the answer. These digits are not significant because the values for the corresponding places in the other measurement are unknown (3240.7??). Consequently, the answer is expressed as 3261.9, with five significant figures. Again, numbers greater than or equal to 5 are rounded up. If our second number in the calculation had been 21.256, then we would have rounded 3261.956 to 3262.0 to complete our calculation.
When we multiply or divide measured values, the answer is limited to the smallest number of significant figures in the calculation; thus,
$42.9 × 8.323 = 357.057 = 357. \nonumber$
Although the second number in the calculation has four significant figures, we are justified in reporting the answer to only three significant figures because the first number in the calculation has only three significant figures. An exception to this rule occurs when multiplying a number by an integer, as in 12.793 × 12. In this case, the number of significant figures in the answer is determined by the number 12.973, because we are in essence adding 12.973 to itself 12 times. The correct answer is therefore 155.516, an increase of one significant figure, not 155.52.
When you use a calculator, it is important to remember that the number shown in the calculator display often shows more digits than can be reported as significant in your answer. When a measurement reported as 5.0 kg is divided by 3.0 L, for example, the display may show 1.666666667 as the answer. We are justified in reporting the answer to only two significant figures, giving 1.7 kg/L as the answer, with the last digit understood to have some uncertainty.
In calculations involving several steps, slightly different answers can be obtained depending on how rounding is handled, specifically whether rounding is performed on intermediate results or postponed until the last step. Rounding to the correct number of significant figures should always be performed at the end of a series of calculations because rounding of intermediate results can sometimes cause the final answer to be significantly in error.
Example $5$
Complete the calculations and report your answers using the correct number of significant figures.
1. 87.25 mL + 3.0201 mL
2. 26.843 g + 12.23 g
3. 6 × 12.011
4. 2(1.008) g + 15.99 g
5. 137.3 + 2(35.45)
6. ${118.7 \over 2} g - 35.5 g$
7. $47.23 g - {207.2 \over 5.92 }g$
8. ${77.604 \over 6.467} −4.8$
9. ${24.86 \over 2.0 } - 3.26 (0.98 )$
10. $(15.9994 \times 9) + 2.0158$
Solution
1. 90.27 mL
2. 39.07 g
3. 72.066 (See rule 5 under “Significant Figures.”)
4. 2(1.008) g + 15.99 g = 2.016 g + 15.99 g = 18.01 g
5. 137.3 + 2(35.45) = 137.3 + 70.90 = 208.2
6. 59.35 g − 35.5 g = 23.9 g
7. 47.23 g − 35.0 g = 12.2 g
8. 12.00 − 4.8 = 7.2
9. 12 − 3.2 = 9
10. 143.9946 + 2.0158 = 146.0104
In practice, chemists generally work with a calculator and carry all digits forward through subsequent calculations. When working on paper, however, we often want to minimize the number of digits we have to write out. Because successive rounding can compound inaccuracies, intermediate roundings need to be handled correctly. When working on paper, always round an intermediate result so as to retain at least one more digit than can be justified and carry this number into the next step in the calculation. The final answer is then rounded to the correct number of significant figures at the very end.
In the worked examples in this text, we will often show the results of intermediate steps in a calculation. In doing so, we will show the results to only the correct number of significant figures allowed for that step, in effect treating each step as a separate calculation. This procedure is intended to reinforce the rules for determining the number of significant figures, but in some cases it may give a final answer that differs in the last digit from that obtained using a calculator, where all digits are carried through to the last step.
Significant Figures: Significant Figures, YouTube(opens in new window) [youtu.be] | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/01%3A_Keys_to_the_Study_of_Chemistry/1.05%3A_Uncertainty_in_Measurement-_Significant_Figures.txt |
In this chapter, you will learn how to describe the composition of chemical compounds. We introduce chemical nomenclature—the language of chemistry—that will enable you to recognize and name the most common kinds of compounds. An understanding of chemical nomenclature not only is essential for your study of chemistry but also has other benefits—for example, it helps you understand the labels on products found in the supermarket and the pharmacy. You will also be better equipped to understand many of the important environmental and medical issues that face society. By the end of this chapter, you will be able to describe what happens chemically when a doctor prepares a cast to stabilize a broken bone, and you will know the composition of common substances such as laundry bleach, the active ingredient in baking powder, and the foul-smelling compound responsible for the odor of spoiled fish. Finally, you will be able to explain the chemical differences among different grades of gasoline.
• 2.1: The Atomic Theory of Matter
This article explains the theories that Dalton used as a basis for his theory: (1) the Law of Conservation of Mass, (2) the Law of Constant Composition, (3) the Law of Multiple Proportions.
• 2.2: The Discovery of Atomic Structure
Atoms, the smallest particles of an element that exhibit the properties of that element, consist of negatively charged electrons around a central nucleus composed of more massive positively charged protons and electrically neutral neutrons. Radioactivity is the emission of energetic particles and rays (radiation) by some substances. Three important kinds of radiation are α particles (helium nuclei), β particles (electrons traveling at high speed), and γ rays.
• 2.3: The Modern View of Atomic Structure
Each atom of an element contains the same number of protons, which is the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. Each isotope of a given element has the same atomic number but a different mass number (A), which is the sum of the numbers of protons and neutrons. The relative masses of atoms are reported using the atomic mass unit (amu).
• 2.4: Atomic Mass
The mass of an atom is a weighted average that is largely determined by the number of its protons and neutrons, and the number of protons and electrons determines its charge. Each atom of an element contains the same number of protons, known as the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. Each isotope of a given element has the same atomic number, but different mass number.
• 2.5: The Periodic Table
The periodic table is used as a predictive tool that arranges of the elements in order of increasing atomic number. Elements that exhibit similar chemistry appear in vertical columns called groups (numbered 1–18 from left to right); the seven horizontal rows are called periods. The elements can be broadly divided into metals, nonmetals, and semimetals. Semimetals exhibit properties intermediate between those of metals and nonmetals.
• 2.6: Molecules and Molecular Compounds
There are two fundamentally different kinds of chemical bonds (covalent and ionic) that cause substances to have very different properties. The atoms in chemical compounds are held together by attractive electrostatic interactions known as chemical bonds. The molecular formula of a covalent compound gives the types and numbers of atoms present. Diatomic molecules contain two atoms, and polyatomic molecules contain more than two.
• 2.7: Ions and Ionic Compounds
The atoms in chemical compounds are held together by attractive electrostatic interactions known as chemical bonds. Ionic compounds contain positively and negatively charged ions in a ratio that results in an overall charge of zero. The ions are held together in a regular spatial arrangement by electrostatic forces. Atoms or groups of atoms that possess a net electrical charge are called ions; they can have either a positive charge (cations) or a negative charge (anions).
• 2.8: Naming Inorganic Compounds
The composition of a compound is represented by an empirical or molecular formula, each consisting of at least one formula unit. Covalent inorganic compounds are named using a procedure similar to that used for ionic compounds, whereas hydrocarbons use a system based on the number of bonds between carbon atoms. Covalent inorganic compounds are named by a procedure similar to that used for ionic compounds, using prefixes to indicate the numbers of atoms in the molecular formula.
• 2.9: Some Simple Organic Compounds
The simplest organic compounds are the hydrocarbons, which contain only carbon and hydrogen. Alkanes contain only carbon–hydrogen and carbon–carbon single bonds, alkenes contain at least one carbon–carbon double bond, and alkynes contain one or more carbon–carbon triple bonds. Hydrocarbons can also be cyclic, with the ends of the chain connected to form a ring. Collectively, alkanes, alkenes, and alkynes are called aliphatic hydrocarbons.
• 2.E: Atoms, Molecules, and Ions (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al.
• 2.S: Atoms, Molecules, and Ions (Summary)
This is the summary Module for the chapter "Atoms, Molecules, and Ions" in the Brown et al. General Chemistry Textmap.
Thumbnail: Spinning Buckminsterfullerene (\(\ce{C60}\)). (CC BY-SA 3.0; unported; Sponk).
02: The Components of Matter
Learning Objectives
• To become familiar with the components and structure of the atom.
Long before the end of the 19th century, it was well known that applying a high voltage to a gas contained at low pressure in a sealed tube (called a gas discharge tube) caused electricity to flow through the gas, which then emitted light (Figure $1$). Researchers trying to understand this phenomenon found that an unusual form of energy was also emitted from the cathode, or negatively charged electrode; this form of energy was called a cathode ray.
In 1897, the British physicist J. J. Thomson (1856–1940) proved that atoms were not the most basic form of matter. He demonstrated that cathode rays could be deflected, or bent, by magnetic or electric fields, which indicated that cathode rays consist of charged particles (Figure $2$). More important, by measuring the extent of the deflection of the cathode rays in magnetic or electric fields of various strengths, Thomson was able to calculate the mass-to-charge ratio of the particles. These particles were emitted by the negatively charged cathode and repelled by the negative terminal of an electric field. Because like charges repel each other and opposite charges attract, Thomson concluded that the particles had a net negative charge; these particles are now called electrons. Most relevant to the field of chemistry, Thomson found that the mass-to-charge ratio of cathode rays is independent of the nature of the metal electrodes or the gas, which suggested that electrons were fundamental components of all atoms.
Subsequently, the American scientist Robert Millikan (1868–1953) carried out a series of experiments using electrically charged oil droplets, which allowed him to calculate the charge on a single electron. With this information and Thomson’s mass-to-charge ratio, Millikan determined the mass of an electron:
$\dfrac {mass}{charge} \times {charge} ={mass} \nonumber$
It was at this point that two separate lines of investigation began to converge, both aimed at determining how and why matter emits energy. The video below shows how JJ Thompson used such a tube to measure the ratio of charge over mass of an electron
Measuring e/m For an Electron. Video from Davidson College demonstrating Thompson's e/m experiment.
Radioactivity
The second line of investigation began in 1896, when the French physicist Henri Becquerel (1852–1908) discovered that certain minerals, such as uranium salts, emitted a new form of energy. Becquerel’s work was greatly extended by Marie Curie (1867–1934) and her husband, Pierre (1854–1906); all three shared the Nobel Prize in Physics in 1903. Marie Curie coined the term radioactivity (from the Latin radius, meaning “ray”) to describe the emission of energy rays by matter. She found that one particular uranium ore, pitchblende, was substantially more radioactive than most, which suggested that it contained one or more highly radioactive impurities. Starting with several tons of pitchblende, the Curies isolated two new radioactive elements after months of work: polonium, which was named for Marie’s native Poland, and radium, which was named for its intense radioactivity. Pierre Curie carried a vial of radium in his coat pocket to demonstrate its greenish glow, a habit that caused him to become ill from radiation poisoning well before he was run over by a horse-drawn wagon and killed instantly in 1906. Marie Curie, in turn, died of what was almost certainly radiation poisoning.
Building on the Curies’ work, the British physicist Ernest Rutherford (1871–1937) performed decisive experiments that led to the modern view of the structure of the atom. While working in Thomson’s laboratory shortly after Thomson discovered the electron, Rutherford showed that compounds of uranium and other elements emitted at least two distinct types of radiation. One was readily absorbed by matter and seemed to consist of particles that had a positive charge and were massive compared to electrons. Because it was the first kind of radiation to be discovered, Rutherford called these substances α particles. Rutherford also showed that the particles in the second type of radiation, β particles, had the same charge and mass-to-charge ratio as Thomson’s electrons; they are now known to be high-speed electrons. A third type of radiation, γ rays, was discovered somewhat later and found to be similar to the lower-energy form of radiation called x-rays, now used to produce images of bones and teeth.
These three kinds of radiation—α particles, β particles, and γ rays—are readily distinguished by the way they are deflected by an electric field and by the degree to which they penetrate matter. As Figure $3$ illustrates, α particles and β particles are deflected in opposite directions; α particles are deflected to a much lesser extent because of their higher mass-to-charge ratio. In contrast, γ rays have no charge, so they are not deflected by electric or magnetic fields. Figure $5$ shows that α particles have the least penetrating power and are stopped by a sheet of paper, whereas β particles can pass through thin sheets of metal but are absorbed by lead foil or even thick glass. In contrast, γ-rays can readily penetrate matter; thick blocks of lead or concrete are needed to stop them.
The Atomic Model
Once scientists concluded that all matter contains negatively charged electrons, it became clear that atoms, which are electrically neutral, must also contain positive charges to balance the negative ones. Thomson proposed that the electrons were embedded in a uniform sphere that contained both the positive charge and most of the mass of the atom, much like raisins in plum pudding or chocolate chips in a cookie (Figure $6$).
In a single famous experiment, however, Rutherford showed unambiguously that Thomson’s model of the atom was incorrect. Rutherford aimed a stream of α particles at a very thin gold foil target (Figure $\PageIndex{7a}$) and examined how the α particles were scattered by the foil. Gold was chosen because it could be easily hammered into extremely thin sheets, minimizing the number of atoms in the target. If Thomson’s model of the atom were correct, the positively-charged α particles should crash through the uniformly distributed mass of the gold target like cannonballs through the side of a wooden house. They might be moving a little slower when they emerged, but they should pass essentially straight through the target (Figure $\PageIndex{7b}$). To Rutherford’s amazement, a small fraction of the α particles were deflected at large angles, and some were reflected directly back at the source (Figure $\PageIndex{7c}$). According to Rutherford, “It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.”
The Nuclear Atom: The Nuclear Atom, YouTube(opens in new window) [youtu.be]
Rutherford’s results were not consistent with a model in which the mass and positive charge are distributed uniformly throughout the volume of an atom. Instead, they strongly suggested that both the mass and positive charge are concentrated in a tiny fraction of the volume of an atom, which Rutherford called the nucleus. It made sense that a small fraction of the α particles collided with the dense, positively charged nuclei in either a glancing fashion, resulting in large deflections, or almost head-on, causing them to be reflected straight back at the source.
Although Rutherford could not explain why repulsions between the positive charges in nuclei that contained more than one positive charge did not cause the nucleus to disintegrate, he reasoned that repulsions between negatively charged electrons would cause the electrons to be uniformly distributed throughout the atom’s volume.Today it is known that strong nuclear forces, which are much stronger than electrostatic interactions, hold the protons and the neutrons together in the nucleus. For this and other insights, Rutherford was awarded the Nobel Prize in Chemistry in 1908. Unfortunately, Rutherford would have preferred to receive the Nobel Prize in Physics because he considered physics superior to chemistry. In his opinion, “All science is either physics or stamp collecting.”
The historical development of the different models of the atom’s structure is summarized in Figure $8$. Rutherford established that the nucleus of the hydrogen atom was a positively charged particle, for which he coined the name proton in 1920. He also suggested that the nuclei of elements other than hydrogen must contain electrically neutral particles with approximately the same mass as the proton. The neutron, however, was not discovered until 1932, when James Chadwick (1891–1974, a student of Rutherford; Nobel Prize in Physics, 1935) discovered it. As a result of Rutherford’s work, it became clear that an α particle contains two protons and neutrons, and is therefore the nucleus of a helium atom.
Rutherford’s model of the atom is essentially the same as the modern model, except that it is now known that electrons are not uniformly distributed throughout an atom’s volume. Instead, they are distributed according to a set of principles described by Quantum Mechanics. Figure $9$ shows how the model of the atom has evolved over time from the indivisible unit of Dalton to the modern view taught today.
Summary
Atoms are the ultimate building blocks of all matter. The modern atomic theory establishes the concepts of atoms and how they compose matter. Atoms, the smallest particles of an element that exhibit the properties of that element, consist of negatively charged electrons around a central nucleus composed of more massive positively charged protons and electrically neutral neutrons. Radioactivity is the emission of energetic particles and rays (radiation) by some substances. Three important kinds of radiation are α particles (helium nuclei), β particles (electrons traveling at high speed), and γ rays (similar to x-rays but higher in energy). | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/02%3A_The_Components_of_Matter/2.02%3A_The_Observations_That_Led_to_an_Atomic_View_of_Matter.txt |
Learning Objectives
• To describe the composition of a chemical compound.
• To name covalent compounds that contain up to three elements.
As with ionic compounds, the system for naming covalent compounds enables chemists to write the molecular formula from the name and vice versa. This and the following section describe the rules for naming simple covalent compounds, beginning with inorganic compounds and then turning to simple organic compounds that contain only carbon and hydrogen.
When chemists synthesize a new compound, they may not yet know its molecular or structural formula. In such cases, they usually begin by determining its empirical formula, the relative numbers of atoms of the elements in a compound, reduced to the smallest whole numbers. Because the empirical formula is based on experimental measurements of the numbers of atoms in a sample of the compound, it shows only the ratios of the numbers of the elements present. The difference between empirical and molecular formulas can be illustrated with butane, a covalent compound used as the fuel in disposable lighters. The molecular formula for butane is \(\ce{C4H10}\). The ratio of carbon atoms to hydrogen atoms in butane is 4:10, which can be reduced to 2:5. The empirical formula for butane is therefore \(\ce{C2H5}\). The formula unit is the absolute grouping of atoms or ions represented by the empirical formula of a compound, either ionic or covalent. Butane has the empirical formula \(\ce{C2H5}\), but it contains two \(\ce{C2H5}\) formula units, giving a molecular formula of \(\ce{C4H10}\).
Because ionic compounds do not contain discrete molecules, empirical formulas are used to indicate their compositions. All compounds, whether ionic or covalent, must be electrically neutral. Consequently, the positive and negative charges in a formula unit must exactly cancel each other. If the cation and the anion have charges of equal magnitude, such as \(\ce{Na^{+}}\) and \(\ce{Cl^{−}}\), then the compound must have a 1:1 ratio of cations to anions, and the empirical formula must be \(\ce{NaCl}\). If the charges are not the same magnitude, then a cation:anion ratio other than 1:1 is needed to produce a neutral compound. In the case of \(\ce{Mg^{2+}}\) and \(\ce{Cl^{−}}\), for example, two Cl ions are needed to balance the two positive charges on each Mg2+ ion, giving an empirical formula of \(\ce{MgCl2}\). Similarly, the formula for the ionic compound that contains Na+ and O2− ions is Na2O.
Ionic compounds do not contain discrete molecules, so empirical formulas are used to indicate their compositions.
Binary Ionic Compounds
An ionic compound that contains only two elements, one present as a cation and one as an anion, is called a binary ionic compound. One example is \(\ce{MgCl_2}\), a coagulant used in the preparation of tofu from soybeans. For binary ionic compounds, the subscripts in the empirical formula can also be obtained by crossing charges: use the absolute value of the charge on one ion as the subscript for the other ion. This method is shown schematically as follows:
When crossing charges, it is sometimes necessary to reduce the subscripts to their simplest ratio to write the empirical formula. Consider, for example, the compound formed by Mg2+ and O2−. Using the absolute values of the charges on the ions as subscripts gives the formula \(\ce{Mg2O2}\):
This simplifies to its correct empirical formula MgO. The empirical formula has one Mg2+ ion and one O2− ion.
Example \(1\): Binary Ionic Compounds
Write the empirical formula for the simplest binary ionic compound formed from each ion or element pair.
1. Ga3+ and As3
2. Eu3+ and O2−
3. calcium and chlorine
Given: ions or elements
Asked for: empirical formula for binary ionic compound
Strategy:
1. If not given, determine the ionic charges based on the location of the elements in the periodic table.
2. Use the absolute value of the charge on each ion as the subscript for the other ion. Reduce the subscripts to the lowest numbers
to write the empirical formula. Check to make sure the empirical formula is electrically neutral.
Solution
a. B Using the absolute values of the charges on the ions as the subscripts gives \(\ce{Ga3As3}\):
Reducing the subscripts to the smallest whole numbers gives the empirical formula GaAs, which is electrically neutral [+3 + (−3) = 0]. Alternatively, we could recognize that Ga3+ and As3 have charges of equal magnitude but opposite signs. One Ga3+ ion balances the charge on one As3 ion, and a 1:1 compound will have no net charge. Because we write subscripts only if the number is greater than 1, the empirical formula is GaAs. GaAs is gallium arsenide, which is widely used in the electronics industry in transistors and other devices.
b. B Because Eu3+ has a charge of +3 and O2− has a charge of −2, a 1:1 compound would have a net charge of +1. We must therefore find multiples of the charges that cancel. We cross charges, using the absolute value of the charge on one ion as the subscript for the other ion:
The subscript for Eu3+ is 2 (from O2−), and the subscript for O2− is 3 (from Eu3+), giving Eu2O3; the subscripts cannot be reduced further. The empirical formula contains a positive charge of 2(+3) = +6 and a negative charge of 3(−2) = −6, for a net charge of 0. The compound Eu2O3 is neutral. Europium oxide is responsible for the red color in television and computer screens.
c. A Because the charges on the ions are not given, we must first determine the charges expected for the most common ions derived from calcium and chlorine. Calcium lies in group 2, so it should lose two electrons to form Ca2+. Chlorine lies in group 17, so it should gain one electron to form Cl.
B Two Cl ions are needed to balance the charge on one Ca2+ ion, which leads to the empirical formula CaCl2. We could also cross charges, using the absolute value of the charge on Ca2+ as the subscript for Cl and the absolute value of the charge on Cl as the subscript for Ca:
The subscripts in CaCl2 cannot be reduced further. The empirical formula is electrically neutral [+2 + 2(−1) = 0]. This compound is calcium chloride, one of the substances used as “salt” to melt ice on roads and sidewalks in winter.
Exercise \(1\)
Write the empirical formula for the simplest binary ionic compound formed from each ion or element pair.
1. Li+ and N3−
2. Al3+ and O2−
3. lithium and oxygen
Answer a
Li3N
Answer b
Al2O3
Answer c
Li2O
Nomenclature of Metals: Nomenclature of Metals(opens in new window) [youtu.be]
Polyatomic Ions
Polyatomic ions are groups of atoms that bear net electrical charges, although the atoms in a polyatomic ion are held together by the same covalent bonds that hold atoms together in molecules. Just as there are many more kinds of molecules than simple elements, there are many more kinds of polyatomic ions than monatomic ions. Two examples of polyatomic cations are the ammonium (NH4+) and the methylammonium (CH3NH3+) ions. Polyatomic anions are much more numerous than polyatomic cations; some common examples are in Table \(1\).
Table \(1\): Common Polyatomic Ions and Their Names
Formula Name of Ion Formula Name of Ion
NH4+ ammonium HPO42 hydrogen phosphate
CH3NH3+ methylammonium H2PO4 dihydrogen phosphate
OH hydroxide ClO hypochlorite
O22 peroxide ClO2 chlorite
CN cyanide ClO3 chlorate
SCN thiocyanate ClO4 perchlorate
NO2 nitrite MnO4 permanganate
NO3 nitrate CrO42 chromate
CO32 carbonate Cr2O72 dichromate
HCO3 hydrogen carbonate, or bicarbonate C2O42 oxalate
SO32 sulfite HCO2 formate
SO42 sulfate CH3CO2 acetate
HSO4 hydrogen sulfate, or bisulfate C6H5CO2 benzoate
PO43 phosphate
The method used to predict the empirical formulas for ionic compounds that contain monatomic ions can also be used for compounds that contain polyatomic ions. The overall charge on the cations must balance the overall charge on the anions in the formula unit. Thus, K+ and NO3 ions combine in a 1:1 ratio to form KNO3 (potassium nitrate or saltpeter), a major ingredient in black gunpowder. Similarly, Ca2+ and SO42 form CaSO4 (calcium sulfate), which combines with varying amounts of water to form gypsum and plaster of Paris. The polyatomic ions NH4+ and NO3 form NH4NO3 (ammonium nitrate), a widely used fertilizer and, in the wrong hands, an explosive. One example of a compound in which the ions have charges of different magnitudes is calcium phosphate, which is composed of Ca2+ and PO43 ions; it is a major component of bones. The compound is electrically neutral because the ions combine in a ratio of three Ca2+ ions [3(+2) = +6] for every two ions [2(−3) = −6], giving an empirical formula of Ca3(PO4)2; the parentheses around PO4 in the empirical formula indicate that it is a polyatomic ion. Writing the formula for calcium phosphate as Ca3P2O8 gives the correct number of each atom in the formula unit, but it obscures the fact that the compound contains readily identifiable PO43 ions.
Example \(2\)
Write the empirical formula for the compound formed from each ion pair.
1. Na+ and HPO42
2. potassium cation and cyanide anion
3. calcium cation and hypochlorite anion
Given: ions
Asked for: empirical formula for ionic compound
Strategy:
1. If it is not given, determine the charge on a monatomic ion from its location in the periodic table. Use Table \(1\) to find the charge on a polyatomic ion.
2. Use the absolute value of the charge on each ion as the subscript for the other ion. Reduce the subscripts to the smallest whole numbers when writing the empirical formula.
Solution:
1. B Because HPO42 has a charge of −2 and Na+ has a charge of +1, the empirical formula requires two Na+ ions to balance the charge of the polyatomic ion, giving Na2HPO4. The subscripts are reduced to the lowest numbers, so the empirical formula is Na2HPO4. This compound is sodium hydrogen phosphate, which is used to provide texture in processed cheese, puddings, and instant breakfasts.
2. A The potassium cation is K+, and the cyanide anion is CN. B Because the magnitude of the charge on each ion is the same, the empirical formula is KCN. Potassium cyanide is highly toxic, and at one time it was used as rat poison. This use has been discontinued, however, because too many people were being poisoned accidentally.
3. A The calcium cation is Ca2+, and the hypochlorite anion is ClO. B Two ClO ions are needed to balance the charge on one Ca2+ ion, giving Ca(ClO)2. The subscripts cannot be reduced further, so the empirical formula is Ca(ClO)2. This is calcium hypochlorite, the “chlorine” used to purify water in swimming pools.
Exercise \(2\)
Write the empirical formula for the compound formed from each ion pair.
1. Ca2+ and H2PO4
2. sodium cation and bicarbonate anion
3. ammonium cation and sulfate anion
Answer a
Ca(H2PO4)2: calcium dihydrogen phosphate is one of the ingredients in baking powder.
Answer b
NaHCO3: sodium bicarbonate is found in antacids and baking powder; in pure form, it is sold as baking soda.
Answer c
(NH4)2SO4: ammonium sulfate is a common source of nitrogen in fertilizers.
Polyatomics: Polyatomics, YouTube(opens in new window) [youtu.be] (opens in new window)
Hydrates
Many ionic compounds occur as hydrates, compounds that contain specific ratios of loosely bound water molecules, called waters of hydration. Waters of hydration can often be removed simply by heating. For example, calcium dihydrogen phosphate can form a solid that contains one molecule of water per \(\ce{Ca(H2PO4)2}\) unit and is used as a leavening agent in the food industry to cause baked goods to rise. The empirical formula for the solid is \(\ce{Ca(H2PO4)2·H2O}\). In contrast, copper sulfate usually forms a blue solid that contains five waters of hydration per formula unit, with the empirical formula \(\ce{CuSO4·5H2O}\). When heated, all five water molecules are lost, giving a white solid with the empirical formula \(\ce{CuSO4}\).
Compounds that differ only in the numbers of waters of hydration can have very different properties. For example, \(\ce{CaSO4·½H2O}\) is plaster of Paris, which was often used to make sturdy casts for broken arms or legs, whereas \(\ce{CaSO4·2H2O}\) is the less dense, flakier gypsum, a mineral used in drywall panels for home construction. When a cast would set, a mixture of plaster of Paris and water crystallized to give solid \(\ce{CaSO4·2H2O}\). Similar processes are used in the setting of cement and concrete.
Binary Acids
Some compounds containing hydrogen are members of an important class of substances known as acids. The chemistry of these compounds is explored in more detail in later chapters of this text, but for now, it will suffice to note that many acids release hydrogen ions, H+, when dissolved in water. To denote this distinct chemical property, a mixture of water with an acid is given a name derived from the compound’s name. If the compound is a binary acid (comprised of hydrogen and one other nonmetallic element):
1. The word “hydrogen” is changed to the prefix hydro-
2. The other nonmetallic element name is modified by adding the suffix -ic
3. The word “acid” is added as a second word
For example, when the gas \(\ce{HCl}\) (hydrogen chloride) is dissolved in water, the solution is called hydrochloric acid. Several other examples of this nomenclature are shown in Table \(2\).
Table \(2\): Names of Some Simple Acids
Name of Gas Name of Acid
HF(g), hydrogen fluoride HF(aq), hydrofluoric acid
HCl(g), hydrogen chloride HCl(aq), hydrochloric acid
HBr(g), hydrogen bromide HBr(aq), hydrobromic acid
HI(g), hydrogen iodide HI(aq), hydroiodic acid
H2S(g), hydrogen sulfide H2S(aq), hydrosulfuric acid
Oxyacids
Many compounds containing three or more elements (such as organic compounds or coordination compounds) are subject to specialized nomenclature rules that you will learn later. However, we will briefly discuss the important compounds known as oxyacids, compounds that contain hydrogen, oxygen, and at least one other element, and are bonded in such a way as to impart acidic properties to the compound (you will learn the details of this in a later chapter). Typical oxyacids consist of hydrogen combined with a polyatomic, oxygen-containing ion. To name oxyacids:
1. Omit “hydrogen”
2. Start with the root name of the anion
3. Replace –ate with –ic, or –ite with –ous
4. Add “acid”
For example, consider H2CO3 (which you might be tempted to call “hydrogen carbonate”). To name this correctly, “hydrogen” is omitted; the –ate of carbonate is replace with –ic; and acid is added—so its name is carbonic acid. Other examples are given in Table \(3\). There are some exceptions to the general naming method (e.g., H2SO4 is called sulfuric acid, not sulfic acid, and H2SO3 is sulfurous, not sulfous, acid).
Table \(3\): Names of Common Oxyacids
Formula Anion Name Acid Name
HC2H3O2 acetate acetic acid
HNO3 nitrate nitric acid
HNO2 nitrite nitrous acid
HClO4 perchlorate perchloric acid
H2CO3 carbonate carbonic acid
H2SO4 sulfate sulfuric acid
H2SO3 sulfite sulfurous acid
H3PO4 phosphate phosphoric acid
Nomenclature of Acids: Nomenclature of Acids, YouTube(opens in new window) [youtu.be]
Bases
We will present more comprehensive definitions of bases in later chapters, but virtually every base you encounter in the meantime will be an ionic compound, such as sodium hydroxide (NaOH) and barium hydroxide [Ba(OH)2], that contain the hydroxide ion and a metal cation. These have the general formula M(OH)n. It is important to recognize that alcohols, with the general formula ROH, are covalent compounds, not ionic compounds; consequently, they do not dissociate in water to form a basic solution (containing OH ions). When a base reacts with any of the acids we have discussed, it accepts a proton (H+). For example, the hydroxide ion (OH) accepts a proton to form H2O. Thus bases are also referred to as proton acceptors.
Concentrated aqueous solutions of ammonia (NH3) contain significant amounts of the hydroxide ion, even though the dissolved substance is not primarily ammonium hydroxide (NH4OH) as is often stated on the label. Thus aqueous ammonia solution is also a common base. Replacing a hydrogen atom of NH3 with an alkyl group results in an amine (RNH2), which is also a base. Amines have pungent odors—for example, methylamine (CH3NH2) is one of the compounds responsible for the foul odor associated with spoiled fish. The physiological importance of amines is suggested in the word vitamin, which is derived from the phrase vital amines. The word was coined to describe dietary substances that were effective at preventing scurvy, rickets, and other diseases because these substances were assumed to be amines. Subsequently, some vitamins have indeed been confirmed to be amines.
Binary Inorganic Compounds
Binary covalent compounds—covalent compounds that contain only two elements—are named using a procedure similar to that used for simple ionic compounds, but prefixes are added as needed to indicate the number of atoms of each kind. The procedure, diagrammed in Figure \(2\) consists of the following steps:
1. Place the elements in their proper order.
• The element farthest to the left in the periodic table is usually named first. If both elements are in the same group, the element closer to the bottom of the column is named first.
• The second element is named as if it were a monatomic anion in an ionic compound (even though it is not), with the suffix -ide attached to the root of the element name.
2. Identify the number of each type of atom present.
1. Prefixes derived from Greek stems are used to indicate the number of each type of atom in the formula unit (Table \(3\)). The prefix mono- (“one”) is used only when absolutely necessary to avoid confusion, just as the subscript 1 is omitted when writing molecular formulas.
To demonstrate steps 1 and 2a, HCl is named hydrogen chloride (because hydrogen is to the left of chlorine in the periodic table), and PCl5 is phosphorus pentachloride. The order of the elements in the name of BrF3, bromine trifluoride, is determined by the fact that bromine lies below fluorine in Group 17.
Table \(3\): Prefixes for Indicating the Number of Atoms in Chemical Names
Prefix Number
mono- 1
di- 2
tri- 3
tetra- 4
penta- 5
hexa- 6
hepta- 7
octa- 8
nona- 9
deca- 10
undeca- 11
dodeca- 12
2. If a molecule contains more than one atom of both elements, then prefixes are used for both. Thus N2O3 is dinitrogen trioxide, as shown in Figure 2.13.
3. In some names, the final a or o of the prefix is dropped to avoid awkward pronunciation. Thus OsO4 is osmium tetroxide rather than osmium tetraoxide.
3. Write the name of the compound.
1. Binary compounds of the elements with oxygen are generally named as “element oxide,” with prefixes that indicate the number of atoms of each element per formula unit. For example, CO is carbon monoxide. The only exception is binary compounds of oxygen with fluorine, which are named as oxygen fluorides.
2. Certain compounds are always called by the common names that were assigned before formulas were used. For example, H2O is water (not dihydrogen oxide); NH3 is ammonia; PH3 is phosphine; SiH4 is silane; and B2H6, a dimer of BH3, is diborane. For many compounds, the systematic name and the common name are both used frequently, requiring familiarity with both. For example, the systematic name for NO is nitrogen monoxide, but it is much more commonly called nitric oxide. Similarly, N2O is usually called nitrous oxide rather than dinitrogen monoxide. Notice that the suffixes -ic and -ous are the same ones used for ionic compounds.
Start with the element at the far left in the periodic table and work to the right. If two or more elements are in the same group, start with the bottom element and work up.
Example \(3\): Binary Covalent Compounds
Write the name of each binary covalent compound.
1. SF6
2. N2O4
3. ClO2
Given: molecular formula
Asked for: name of compound
Strategy:
1. List the elements in order according to their positions in the periodic table. Identify the number of each type of atom in the chemical formula and then use Table \(2\) to determine the prefixes needed.
2. If the compound contains oxygen, follow step 3a. If not, decide whether to use the common name or the systematic name.
Solution:
1. A Because sulfur is to the left of fluorine in the periodic table, sulfur is named first. Because there is only one sulfur atom in the formula, no prefix is needed. B There are, however, six fluorine atoms, so we use the prefix for six: hexa- (Table \(2\)). The compound is sulfur hexafluoride.
2. A Because nitrogen is to the left of oxygen in the periodic table, nitrogen is named first. Because more than one atom of each element is present, prefixes are needed to indicate the number of atoms of each. According to Table \(2\) "Prefixes for Indicating the Number of Atoms in Chemical Names", the prefix for two is di-, and the prefix for four is tetra-. B The compound is dinitrogen tetroxide (omitting the a in tetra- according to step 2c) and is used as a component of some rocket fuels.
3. A Although oxygen lies to the left of chlorine in the periodic table, it is not named first because ClO2 is an oxide of an element other than fluorine (step 3a). Consequently, chlorine is named first, but a prefix is not necessary because each molecule has only one atom of chlorine. B Because there are two oxygen atoms, the compound is a dioxide. Thus the compound is chlorine dioxide. It is widely used as a substitute for chlorine in municipal water treatment plants because, unlike chlorine, it does not react with organic compounds in water to produce potentially toxic chlorinated compounds.
Example \(3\)
Write the name of each binary covalent compound.
1. IF7
2. N2O5
3. OF2
Answer a
iodine heptafluoride
Answer b
dinitrogen pentoxide
Answer c
oxygen difluoride
Example \(4\)
Write the formula for each binary covalent compound.
1. sulfur trioxide
2. diiodine pentoxide
Given: name of compound
Asked for: formula
Strategy:
List the elements in the same order as in the formula, use Table \(2\) to identify the number of each type of atom present, and then indicate this quantity as a subscript to the right of that element when writing the formula.
Solution:
1. Sulfur has no prefix, which means that each molecule has only one sulfur atom. The prefix tri- indicates that there are three oxygen atoms. The formula is therefore SO3. Sulfur trioxide is produced industrially in huge amounts as an intermediate in the synthesis of sulfuric acid.
2. The prefix di- tells you that each molecule has two iodine atoms, and the prefix penta- indicates that there are five oxygen atoms. The formula is thus I2O5, a compound used to remove carbon monoxide from air in respirators.
Exercise \(4\)
Write the formula for each binary covalent compound.
1. silicon tetrachloride
2. disulfur decafluoride
Answer a
SiCl4
Answer b
S2F10
The structures of some of the compounds in Examples \(3\) and \(4\) are shown in Figure \(2\) along with the location of the “central atom” of each compound in the periodic table. It may seem that the compositions and structures of such compounds are entirely random, but this is not true. After mastering the material discussed later on this course, one is able to predict the compositions and structures of compounds of this type with a high degree of accuracy.
Nomenclature of Nonmetals: Nomenclature of Nonmetals, YouTube(opens in new window) [youtu.be] (opens in new window)
Summary
The composition of a compound is represented by an empirical or molecular formula, each consisting of at least one formula unit. Covalent inorganic compounds are named using a procedure similar to that used for ionic compounds, whereas hydrocarbons use a system based on the number of bonds between carbon atoms. Covalent inorganic compounds are named by a procedure similar to that used for ionic compounds, using prefixes to indicate the numbers of atoms in the molecular formula. An empirical formula gives the relative numbers of atoms of the elements in a compound, reduced to the lowest whole numbers. The formula unit is the absolute grouping represented by the empirical formula of a compound, either ionic or covalent. Empirical formulas are particularly useful for describing the composition of ionic compounds, which do not contain readily identifiable molecules. Some ionic compounds occur as hydrates, which contain specific ratios of loosely bound water molecules called waters of hydration. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/02%3A_The_Components_of_Matter/2.07%3A_Compounds_-_Formulas_Names_and_Masses.txt |
Stoichiometry is the calculation of relative quantities of reactants and products in chemical reactions. Stoichiometry is founded on the law of conservation of mass where the total mass of the reactants equals the total mass of the products leading to the insight that the relations among quantities of reactants and products typically form a ratio of positive integers. This means that if the amounts of the separate reactants are known, then the amount of the product can be calculated. Conversely, if one reactant has a known quantity and the quantity of product can be empirically determined, then the amount of the other reactants can also be calculated.
We begin this chapter by describing the relationship between the mass of a sample of a substance and its composition. We then develop methods for determining the quantities of compounds produced or consumed in chemical reactions, and we describe some fundamental types of chemical reactions. By applying the concepts and skills introduced in this chapter, you will be able to explain what happens to the sugar in a candy bar you eat, what reaction occurs in a battery when you start your car, what may be causing the “ozone hole” over Antarctica, and how we might prevent the hole’s growth.
• 3.1: Chemical Equations
A chemical reaction is described by a chemical equation that gives the identities and quantities of the reactants and the products. In a chemical reaction, one or more substances are transformed to new substances. A chemical reaction is described by a chemical equation, an expression that gives the identities and quantities of the substances involved in a reaction. A chemical equation shows the starting compound(s)—the reactants—on the left and the final compound(s)—the products—on the right.
• 3.2: Some Simple Patterns of Chemical Reactivity
By recognizing general patterns of chemical reactivity, you will be able to successfully predict the products formed by a given combination of reactants We can often predict a reaction if we have seen a similar reaction before.
• 3.3: Formula Masses
The empirical formula of a substance can be calculated from its percent composition, and the molecular formula can be determined from the empirical formula and the compound’s molar mass. The empirical formula of a substance can be calculated from the experimentally determined percent composition, the percentage of each element present in a pure substance by mass. In many cases, these percentages can be determined by combustion analysis.
• 3.4: Avogadro's Number and the Mole
The molecular mass and the formula mass of a compound are obtained by adding together the atomic masses of the atoms present in the molecular formula or empirical formula, respectively; the units of both are atomic mass units (amu). The mole is defined as the amount of substance that contains the number of carbon atoms in exactly 12 g of carbon-12, Avogadro’s number ($6.022 \times 10^{23}$) of atoms of carbon-12.
• 3.5: Empirical Formulas from Analysis
Chemical formulas tell you how many atoms of each element are in a compound, and empirical formulas tell you the simplest or most reduced ratio of elements in a compound. If a compound's molecular formula cannot be reduced any more, then the empirical formula is the same as the chemical formula. Combustion analysis can determine the empirical formula of a compound, but cannot determine the chemical formula (other techniques can though).
• 3.6: Quantitative Information from Balanced Equations
Either the masses or the volumes of solutions of reactants and products can be used to determine the amounts of other species in the balanced chemical equation. Quantitative calculations that involve the stoichiometry of reactions in solution use volumes of solutions of known concentration instead of masses of reactants or products. The coefficients in the balanced chemical equation tell how many moles of reactants are needed and how many moles of product can be produced.
• 3.7: Limiting Reactants
The stoichiometry of a balanced chemical equation identifies the maximum amount of product that can be obtained. The stoichiometry of a reaction describes the relative amounts of reactants and products in a balanced chemical equation. A stoichiometric quantity of a reactant is the amount necessary to react completely with the other reactant(s). If a reactant remains unconsumed after complete reaction has occurred, it is in excess. The reactant that is consumed first is the limiting reagent.
• 3.E: Stoichiometry (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al.
• 3.S: Stoichiometry (Summary)
This is the summary Module for the chapter "Stoichiometry" in the Brown et al. General Chemistry Textmap.
Thumbnail: The combustion of methane, a hydrocarbon. (CC BY-SA 4.0 International; Yulo1985 (cropper) via Wikipedia)
03: Stoichiometry of Formulas and Equation
Learning Objectives
• To calculate the molecular mass of a covalent compound and the formula mass of an ionic compound, and to calculate the number of atoms, molecules, or formula units in a sample of a substances.
As discussed previosuly, the mass number is the sum of the numbers of protons and neutrons present in the nucleus of an atom. The mass number is an integer that is approximately equal to the numerical value of the atomic mass. Although the mass number is unitless, it is assigned units called atomic mass units (amu). Because a molecule or a polyatomic ion is an assembly of atoms whose identities are given in its molecular or ionic formula, the average atomic mass of any molecule or polyatomic ion can be calculated from its composition by adding together the masses of the constituent atoms. The average mass of a monatomic ion is the same as the average mass of an atom of the element because the mass of electrons is so small that it is insignificant in most calculations.
Molecular and Formula Masses
The molecular mass of a substance is the sum of the average masses of the atoms in one molecule of a substance. It is calculated by adding together the atomic masses of the elements in the substance, each multiplied by its subscript (written or implied) in the molecular formula. Because the units of atomic mass are atomic mass units, the units of molecular mass are also atomic mass units. The procedure for calculating molecular masses is illustrated in Example $1$.
Example $1$: Molecular Mass of Ethanol
Calculate the molecular mass of ethanol, whose condensed structural formula is $\ce{CH3CH2OH}$. Among its many uses, ethanol is a fuel for internal combustion engines.
Given: molecule
Asked for: molecular mass
Strategy:
1. Determine the number of atoms of each element in the molecule.
2. Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element.
3. Add together the masses to give the molecular mass.
Solution:
A The molecular formula of ethanol may be written in three different ways: $\ce{CH3CH2OH}$ (which illustrates the presence of an ethyl group, CH3CH2, and an −OH group), $\ce{C2H5OH}$, and $\ce{C2H6O}$; all show that ethanol has two carbon atoms, six hydrogen atoms, and one oxygen atom.
B Taking the atomic masses from the periodic table, we obtain
\begin{align*} 2 \times \text { atomic mass of carbon} &= 2 \, atoms \left ( {12.011 \, amu \over atoms } \right ) \[4pt] &= 24.022 \,amu \end{align*}\nonumber
\begin{align*} 6 \times \text { atomic mass of hydrogen} &= 2 \, atoms \left ( {1.0079 \, amu \over atoms } \right ) \[4pt] &= 6.0474 \,amu \end{align*}\nonumber
\begin{align*} 1 \times \text { atomic mass of oxygen} &= 1 \, atoms \left ( {15.9994 \, amu \over atoms } \right ) \[4pt] &= 15.994 \,amu \end{align*}\nonumber
C Adding together the masses gives the molecular mass:
$24.022 \,amu + 6.0474 \,amu + 15.9994 \,amu = 46.069 \,amu \nonumber \nonumber$
Alternatively, we could have used unit conversions to reach the result in one step:
$\left [ 2 \, atoms\, C \left ( {12.011 \, amu \over 1 \, atom C} \right ) \right ] + \left [ 6 \, atoms\, H \left ( {1.0079 \, amu \over 1 \, atom H} \right ) \right ] + \left [ 1 \, atoms\, C \left ( {15.9994 \, amu \over 1 \, atom 0} \right ) \right ] = 46.069 \, amu \nonumber$
The same calculation can also be done in a tabular format, which is especially helpful for more complex molecules:
$2 \times C \, \, \, (2\, atoms) (12.011 \, amu/atom ) = 24.022 \, amu \nonumber$
$6 \times H \, \, \, (6\, atoms) (1.0079 \, amu/atom ) = 6.0474 \, amu \nonumber$
$1 \times \,O \, \, \, (1\, atoms) (15.9994 \, amu/atom ) = 15.9994 \, amu \nonumber$
$C_2H_6O \, \, \, \, \, \text {molecular mass of ethanol} = 46.069 \, amu \nonumber$
Exercise $1$: Molecular Mass of Freon
Calculate the molecular mass of trichlorofluoromethane, also known as Freon-11, whose condensed structural formula is $\ce{CCl_3F}$. Until recently, it was used as a refrigerant. The structure of a molecule of Freon-11 is as follows:
Answer
137.368 amu
Unlike molecules, which form covalent bonds, ionic compounds do not have a readily identifiable molecular unit. Therefore, for ionic compounds, the formula mass (also called the empirical formula mass) of the compound is used instead of the molecular mass. The formula mass is the sum of the atomic masses of all the elements in the empirical formula, each multiplied by its subscript (written or implied). It is directly analogous to the molecular mass of a covalent compound. The units are atomic mass units.
Atomic mass, molecular mass, and formula mass all have the same units: atomic mass units.
Example $2$: Formula Mass of Calcium Phosphate
Calculate the formula mass of $\ce{Ca3(PO4)2}$, commonly called calcium phosphate. This compound is the principal source of calcium found in bovine milk.
Given: ionic compound
Asked for: formula mass
Strategy:
1. Determine the number of atoms of each element in the empirical formula.
2. Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element.
3. Add together the masses to give the formula mass.
Solution:
A The empirical formula—Ca3(PO4)2—indicates that the simplest electrically neutral unit of calcium phosphate contains three Ca2+ ions and two PO43 ions. The formula mass of this molecular unit is calculated by adding together the atomic masses of three calcium atoms, two phosphorus atoms, and eight oxygen atoms.
B Taking atomic masses from the periodic table, we obtain
$3 \times \text {atomic mass of calcium} = 3 \, atoms \left ( {40.078 \, amu \over atom } \right ) = 120.234 \, amu \nonumber \nonumber$
$2 \times \text {atomic mass of phosphorus} = 2 \, atoms \left ( {30.973761 \, amu \over atom } \right ) = 61.947522 \, amu \nonumber \nonumber$
$8 \times \text {atomic mass of oxygen} = 8 \, atoms \left ( {15.9994 \, amu \over atom } \right ) = 127.9952 \, amu \nonumber \nonumber$
C Adding together the masses gives the formula mass of $\ce{Ca3(PO4)2}$:
$120.234 \,amu + 61.947522 \, amu + 127.9952 \, amu = 310.177 \, amu \nonumber \nonumber$
We could also find the formula mass of $\ce{Ca3(PO4)2}$ in one step by using unit conversions or a tabular format:
$\left [ 3 \, atoms Ca \left ({40.078 \, amu \over 1 \, atom Ca } \right ) \right ] + \left [ 2 \, atoms P \left ({30.973761 \, amu \over 1 \, atom P } \right ) \right ] + \left [ 8 \, atoms O \left ({15.9994 \, amu \over 1 \, atom O } \right ) \right ] = 310.177 \,amu \nonumber$
$3Ca \, \, \, \, (3\, atoms)(40.078 \, amu/atom) = 120.234 \, amu \nonumber$
$2P \, \, \, \, (2\, atoms)(30.973761 \, amu/atom) = 61.947522 \, amu \nonumber$
$+ 8O \, \, \, \, (8\, atoms)(15.9994 \, amu/atom) = 127.9952 \, amu \nonumber$
$Ca_3P_2O_8 \, \, \, \, \text {formula mass of Ca}_3(PO_4)_2 = 310.177 \, amu \nonumber$
Exercise $2$: Formula Mass of Silicon Nitride
Calculate the formula mass of $\ce{Si3N4}$, commonly called silicon nitride. It is an extremely hard and inert material that is used to make cutting tools for machining hard metal alloys.
Answer
$140.29 \,amu$
Molar Masses of Compounds: Molar Masses of Compounds, YouTube(opens in new window) [youtu.be]
The Mole
Dalton’s theory that each chemical compound has a particular combination of atoms and that the ratios of the numbers of atoms of the elements present are usually small whole numbers. It also describes the law of multiple proportions, which states that the ratios of the masses of elements that form a series of compounds are small whole numbers. The problem for Dalton and other early chemists was to discover the quantitative relationship between the number of atoms in a chemical substance and its mass. Because the masses of individual atoms are so minuscule (on the order of 10−23 g/atom), chemists do not measure the mass of individual atoms or molecules. In the laboratory, for example, the masses of compounds and elements used by chemists typically range from milligrams to grams, while in industry, chemicals are bought and sold in kilograms and tons. To analyze the transformations that occur between individual atoms or molecules in a chemical reaction, it is therefore essential for chemists to know how many atoms or molecules are contained in a measurable quantity in the laboratory—a given mass of sample. The unit that provides this link is the mole (mol), from the Latin moles, meaning “pile” or “heap.”
Many familiar items are sold in numerical quantities with distinct names. For example, cans of soda come in a six-pack, eggs are sold by the dozen (12), and pencils often come in a gross (12 dozen, or 144). Sheets of printer paper are packaged in reams of 500, a seemingly large number. Atoms are so small, however, that even 500 atoms are too small to see or measure by most common techniques. Any readily measurable mass of an element or compound contains an extraordinarily large number of atoms, molecules, or ions, so an extremely large numerical unit is needed to count them. The mole is used for this purpose.
A mole is defined as the amount of a substance that contains the number of carbon atoms in exactly 12 g of isotopically pure carbon-12. According to the most recent experimental measurements, this mass of carbon-12 contains 6.022142 × 1023 atoms, but for most purposes 6.022 × 1023 provides an adequate number of significant figures. Just as 1 mole of atoms contains 6.022 × 1023 atoms, 1 mole of eggs contains 6.022 × 1023 eggs. This number is called Avogadro’s number, after the 19th-century Italian scientist who first proposed a relationship between the volumes of gases and the numbers of particles they contain.
It is not obvious why eggs come in dozens rather than 10s or 14s, or why a ream of paper contains 500 sheets rather than 400 or 600. The definition of a mole—that is, the decision to base it on 12 g of carbon-12—is also arbitrary. The important point is that 1 mole of carbon—or of anything else, whether atoms, compact discs, or houses—always has the same number of objects: 6.022 × 1023.
One mole always has the same number of objects: 6.022 × 1023.
To appreciate the magnitude of Avogadro’s number, consider a mole of pennies. Stacked vertically, a mole of pennies would be 4.5 × 1017 mi high, or almost six times the diameter of the Milky Way galaxy. If a mole of pennies were distributed equally among the entire population on Earth, each person would have more than one trillion dollars. The mole is so large that it is useful only for measuring very small objects, such as atoms.
The concept of the mole allows scientists to count a specific number of individual atoms and molecules by weighing measurable quantities of elements and compounds. To obtain 1 mol of carbon-12 atoms, one weighs out 12 g of isotopically pure carbon-12. Because each element has a different atomic mass, however, a mole of each element has a different mass, even though it contains the same number of atoms (6.022 × 1023). This is analogous to the fact that a dozen extra large eggs weighs more than a dozen small eggs, or that the total weight of 50 adult humans is greater than the total weight of 50 children. Because of the way the mole is defined, for every element the number of grams in a mole is the same as the number of atomic mass units in the atomic mass of the element. For example, the mass of 1 mol of magnesium (atomic mass = 24.305 amu) is 24.305 g. Because the atomic mass of magnesium (24.305 amu) is slightly more than twice that of a carbon-12 atom (12 amu), the mass of 1 mol of magnesium atoms (24.305 g) is slightly more than twice that of 1 mol of carbon-12 (12 g). Similarly, the mass of 1 mol of helium (atomic mass = 4.002602 amu) is 4.002602 g, which is about one-third that of 1 mol of carbon-12. Using the concept of the mole, Dalton’s theory can be restated: 1 mol of a compound is formed by combining elements in amounts whose mole ratios are small whole numbers. For example, 1 mol of water (H2O) has 2 mol of hydrogen atoms and 1 mol of oxygen atoms.
Molar Mass
The molar mass of a substance is defined as the mass in grams of 1 mole of that substance. One mole of isotopically pure carbon-12 has a mass of 12 g. For an element, the molar mass is the mass of 1 mol of atoms of that element; for a covalent molecular compound, it is the mass of 1 mol of molecules of that compound; for an ionic compound, it is the mass of 1 mol of formula units. That is, the molar mass of a substance is the mass (in grams per mole) of 6.022 × 1023 atoms, molecules, or formula units of that substance. In each case, the number of grams in 1 mol is the same as the number of atomic mass units that describe the atomic mass, the molecular mass, or the formula mass, respectively.
The molar mass of any substance is its atomic mass, molecular mass, or formula mass in grams per mole.
The periodic table lists the atomic mass of carbon as 12.011 amu; the average molar mass of carbon—the mass of 6.022 × 1023 carbon atoms—is therefore 12.011 g/mol:
Atomic mass of carbon as 12.011 amu; the average molar mass of carbon
Substance (formula) Atomic, Molecular, or Formula Mass (amu) Molar Mass (g/mol)
carbon (C) 12.011 (atomic mass) 12.011
ethanol (C2H5OH) 46.069 (molecular mass) 46.069
calcium phosphate [Ca3(PO4)2] 310.177 (formula mass) 310.177
Determining the Molar Mass of a Molecule, YouTube: Determining the Molar Mass of a Molecule(opens in new window) [youtu.be]
The molar mass of naturally-occurring carbon is different from that of carbon-12, and is not an integer because carbon occurs as a mixture of carbon-12, carbon-13, and carbon-14. One mole of carbon still has 6.022 × 1023 carbon atoms, but 98.89% of those atoms are carbon-12, 1.11% are carbon-13, and a trace (about 1 atom in 1012) are carbon-14. (For more information, see Section 1.6.) Similarly, the molar mass of uranium is 238.03 g/mol, and the molar mass of iodine is 126.90 g/mol. When dealing with elements such as iodine and sulfur, which occur as a diatomic molecule (I2) and a polyatomic molecule (S8), respectively, molar mass usually refers to the mass of 1 mol of atoms of the element—in this case I and S, not to the mass of 1 mol of molecules of the element (I2 and S8).
The molar mass of ethanol is the mass of ethanol (C2H5OH) that contains 6.022 × 1023 ethanol molecules. As in Example $1$, the molecular mass of ethanol is 46.069 amu. Because 1 mol of ethanol contains 2 mol of carbon atoms (2 × 12.011 g), 6 mol of hydrogen atoms (6 × 1.0079 g), and 1 mol of oxygen atoms (1 × 15.9994 g), its molar mass is 46.069 g/mol. Similarly, the formula mass of calcium phosphate [Ca3(PO4)2] is 310.177 amu, so its molar mass is 310.177 g/mol. This is the mass of calcium phosphate that contains 6.022 × 1023 formula units.
The mole is the basis of quantitative chemistry. It provides chemists with a way to convert easily between the mass of a substance and the number of individual atoms, molecules, or formula units of that substance. Conversely, it enables chemists to calculate the mass of a substance needed to obtain a desired number of atoms, molecules, or formula units. For example, to convert moles of a substance to mass, the following relationship is used:
$(\text{moles}) \times (\text{molar mass}) \rightarrow \text{mass} \label{3.4.1}$
or, more specifically,
$\cancel{\text{moles}} \times \left ( {\text{grams} \over \cancel{\text{mole}} } \right ) = \text{grams} \nonumber$
Conversely, to convert the mass of a substance to moles:
$\left ( {\text{grams} \over \text{molar mass} } \right ) \rightarrow \text{moles} \label{3.4.2A}$
$\left ( { \text{grams} \over \text{grams/mole}} \right ) = \cancel{\text{grams}} \left ( {\text{mole} \over \cancel{\text{grams}} } \right ) = \text{moles} \label{3.4.2B}$
The coefficients in a balanced chemical equation can be interpreted both as the relative numbers of molecules involved in the reaction and as the relative number of moles. For example, in the balanced equation:
$\ce{2H2(g) + O2(g) \rightarrow 2H2O(\ell) } \nonumber$
the production of two moles of water would require the consumption of 2 moles of $\ce{H_2}$ and one mole of $\ce{O_2}$. Therefore, when considering this particular reaction
• 2 moles of $\ce{H2}$
• 1 mole of $\ce{O2}$ and
• 2 moles of $\ce{H2O}$
would be considered to be stoichiometrically equivalent quantitites.
These stoichiometric relationships, derived from balanced equations, can be used to determine expected amounts of products given amounts of reactants. For example, how many moles of $H_2O$ would be produced from 1.57 moles of $O_2$?
$(1.57\; mol\; \ce{O_2}) \left( \dfrac{2\; mol\, \ce{H_2O}}{1\;mol\; \ce{O_2}} \right) = 3.14\; mol\; \ce{H_2O}\nonumber$
The ratio $\left( \dfrac{2\; mol\; \ce{H_2O}}{1\;mol\;\ce{O_2}} \right)$ is the stoichiometric relationship between $\ce{H_2O}$ and $\ce{O_2}$ from the balanced equation for this reaction.
Be sure to pay attention to the units when converting between mass and moles. Figure $1$ is a flowchart for converting between mass; the number of moles; and the number of atoms, molecules, or formula units. The use of these conversions is illustrated in Examples $3$ and $4$.
Example $3$: Combustion of Butane
For the combustion of butane ($\ce{C_4H_{10}}$) the balanced equation is:
$\ce{2C4H_{10}(l) + 13O2(g) \rightarrow 8CO2(g) + 10H2O(l) }\nonumber$
Calculate the mass of $\ce{CO_2}$ that is produced in burning 1.00 gram of $\ce{C_4H_{10}}$.
Solution
Thus, the overall sequence of steps to solve this problem is:
First of all we need to calculate how many moles of butane we have in a 1.00 gram sample:
$(1.00\; g\; \ce{C_4H_{10}}) \left(\dfrac{1\; mol\; \ce{C_4H_{10}}}{58.0\;g\; \ce{C_4H_{10}}}\right) = 1.72 \times 10^{-2} \; mol\; \ce{C_4H_{10}}\nonumber$
Now, the stoichiometric relationship between $\ce{C_4H_{10}}$ and $\ce{CO_2}$ is:
$\left( \dfrac{8\; mol\; \ce{CO_2}}{2\; mol\; \ce{C_4H_{10}}}\right)\nonumber$
Therefore:
$\left(\dfrac{8\; mol\; \ce{CO_2}}{2\; mol\; \ce{C_4H_{10}}} \right) \times 1.72 \times 10^{-2} \; mol\; \ce{C_4H_{10}} = 6.88 \times 10^{-2} \; mol\; \ce{CO_2} \nonumber$
The question called for the determination of the mass of $\ce{CO_2}$ produced, thus we have to convert moles of $\ce{CO_2}$ into grams (by using the molecular weight of $\ce{CO_2}$):
$6.88 \times 10^{-2} \; mol\; \ce{CO_2} \left( \dfrac{44.0\; g\; \ce{CO_2}}{1\; mol\; \ce{CO_2}} \right) = 3.03\;g \; \ce{CO_2}\nonumber$
Example $4$: Ethylene Glycol
For 35.00 g of ethylene glycol (\ce{HOCH2CH2OH}), which is used in inks for ballpoint pens, calculate the number of
1. moles.
2. molecules.
Given: mass and molecular formula
Asked for: number of moles and number of molecules
Strategy:
1. Use the molecular formula of the compound to calculate its molecular mass in grams per mole.
2. Convert from mass to moles by dividing the mass given by the compound’s molar mass.
3. Convert from moles to molecules by multiplying the number of moles by Avogadro’s number.
Solution:
A The molecular mass of ethylene glycol can be calculated from its molecular formula using the method illustrated in Example $1$:
$2C (2 \,atoms )(12.011 \, amu/atom) = 24.022 \, amu \nonumber$
$6H (6 \,atoms )(1.0079 \, amu/atom) = 6.0474 \, amu \nonumber$
$2O (2 \,atoms )(15.9994 \, amu/atom) = 31.9988 \, amu \nonumber$
$C_2H_6O_2 \text {molecular mass of ethylene glycol} = 62.068 \, amu \nonumber$
The molar mass of ethylene glycol is 62.068 g/mol.
B The number of moles of ethylene glycol present in 35.00 g can be calculated by dividing the mass (in grams) by the molar mass (in grams per mole):
${ \text {mass of ethylene glycol (g)} \over \text {molar mass (g/mol)} } = \text {moles ethylene glycol (mol) }\nonumber$
So
$35.00 \, g \text {ethylene glycol} \left ( {1 \, mole\, \text {ethylene glycol} \over 62.068 \, g \, \text {ethylene glycol } } \right ) = 0.5639 \,mol \text {ethylene glycol} \nonumber$
It is always a good idea to estimate the answer before you do the actual calculation. In this case, the mass given (35.00 g) is less than the molar mass, so the answer should be less than 1 mol. The calculated answer (0.5639 mol) is indeed less than 1 mol, so we have probably not made a major error in the calculations.
C To calculate the number of molecules in the sample, we multiply the number of moles by Avogadro’s number:
\begin{align*} \text {molecules of ethylene glycol} &= 0.5639 \, \cancel{mol} \left ( {6.022 \times 10^{23} \, molecules \over 1 \, \cancel{mol} } \right ) \[4pt] &= 3.396 \times 10^{23} \, molecules \end{align*} \nonumber
Because we are dealing with slightly more than 0.5 mol of ethylene glycol, we expect the number of molecules present to be slightly more than one-half of Avogadro’s number, or slightly more than 3 × 1023 molecules, which is indeed the case.
Exercise $4$: Freon-11
For 75.0 g of CCl3F (Freon-11), calculate the number of
1. moles.
2. molecules.
Answer a
0.546 mol
Answer b
3.29 × 1023 molecules
Example $5$
Calculate the mass of 1.75 mol of each compound.
1. S2Cl2 (common name: sulfur monochloride; systematic name: disulfur dichloride)
2. Ca(ClO)2 (calcium hypochlorite)
Given: number of moles and molecular or empirical formula
Asked for: mass
Strategy:
A Calculate the molecular mass of the compound in grams from its molecular formula (if covalent) or empirical formula (if ionic).
B Convert from moles to mass by multiplying the moles of the compound given by its molar mass.
Solution:
We begin by calculating the molecular mass of S2Cl2 and the formula mass of Ca(ClO)2.
A The molar mass of S2Cl2 is obtained from its molecular mass as follows:
$2S (2 \, atoms)(32.065 \, amu/atom ) = 64.130 \, amu \nonumber$
$+ 2Cl (2 \, atoms )(35.453 \, amu/atom ) = 70.906 \, amu \nonumber$
$S_2 Cl_2 \text {molecular mass of } S_2Cl_2 = 135.036 \, amu \nonumber$
The molar mass of S2Cl2 is 135.036 g/mol.
B The mass of 1.75 mol of S2Cl2 is calculated as follows:
$moles S_2Cl_2 \left [\text {molar mass}\left ({ g \over mol} \right )\right ] \rightarrow mass of S_2Cl_2 \, (g) \nonumber$
$1.75 \, mol S_2Cl_2\left ({135.036 \, g S_2Cl_2 \over 1 \, mol S_2Cl_2 } \right ) = 236 \, g S_2Cl_2 \nonumber$
A The formula mass of Ca(ClO)2 is obtained as follows:
$1Ca (1 \, atom)(40.078 \, amu/atom) = 40.078 \, amu \nonumber$
$2Cl (2 \, atoms)(35.453 \, amu/atom) = 70.906 \, amu \nonumber$
$+ 2O (2 \, atoms)(15.9994 \, amu/atom) = 31.9988 \, amu \nonumber$
$Ca (ClO)_2 \text { formula mass of } Ca (ClO)_2 = 142.983 \, amu\nonumber$
The molar mass of Ca(ClO)2 142.983 g/mol.
B The mass of 1.75 mol of Ca(ClO)2 is calculated as follows:
$moles Ca(ClO)_2 \left [{\text {molar mass} Ca(ClO)_2 \over 1 \, mol Ca(ClO)_2} \right ]=mass Ca(ClO)_2\nonumber$
$1.75 \, mol Ca(ClO)_2 \left [ {142.983 \, g Ca(ClO)_2 \over 1 \, mol Ca(ClO)_2 } \right ] = 250 \, g Ca(ClO)_2 \nonumber$
Because 1.75 mol is less than 2 mol, the final quantity in grams in both cases should be less than twice the molar mass, which it is.
Exercise $5$
Calculate the mass of 0.0122 mol of each compound.
1. Si3N4 (silicon nitride), used as bearings and rollers
2. (CH3)3N (trimethylamine), a corrosion inhibitor
Answer a
1.71 g
Answer b
0.721 g
Conversions Between Grams, Mol, & Atoms: Conversions Between Grams, Mol & Atoms, YouTube(opens in new window) [youtu.be]
The coefficients in a balanced chemical equation can be interpreted both as the relative numbers of molecules involved in the reaction and as the relative number of moles. For example, in the balanced equation:
$\ce{2H2(g) + O2(g) \rightarrow 2H_2O(l)} \nonumber \nonumber$
the production of two moles of water would require the consumption of 2 moles of $H_2$ and one mole of $O_2$. Therefore, when considering this particular reaction
• 2 moles of H2
• 1 mole of O2 and
• 2 moles of H2O
would be considered to be stoichiometrically equivalent quantitites.
These stoichiometric relationships, derived from balanced equations, can be used to determine expected amounts of products given amounts of reactants. For example, how many moles of $H_2O$ would be produced from 1.57 moles of $O_2$?
$(1.57\; mol\; O_2) \left( \dfrac{2\; mol H_2O}{1\;mol\;O_2} \right) = 3.14\; mol\; H_2O \nonumber \nonumber$
The ratio $\left( \dfrac{2\; mol\l H_2O}{1\;mol\;O_2} \right)$ is the stoichiometric relationship between $H_2O$ and $O_2$ from the balanced equation for this reaction.
Example $6$
For the combustion of butane ($C_4H_{10}$) the balanced equation is:
$\ce{2C4H_{10} (l) + 13O2(g) \rightarrow 8CO2(g) + 10H2O(l)} \nonumber \nonumber$
Calculate the mass of $CO_2$ that is produced in burning 1.00 gram of $C_4H_{10}$.
Solution
First of all we need to calculate how many moles of butane we have in a 1.00 gram sample:
$(1.00\; g\; C_4H_{10}) \left(\dfrac{1\; mol\; C_4H_{10}}{58.0\;g\; C_4H_{10}}\right) = 1.72 \times 10^{-2} \; mol\; C_4H_{10} \nonumber \nonumber$
Now, the stoichiometric relationship between $C_4H_{10}$ and $CO_2$ is:
$\left( \dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}}\right) \nonumber \nonumber$
Therefore:
$\left(\dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}} \right) \times 1.72 \times 10^{-2} \; mol\; C_4H_{10} = 6.88 \times 10^{-2} \; mol\; CO_2 \nonumber \nonumber$
The question called for the determination of the mass of $CO_2$ produced, thus we have to convert moles of $CO_2$ into grams (by using the molecular weight of $CO_2$):
$6.88 \times 10^{-2} \; mol\; CO_2 \left( \dfrac{44.0\; g\; CO_2}{1\; mol\; CO_2} \right) = 3.03\;g \; CO_2 \nonumber \nonumber$
Thus, the overall sequence of steps to solve this problem were:
In a similar way we could determine the mass of water produced, or oxygen consumed, etc.
Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios): Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios), YouTube(opens in new window) [youtu.be]
Summary
To analyze chemical transformations, it is essential to use a standardized unit of measure called the mole. The molecular mass and the formula mass of a compound are obtained by adding together the atomic masses of the atoms present in the molecular formula or empirical formula, respectively; the units of both are atomic mass units (amu). The mole is a unit used to measure the number of atoms, molecules, or (in the case of ionic compounds) formula units in a given mass of a substance. The mole is defined as the amount of substance that contains the number of carbon atoms in exactly 12 g of carbon-12, Avogadro’s number (6.022 × 1023) of atoms of carbon-12. The molar mass of a substance is defined as the mass of 1 mol of that substance, expressed in grams per mole, and is equal to the mass of 6.022 × 1023 atoms, molecules, or formula units of that substance. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/03%3A_Stoichiometry_of_Formulas_and_Equation/3.01%3A_The_Mole.txt |
Learning Objectives
• To describe the concentrations of solutions quantitatively
Many people have a qualitative idea of what is meant by concentration. Anyone who has made instant coffee or lemonade knows that too much powder gives a strongly flavored, highly concentrated drink, whereas too little results in a dilute solution that may be hard to distinguish from water. In chemistry, the concentration of a solution is the quantity of a solute that is contained in a particular quantity of solvent or solution. Knowing the concentration of solutes is important in controlling the stoichiometry of reactants for solution reactions. Chemists use many different methods to define concentrations, some of which are described in this section.
Molarity
The most common unit of concentration is molarity, which is also the most useful for calculations involving the stoichiometry of reactions in solution. The molarity (M) is defined as the number of moles of solute present in exactly 1 L of solution. It is, equivalently, the number of millimoles of solute present in exactly 1 mL of solution:
$molarity = \dfrac{moles\: of\: solute}{liters\: of\: solution} = \dfrac{mmoles\: of\: solute} {milliliters\: of\: solution} \label{4.5.1}$
The units of molarity are therefore moles per liter of solution (mol/L), abbreviated as $M$. An aqueous solution that contains 1 mol (342 g) of sucrose in enough water to give a final volume of 1.00 L has a sucrose concentration of 1.00 mol/L or 1.00 M. In chemical notation, square brackets around the name or formula of the solute represent the molar concentration of a solute. Therefore,
$[\rm{sucrose}] = 1.00\: M \nonumber$
is read as “the concentration of sucrose is 1.00 molar.” The relationships between volume, molarity, and moles may be expressed as either
$V_L M_{mol/L} = \cancel{L} \left( \dfrac{mol}{\cancel{L}} \right) = moles \label{4.5.2}$
or
$V_{mL} M_{mmol/mL} = \cancel{mL} \left( \dfrac{mmol} {\cancel{mL}} \right) = mmoles \label{4.5.3}$
Figure $1$ illustrates the use of Equations $\ref{4.5.2}$ and $\ref{4.5.3}$.
Example $1$: Calculating Moles from Concentration of NaOH
Calculate the number of moles of sodium hydroxide (NaOH) in 2.50 L of 0.100 M NaOH.
Given: identity of solute and volume and molarity of solution
Asked for: amount of solute in moles
Strategy:
Use either Equation \ref{4.5.2} or Equation \ref{4.5.3}, depending on the units given in the problem.
Solution:
Because we are given the volume of the solution in liters and are asked for the number of moles of substance, Equation \ref{4.5.2} is more useful:
$moles\: NaOH = V_L M_{mol/L} = (2 .50\: \cancel{L} ) \left( \dfrac{0.100\: mol } {\cancel{L}} \right) = 0 .250\: mol\: NaOH$
Exercise $1$: Calculating Moles from Concentration of Alanine
Calculate the number of millimoles of alanine, a biologically important molecule, in 27.2 mL of 1.53 M alanine.
Answer
41.6 mmol
Calculations Involving Molarity (M): Calculations Involving Molarity (M), YouTube(opens in new window) [youtu.be]
Concentrations are also often reported on a mass-to-mass (m/m) basis or on a mass-to-volume (m/v) basis, particularly in clinical laboratories and engineering applications. A concentration expressed on an m/m basis is equal to the number of grams of solute per gram of solution; a concentration on an m/v basis is the number of grams of solute per milliliter of solution. Each measurement can be expressed as a percentage by multiplying the ratio by 100; the result is reported as percent m/m or percent m/v. The concentrations of very dilute solutions are often expressed in parts per million (ppm), which is grams of solute per 106 g of solution, or in parts per billion (ppb), which is grams of solute per 109 g of solution. For aqueous solutions at 20°C, 1 ppm corresponds to 1 μg per milliliter, and 1 ppb corresponds to 1 ng per milliliter. These concentrations and their units are summarized in Table $1$.
Table $1$: Common Units of Concentration
Concentration Units
m/m g of solute/g of solution
m/v g of solute/mL of solution
ppm g of solute/106 g of solution
μg/mL
ppb g of solute/109 g of solution
ng/mL
The Preparation of Solutions
To prepare a solution that contains a specified concentration of a substance, it is necessary to dissolve the desired number of moles of solute in enough solvent to give the desired final volume of solution. Figure $1$ illustrates this procedure for a solution of cobalt(II) chloride dihydrate in ethanol. Note that the volume of the solvent is not specified. Because the solute occupies space in the solution, the volume of the solvent needed is almost always less than the desired volume of solution. For example, if the desired volume were 1.00 L, it would be incorrect to add 1.00 L of water to 342 g of sucrose because that would produce more than 1.00 L of solution. As shown in Figure $2$, for some substances this effect can be significant, especially for concentrated solutions.
Example $2$
The solution contains 10.0 g of cobalt(II) chloride dihydrate, CoCl2•2H2O, in enough ethanol to make exactly 500 mL of solution. What is the molar concentration of $\ce{CoCl2•2H2O}$?
Given: mass of solute and volume of solution
Asked for: concentration (M)
Strategy:
To find the number of moles of $\ce{CoCl2•2H2O}$, divide the mass of the compound by its molar mass. Calculate the molarity of the solution by dividing the number of moles of solute by the volume of the solution in liters.
Solution:
The molar mass of CoCl2•2H2O is 165.87 g/mol. Therefore,
$moles\: CoCl_2 \cdot 2H_2O = \left( \dfrac{10.0 \: \cancel{g}} {165 .87\: \cancel{g} /mol} \right) = 0 .0603\: mol \nonumber$
The volume of the solution in liters is
$volume = 500\: \cancel{mL} \left( \dfrac{1\: L} {1000\: \cancel{mL}} \right) = 0 .500\: L \nonumber$
Molarity is the number of moles of solute per liter of solution, so the molarity of the solution is
$molarity = \dfrac{0.0603\: mol} {0.500\: L} = 0.121\: M = CoCl_2 \cdot H_2O \nonumber$
Exercise $2$
The solution shown in Figure $2$ contains 90.0 g of (NH4)2Cr2O7 in enough water to give a final volume of exactly 250 mL. What is the molar concentration of ammonium dichromate?
Answer
$(NH_4)_2Cr_2O_7 = 1.43\: M \nonumber$
To prepare a particular volume of a solution that contains a specified concentration of a solute, we first need to calculate the number of moles of solute in the desired volume of solution using the relationship shown in Equation $\ref{4.5.2}$. We then convert the number of moles of solute to the corresponding mass of solute needed. This procedure is illustrated in Example $3$.
Example $3$: D5W Solution
The so-called D5W solution used for the intravenous replacement of body fluids contains 0.310 M glucose. (D5W is an approximately 5% solution of dextrose [the medical name for glucose] in water.) Calculate the mass of glucose necessary to prepare a 500 mL pouch of D5W. Glucose has a molar mass of 180.16 g/mol.
Given: molarity, volume, and molar mass of solute
Asked for: mass of solute
Strategy:
1. Calculate the number of moles of glucose contained in the specified volume of solution by multiplying the volume of the solution by its molarity.
2. Obtain the mass of glucose needed by multiplying the number of moles of the compound by its molar mass.
Solution:
A We must first calculate the number of moles of glucose contained in 500 mL of a 0.310 M solution:
$V_L M_{mol/L} = moles$
$500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .310\: mol\: glucose} {1\: \cancel{L}} \right) = 0 .155\: mol\: glucose$
B We then convert the number of moles of glucose to the required mass of glucose:
$mass \: of \: glucose = 0.155 \: \cancel{mol\: glucose} \left( \dfrac{180.16 \: g\: glucose} {1\: \cancel{mol\: glucose}} \right) = 27.9 \: g \: glucose$
Exercise $3$
Another solution commonly used for intravenous injections is normal saline, a 0.16 M solution of sodium chloride in water. Calculate the mass of sodium chloride needed to prepare 250 mL of normal saline solution.
Answer
2.3 g NaCl
A solution of a desired concentration can also be prepared by diluting a small volume of a more concentrated solution with additional solvent. A stock solution is a commercially prepared solution of known concentration and is often used for this purpose. Diluting a stock solution is preferred because the alternative method, weighing out tiny amounts of solute, is difficult to carry out with a high degree of accuracy. Dilution is also used to prepare solutions from substances that are sold as concentrated aqueous solutions, such as strong acids.
The procedure for preparing a solution of known concentration from a stock solution is shown in Figure $3$. It requires calculating the number of moles of solute desired in the final volume of the more dilute solution and then calculating the volume of the stock solution that contains this amount of solute. Remember that diluting a given quantity of stock solution with solvent does not change the number of moles of solute present. The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution is therefore
$(V_s)(M_s) = moles\: of\: solute = (V_d)(M_d)\label{4.5.4}$
where the subscripts s and d indicate the stock and dilute solutions, respectively. Example $4$ demonstrates the calculations involved in diluting a concentrated stock solution.
Example $4$
What volume of a 3.00 M glucose stock solution is necessary to prepare 2500 mL of the D5W solution in Example $3$?
Given: volume and molarity of dilute solution
Asked for: volume of stock solution
Strategy:
1. Calculate the number of moles of glucose contained in the indicated volume of dilute solution by multiplying the volume of the solution by its molarity.
2. To determine the volume of stock solution needed, divide the number of moles of glucose by the molarity of the stock solution.
Solution:
A The D5W solution in Example 4.5.3 was 0.310 M glucose. We begin by using Equation 4.5.4 to calculate the number of moles of glucose contained in 2500 mL of the solution:
$moles\: glucose = 2500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .310\: mol\: glucose} {1\: \cancel{L}} \right) = 0 .775\: mol\: glucose \nonumber$
B We must now determine the volume of the 3.00 M stock solution that contains this amount of glucose:
$volume\: of\: stock\: soln = 0 .775\: \cancel{mol\: glucose} \left( \dfrac{1\: L} {3 .00\: \cancel{mol\: glucose}} \right) = 0 .258\: L\: or\: 258\: mL \nonumber$
In determining the volume of stock solution that was needed, we had to divide the desired number of moles of glucose by the concentration of the stock solution to obtain the appropriate units. Also, the number of moles of solute in 258 mL of the stock solution is the same as the number of moles in 2500 mL of the more dilute solution; only the amount of solvent has changed. The answer we obtained makes sense: diluting the stock solution about tenfold increases its volume by about a factor of 10 (258 mL → 2500 mL). Consequently, the concentration of the solute must decrease by about a factor of 10, as it does (3.00 M → 0.310 M).
We could also have solved this problem in a single step by solving Equation 4.5.4 for Vs and substituting the appropriate values:
$V_s = \dfrac{( V_d )(M_d )}{M_s} = \dfrac{(2 .500\: L)(0 .310\: \cancel{M} )} {3 .00\: \cancel{M}} = 0 .258\: L \nonumber$
As we have noted, there is often more than one correct way to solve a problem.
Exercise $4$
What volume of a 5.0 M NaCl stock solution is necessary to prepare 500 mL of normal saline solution (0.16 M NaCl)?
Answer
16 mL
Ion Concentrations in Solution
In Example $2$, the concentration of a solution containing 90.00 g of ammonium dichromate in a final volume of 250 mL were calculated to be 1.43 M. Let’s consider in more detail exactly what that means. Ammonium dichromate is an ionic compound that contains two NH4+ ions and one Cr2O72 ion per formula unit. Like other ionic compounds, it is a strong electrolyte that dissociates in aqueous solution to give hydrated NH4+ and Cr2O72 ions:
$(NH_4 )_2 Cr_2 O_7 (s) \xrightarrow {H_2 O(l)} 2NH_4^+ (aq) + Cr_2 O_7^{2-} (aq)\label{4.5.5}$
Thus 1 mol of ammonium dichromate formula units dissolves in water to produce 1 mol of Cr2O72 anions and 2 mol of NH4+ cations (see Figure $4$).
When carrying out a chemical reaction using a solution of a salt such as ammonium dichromate, it is important to know the concentration of each ion present in the solution. If a solution contains 1.43 M (NH4)2Cr2O7, then the concentration of Cr2O72 must also be 1.43 M because there is one Cr2O72 ion per formula unit. However, there are two NH4+ ions per formula unit, so the concentration of NH4+ ions is 2 × 1.43 M = 2.86 M. Because each formula unit of (NH4)2Cr2O7 produces three ions when dissolved in water (2NH4+ + 1Cr2O72), the total concentration of ions in the solution is 3 × 1.43 M = 4.29 M.
Concentration of Ions in Solution from a Soluble Salt: Concentration of Ions in Solution from a Soluble Salt, YouTube(opens in new window) [youtu.be]
Example $5$
What are the concentrations of all species derived from the solutes in these aqueous solutions?
1. 0.21 M NaOH
2. 3.7 M (CH3)2CHOH
3. 0.032 M In(NO3)3
Given: molarity
Asked for: concentrations
Strategy:
A Classify each compound as either a strong electrolyte or a nonelectrolyte.
B If the compound is a nonelectrolyte, its concentration is the same as the molarity of the solution. If the compound is a strong electrolyte, determine the number of each ion contained in one formula unit. Find the concentration of each species by multiplying the number of each ion by the molarity of the solution.
Solution:
1. Sodium hydroxide is an ionic compound that is a strong electrolyte (and a strong base) in aqueous solution: $NaOH(s) \xrightarrow {H_2 O(l)} Na^+ (aq) + OH^- (aq)$
B Because each formula unit of NaOH produces one Na+ ion and one OH ion, the concentration of each ion is the same as the concentration of NaOH: [Na+] = 0.21 M and [OH] = 0.21 M.
2. A The formula (CH3)2CHOH represents 2-propanol (isopropyl alcohol) and contains the –OH group, so it is an alcohol. Recall from Section 4.1 that alcohols are covalent compounds that dissolve in water to give solutions of neutral molecules. Thus alcohols are nonelectrolytes.
B The only solute species in solution is therefore (CH3)2CHOH molecules, so [(CH3)2CHOH] = 3.7 M.
3. A Indium nitrate is an ionic compound that contains In3+ ions and NO3 ions, so we expect it to behave like a strong electrolyte in aqueous solution:
$In(NO _3 ) _3 (s) \xrightarrow {H_ 2 O(l)} In ^{3+} (aq) + 3NO _3^- (aq)$
B One formula unit of In(NO3)3 produces one In3+ ion and three NO3 ions, so a 0.032 M In(NO3)3 solution contains 0.032 M In3+ and 3 × 0.032 M = 0.096 M NO3—that is, [In3+] = 0.032 M and [NO3] = 0.096 M.
Exercise $5$
What are the concentrations of all species derived from the solutes in these aqueous solutions?
1. 0.0012 M Ba(OH)2
2. 0.17 M Na2SO4
3. 0.50 M (CH3)2CO, commonly known as acetone
Answer a
$[Ba^{2+}] = 0.0012\: M; \: [OH^-] = 0.0024\: M$
Answer b
$[Na^+] = 0.34\: M; \: [SO_4^{2-}] = 0.17\: M$
Answer c
$[(CH_3)_2CO] = 0.50\: M$
Summary
Solution concentrations are typically expressed as molarities and can be prepared by dissolving a known mass of solute in a solvent or diluting a stock solution.
• definition of molarity: $molarity = \dfrac{moles\: of\: solute}{liters\: of\: solution} = \dfrac{mmoles\: of\: solute} {milliliters\: of\: solution} \nonumber$
• relationship among volume, molarity, and moles: $V_L M_{mol/L} = \cancel{L} \left( \dfrac{mol}{\cancel{L}} \right) = moles \nonumber$
• relationship between volume and concentration of stock and dilute solutions: $(V_s)(M_s) = moles\: of\: solute = (V_d)(M_d) \nonumber$
The concentration of a substance is the quantity of solute present in a given quantity of solution. Concentrations are usually expressed in terms of molarity, defined as the number of moles of solute in 1 L of solution. Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/04%3A_Three_Major_Classes_of_Chemical_Reactions/4.01%3A_Solution_Concentration_and_the_Role_of_Water_as_a_Solvent.txt |
Learning Objectives
• To identify a precipitation reaction and predict solubilities.
Exchange (Double-Displacement) Reactions
A precipitation reaction is a reaction that yields an insoluble product—a precipitate—when two solutions are mixed. We described a precipitation reaction in which a colorless solution of silver nitrate was mixed with a yellow-orange solution of potassium dichromate to give a reddish precipitate of silver dichromate:
$\ce{AgNO_3(aq) + K_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + KNO_3(aq)} \label{4.2.1}$
This unbalanced equation has the general form of an exchange reaction:
$\overbrace{\ce{AC}}^{\text{soluble}} + \overbrace{\ce{BD}}^{\text{soluble}} \rightarrow \underbrace{\ce{AD}}_{\text{insoluble}} + \overbrace{\ce{BC}}^{\text{soluble}} \label{4.2.2}$
The solubility and insoluble annotations are specific to the reaction in Equation \ref{4.2.1} and not characteristic of all exchange reactions (e.g., both products can be soluble or insoluble). Precipitation reactions are a subclass of exchange reactions that occur between ionic compounds when one of the products is insoluble. Because both components of each compound change partners, such reactions are sometimes called double-displacement reactions. Two important uses of precipitation reactions are to isolate metals that have been extracted from their ores and to recover precious metals for recycling.
Video $1$: Mixing Potassium Chromate and Silver Nitrate together to initiate a precipitation reaction (Equation $\ref{4.2.1}$).
While full chemical equations show the identities of the reactants and the products and give the stoichiometries of the reactions, they are less effective at describing what is actually occurring in solution. In contrast, equations that show only the hydrated species focus our attention on the chemistry that is taking place and allow us to see similarities between reactions that might not otherwise be apparent.
Let’s consider the reaction of silver nitrate with potassium dichromate above. When aqueous solutions of silver nitrate and potassium dichromate are mixed, silver dichromate forms as a red solid. The overall balanced chemical equation for the reaction shows each reactant and product as undissociated, electrically neutral compounds:
$\ce{2AgNO_3(aq)} + \ce{K_2Cr_2O_7(aq)} \rightarrow \ce{Ag_2Cr_2O_7(s) }+ \ce{2KNO_3(aq)} \label{4.2.1a}$
Although Equation $\ref{4.2.1a}$ gives the identity of the reactants and the products, it does not show the identities of the actual species in solution. Because ionic substances such as $\ce{AgNO3}$ and $\ce{K2Cr2O7}$ are strong electrolytes (i.e., they dissociate completely in aqueous solution to form ions). In contrast, because $\ce{Ag2Cr2O7}$ is not very soluble, it separates from the solution as a solid. To find out what is actually occurring in solution, it is more informative to write the reaction as a complete ionic equation showing which ions and molecules are hydrated and which are present in other forms and phases:
$\ce{2Ag^{+}(aq) + 2NO_3^{-} (aq) + 2K^{+}(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s) + 2K^{+}(aq) + 2NO_3^{-}(aq)}\label{4.2.2a}$
Note that $\ce{K^+ (aq)}$ and $\ce{NO3^{−} (aq)}$ ions are present on both sides of Equation $\ref{4.2.2a}$ and their coefficients are the same on both sides. These ions are called spectator ions because they do not participate in the actual reaction. Canceling the spectator ions gives the net ionic equation, which shows only those species that participate in the chemical reaction:
$2Ag^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s)\label{4.2.3}$
Both mass and charge must be conserved in chemical reactions because the numbers of electrons and protons do not change. For charge to be conserved, the sum of the charges of the ions multiplied by their coefficients must be the same on both sides of the equation. In Equation $\ref{4.2.3}$, the charge on the left side is 2(+1) + 1(−2) = 0, which is the same as the charge of a neutral $\ce{Ag2Cr2O7}$ formula unit on the right side.
By eliminating the spectator ions, we can focus on the chemistry that takes place in a solution. For example, the overall chemical equation for the reaction between silver fluoride and ammonium dichromate is as follows:
$2AgF(aq) + (NH_4)_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + 2NH_4F(aq)\label{4.2.4}$
The complete ionic equation for this reaction is as follows:
$\ce{2Ag^{+}(aq)} + \cancel{\ce{2F^{-}(aq)}} + \cancel{\ce{2NH_4^{+}(aq)}} + \ce{Cr_2O_7^{2-}(aq)} \rightarrow \ce{Ag_2Cr_2O_7(s)} + \cancel{\ce{2NH_4^{+}(aq)}} + \cancel{\ce{2F^{-}(aq)}} \label{4.2.5}$
Because two $\ce{NH4^{+}(aq)}$ and two $\ce{F^{−} (aq)}$ ions appear on both sides of Equation $\ref{4.2.5}$, they are spectator ions. They can therefore be canceled to give the net ionic equation (Equation $\ref{4.2.6}$), which is identical to Equation $\ref{4.2.3}$:
$\ce{2Ag^{+}(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s)} \label{4.2.6}$
If we look at net ionic equations, it becomes apparent that many different combinations of reactants can result in the same net chemical reaction. For example, we can predict that silver fluoride could be replaced by silver nitrate in the preceding reaction without affecting the outcome of the reaction.
Determining the Products for Precipitation Reactions: Determining the Products for Precipitation Reactions, YouTube(opens in new window) [youtu.be]
Example $1$: Balancing Precipitation Equations
Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous barium nitrate with aqueous sodium phosphate to give solid barium phosphate and a solution of sodium nitrate.
Given: reactants and products
Asked for: overall, complete ionic, and net ionic equations
Strategy:
Write and balance the overall chemical equation. Write all the soluble reactants and products in their dissociated form to give the complete ionic equation; then cancel species that appear on both sides of the complete ionic equation to give the net ionic equation.
Solution:
From the information given, we can write the unbalanced chemical equation for the reaction:
$\ce{Ba(NO_3)_2(aq) + Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s) + NaNO_3(aq)} \nonumber$
Because the product is Ba3(PO4)2, which contains three Ba2+ ions and two PO43 ions per formula unit, we can balance the equation by inspection:
$\ce{3Ba(NO_3)_2(aq) + 2Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s) + 6NaNO_3(aq)} \nonumber$
This is the overall balanced chemical equation for the reaction, showing the reactants and products in their undissociated form. To obtain the complete ionic equation, we write each soluble reactant and product in dissociated form:
$\ce{3Ba^{2+}(aq)} + \cancel{\ce{6NO_3^{-}(aq)}} + \cancel{\ce{6Na^{+} (aq)}} + \ce{2PO_4^{3-} (aq)} \rightarrow \ce{Ba_3(PO_4)_2(s)} + \cancel{\ce{6Na^+(aq)}} + \cancel{\ce{6NO_3^{-}(aq)}} \nonumber$
The six NO3(aq) ions and the six Na+(aq) ions that appear on both sides of the equation are spectator ions that can be canceled to give the net ionic equation:
$\ce{3Ba^{2+}(aq) + 2PO_4^{3-}(aq) \rightarrow Ba_3(PO_4)_2(s)} \nonumber$
Exercise $1$: Mixing Silver Fluoride with Sodium Phosphate
Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous silver fluoride with aqueous sodium phosphate to give solid silver phosphate and a solution of sodium fluoride.
Answer
overall chemical equation:
$\ce{3AgF(aq) + Na_3PO_4(aq) \rightarrow Ag_3PO_4(s) + 3NaF(aq) } \nonumber$
complete ionic equation:
$\ce{3Ag^+(aq) + 3F^{-}(aq) + 3Na^{+}(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s) + 3Na^{+}(aq) + 3F^{-}(aq) } \nonumber$
net ionic equation:
$\ce{3Ag^{+}(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s)} \nonumber$
So far, we have always indicated whether a reaction will occur when solutions are mixed and, if so, what products will form. As you advance in chemistry, however, you will need to predict the results of mixing solutions of compounds, anticipate what kind of reaction (if any) will occur, and predict the identities of the products. Students tend to think that this means they are supposed to “just know” what will happen when two substances are mixed. Nothing could be further from the truth: an infinite number of chemical reactions is possible, and neither you nor anyone else could possibly memorize them all. Instead, you must begin by identifying the various reactions that could occur and then assessing which is the most probable (or least improbable) outcome.
The most important step in analyzing an unknown reaction is to write down all the species—whether molecules or dissociated ions—that are actually present in the solution (not forgetting the solvent itself) so that you can assess which species are most likely to react with one another. The easiest way to make that kind of prediction is to attempt to place the reaction into one of several familiar classifications, refinements of the five general kinds of reactions (acid–base, exchange, condensation, cleavage, and oxidation–reduction reactions). In the sections that follow, we discuss three of the most important kinds of reactions that occur in aqueous solutions: precipitation reactions (also known as exchange reactions), acid–base reactions, and oxidation–reduction reactions.
Predicting Solubilities
Table $1$ gives guidelines for predicting the solubility of a wide variety of ionic compounds. To determine whether a precipitation reaction will occur, we identify each species in the solution and then refer to Table $1$ to see which, if any, combination(s) of cation and anion are likely to produce an insoluble salt. In doing so, it is important to recognize that soluble and insoluble are relative terms that span a wide range of actual solubilities. We will discuss solubilities in more detail later, where you will learn that very small amounts of the constituent ions remain in solution even after precipitation of an “insoluble” salt. For our purposes, however, we will assume that precipitation of an insoluble salt is complete.
Table $1$: Guidelines for Predicting the Solubility of Ionic Compounds in Water
Soluble Exceptions
Rule 1 most salts that contain an alkali metal (Li+, Na+, K+, Rb+, and Cs+) and ammonium (NH4+)
Rule 2 most salts that contain the nitrate (NO3) anion
Rule 3 most salts of anions derived from monocarboxylic acids (e.g., CH3CO2) but not silver acetate and salts of long-chain carboxylates
Rule 4 most chloride, bromide, and iodide salts but not salts of metal ions located on the lower right side of the periodic table (e.g., Cu+, Ag+, Pb2+, and Hg22+).
Insoluble Exceptions
Rule 5 most salts that contain the hydroxide (OH) and sulfide (S2−) anions but not salts of the alkali metals (group 1), the heavier alkaline earths (Ca2+, Sr2+, and Ba2+ in group 2), and the NH4+ ion.
Rule 6 most carbonate (CO32) and phosphate (PO43) salts but not salts of the alkali metals or the NH4+ ion.
Rule 7 most sulfate (SO42) salts that contain main group cations with a charge ≥ +2 but not salts of +1 cations, Mg2+, and dipositive transition metal cations (e.g., Ni2+)
Just as important as predicting the product of a reaction is knowing when a chemical reaction will not occur. Simply mixing solutions of two different chemical substances does not guarantee that a reaction will take place. For example, if 500 mL of a 1.0 M aqueous NaCl solution is mixed with 500 mL of a 1.0 M aqueous KBr solution, the final solution has a volume of 1.00 L and contains 0.50 M Na+(aq), 0.50 M Cl(aq), 0.50 M K+(aq), and 0.50 M Br(aq). As you will see in the following sections, none of these species reacts with any of the others. When these solutions are mixed, the only effect is to dilute each solution with the other (Figure $1$).
Example $2$
Using the information in Table $1$, predict what will happen in each case involving strong electrolytes. Write the net ionic equation for any reaction that occurs.
1. Aqueous solutions of barium chloride and lithium sulfate are mixed.
2. Aqueous solutions of rubidium hydroxide and cobalt(II) chloride are mixed.
3. Aqueous solutions of strontium bromide and aluminum nitrate are mixed.
4. Solid lead(II) acetate is added to an aqueous solution of ammonium iodide.
Given: reactants
Asked for: reaction and net ionic equation
Strategy:
1. Identify the ions present in solution and write the products of each possible exchange reaction.
2. Refer to Table $1$ to determine which, if any, of the products is insoluble and will therefore form a precipitate. If a precipitate forms, write the net ionic equation for the reaction.
Solution:
A Because barium chloride and lithium sulfate are strong electrolytes, each dissociates completely in water to give a solution that contains the constituent anions and cations. Mixing the two solutions initially gives an aqueous solution that contains Ba2+, Cl, Li+, and SO42 ions. The only possible exchange reaction is to form LiCl and BaSO4:
B We now need to decide whether either of these products is insoluble. Table $1$ shows that LiCl is soluble in water (rules 1 and 4), but BaSO4 is not soluble in water (rule 5). Thus BaSO4 will precipitate according to the net ionic equation
$Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s) \nonumber$
Although soluble barium salts are toxic, BaSO4 is so insoluble that it can be used to diagnose stomach and intestinal problems without being absorbed into tissues. An outline of the digestive organs appears on x-rays of patients who have been given a “barium milkshake” or a “barium enema”—a suspension of very fine BaSO4 particles in water.
1. A Rubidium hydroxide and cobalt(II) chloride are strong electrolytes, so when aqueous solutions of these compounds are mixed, the resulting solution initially contains Rb+, OH, Co2+, and Cl ions. The possible products of an exchange reaction are rubidium chloride and cobalt(II) hydroxide):
B According to Table $1$, RbCl is soluble (rules 1 and 4), but Co(OH)2 is not soluble (rule 5). Hence Co(OH)2 will precipitate according to the following net ionic equation:
$Co^{2+}(aq) + 2OH^-(aq) \rightarrow Co(OH)_2(s)$
2. A When aqueous solutions of strontium bromide and aluminum nitrate are mixed, we initially obtain a solution that contains Sr2+, Br, Al3+, and NO3 ions. The two possible products from an exchange reaction are aluminum bromide and strontium nitrate:
B According to Table $1$, both AlBr3 (rule 4) and Sr(NO3)2 (rule 2) are soluble. Thus no net reaction will occur.
1. A According to Table $1$, lead acetate is soluble (rule 3). Thus solid lead acetate dissolves in water to give Pb2+ and CH3CO2 ions. Because the solution also contains NH4+ and I ions, the possible products of an exchange reaction are ammonium acetate and lead(II) iodide:
B According to Table $1$, ammonium acetate is soluble (rules 1 and 3), but PbI2 is insoluble (rule 4). Thus Pb(C2H3O2)2 will dissolve, and PbI2 will precipitate. The net ionic equation is as follows:
$Pb^{2+} (aq) + 2I^-(aq) \rightarrow PbI_2(s)$
Exercise $2$
Using the information in Table $1$, predict what will happen in each case involving strong electrolytes. Write the net ionic equation for any reaction that occurs.
1. An aqueous solution of strontium hydroxide is added to an aqueous solution of iron(II) chloride.
2. Solid potassium phosphate is added to an aqueous solution of mercury(II) perchlorate.
3. Solid sodium fluoride is added to an aqueous solution of ammonium formate.
4. Aqueous solutions of calcium bromide and cesium carbonate are mixed.
Answer a
$Fe^{2+}(aq) + 2OH^-(aq) \rightarrow Fe(OH)_2(s)$
Answer b
$2PO_4^{3-}(aq) + 3Hg^{2+}(aq) \rightarrow Hg_3(PO_4)_2(s)$
Answer c
$NaF(s)$ dissolves; no net reaction
Answer d
$Ca^{2+}(aq) + CO_3^{2-}(aq) \rightarrow CaCO_3(s)$
Predicting the Solubility of Ionic Compounds: Predicting the Solubility of Ionic Compounds, YouTube(opens in new window) [youtu.be] (opens in new window) | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/04%3A_Three_Major_Classes_of_Chemical_Reactions/4.03%3A_Precipitation_Reactions.txt |
Learning Objectives
• To know the characteristic properties of acids and bases.
Acid–base reactions are essential in both biochemistry and industrial chemistry. Moreover, many of the substances we encounter in our homes, the supermarket, and the pharmacy are acids or bases. For example, aspirin is an acid (acetylsalicylic acid), and antacids are bases. In fact, every amateur chef who has prepared mayonnaise or squeezed a wedge of lemon to marinate a piece of fish has carried out an acid–base reaction. Before we discuss the characteristics of such reactions, let’s first describe some of the properties of acids and bases.
Definitions of Acids and Bases
We can define acids as substances that dissolve in water to produce H+ ions, whereas bases are defined as substances that dissolve in water to produce OH ions. In fact, this is only one possible set of definitions. Although the general properties of acids and bases have been known for more than a thousand years, the definitions of acid and base have changed dramatically as scientists have learned more about them. In ancient times, an acid was any substance that had a sour taste (e.g., vinegar or lemon juice), caused consistent color changes in dyes derived from plants (e.g., turning blue litmus paper red), reacted with certain metals to produce hydrogen gas and a solution of a salt containing a metal cation, and dissolved carbonate salts such as limestone (CaCO3) with the evolution of carbon dioxide. In contrast, a base was any substance that had a bitter taste, felt slippery to the touch, and caused color changes in plant dyes that differed diametrically from the changes caused by acids (e.g., turning red litmus paper blue). Although these definitions were useful, they were entirely descriptive.
The Arrhenius Definition of Acids and Bases
The first person to define acids and bases in detail was the Swedish chemist Svante Arrhenius (1859–1927; Nobel Prize in Chemistry, 1903). According to the Arrhenius definition, an acid is a substance like hydrochloric acid that dissolves in water to produce H+ ions (protons; Equation $\ref{4.3.1}$), and a base is a substance like sodium hydroxide that dissolves in water to produce hydroxide (OH) ions (Equation $\ref{4.3.2}$):
$\underset{an\: Arrhenius\: acid}{HCl_{(g)}} \xrightarrow {H_2 O_{(l)}} H^+_{(aq)} + Cl^-_{(aq)} \label{4.3.1}$
$\underset{an\: Arrhenius\: base}{NaOH_{(s)}} \xrightarrow {H_2O_{(l)}} Na^+_{(aq)} + OH^-_{(aq)} \label{4.3.2}$
According to Arrhenius, the characteristic properties of acids and bases are due exclusively to the presence of H+ and OH ions, respectively, in solution. Although Arrhenius’s ideas were widely accepted, his definition of acids and bases had two major limitations:
1. First, because acids and bases were defined in terms of ions obtained from water, the Arrhenius concept applied only to substances in aqueous solution.
2. Second, and more important, the Arrhenius definition predicted that only substances that dissolve in water to produce $H^+$ and $OH^−$ ions should exhibit the properties of acids and bases, respectively. For example, according to the Arrhenius definition, the reaction of ammonia (a base) with gaseous HCl (an acid) to give ammonium chloride (Equation $\ref{4.3.3}$) is not an acid–base reaction because it does not involve $H^+$ and $OH^−$:
$NH_{3\;(g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)} \label{4.3.3}$
The Brønsted–Lowry Definition of Acids and Bases
Because of the limitations of the Arrhenius definition, a more general definition of acids and bases was needed. One was proposed independently in 1923 by the Danish chemist J. N. Brønsted (1879–1947) and the British chemist T. M. Lowry (1874–1936), who defined acid–base reactions in terms of the transfer of a proton (H+ ion) from one substance to another.
According to Brønsted and Lowry, an acid (A substance with at least one hydrogen atom that can dissociate to form an anion and an $H^+$ ion (a proton) in aqueous solution, thereby forming an acidic solution) is any substance that can donate a proton, and a base (a substance that produces one or more hydroxide ions ($OH^-$ and a cation when dissolved in aqueous solution, thereby forming a basic solution) is any substance that can accept a proton. The Brønsted–Lowry definition of an acid is essentially the same as the Arrhenius definition, except that it is not restricted to aqueous solutions. The Brønsted–Lowry definition of a base, however, is far more general because the hydroxide ion is just one of many substances that can accept a proton. Ammonia, for example, reacts with a proton to form $NH_4^+$, so in Equation $\ref{4.3.3}$, $NH_3$ is a Brønsted–Lowry base and $HCl$ is a Brønsted–Lowry acid. Because of its more general nature, the Brønsted–Lowry definition is used throughout this text unless otherwise specified.
Polyprotic Acids
Acids differ in the number of protons they can donate. For example, monoprotic acids (a compound that is capable of donating one proton per molecule) are compounds that are capable of donating a single proton per molecule. Monoprotic acids include HF, HCl, HBr, HI, HNO3, and HNO2. All carboxylic acids that contain a single −CO2H group, such as acetic acid (CH3CO2H), are monoprotic acids, dissociating to form RCO2 and H+. A compound that can donate more than one proton per molecule is known as a polyprotic acid. For example, H2SO4 can donate two H+ ions in separate steps, so it is a diprotic acid (a compound that can donate two protons per molecule in separate steps) and H3PO4, which is capable of donating three protons in successive steps, is a triprotic acid (a compound that can donate three protons per molecule in separate steps), (Equation $\ref{4.3.4}$, Equation $\ref{4.3.5}$, and Equation $\ref{4.3.6}$):
$H_3 PO_4 (l) \overset{H_2 O(l)}{\rightleftharpoons} H ^+ ( a q ) + H_2 PO_4 ^- (aq) \label{4.3.4}$
$H_2 PO_4 ^- (aq) \rightleftharpoons H ^+ (aq) + HPO_4^{2-} (aq) \label{4.3.5}$
$HPO_4^{2-} (aq) \rightleftharpoons H^+ (aq) + PO_4^{3-} (aq) \label{4.3.6}$
In chemical equations such as these, a double arrow is used to indicate that both the forward and reverse reactions occur simultaneously, so the forward reaction does not go to completion. Instead, the solution contains significant amounts of both reactants and products. Over time, the reaction reaches a state in which the concentration of each species in solution remains constant. The reaction is then said to be in equilibrium (the point at which the rates of the forward and reverse reactions become the same, so that the net composition of the system no longer changes with time).
Strengths of Acids and Bases
We will not discuss the strengths of acids and bases quantitatively until next semester. Qualitatively, however, we can state that strong acids react essentially completely with water to give $H^+$ and the corresponding anion. Similarly, strong bases dissociate essentially completely in water to give $OH^−$ and the corresponding cation. Strong acids and strong bases are both strong electrolytes. In contrast, only a fraction of the molecules of weak acids and weak bases react with water to produce ions, so weak acids and weak bases are also weak electrolytes. Typically less than 5% of a weak electrolyte dissociates into ions in solution, whereas more than 95% is present in undissociated form.
In practice, only a few strong acids are commonly encountered: HCl, HBr, HI, HNO3, HClO4, and H2SO4 (H3PO4 is only moderately strong). The most common strong bases are ionic compounds that contain the hydroxide ion as the anion; three examples are NaOH, KOH, and Ca(OH)2. Common weak acids include HCN, H2S, HF, oxoacids such as HNO2 and HClO, and carboxylic acids such as acetic acid. The ionization reaction of acetic acid is as follows:
$CH_3 CO_2 H(l) \overset{H_2 O(l)}{\rightleftharpoons} H^+ (aq) + CH_3 CO_2^- (aq) \label{4.3.7}$
Although acetic acid is very soluble in water, almost all of the acetic acid in solution exists in the form of neutral molecules (less than 1% dissociates). Sulfuric acid is unusual in that it is a strong acid when it donates its first proton (Equation $\ref{4.3.8}$) but a weak acid when it donates its second proton (Equation $\ref{4.3.9}$) as indicated by the single and double arrows, respectively:
$\underset{strong\: acid}{H_2 SO_4 (l)} \xrightarrow {H_2 O(l)} H ^+ (aq) + HSO_4 ^- (aq) \label{4.3.8}$
$\underset{weak\: acid}{HSO_4^- (aq)} \rightleftharpoons H^+ (aq) + SO_4^{2-} (aq) \label{4.3.9}$
Consequently, an aqueous solution of sulfuric acid contains $H^+_{(aq)}$ ions and a mixture of $HSO^-_{4\;(aq)}$ and $SO^{2−}_{4\;(aq)}$ ions, but no $H_2SO_4$ molecules. All other polyprotic acids, such as H3PO4, are weak acids.
The most common weak base is ammonia, which reacts with water to form small amounts of hydroxide ion:
$NH_3 (g) + H_2 O(l) \rightleftharpoons NH_4^+ (aq) + OH^- (aq) \label{4.3.10}$
Most of the ammonia (>99%) is present in the form of NH3(g). Amines, which are organic analogues of ammonia, are also weak bases, as are ionic compounds that contain anions derived from weak acids (such as S2−).
There is no correlation between the solubility of a substance and whether it is a strong electrolyte, a weak electrolyte, or a nonelectrolyte.
Definition of Strong/Weak Acids & Bases: Definition of Strong/Weak Acids & Bases, YouTube (opens in new window) [Definition of Strong] [Definition of Strong] [youtu.be] (opens in new window)
Table $1$ lists some common strong acids and bases. Acids other than the six common strong acids are almost invariably weak acids. The only common strong bases are the hydroxides of the alkali metals and the heavier alkaline earths (Ca, Sr, and Ba); any other bases you encounter are most likely weak. Remember that there is no correlation between solubility and whether a substance is a strong or a weak electrolyte! Many weak acids and bases are extremely soluble in water.
Table $1$: Common Strong Acids and Bases
Strong Acids Strong Bases
Hydrogen Halides Oxoacids Group 1 Hydroxides Hydroxides of Heavy Group 2 Elements
HCl HNO3 LiOH Ca(OH)2
HBr H2SO4 NaOH Sr(OH)2
HI HClO4 KOH Ba(OH)2
RbOH
CsOH
Example $1$: Acid Strength
Classify each compound as a strong acid, a weak acid, a strong base, a weak base, or none of these.
1. CH3CH2CO2H
2. CH3OH
3. Sr(OH)2
4. CH3CH2NH2
5. HBrO4
Given: compound
Asked for: acid or base strength
Strategy:
A Determine whether the compound is organic or inorganic.
B If inorganic, determine whether the compound is acidic or basic by the presence of dissociable H+ or OH ions, respectively. If organic, identify the compound as a weak base or a weak acid by the presence of an amine or a carboxylic acid group, respectively. Recall that all polyprotic acids except H2SO4 are weak acids.
Solution:
1. A This compound is propionic acid, which is organic. B It contains a carboxylic acid group analogous to that in acetic acid, so it must be a weak acid.
2. A CH3OH is methanol, an organic compound that contains the −OH group. B As a covalent compound, it does not dissociate to form the OH ion. Because it does not contain a carboxylic acid (−CO2H) group, methanol also cannot dissociate to form H+(aq) ions. Thus we predict that in aqueous solution methanol is neither an acid nor a base.
3. A Sr(OH)2 is an inorganic compound that contains one Sr2+ and two OH ions per formula unit. B We therefore expect it to be a strong base, similar to Ca(OH)2.
4. A CH3CH2NH2 is an amine (ethylamine), an organic compound in which one hydrogen of ammonia has been replaced by an R group. B Consequently, we expect it to behave similarly to ammonia (Equation $\ref{4.3.7}$), reacting with water to produce small amounts of the OH ion. Ethylamine is therefore a weak base.
5. A HBrO4 is perbromic acid, an inorganic compound. B It is not listed in Table $1$ as one of the common strong acids, but that does not necessarily mean that it is a weak acid. If you examine the periodic table, you can see that Br lies directly below Cl in group 17. We might therefore expect that HBrO4 is chemically similar to HClO4, a strong acid—and, in fact, it is.
Exercise $1$: Acid Strength
Classify each compound as a strong acid, a weak acid, a strong base, a weak base, or none of these.
1. Ba(OH)2
2. HIO4
3. CH3CH2CH2CO2H
4. (CH3)2NH
5. CH2O
Answer a
strong base
Answer b
strong acid
Answer c
weak acid
Answer d
weak base
Answer e
none of these; formaldehyde is a neutral molecule | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/04%3A_Three_Major_Classes_of_Chemical_Reactions/4.04%3A_Acid-Base_Reactions.txt |
Learning Objectives
• To identify oxidation–reduction reactions in solution.
The term oxidation was first used to describe reactions in which metals react with oxygen in air to produce metal oxides. When iron is exposed to air in the presence of water, for example, the iron turns to rust—an iron oxide. When exposed to air, aluminum metal develops a continuous, transparent layer of aluminum oxide on its surface. In both cases, the metal acquires a positive charge by transferring electrons to the neutral oxygen atoms of an oxygen molecule. As a result, the oxygen atoms acquire a negative charge and form oxide ions (O2−). Because the metals have lost electrons to oxygen, they have been oxidized; oxidation is therefore the loss of electrons. Conversely, because the oxygen atoms have gained electrons, they have been reduced, so reduction is the gain of electrons. For every oxidation, there must be an associated reduction. Therefore, these reactions are known as oxidation-reduction reactions, or "redox" reactions for short.
Any oxidation must ALWAYS be accompanied by a reduction and vice versa.
Originally, the term reduction referred to the decrease in mass observed when a metal oxide was heated with carbon monoxide, a reaction that was widely used to extract metals from their ores. When solid copper(I) oxide is heated with hydrogen, for example, its mass decreases because the formation of pure copper is accompanied by the loss of oxygen atoms as a volatile product (water vapor). The reaction is as follows:
$\ce{Cu_2O (s) + H_2 (g) \rightarrow 2Cu (s) + H_2O (g)} \label{4.4.1}$
Oxidation-reduction reactions are now defined as reactions that exhibit a change in the oxidation states of one or more elements in the reactants by a transfer of electrons, which follows the mnemonic "oxidation is loss, reduction is gain", or "oil rig". The oxidation state of each atom in a compound is the charge an atom would have if all its bonding electrons were transferred to the atom with the greater attraction for electrons. Atoms in their elemental form, such as O2 or H2, are assigned an oxidation state of zero. For example, the reaction of aluminum with oxygen to produce aluminum oxide is
$\ce{ 4 Al (s) + 3O_2 \rightarrow 2Al_2O_3 (s)} \label{4.4.2}$
Each neutral oxygen atom gains two electrons and becomes negatively charged, forming an oxide ion; thus, oxygen has an oxidation state of −2 in the product and has been reduced. Each neutral aluminum atom loses three electrons to produce an aluminum ion with an oxidation state of +3 in the product, so aluminum has been oxidized. In the formation of Al2O3, electrons are transferred as follows (the small overset number emphasizes the oxidation state of the elements):
$4 \overset{0}{\ce{Al}} + 3 \overset{0}{\ce{O2}} \rightarrow \ce{4 Al^{3+} + 6 O^{2-} }\label{4.4.3}$
Equation $\ref{4.4.1}$ and Equation $\ref{4.4.2}$ are examples of oxidation–reduction (redox) reactions. In redox reactions, there is a net transfer of electrons from one reactant to another. In any redox reaction, the total number of electrons lost must equal the total of electrons gained to preserve electrical neutrality. In Equation $\ref{4.4.3}$, for example, the total number of electrons lost by aluminum is equal to the total number gained by oxygen:
\begin{align*} \text{electrons lost} &= \ce{4 Al} \, \text{atoms} \times {3 \, e^- \, \text{lost} \over \ce{Al} \, \text{atom} } \[4pt] &= 12 \, e^- \, \text{lost} \label{4.4.4a} \end{align*}
\begin{align*} \text{electrons gained} &= \ce{6 O} \, \text{atoms} \times {2 \, e^- \, \text{gained} \over \ce{O} \, \text{atom}} \[4pt] &= 12 \, e^- \, \text{gained} \label{4.4.4b}\end{align*}
The same pattern is seen in all oxidation–reduction reactions: the number of electrons lost must equal the number of electrons gained. An additional example of a redox reaction, the reaction of sodium metal with chlorine is illustrated in Figure $1$.
In all oxidation–reduction (redox) reactions, the number of electrons lost equals the number of electrons gained.
Assigning Oxidation States
Assigning oxidation states to the elements in binary ionic compounds is straightforward: the oxidation states of the elements are identical to the charges on the monatomic ions. Previously, you learned how to predict the formulas of simple ionic compounds based on the sign and magnitude of the charge on monatomic ions formed by the neutral elements. Examples of such compounds are sodium chloride (NaCl; Figure $1$), magnesium oxide (MgO), and calcium chloride (CaCl2). In covalent compounds, in contrast, atoms share electrons. However, we can still assign oxidation states to the elements involved by treating them as if they were ionic (that is, as if all the bonding electrons were transferred to the more attractive element). Oxidation states in covalent compounds are somewhat arbitrary, but they are useful bookkeeping devices to help you understand and predict many reactions.
A set of rules for assigning oxidation states to atoms in chemical compounds follows.
Rules for Assigning Oxidation States
1. The oxidation state of an atom in any pure element, whether monatomic, diatomic, or polyatomic, is zero.
2. The oxidation state of a monatomic ion is the same as its charge—for example, Na+ = +1, Cl = −1.
3. The oxidation state of fluorine in chemical compounds is always −1. Other halogens usually have oxidation states of −1 as well, except when combined with oxygen or other halogens.
4. Hydrogen is assigned an oxidation state of +1 in its compounds with nonmetals and −1 in its compounds with metals.
5. Oxygen is normally assigned an oxidation state of −2 in compounds, with two exceptions: in compounds that contain oxygen–fluorine or oxygen–oxygen bonds, the oxidation state of oxygen is determined by the oxidation states of the other elements present.
6. The sum of the oxidation states of all the atoms in a neutral molecule or ion must equal the charge on the molecule or ion.
Nonintegral (fractional) oxidation states are encountered occasionally. They are usually due to the presence of two or more atoms of the same element with different oxidation states.
In any chemical reaction, the net charge must be conserved; that is, in a chemical reaction, the total number of electrons is constant, just like the total number of atoms. Consistent with this, rule 1 states that the sum of the individual oxidation states of the atoms in a molecule or ion must equal the net charge on that molecule or ion. In NaCl, for example, Na has an oxidation state of +1 and Cl is −1. The net charge is zero, as it must be for any compound.
Rule 3 is required because fluorine attracts electrons more strongly than any other element, for reasons you will discover in Chapter 6. Hence fluorine provides a reference for calculating the oxidation states of other atoms in chemical compounds. Rule 4 reflects the difference in chemistry observed for compounds of hydrogen with nonmetals (such as chlorine) as opposed to compounds of hydrogen with metals (such as sodium). For example, NaH contains the H ion, whereas HCl forms H+ and Cl ions when dissolved in water. Rule 5 is necessary because fluorine has a greater attraction for electrons than oxygen does; this rule also prevents violations of rule 2. So the oxidation state of oxygen is +2 in OF2 but −½ in KO2. Note that an oxidation state of −½ for O in KO2 is perfectly acceptable.
The reduction of copper(I) oxide shown in Equation $\ref{4.4.5}$ demonstrates how to apply these rules. Rule 1 states that atoms in their elemental form have an oxidation state of zero, which applies to H2 and Cu. From rule 4, hydrogen in H2O has an oxidation state of +1, and from rule 5, oxygen in both Cu2O and H2O has an oxidation state of −2. Rule 6 states that the sum of the oxidation states in a molecule or formula unit must equal the net charge on that compound. This means that each Cu atom in Cu2O must have a charge of +1: 2(+1) + (−2) = 0. So the oxidation states are as follows:
$\overset {\color{ref}{+1}}{\ce{Cu_2}} \overset {\color{ref}-2}{\ce{O}} (s) + \overset {\color{ref}0}{\ce{H_2}} (g) \rightarrow 2 \overset {\color{ref}0}{\ce{Cu}} (s) + \overset {\color{ref}+1}{\ce{H}}_2 \overset {\color{ref}-2}{\ce{O}} (g) \label{4.4.5}$
Assigning oxidation states allows us to see that there has been a net transfer of electrons from hydrogen (0 → +1) to copper (+1 → 0). Thus, this is a redox reaction. Once again, the number of electrons lost equals the number of electrons gained, and there is a net conservation of charge:
$\text{electrons lost} = 2 \, H \, \text{atoms} \times {1 \, e^- \, \text{lost} \over H \, \text{atom} } = 2 \, e^- \, \text{lost} \label{4.4.6a}$
$\text{electrons gained} = 2 \, Cu \, \text{atoms} \times {1 \, e^- \, \text{gained} \over Cu \, \text{atom}} = 2 \, e^- \, \text{gained} \label{4.4.6b}$
Remember that oxidation states are useful for visualizing the transfer of electrons in oxidation–reduction reactions, but the oxidation state of an atom and its actual charge are the same only for simple ionic compounds. Oxidation states are a convenient way of assigning electrons to atoms, and they are useful for predicting the types of reactions that substances undergo.
Example $1$: Oxidation States
Assign oxidation states to all atoms in each compound.
1. sulfur hexafluoride (SF6)
2. methanol (CH3OH)
3. ammonium sulfate [(NH4)2SO4]
4. magnetite (Fe3O4)
5. ethanoic (acetic) acid (CH3CO2H)
Given: molecular or empirical formula
Asked for: oxidation states
Strategy:
Begin with atoms whose oxidation states can be determined unambiguously from the rules presented (such as fluorine, other halogens, oxygen, and monatomic ions). Then determine the oxidation states of other atoms present according to rule 1.
Solution:
a. We know from rule 3 that fluorine always has an oxidation state of −1 in its compounds. The six fluorine atoms in sulfur hexafluoride give a total negative charge of −6. Because rule 1 requires that the sum of the oxidation states of all atoms be zero in a neutral molecule (here SF6), the oxidation state of sulfur must be +6:
[(6 F atoms)(−1)] + [(1 S atom) (+6)] = 0
b. According to rules 4 and 5, hydrogen and oxygen have oxidation states of +1 and −2, respectively. Because methanol has no net charge, carbon must have an oxidation state of −2:
[(4 H atoms)(+1)] + [(1 O atom)(−2)] + [(1 C atom)(−2)] = 0
c. Note that (NH4)2SO4 is an ionic compound that consists of both a polyatomic cation (NH4+) and a polyatomic anion (SO42) (see Table 2.4). We assign oxidation states to the atoms in each polyatomic ion separately. For NH4+, hydrogen has an oxidation state of +1 (rule 4), so nitrogen must have an oxidation state of −3:
[(4 H atoms)(+1)] + [(1 N atom)(−3)] = +1, the charge on the NH4+ ion
For SO42−, oxygen has an oxidation state of −2 (rule 5), so sulfur must have an oxidation state of +6:
[(4 O atoms) (−2)] + [(1 S atom)(+6)] = −2, the charge on the sulfate ion
d. Oxygen has an oxidation state of −2 (rule 5), giving an overall charge of −8 per formula unit. This must be balanced by the positive charge on three iron atoms, giving an oxidation state of +8/3 for iron:
[(4 O atoms)(−2)]+[(3 Fe atoms)$\left (+{8 \over 3} \right )$]= 0
Fractional oxidation states are allowed because oxidation states are a somewhat arbitrary way of keeping track of electrons. In fact, Fe3O4 can be viewed as having two Fe3+ ions and one Fe2+ ion per formula unit, giving a net positive charge of +8 per formula unit. Fe3O4 is a magnetic iron ore commonly called magnetite. In ancient times, magnetite was known as lodestone because it could be used to make primitive compasses that pointed toward Polaris (the North Star), which was called the “lodestar.”
e. Initially, we assign oxidation states to the components of CH3CO2H in the same way as any other compound. Hydrogen and oxygen have oxidation states of +1 and −2 (rules 4 and 5, respectively), resulting in a total charge for hydrogen and oxygen of
[(4 H atoms)(+1)] + [(2 O atoms)(−2)] = 0
So the oxidation state of carbon must also be zero (rule 6). This is, however, an average oxidation state for the two carbon atoms present. Because each carbon atom has a different set of atoms bonded to it, they are likely to have different oxidation states. To determine the oxidation states of the individual carbon atoms, we use the same rules as before but with the additional assumption that bonds between atoms of the same element do not affect the oxidation states of those atoms. The carbon atom of the methyl group (−CH3) is bonded to three hydrogen atoms and one carbon atom. We know from rule 4 that hydrogen has an oxidation state of +1, and we have just said that the carbon–carbon bond can be ignored in calculating the oxidation state of the carbon atom. For the methyl group to be electrically neutral, its carbon atom must have an oxidation state of −3. Similarly, the carbon atom of the carboxylic acid group (−CO2H) is bonded to one carbon atom and two oxygen atoms. Again ignoring the bonded carbon atom, we assign oxidation states of −2 and +1 to the oxygen and hydrogen atoms, respectively, leading to a net charge of
[(2 O atoms)(−2)] + [(1 H atom)(+1)] = −3
To obtain an electrically neutral carboxylic acid group, the charge on this carbon must be +3. The oxidation states of the individual atoms in acetic acid are thus
$\underset {-3}{C} \overset {+1}{H_3} \overset {+3}{C} \underset {-2}{O_2} \overset {+1}{H} \nonumber$
Thus the sum of the oxidation states of the two carbon atoms is indeed zero.
Exercise $1$: Oxidation States
Assign oxidation states to all atoms in each compound.
1. barium fluoride (BaF2)
2. formaldehyde (CH2O)
3. potassium dichromate (K2Cr2O7)
4. cesium oxide (CsO2)
5. ethanol (CH3CH2OH)
Answer a
Ba, +2; F, −1
Answer b
C, 0; H, +1; O, −2
Answer c
K, +1; Cr, +6; O, −2
Answer d
Cs, +1; O, −½
Answer e
C, −3; H, +1; C, −1; H, +1; O, −2; H, +1
Types of Redox Reactions
Many types of chemical reactions are classified as redox reactions, and it would be impossible to memorize all of them. However, there are a few important types of redox reactions that you are likely to encounter and should be familiar with. These include:
• Synthesis reactions: The formation of any compound directly from the elements is a redox reaction, for example, the formation of water from hydrogen and oxygen: $\ce{ 2H_2(g) + O_2(g) \rightarrow 2H_2O(g)} \nonumber$
• Decomposition reactions: Conversely, the decomposition of a compound to its elements is also a redox reaction, as in the electrolysis of water: $\ce{2H_2O(l) \rightarrow 2H_2(g) + O_2(g)} \nonumber$
• Combustion reactions: Many chemicals combust (burn) with oxygen. In particular, organic chemicals such as hydrocarbons burn in the presence of oxygen to produce carbon dioxide and water as the products: $\ce{ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)} \nonumber$
The following sections describe another important class of redox reactions: single-displacement reactions of metals in solution.
Redox Reactions of Solid Metals in Aqueous Solution
• A widely encountered class of oxidation–reduction reactions is the reaction of aqueous solutions of acids or metal salts with solid metals. An example is the corrosion of metal objects, such as the rusting of an automobile (Figure $2$). Rust is formed from a complex oxidation–reduction reaction involving dilute acid solutions that contain Cl ions (effectively, dilute HCl), iron metal, and oxygen. When an object rusts, iron metal reacts with HCl(aq) to produce iron(II) chloride and hydrogen gas:
$\ce{ Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g)} \label{4.4.81}$
In subsequent steps, $\ce{FeCl2}$ undergoes oxidation to form a reddish-brown precipitate of $\ce{Fe(OH)3}$.
Many metals dissolve through reactions of this type, which have the general form
$\text{metal} + \text{acid} \rightarrow \text{salt} + \text{hydrogen} \label{4.4.82}$
Some of these reactions have important consequences. For example, it has been proposed that one factor that contributed to the fall of the Roman Empire was the widespread use of lead in cooking utensils and pipes that carried water. Rainwater, as we have seen, is slightly acidic, and foods such as fruits, wine, and vinegar contain organic acids. In the presence of these acids, lead dissolves:
$\ce{Pb(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2(g) } \label{4.4.83}$
Consequently, it has been speculated that both the water and the food consumed by Romans contained toxic levels of lead, which resulted in widespread lead poisoning and eventual madness. Perhaps this explains why the Roman Emperor Caligula appointed his favorite horse as consul!
• Single-Displacement Reactions
Certain metals are oxidized by aqueous acid, whereas others are oxidized by aqueous solutions of various metal salts. Both types of reactions are called single-displacement reactions, in which the ion in solution is displaced through oxidation of the metal. Two examples of single-displacement reactions are the reduction of iron salts by zinc (Equation $\ref{4.4.84}$) and the reduction of silver salts by copper (Equation $\ref{4.4.85}$ and Figure $3$):
$\ce{Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s)} \label{4.4.84}$
$\ce{ Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)} \label{4.4.85}$
The reaction in Equation $\ref{4.4.84}$ is widely used to prevent (or at least postpone) the corrosion of iron or steel objects, such as nails and sheet metal. The process of “galvanizing” consists of applying a thin coating of zinc to the iron or steel, thus protecting it from oxidation as long as zinc remains on the object.
• The Activity Series
By observing what happens when samples of various metals are placed in contact with solutions of other metals, chemists have arranged the metals according to the relative ease or difficulty with which they can be oxidized in a single-displacement reaction. For example, metallic zinc reacts with iron salts, and metallic copper reacts with silver salts. Experimentally, it is found that zinc reacts with both copper salts and silver salts, producing $\ce{Zn2+}$. Zinc therefore has a greater tendency to be oxidized than does iron, copper, or silver. Although zinc will not react with magnesium salts to give magnesium metal, magnesium metal will react with zinc salts to give zinc metal:
$\ce{Zn(s) + Mg^{2+}(aq) \xcancel{\rightarrow} Zn^{2+}(aq) + Mg(s)} \label{4.4.10}$
$\ce{Mg(s) + Zn^{2+}(aq) \rightarrow Mg^{2+}(aq) + Zn(s)} \label{4.4.11}$
Magnesium has a greater tendency to be oxidized than zinc does.
Pairwise reactions of this sort are the basis of the activity series (Figure $4$), which lists metals and hydrogen in order of their relative tendency to be oxidized. The metals at the top of the series, which have the greatest tendency to lose electrons, are the alkali metals (group 1), the alkaline earth metals (group 2), and Al (group 13). In contrast, the metals at the bottom of the series, which have the lowest tendency to be oxidized, are the precious metals or coinage metals—platinum, gold, silver, and copper, and mercury, which are located in the lower right portion of the metals in the periodic table. You should be generally familiar with which kinds of metals are active metals, which have the greatest tendency to be oxidized. (located at the top of the series) and which are inert metals, which have the least tendency to be oxidized. (at the bottom of the series).
When using the activity series to predict the outcome of a reaction, keep in mind that any element will reduce compounds of the elements below it in the series. Because magnesium is above zinc in Figure $4$, magnesium metal will reduce zinc salts but not vice versa. Similarly, the precious metals are at the bottom of the activity series, so virtually any other metal will reduce precious metal salts to the pure precious metals. Hydrogen is included in the series, and the tendency of a metal to react with an acid is indicated by its position relative to hydrogen in the activity series. Only those metals that lie above hydrogen in the activity series dissolve in acids to produce H2. Because the precious metals lie below hydrogen, they do not dissolve in dilute acid and therefore do not corrode readily. Example $2$ demonstrates how a familiarity with the activity series allows you to predict the products of many single-displacement reactions.
Example $2$: Activity
Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.
1. A strip of aluminum foil is placed in an aqueous solution of silver nitrate.
2. A few drops of liquid mercury are added to an aqueous solution of lead(II) acetate.
3. Some sulfuric acid from a car battery is accidentally spilled on the lead cable terminals.
Given: reactants
Asked for: overall reaction and net ionic equation
Strategy:
1. Locate the reactants in the activity series in Figure $4$ and from their relative positions, predict whether a reaction will occur. If a reaction does occur, identify which metal is oxidized and which is reduced.
2. Write the net ionic equation for the redox reaction.
Solution:
1. A Aluminum is an active metal that lies above silver in the activity series, so we expect a reaction to occur. According to their relative positions, aluminum will be oxidized and dissolve, and silver ions will be reduced to silver metal. B The net ionic equation is as follows:
$\ce{ Al(s) + 3Ag^+(aq) \rightarrow Al^{3+}(aq) + 3Ag(s)} \nonumber$
Recall from our discussion of solubilities that most nitrate salts are soluble. In this case, the nitrate ions are spectator ions and are not involved in the reaction.
2. A Mercury lies below lead in the activity series, so no reaction will occur.
3. A Lead is above hydrogen in the activity series, so the lead terminals will be oxidized, and the acid will be reduced to form H2. B From our discussion of solubilities, recall that Pb2+ and SO42 form insoluble lead(II) sulfate. In this case, the sulfate ions are not spectator ions, and the reaction is as follows:
$\ce{Pb(s) + 2H^+(aq) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + H_2(g) } \nonumber$
Lead(II) sulfate is the white solid that forms on corroded battery terminals.
Corroded battery terminals. The white solid is lead(II) sulfate, formed from the reaction of solid lead with a solution of sulfuric acid.
Exercise $2$
Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.
1. A strip of chromium metal is placed in an aqueous solution of aluminum chloride.
2. A strip of zinc is placed in an aqueous solution of chromium(III) nitrate.
3. A piece of aluminum foil is dropped into a glass that contains vinegar (the active ingredient is acetic acid).
Answer a
$no\: reaction$
Answer b
$3Zn(s) + 2Cr^{3+}(aq) \rightarrow 3Zn^{2+}(aq) + 2Cr(s)$
Answer c | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/04%3A_Three_Major_Classes_of_Chemical_Reactions/4.05%3A_Oxidation-Reduction_%28Redox%29_Reactions.txt |
• 5.1: An Overview of the Physical States of Matter
Bulk matter can exist in three states: gas, liquid, and solid. Gases have the lowest density of the three, are highly compressible, and fill their containers completely. Elements that exist as gases at room temperature and pressure are clustered on the right side of the periodic table; they occur as either monatomic gases (the noble gases) or diatomic molecules (some halogens, N2, O2).
• 5.2: Gas Pressure and Its Measurement
Pressure is defined as the force exerted per unit area; it can be measured using a barometer or manometer. Four quantities must be known for a complete physical description of a sample of a gas: temperature, volume, amount, and pressure. Pressure is force per unit area of surface; the SI unit for pressure is the pascal (Pa), defined as 1 newton per square meter (N/m2). The pressure exerted by an object is proportional to the force it exerts and inversely proportional to the area.
• 5.3: The Gas Laws and Their Experimental Foundations
The volume of a gas is inversely proportional to its pressure and directly proportional to its temperature and the amount of gas. Boyle showed that the volume of a sample of a gas is inversely proportional to pressure (Boyle’s law), Charles and Gay-Lussac demonstrated that the volume of a gas is directly proportional to its temperature at constant pressure (Charles’s law), and Avogadro showed that the volume of a gas is directly proportional to the number of moles of gas (Avogadro’s law).
• 5.4: Rearrangements of the Ideal Gas Law
• 5.5: The Kinetic-Molecular Theory - A Model for Gas Behavior
The behavior of ideal gases is explained by the kinetic molecular theory of gases. Molecular motion, which leads to collisions between molecules and the container walls, explains pressure, and the large intermolecular distances in gases explain their high compressibility. Although all gases have the same average kinetic energy at a given temperature, they do not all possess the same root mean square speed. The actual values of speed and kinetic energy are not the same for all gas particles.
• 5.6: Real Gases - Deviations from Ideal Behavior
No real gas exhibits ideal gas behavior, although many real gases approximate it over a range of conditions. Gases most closely approximate ideal gas behavior at high temperatures and low pressures. Deviations from ideal gas law behavior can be described by the van der Waals equation, which includes empirical constants to correct for the actual volume of the gaseous molecules and quantify the reduction in pressure due to intermolecular attractive forces.
• 5.E: Gases and the Kinetic-Molecular Theory (Exercises)
05: Gases and the Kinetic-Molecular Theory
Learning Objectives
• To describe the characteristics of a gas.
The three common phases (or states) of matter are gases, liquids, and solids. Gases have the lowest density of the three, are highly compressible, and completely fill any container in which they are placed. Gases behave this way because their intermolecular forces are relatively weak, so their molecules are constantly moving independently of the other molecules present. Solids, in contrast, are relatively dense, rigid, and incompressible because their intermolecular forces are so strong that the molecules are essentially locked in place. Liquids are relatively dense and incompressible, like solids, but they flow readily to adapt to the shape of their containers, like gases. We can therefore conclude that the sum of the intermolecular forces in liquids are between those of gases and solids. Figure \(1\) compares the three states of matter and illustrates the differences at the molecular level.
The state of a given substance depends strongly on conditions. For example, H2O is commonly found in all three states: solid ice, liquid water, and water vapor (its gaseous form). Under most conditions, we encounter water as the liquid that is essential for life; we drink it, cook with it, and bathe in it. When the temperature is cold enough to transform the liquid to ice, we can ski or skate on it, pack it into a snowball or snow cone, and even build dwellings with it. Water vapor (the term vapor refers to the gaseous form of a substance that is a liquid or a solid under normal conditions so nitrogen (N2) and oxygen (O2) are referred to as gases, but gaseous water in the atmosphere is called water vapor) is a component of the air we breathe, and it is produced whenever we heat water for cooking food or making coffee or tea. Water vapor at temperatures greater than 100°C is called steam. Steam is used to drive large machinery, including turbines that generate electricity. The properties of the three states of water are summarized in Table 10.1.
Table \(1\): Properties of Water at 1.0 atm
Temperature State Density (g/cm3)
≤0°C solid (ice) 0.9167 (at 0.0°C)
0°C–100°C liquid (water) 0.9997 (at 4.0°C)
≥100°C vapor (steam) 0.005476 (at 127°C)
The geometric structure and the physical and chemical properties of atoms, ions, and molecules usually do not depend on their physical state; the individual water molecules in ice, liquid water, and steam, for example, are all identical. In contrast, the macroscopic properties of a substance depend strongly on its physical state, which is determined by intermolecular forces and conditions such as temperature and pressure.
Figure \(2\) shows the locations in the periodic table of those elements that are commonly found in the gaseous, liquid, and solid states. Except for hydrogen, the elements that occur naturally as gases are on the right side of the periodic table. Of these, all the noble gases (group 18) are monatomic gases, whereas the other gaseous elements are diatomic molecules (H2, N2, O2, F2, and Cl2). Oxygen can also form a second allotrope, the highly reactive triatomic molecule ozone (O3), which is also a gas. In contrast, bromine (as Br2) and mercury (Hg) are liquids under normal conditions (25°C and 1.0 atm, commonly referred to as “room temperature and pressure”). Gallium (Ga), which melts at only 29.76°C, can be converted to a liquid simply by holding a container of it in your hand or keeping it in a non-air-conditioned room on a hot summer day. The rest of the elements are all solids under normal conditions.
All of the gaseous elements (other than the monatomic noble gases) are molecules. Within the same group (1, 15, 16 and 17), the lightest elements are gases. All gaseous substances are characterized by weak interactions between the constituent molecules or atoms.
Summary
Bulk matter can exist in three states: gas, liquid, and solid. Gases have the lowest density of the three, are highly compressible, and fill their containers completely. Elements that exist as gases at room temperature and pressure are clustered on the right side of the periodic table; they occur as either monatomic gases (the noble gases) or diatomic molecules (some halogens, N2, O2). | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/05%3A_Gases_and_the_Kinetic-Molecular_Theory/5.01%3A_An_Overview_of_the_Physical_States_of_Matter.txt |
Learning Objectives
• to describe and measure the pressure of a gas.
At the macroscopic level, a complete physical description of a sample of a gas requires four quantities:
• temperature (expressed in kelvins),
• volume (expressed in liters),
• amount (expressed in moles), and
• pressure (in atmospheres).
As we demonstrated below, these variables are not independent (i.e., they cannot be arbitrarily be varied). If we know the values of any three of these quantities, we can calculate the fourth and thereby obtain a full physical description of the gas. Temperature, volume, and amount have been discussed in previous chapters. We now discuss pressure and its units of measurement.
Units of Pressure
Any object, whether it is your computer, a person, or a sample of gas, exerts a force on any surface with which it comes in contact. The air in a balloon, for example, exerts a force against the interior surface of the balloon, and a liquid injected into a mold exerts a force against the interior surface of the mold, just as a chair exerts a force against the floor because of its mass and the effects of gravity. If the air in a balloon is heated, the increased kinetic energy of the gas eventually causes the balloon to burst because of the increased pressure ($P$) of the gas, the force ($F$) per unit area ($A$) of surface:
$P=\dfrac{\rm Force}{\rm Area}=\dfrac{F}{A}\label{10.2.1}$
Pressure is dependent on both the force exerted and the size of the area to which the force is applied. We know from Equation $\ref{10.2.1}$ that applying the same force to a smaller area produces a higher pressure. When we use a hose to wash a car, for example, we can increase the pressure of the water by reducing the size of the opening of the hose with a thumb.
The units of pressure are derived from the units used to measure force and area. The SI unit for pressure, derived from the SI units for force (newtons) and area (square meters), is the newton per square meter ($N/m^2$), which is called the Pascal (Pa), after the French mathematician Blaise Pascal (1623–1662):
$\rm 1 \;Pa=1\;N/m^2 \label{10.2.2}$
Example $1$
Assuming a paperback book has a mass of 2.00 kg, a length of 27.0 cm, a width of 21.0 cm, and a thickness of 4.5 cm, what pressure does it exert on a surface if it is
1. lying flat?
2. standing on edge in a bookcase?
Given: mass and dimensions of object
Asked for: pressure
Strategy:
1. Calculate the force exerted by the book and then compute the area that is in contact with a surface.
2. Substitute these two values into Equation $\ref{10.2.1}$ to find the pressure exerted on the surface in each orientation.
Solution:
The force exerted by the book does not depend on its orientation. Recall that the force exerted by an object is F = ma, where m is its mass and a is its acceleration. In Earth’s gravitational field, the acceleration is due to gravity (9.8067 m/s2 at Earth’s surface). In SI units, the force exerted by the book is therefore
$F = ma = 2.00 \;\rm kg\times 9.8067 \dfrac{\rm m}{\rm s^2} = 19.6 \dfrac{\rm kg·m}{\rm s^2} = 19.6\;\rm N \nonumber$
A We calculated the force as 19.6 N. When the book is lying flat, the area is
$A=\rm0.270 \;m\times0.210 \;m= 0.0567 \;m^2. \nonumber$
B The pressure exerted by the text lying flat is thus
$P=\dfrac{F}{A}=\dfrac{19.6\;\rm N}{0.0567\;\rm m^2}=3.46\times10^2 \rm Pa \nonumber$
A If the book is standing on its end, the force remains the same, but the area decreases:
$\rm A=\rm21.0 \;cm\times4.5 \;cm = 0.210 \;m\times0.045 \;m = 9.5 \times 10^{−3} \;\rm m^2 \nonumber$
B The pressure exerted by the text lying flat is thus
$P=\dfrac{19.6\;\rm N}{9.5\times10^{-3}\;\rm m^2}=2.06\times10^3 \;\rm Pa \nonumber$
Exercise $1$
What pressure does a 60.0 kg student exert on the floor
1. when standing flat-footed in the laboratory in a pair of tennis shoes (the surface area of the soles is approximately 180 cm2)?
2. as she steps heel-first onto a dance floor wearing high-heeled shoes (the area of the heel = 1.0 cm2)?
Answer a
3.27 × 104 Pa
Answer b
5.9 × 106 Pa
Barometric Pressure
Just as we exert pressure on a surface because of gravity, so does our atmosphere. We live at the bottom of an ocean of gases that becomes progressively less dense with increasing altitude. Approximately 99% of the mass of the atmosphere lies within 30 km of Earth’s surface (Figure $1$). Every point on Earth’s surface experiences a net pressure called barometric pressure. The pressure exerted by the atmosphere is considerable: a 1 m2 column, measured from sea level to the top of the atmosphere, has a mass of about 10,000 kg, which gives a pressure of about 101 kPa:
Barometric pressure can be measured using a barometer, a device invented in 1643 by one of Galileo’s students, Evangelista Torricelli (1608–1647). A barometer may be constructed from a long glass tube that is closed at one end. It is filled with mercury and placed upside down in a dish of mercury without allowing any air to enter the tube. Some of the mercury will run out of the tube, but a relatively tall column remains inside (Figure $2$). Why doesn’t all the mercury run out? Gravity is certainly exerting a downward force on the mercury in the tube, but it is opposed by the pressure of the atmosphere pushing down on the surface of the mercury in the dish, which has the net effect of pushing the mercury up into the tube. Because there is no air above the mercury inside the tube in a properly filled barometer (it contains a vacuum), there is no pressure pushing down on the column. Thus the mercury runs out of the tube until the pressure exerted by the mercury column itself exactly balances the pressure of the atmosphere. The pressure exerted by the mercury column can be expressed as:
\begin{align} P&=\dfrac{F}{A} \[4pt] &= \dfrac{mg}{A} \[4pt] &= \dfrac{\rho V\cdot g}{A} \[4pt] &= \dfrac{ \rho \cdot Ah\cdot g}{A} \[4pt] &= \rho gh \end{align} \nonumber
with
• $g$ is the gravitational acceleration,
• $m$ is the mass,
• $\rho$ is the density,
• $V$ is the volume,
• $A$ is the bottom area, and
• $h$ is height of the mercury column.
Under normal weather conditions at sea level, the two forces are balanced when the top of the mercury column is approximately 760 mm above the level of the mercury in the dish, as shown in Figure $2$. This value varies with meteorological conditions and altitude. In Denver, Colorado, for example, at an elevation of about 1 mile, or 1609 m (5280 ft), the height of the mercury column is 630 mm rather than 760 mm.
Mercury barometers have been used to measure barometric pressure for so long that they have their own unit for pressure: the millimeter of mercury (mmHg), often called the torr, after Torricelli. Standard barometric pressure is the barometric pressure required to support a column of mercury exactly 760 mm tall; this pressure is also referred to as 1 atmosphere (atm). These units are also related to the pascal:
\begin{align} \rm 1\; atm &= 760 \; mmHg \[4pt] &= 760 \; torr \[4pt] &= 1.01325 \times 10^5 \; Pa \[4pt] &= 101.325 \; kPa\label{10.2.3} \end{align}
Thus a pressure of 1 atm equals 760 mmHg exactly.
We are so accustomed to living under this pressure that we never notice it. Instead, what we notice are changes in the pressure, such as when our ears pop in fast elevators in skyscrapers or in airplanes during rapid changes in altitude. We make use of barometric pressure in many ways. We can use a drinking straw because sucking on it removes air and thereby reduces the pressure inside the straw. The barometric pressure pushing down on the liquid in the glass then forces the liquid up the straw.
Example $2$: Barometric Pressure
One of the authors visited Rocky Mountain National Park several years ago. After departing from an airport at sea level in the eastern United States, he arrived in Denver (altitude 5280 ft), rented a car, and drove to the top of the highway outside Estes Park (elevation 14,000 ft). He noticed that even slight exertion was very difficult at this altitude, where the barometric pressure is only 454 mmHg. Convert this pressure to
1. atmospheres (atm).
2. bar.
Given: pressure in millimeters of mercury
Asked for: pressure in atmospheres and bar
Strategy:
Use the conversion factors in Equation $\ref{10.2.3}$ to convert from millimeters of mercury to atmospheres and kilopascals.
Solution:
From Equation $\ref{10.2.3}$, we have 1 atm = 760 mmHg = 101.325 kPa. The pressure at 14,000 ft in atm is thus
\begin{align} P &=\rm 454 \;mmHg\times\dfrac{1\;atm}{760\;mmHg} \[4pt] &= 0.597\;atm \nonumber \end{align} \nonumber
The pressure in bar is given by
\begin{align} P&=\rm 0.597\;atm\times\dfrac{1.01325\;bar}{1\;atm}\[4pt] &= 0.605\;bar \nonumber \end{align} \nonumber
Exercise $2$: Barometric Pressure
Mt. Everest, at 29,028 ft above sea level, is the world’s tallest mountain. The normal barometric pressure at this altitude is about 0.308 atm. Convert this pressure to
1. millimeters of mercury.
2. bar.
Answer a
234 mmHg;
Answer b
0.312 bar
Manometers
Barometers measure barometric pressure, but manometers measure the pressures of samples of gases contained in an apparatus. The key feature of a manometer is a U-shaped tube containing mercury (or occasionally another nonvolatile liquid). A closed-end manometer is shown schematically in part (a) in Figure $3$. When the bulb contains no gas (i.e., when its interior is a near vacuum), the heights of the two columns of mercury are the same because the space above the mercury on the left is a near vacuum (it contains only traces of mercury vapor). If a gas is released into the bulb on the right, it will exert a pressure on the mercury in the right column, and the two columns of mercury will no longer be the same height. The difference between the heights of the two columns is equal to the pressure of the gas.
If the tube is open to the atmosphere instead of closed, as in the open-end manometer shown in part (b) in Figure $3$, then the two columns of mercury have the same height only if the gas in the bulb has a pressure equal to the barometric pressure. If the gas in the bulb has a higher pressure, the mercury in the open tube will be forced up by the gas pushing down on the mercury in the other arm of the U-shaped tube. The pressure of the gas in the bulb is therefore the sum of the barometric pressure (measured with a barometer) and the difference in the heights of the two columns. If the gas in the bulb has a pressure less than that of the atmosphere, then the height of the mercury will be greater in the arm attached to the bulb. In this case, the pressure of the gas in the bulb is the barometric pressure minus the difference in the heights of the two columns.
Example $3$
Suppose you want to construct a closed-end manometer to measure gas pressures in the range 0.000–0.200 atm. Because of the toxicity of mercury, you decide to use water rather than mercury. How tall a column of water do you need? (The density of water is 1.00 g/cm3; the density of mercury is 13.53 g/cm3.)
Given: pressure range and densities of water and mercury
Asked for: column height
Strategy:
1. Calculate the height of a column of mercury corresponding to 0.200 atm in millimeters of mercury. This is the height needed for a mercury-filled column.
2. From the given densities, use a proportion to compute the height needed for a water-filled column.
Solution:
A In millimeters of mercury, a gas pressure of 0.200 atm is
$P=\rm 0.200\;atm\times\dfrac{760\;mmHg}{1\;atm}=152\;mmHg \nonumber$
Using a mercury manometer, you would need a mercury column at least 152 mm high.
B Because water is less dense than mercury, you need a taller column of water to achieve the same pressure as a given column of mercury. The height needed for a water-filled column corresponding to a pressure of 0.200 atm is proportional to the ratio of the density of mercury to the density of water
$P=d_{\rm wat}gh_{\rm wat}=d_{\rm Hg}gh_{\rm Hg} \nonumber$
$h_{\rm wat}=h_{\rm Hg}\times\dfrac{d_{\rm Hg}}{g_{\rm wat}}=\rm152\;mm\times\dfrac{13.53\;g/cm^3}{1.00\;g/cm^3}=2070\;mm \nonumber$
The answer makes sense: it takes a taller column of a less dense liquid to achieve the same pressure.
Exercise $3$
Suppose you want to design a barometer to measure barometric pressure in an environment that is always hotter than 30°C. To avoid using mercury, you decide to use gallium, which melts at 29.76°C; the density of liquid gallium at 25°C is 6.114 g/cm3. How tall a column of gallium do you need if P = 1.00 atm?
Answer
1.68 m
The answer to Example $3$ also tells us the maximum depth of a farmer’s well if a simple suction pump will be used to get the water out. The 1.00 atm corresponds to a column height of
\begin{align} h_{\rm wat} &=h_{\rm Hg}\times\dfrac{d_{\rm Hg}}{g_{\rm wat}} \nonumber \[4pt] &=\rm760\;mm\times\dfrac{13.53\;g/cm^3}{1.00\;g/cm^3} \nonumber \[4pt] &= 1.03\times10^4\;mm \nonumber \[4pt] &= 10.3\;m \nonumber \end{align} \nonumber
A suction pump is just a more sophisticated version of a straw: it creates a vacuum above a liquid and relies on barometric pressure to force the liquid up a tube. If 1 atm pressure corresponds to a 10.3 m (33.8 ft) column of water, then it is physically impossible for barometric pressure to raise the water in a well higher than this. Until electric pumps were invented to push water mechanically from greater depths, this factor greatly limited where people could live because obtaining water from wells deeper than about 33 ft was difficult.
Summary
Pressure is defined as the force exerted per unit area; it can be measured using a barometer or manometer. Four quantities must be known for a complete physical description of a sample of a gas: temperature, volume, amount, and pressure. Pressure is force per unit area of surface; the SI unit for pressure is the pascal (Pa), defined as 1 newton per square meter (N/m2). The pressure exerted by an object is proportional to the force it exerts and inversely proportional to the area on which the force is exerted. The pressure exerted by Earth’s atmosphere, called barometric pressure, is about 101 kPa or 14.7 lb/in.2 at sea level. barometric pressure can be measured with a barometer, a closed, inverted tube filled with mercury. The height of the mercury column is proportional to barometric pressure, which is often reported in units of millimeters of mercury (mmHg), also called torr. Standard barometric pressure, the pressure required to support a column of mercury 760 mm tall, is yet another unit of pressure: 1 atmosphere (atm). A manometer is an apparatus used to measure the pressure of a sample of a gas. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/05%3A_Gases_and_the_Kinetic-Molecular_Theory/5.02%3A_Gas_Pressure_and_Its_Measurement.txt |
Learning Objectives
• To understand the relationships among pressure, temperature, volume, and the amount of a gas.
Early scientists explored the relationships among the pressure of a gas (P) and its temperature (T), volume (V), and amount (n) by holding two of the four variables constant (amount and temperature, for example), varying a third (such as pressure), and measuring the effect of the change on the fourth (in this case, volume). The history of their discoveries provides several excellent examples of the scientific method.
The Relationship between Pressure and Volume: Boyle's Law
As the pressure on a gas increases, the volume of the gas decreases because the gas particles are forced closer together. Conversely, as the pressure on a gas decreases, the gas volume increases because the gas particles can now move farther apart. Weather balloons get larger as they rise through the atmosphere to regions of lower pressure because the volume of the gas has increased; that is, the atmospheric gas exerts less pressure on the surface of the balloon, so the interior gas expands until the internal and external pressures are equal.
The Irish chemist Robert Boyle (1627–1691) carried out some of the earliest experiments that determined the quantitative relationship between the pressure and the volume of a gas. Boyle used a J-shaped tube partially filled with mercury, as shown in Figure $1$. In these experiments, a small amount of a gas or air is trapped above the mercury column, and its volume is measured at atmospheric pressure and constant temperature. More mercury is then poured into the open arm to increase the pressure on the gas sample. The pressure on the gas is atmospheric pressure plus the difference in the heights of the mercury columns, and the resulting volume is measured. This process is repeated until either there is no more room in the open arm or the volume of the gas is too small to be measured accurately. Data such as those from one of Boyle’s own experiments may be plotted in several ways (Figure $2$). A simple plot of $V$ versus $P$ gives a curve called a hyperbola and reveals an inverse relationship between pressure and volume: as the pressure is doubled, the volume decreases by a factor of two. This relationship between the two quantities is described as follows:
$PV = \rm constant \label{10.3.1}$
Dividing both sides by $P$ gives an equation illustrating the inverse relationship between $P$ and $V$:
$V=\dfrac{\rm const.}{P} = {\rm const.}\left(\dfrac{1}{P}\right) \label{10.3.2}$
or
$V \propto \dfrac{1}{P} \label{10.3.3}$
where the ∝ symbol is read “is proportional to.” A plot of V versus 1/P is thus a straight line whose slope is equal to the constant in Equations $\ref{10.3.1}$ and $\ref{10.3.3}$. Dividing both sides of Equation $\ref{10.3.1}$ by V instead of P gives a similar relationship between P and 1/V. The numerical value of the constant depends on the amount of gas used in the experiment and on the temperature at which the experiments are carried out. This relationship between pressure and volume is known as Boyle’s law, after its discoverer, and can be stated as follows: At constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure. This law in practice is shown in Figure $2$.
At constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure
The Relationship between Temperature and Volume: Charles's Law
Hot air rises, which is why hot-air balloons ascend through the atmosphere and why warm air collects near the ceiling and cooler air collects at ground level. Because of this behavior, heating registers are placed on or near the floor, and vents for air-conditioning are placed on or near the ceiling. The fundamental reason for this behavior is that gases expand when they are heated. Because the same amount of substance now occupies a greater volume, hot air is less dense than cold air. The substance with the lower density—in this case hot air—rises through the substance with the higher density, the cooler air.
The first experiments to quantify the relationship between the temperature and the volume of a gas were carried out in 1783 by an avid balloonist, the French chemist Jacques Alexandre César Charles (1746–1823). Charles’s initial experiments showed that a plot of the volume of a given sample of gas versus temperature (in degrees Celsius) at constant pressure is a straight line. Similar but more precise studies were carried out by another balloon enthusiast, the Frenchman Joseph-Louis Gay-Lussac (1778–1850), who showed that a plot of V versus T was a straight line that could be extrapolated to a point at zero volume, a theoretical condition now known to correspond to −273.15°C (Figure $3$).A sample of gas cannot really have a volume of zero because any sample of matter must have some volume. Furthermore, at 1 atm pressure all gases liquefy at temperatures well above −273.15°C. Note from part (a) in Figure $3$ that the slope of the plot of V versus T varies for the same gas at different pressures but that the intercept remains constant at −273.15°C. Similarly, as shown in part (b) in Figure $3$, plots of V versus T for different amounts of varied gases are straight lines with different slopes but the same intercept on the T axis.
The significance of the invariant T intercept in plots of V versus T was recognized in 1848 by the British physicist William Thomson (1824–1907), later named Lord Kelvin. He postulated that −273.15°C was the lowest possible temperature that could theoretically be achieved, for which he coined the term absolute zero (0 K).
We can state Charles’s and Gay-Lussac’s findings in simple terms: At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature (in kelvins). This relationship, illustrated in part (b) in Figure $3$ is often referred to as Charles’s law and is stated mathematically as
$V ={\rm const.}\; T \label{10.3.4}$
or
$V \propto T \label{10.3.5}$
with temperature expressed in kelvins, not in degrees Celsius. Charles’s law is valid for virtually all gases at temperatures well above their boiling points.
The Relationship between Amount and Volume: Avogadro's Law
We can demonstrate the relationship between the volume and the amount of a gas by filling a balloon; as we add more gas, the balloon gets larger. The specific quantitative relationship was discovered by the Italian chemist Amedeo Avogadro, who recognized the importance of Gay-Lussac’s work on combining volumes of gases. In 1811, Avogadro postulated that, at the same temperature and pressure, equal volumes of gases contain the same number of gaseous particles (Figure $4$). This is the historic “Avogadro’s hypothesis.”
A logical corollary to Avogadro's hypothesis (sometimes called Avogadro’s law) describes the relationship between the volume and the amount of a gas: At constant temperature and pressure, the volume of a sample of gas is directly proportional to the number of moles of gas in the sample. Stated mathematically,
$V ={\rm const.} \; (n) \label{10.3.6}$
or
$V \propto.n \text{@ constant T and P} \label{10.3.7}$
This relationship is valid for most gases at relatively low pressures, but deviations from strict linearity are observed at elevated pressures.
For a sample of gas,
• V increases as P decreases (and vice versa)
• V increases as T increases (and vice versa)
• V increases as n increases (and vice versa)
The relationships among the volume of a gas and its pressure, temperature, and amount are summarized in Figure $5$. Volume increases with increasing temperature or amount, but decreases with increasing pressure.
Summary
The volume of a gas is inversely proportional to its pressure and directly proportional to its temperature and the amount of gas. Boyle showed that the volume of a sample of a gas is inversely proportional to its pressure (Boyle’s law), Charles and Gay-Lussac demonstrated that the volume of a gas is directly proportional to its temperature (in kelvins) at constant pressure (Charles’s law), and Avogadro postulated that the volume of a gas is directly proportional to the number of moles of gas present (Avogadro’s law). Plots of the volume of gases versus temperature extrapolate to zero volume at −273.15°C, which is absolute zero (0 K), the lowest temperature possible. Charles’s law implies that the volume of a gas is directly proportional to its absolute temperature. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/05%3A_Gases_and_the_Kinetic-Molecular_Theory/5.03%3A_The_Gas_Laws_and_Their_Experimental_Foundations.txt |
Learning Objectives
• To understand the significance of the kinetic molecular theory of gases.
The laws that describe the behavior of gases were well established long before anyone had developed a coherent model of the properties of gases. In this section, we introduce a theory that describes why gases behave the way they do. The theory we introduce can also be used to derive laws such as the ideal gas law from fundamental principles and the properties of individual particles.
A Molecular Description
The kinetic molecular theory of gases explains the laws that describe the behavior of gases. Developed during the mid-19th century by several physicists, including the Austrian Ludwig Boltzmann (1844–1906), the German Rudolf Clausius (1822–1888), and the Englishman James Clerk Maxwell (1831–1879), who is also known for his contributions to electricity and magnetism, this theory is based on the properties of individual particles as defined for an ideal gas and the fundamental concepts of physics. Thus the kinetic molecular theory of gases provides a molecular explanation for observations that led to the development of the ideal gas law. The kinetic molecular theory of gases is based on the following five postulates:
five postulates of Kinetic Molecular Theory
1. A gas is composed of a large number of particles called molecules (whether monatomic or polyatomic) that are in constant random motion.
2. Because the distance between gas molecules is much greater than the size of the molecules, the volume of the molecules is negligible.
3. Intermolecular interactions, whether repulsive or attractive, are so weak that they are also negligible.
4. Gas molecules collide with one another and with the walls of the container, but these collisions are perfectly elastic; that is, they do not change the average kinetic energy of the molecules.
5. The average kinetic energy of the molecules of any gas depends on only the temperature, and at a given temperature, all gaseous molecules have exactly the same average kinetic energy.
Although the molecules of real gases have nonzero volumes and exert both attractive and repulsive forces on one another, for the moment we will focus on how the kinetic molecular theory of gases relates to the properties of gases we have been discussing. In the following sections, we explain how this theory must be modified to account for the behavior of real gases.
Postulates 1 and 4 state that gas molecules are in constant motion and collide frequently with the walls of their containers. The collision of molecules with their container walls results in a momentum transfer (impulse) from molecules to the walls (Figure $2$).
The momentum transfer to the wall perpendicular to $x$ axis as a molecule with an initial velocity $u_x$ in $x$ direction hits is expressed as:
$\Delta p_x=2mu_x \label{10.7.1}$
The collision frequency, a number of collisions of the molecules to the wall per unit area and per second, increases with the molecular speed and the number of molecules per unit volume.
$f\propto (u_x) \times \Big(\dfrac{N}{V}\Big) \label{10.7.2}$
The pressure the gas exerts on the wall is expressed as the product of impulse and the collision frequency.
$P\propto (2mu_x)\times(u_x)\times\Big(\dfrac{N}{V}\Big)\propto \Big(\dfrac{N}{V}\Big)mu_x^2 \label{10.7.3}$
At any instant, however, the molecules in a gas sample are traveling at different speed. Therefore, we must replace $u_x^2$ in the expression above with the average value of $u_x^2$, which is denoted by $\overline{u_x^2}$. The overbar designates the average value over all molecules.
The exact expression for pressure is given as :
$P=\dfrac{N}{V}m\overline{u_x^2} \label{10.7.4}$
Finally, we must consider that there is nothing special about $x$ direction. We should expect that
$\overline{u_x^2}= \overline{u_y^2}=\overline{u_z^2}=\dfrac{1}{3}\overline{u^2}. \nonumber$
Here the quantity $\overline{u^2}$ is called the mean-square speed defined as the average value of square-speed ($u^2$) over all molecules. Since
$u^2=u_x^2+u_y^2+u_z^2 \nonumber$
for each molecule, then
$\overline{u^2}=\overline{u_x^2}+\overline{u_y^2}+\overline{u_z^2}. \nonumber$
By substituting $\dfrac{1}{3}\overline{u^2}$ for $\overline{u_x^2}$ in the expression above, we can get the final expression for the pressure:
$P=\dfrac{1}{3}\dfrac{N}{V}m\overline{u^2} \label{10.7.5}$
Because volumes and intermolecular interactions are negligible, postulates 2 and 3 state that all gaseous particles behave identically, regardless of the chemical nature of their component molecules. This is the essence of the ideal gas law, which treats all gases as collections of particles that are identical in all respects except mass. Postulate 2 also explains why it is relatively easy to compress a gas; you simply decrease the distance between the gas molecules.
Postulate 5 provides a molecular explanation for the temperature of a gas. Postulate 5 refers to the average translational kinetic energy of the molecules of a gas $(\overline{e_K})$, which can be represented as and states that at a given Kelvin temperature $(T)$, all gases have the same value of
$\overline{e_K}=\dfrac{1}{2}m\overline{u^2}=\dfrac{3}{2}\dfrac{R}{N_A}T \label{10.7.6}$
where $N_A$ is the Avogadro's constant. The total translational kinetic energy of 1 mole of molecules can be obtained by multiplying the equation by $N_A$:
$N_A\overline{e_K}=\dfrac{1}{2}M\overline{u^2}=\dfrac{3}{2}RT \label{10.7.7}$
where $M$ is the molar mass of the gas molecules and is related to the molecular mass by $M=N_Am$. By rearranging the equation, we can get the relationship between the root-mean square speed ($u_{\rm rms}$) and the temperature. The rms speed ($u_{\rm rms}$) is the square root of the sum of the squared speeds divided by the number of particles:
$u_{\rm rms}=\sqrt{\overline{u^2}}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}} \label{10.7.8}$
where $N$ is the number of particles and $u_i$ is the speed of particle $i$.
The relationship between $u_{\rm rms}$ and the temperature is given by:
$u_{\rm rms}=\sqrt{\dfrac{3RT}{M}} \label{10.7.9}$
In Equation $\ref{10.7.9}$, $u_{\rm rms}$ has units of meters per second; consequently, the units of molar mass $M$ are kilograms per mole, temperature $T$ is expressed in kelvins, and the ideal gas constant $R$ has the value 8.3145 J/(K•mol). Equation $\ref{10.7.9}$ shows that $u_{\rm rms}$ of a gas is proportional to the square root of its Kelvin temperature and inversely proportional to the square root of its molar mass. The root mean-square speed of a gas increase with increasing temperature. At a given temperature, heavier gas molecules have slower speeds than do lighter ones.
The rms speed and the average speed do not differ greatly (typically by less than 10%). The distinction is important, however, because the rms speed is the speed of a gas particle that has average kinetic energy. Particles of different gases at the same temperature have the same average kinetic energy, not the same average speed. In contrast, the most probable speed (vp) is the speed at which the greatest number of particles is moving. If the average kinetic energy of the particles of a gas increases linearly with increasing temperature, then Equation $\ref{10.7.8}$ tells us that the rms speed must also increase with temperature because the mass of the particles is constant. At higher temperatures, therefore, the molecules of a gas move more rapidly than at lower temperatures, and vp increases.
At a given temperature, all gaseous particles have the same average kinetic energy but not the same average speed.
Example $1$
The speeds of eight particles were found to be 1.0, 4.0, 4.0, 6.0, 6.0, 6.0, 8.0, and 10.0 m/s. Calculate their average speed ($v_{\rm av}$) root mean square speed ($v_{\rm rms}$), and most probable speed ($v_{\rm m}$).
Given: particle speeds
Asked for: average speed ($v_{\rm av}$), root mean square speed ($v_{\rm rms}$), and most probable speed ($v_{\rm m}$)
Strategy:
Use Equation $\ref{10.7.6}$ to calculate the average speed and Equation $\ref{10.7.8}$ to calculate the rms speed. Find the most probable speed by determining the speed at which the greatest number of particles is moving.
Solution:
The average speed is the sum of the speeds divided by the number of particles:
$v_{\rm av}=\rm\dfrac{(1.0+4.0+4.0+6.0+6.0+6.0+8.0+10.0)\;m/s}{8}=5.6\;m/s \nonumber$
The rms speed is the square root of the sum of the squared speeds divided by the number of particles:
$v_{\rm rms}=\rm\sqrt{\dfrac{(1.0^2+4.0^2+4.0^2+6.0^2+6.0^2+6.0^2+8.0^2+10.0^2)\;m^2/s^2}{8}}=6.2\;m/s \nonumber$
The most probable speed is the speed at which the greatest number of particles is moving. Of the eight particles, three have speeds of 6.0 m/s, two have speeds of 4.0 m/s, and the other three particles have different speeds. Hence $v_{\rm m}=6.0$ m/s. The $v_{\rm rms}$ of the particles, which is related to the average kinetic energy, is greater than their average speed.
Boltzmann Distributions
At any given time, what fraction of the molecules in a particular sample has a given speed? Some of the molecules will be moving more slowly than average, and some will be moving faster than average, but how many in each situation? Answers to questions such as these can have a substantial effect on the amount of product formed during a chemical reaction. This problem was solved mathematically by Maxwell in 1866; he used statistical analysis to obtain an equation that describes the distribution of molecular speeds at a given temperature. Typical curves showing the distributions of speeds of molecules at several temperatures are displayed in Figure $3$. Increasing the temperature has two effects. First, the peak of the curve moves to the right because the most probable speed increases. Second, the curve becomes broader because of the increased spread of the speeds. Thus increased temperature increases the value of the most probable speed but decreases the relative number of molecules that have that speed. Although the mathematics behind curves such as those in Figure $3$ were first worked out by Maxwell, the curves are almost universally referred to as Boltzmann distributions, after one of the other major figures responsible for the kinetic molecular theory of gases.
The Relationships among Pressure, Volume, and Temperature
We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously.
• Pressure versus Volume: At constant temperature, the kinetic energy of the molecules of a gas and hence the rms speed remain unchanged. If a given gas sample is allowed to occupy a larger volume, then the speed of the molecules does not change, but the density of the gas (number of particles per unit volume) decreases, and the average distance between the molecules increases. Hence the molecules must, on average, travel farther between collisions. They therefore collide with one another and with the walls of their containers less often, leading to a decrease in pressure. Conversely, increasing the pressure forces the molecules closer together and increases the density, until the collective impact of the collisions of the molecules with the container walls just balances the applied pressure.
• Volume versus Temperature: Raising the temperature of a gas increases the average kinetic energy and therefore the rms speed (and the average speed) of the gas molecules. Hence as the temperature increases, the molecules collide with the walls of their containers more frequently and with greater force. This increases the pressure, unless the volume increases to reduce the pressure, as we have just seen. Thus an increase in temperature must be offset by an increase in volume for the net impact (pressure) of the gas molecules on the container walls to remain unchanged.
• Pressure of Gas Mixtures: Postulate 3 of the kinetic molecular theory of gases states that gas molecules exert no attractive or repulsive forces on one another. If the gaseous molecules do not interact, then the presence of one gas in a gas mixture will have no effect on the pressure exerted by another, and Dalton’s law of partial pressures holds.
Example $2$
The temperature of a 4.75 L container of N2 gas is increased from 0°C to 117°C. What is the qualitative effect of this change on the
1. average kinetic energy of the N2 molecules?
2. rms speed of the N2 molecules?
3. average speed of the N2 molecules?
4. impact of each N2 molecule on the wall of the container during a collision with the wall?
5. total number of collisions per second of N2 molecules with the walls of the entire container?
6. number of collisions per second of N2 molecules with each square centimeter of the container wall?
7. pressure of the N2 gas?
Given: temperatures and volume
Asked for: effect of increase in temperature
Strategy:
Use the relationships among pressure, volume, and temperature to predict the qualitative effect of an increase in the temperature of the gas.
Solution:
1. Increasing the temperature increases the average kinetic energy of the N2 molecules.
2. An increase in average kinetic energy can be due only to an increase in the rms speed of the gas particles.
3. If the rms speed of the N2 molecules increases, the average speed also increases.
4. If, on average, the particles are moving faster, then they strike the container walls with more energy.
5. Because the particles are moving faster, they collide with the walls of the container more often per unit time.
6. The number of collisions per second of N2 molecules with each square centimeter of container wall increases because the total number of collisions has increased, but the volume occupied by the gas and hence the total area of the walls are unchanged.
7. The pressure exerted by the N2 gas increases when the temperature is increased at constant volume, as predicted by the ideal gas law.
Exercise $2$
A sample of helium gas is confined in a cylinder with a gas-tight sliding piston. The initial volume is 1.34 L, and the temperature is 22°C. The piston is moved to allow the gas to expand to 2.12 L at constant temperature. What is the qualitative effect of this change on the
1. average kinetic energy of the He atoms?
2. rms speed of the He atoms?
3. average speed of the He atoms?
4. impact of each He atom on the wall of the container during a collision with the wall?
5. total number of collisions per second of He atoms with the walls of the entire container?
6. number of collisions per second of He atoms with each square centimeter of the container wall?
7. pressure of the He gas?
Answer a
no change
Answer b
no change
Answer c
no change
Answer d
no change
Answer e
decreases
Answer f
decreases
Answer g
decreases
Summary
• The kinetic molecular theory of gases provides a molecular explanation for the observations that led to the development of the ideal gas law.
• Average kinetic energy:$\overline{e_K}=\dfrac{1}{2}m{u_{\rm rms}}^2=\dfrac{3}{2}\dfrac{R}{N_A}T n\nonumber$
• Root mean square speed: $u_{\rm rms}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}} \nonumber$
• Kinetic molecular theory of gases: $u_{\rm rms}=\sqrt{\dfrac{3RT}{M}} \nonumber$
The behavior of ideal gases is explained by the kinetic molecular theory of gases. Molecular motion, which leads to collisions between molecules and the container walls, explains pressure, and the large intermolecular distances in gases explain their high compressibility. Although all gases have the same average kinetic energy at a given temperature, they do not all possess the same root mean square (rms) speed (vrms). The actual values of speed and kinetic energy are not the same for all particles of a gas but are given by a Boltzmann distribution, in which some molecules have higher or lower speeds (and kinetic energies) than average. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/05%3A_Gases_and_the_Kinetic-Molecular_Theory/5.05%3A_The_Kinetic-Molecular_Theory_-_A_Model_for_Gas_Behavior.txt |
Learning Objectives
• To recognize the differences between the behavior of an ideal gas and a real gas
• To understand how molecular volumes and intermolecular attractions cause the properties of real gases to deviate from those predicted by the ideal gas law.
The postulates of the kinetic molecular theory of gases ignore both the volume occupied by the molecules of a gas and all interactions between molecules, whether attractive or repulsive. In reality, however, all gases have nonzero molecular volumes. Furthermore, the molecules of real gases interact with one another in ways that depend on the structure of the molecules and therefore differ for each gaseous substance. In this section, we consider the properties of real gases and how and why they differ from the predictions of the ideal gas law. We also examine liquefaction, a key property of real gases that is not predicted by the kinetic molecular theory of gases.
Pressure, Volume, and Temperature Relationships in Real Gases
For an ideal gas, a plot of $PV/nRT$ versus $P$ gives a horizontal line with an intercept of 1 on the $PV/nRT$ axis. Real gases, however, show significant deviations from the behavior expected for an ideal gas, particularly at high pressures (Figure $\PageIndex{1a}$). Only at relatively low pressures (less than 1 atm) do real gases approximate ideal gas behavior (Figure $\PageIndex{1b}$).
Real gases also approach ideal gas behavior more closely at higher temperatures, as shown in Figure $2$ for $N_2$. Why do real gases behave so differently from ideal gases at high pressures and low temperatures? Under these conditions, the two basic assumptions behind the ideal gas law—namely, that gas molecules have negligible volume and that intermolecular interactions are negligible—are no longer valid.
Because the molecules of an ideal gas are assumed to have zero volume, the volume available to them for motion is always the same as the volume of the container. In contrast, the molecules of a real gas have small but measurable volumes. At low pressures, the gaseous molecules are relatively far apart, but as the pressure of the gas increases, the intermolecular distances become smaller and smaller (Figure $3$). As a result, the volume occupied by the molecules becomes significant compared with the volume of the container. Consequently, the total volume occupied by the gas is greater than the volume predicted by the ideal gas law. Thus at very high pressures, the experimentally measured value of PV/nRT is greater than the value predicted by the ideal gas law.
Moreover, all molecules are attracted to one another by a combination of forces. These forces become particularly important for gases at low temperatures and high pressures, where intermolecular distances are shorter. Attractions between molecules reduce the number of collisions with the container wall, an effect that becomes more pronounced as the number of attractive interactions increases. Because the average distance between molecules decreases, the pressure exerted by the gas on the container wall decreases, and the observed pressure is less than expected (Figure $4$). Thus as shown in Figure $2$, at low temperatures, the ratio of $PV/nRT$ is lower than predicted for an ideal gas, an effect that becomes particularly evident for complex gases and for simple gases at low temperatures. At very high pressures, the effect of nonzero molecular volume predominates. The competition between these effects is responsible for the minimum observed in the $PV/nRT$ versus $P$ plot for many gases.
Nonzero molecular volume makes the actual volume greater than predicted at high pressures; intermolecular attractions make the pressure less than predicted.
At high temperatures, the molecules have sufficient kinetic energy to overcome intermolecular attractive forces, and the effects of nonzero molecular volume predominate. Conversely, as the temperature is lowered, the kinetic energy of the gas molecules decreases. Eventually, a point is reached where the molecules can no longer overcome the intermolecular attractive forces, and the gas liquefies (condenses to a liquid).
The van der Waals Equation
The Dutch physicist Johannes van der Waals (1837–1923; Nobel Prize in Physics, 1910) modified the ideal gas law to describe the behavior of real gases by explicitly including the effects of molecular size and intermolecular forces. In his description of gas behavior, the so-called van der Waals equation,
$\underbrace{ \left(P + \dfrac{an^2}{V^2}\right)}_{\text{Pressure Term}} \overbrace{(V − nb)}^{\text{Pressure Term}} =nRT \label{10.9.1}$
a and b are empirical constants that are different for each gas. The values of $a$ and $b$ are listed in Table $1$ for several common gases.
Table $1$:: van der Waals Constants for Some Common Gases (see Table A8 for more complete list)
Gas a ((L2·atm)/mol2) b (L/mol)
He 0.03410 0.0238
Ne 0.205 0.0167
Ar 1.337 0.032
H2 0.2420 0.0265
N2 1.352 0.0387
O2 1.364 0.0319
Cl2 6.260 0.0542
NH3 4.170 0.0371
CH4 2.273 0.0430
CO2 3.610 0.0429
The pressure term in Equation $\ref{10.9.1}$ corrects for intermolecular attractive forces that tend to reduce the pressure from that predicted by the ideal gas law. Here, $n^2/V^2$ represents the concentration of the gas ($n/V$) squared because it takes two particles to engage in the pairwise intermolecular interactions of the type shown in Figure $4$. The volume term corrects for the volume occupied by the gaseous molecules.
The correction for volume is negative, but the correction for pressure is positive to reflect the effect of each factor on V and P, respectively. Because nonzero molecular volumes produce a measured volume that is larger than that predicted by the ideal gas law, we must subtract the molecular volumes to obtain the actual volume available. Conversely, attractive intermolecular forces produce a pressure that is less than that expected based on the ideal gas law, so the $an^2/V^2$ term must be added to the measured pressure to correct for these effects.
Example $1$
You are in charge of the manufacture of cylinders of compressed gas at a small company. Your company president would like to offer a 4.00 L cylinder containing 500 g of chlorine in the new catalog. The cylinders you have on hand have a rupture pressure of 40 atm. Use both the ideal gas law and the van der Waals equation to calculate the pressure in a cylinder at 25°C. Is this cylinder likely to be safe against sudden rupture (which would be disastrous and certainly result in lawsuits because chlorine gas is highly toxic)?
Given: volume of cylinder, mass of compound, pressure, and temperature
Asked for: safety
Strategy:
A Use the molar mass of chlorine to calculate the amount of chlorine in the cylinder. Then calculate the pressure of the gas using the ideal gas law.
B Obtain a and b values for Cl2 from Table $1$. Use the van der Waals equation ($\ref{10.9.1}$) to solve for the pressure of the gas. Based on the value obtained, predict whether the cylinder is likely to be safe against sudden rupture.
Solution:
A We begin by calculating the amount of chlorine in the cylinder using the molar mass of chlorine (70.906 g/mol):
\begin{align} n &=\dfrac{m}{M} \[4pt] &= \rm\dfrac{500\;g}{70.906\;g/mol} \[4pt] &=7.052\;mol\nonumber \end{align} \nonumber
Using the ideal gas law and the temperature in kelvin (298 K), we calculate the pressure:
\begin{align} P &=\dfrac{nRT}{V} \[4pt] &=\rm\dfrac{7.052\;mol\times 0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L} \[4pt] &= 43.1\;atm \end{align} \nonumber
If chlorine behaves like an ideal gas, you have a real problem!
B Now let’s use the van der Waals equation with the a and b values for Cl2 from Table $1$. Solving for $P$ gives
\begin{align}P&=\dfrac{nRT}{V-nb}-\dfrac{an^2}{V^2}\&=\rm\dfrac{7.052\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L-7.052\;mol\times0.0542\dfrac{L}{mol}}-\dfrac{6.260\dfrac{L^2atm}{mol^2}\times(7.052\;mol)^2}{(4.00\;L)^2}\&=\rm28.2\;atm\end{align} \nonumber
This pressure is well within the safety limits of the cylinder. The ideal gas law predicts a pressure 15 atm higher than that of the van der Waals equation.
Exercise $1$
A 10.0 L cylinder contains 500 g of methane. Calculate its pressure to two significant figures at 27°C using the
1. ideal gas law.
2. van der Waals equation.
Answer a
77 atm
Answer b
67 atm
Liquefaction of Gases
Liquefaction of gases is the condensation of gases into a liquid form, which is neither anticipated nor explained by the kinetic molecular theory of gases. Both the theory and the ideal gas law predict that gases compressed to very high pressures and cooled to very low temperatures should still behave like gases, albeit cold, dense ones. As gases are compressed and cooled, however, they invariably condense to form liquids, although very low temperatures are needed to liquefy light elements such as helium (for He, 4.2 K at 1 atm pressure).
Liquefaction can be viewed as an extreme deviation from ideal gas behavior. It occurs when the molecules of a gas are cooled to the point where they no longer possess sufficient kinetic energy to overcome intermolecular attractive forces. The precise combination of temperature and pressure needed to liquefy a gas depends strongly on its molar mass and structure, with heavier and more complex molecules usually liquefying at higher temperatures. In general, substances with large van der Waals $a$ coefficients are relatively easy to liquefy because large a coefficients indicate relatively strong intermolecular attractive interactions. Conversely, small molecules with only light elements have small a coefficients, indicating weak intermolecular interactions, and they are relatively difficult to liquefy. Gas liquefaction is used on a massive scale to separate O2, N2, Ar, Ne, Kr, and Xe. After a sample of air is liquefied, the mixture is warmed, and the gases are separated according to their boiling points.
A large value of a in the van der Waals equation indicates the presence of relatively strong intermolecular attractive interactions.
The ultracold liquids formed from the liquefaction of gases are called cryogenic liquids, from the Greek kryo, meaning “cold,” and genes, meaning “producing.” They have applications as refrigerants in both industry and biology. For example, under carefully controlled conditions, the very cold temperatures afforded by liquefied gases such as nitrogen (boiling point = 77 K at 1 atm) can preserve biological materials, such as semen for the artificial insemination of cows and other farm animals. These liquids can also be used in a specialized type of surgery called cryosurgery, which selectively destroys tissues with a minimal loss of blood by the use of extreme cold.
Moreover, the liquefaction of gases is tremendously important in the storage and shipment of fossil fuels (Figure $5$). Liquefied natural gas (LNG) and liquefied petroleum gas (LPG) are liquefied forms of hydrocarbons produced from natural gas or petroleum reserves. LNG consists mostly of methane, with small amounts of heavier hydrocarbons; it is prepared by cooling natural gas to below about −162°C. It can be stored in double-walled, vacuum-insulated containers at or slightly above atmospheric pressure. Because LNG occupies only about 1/600 the volume of natural gas, it is easier and more economical to transport. LPG is typically a mixture of propane, propene, butane, and butenes and is primarily used as a fuel for home heating. It is also used as a feedstock for chemical plants and as an inexpensive and relatively nonpolluting fuel for some automobiles.
Summary
No real gas exhibits ideal gas behavior, although many real gases approximate it over a range of conditions. Deviations from ideal gas behavior can be seen in plots of PV/nRT versus P at a given temperature; for an ideal gas, PV/nRT versus P = 1 under all conditions. At high pressures, most real gases exhibit larger PV/nRT values than predicted by the ideal gas law, whereas at low pressures, most real gases exhibit PV/nRT values close to those predicted by the ideal gas law. Gases most closely approximate ideal gas behavior at high temperatures and low pressures. Deviations from ideal gas law behavior can be described by the van der Waals equation, which includes empirical constants to correct for the actual volume of the gaseous molecules and quantify the reduction in pressure due to intermolecular attractive forces. If the temperature of a gas is decreased sufficiently, liquefaction occurs, in which the gas condenses into a liquid form. Liquefied gases have many commercial applications, including the transport of large amounts of gases in small volumes and the uses of ultracold cryogenic liquids. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/05%3A_Gases_and_the_Kinetic-Molecular_Theory/5.06%3A_Real_Gases_-_Deviations_from_Ideal_Behavior.txt |
with various forms of radiant, or transmitted, energy, such as the energy associated with the visible light we detect with our eyes, the infrared radiation we feel as heat, the that causes sunburn, and the x-rays that produce images of our teeth or bones. All these forms of radiant energy should be familiar to you. We begin our discussion of the development of our current atomic model by describing the properties of waves and the various forms of electromagnetic radiation. is a periodic oscillation that transmits energy through space. Anyone who has visited a beach or dropped a stone into a puddle has observed waves traveling through water (Figure $1$). These waves are produced when wind, a stone, or some other disturbance, such as a passing boat, transfers energy to the water, causing the surface to oscillate up and down as the energy travels outward from its point of origin. As a passes a particular point on the surface of the water, anything floating there moves up and down. —between the midpoints of two peaks, for example, or two troughs—is the ($λ$, lowercase Greek lambda). Wavelengths are described by a unit of distance, typically meters. The ($u$, lowercase Greek nu) of a is the number of oscillations that pass a particular point in a given period of time. The usual units are oscillations per second (1/s = s), which in the SI system is called the hertz (Hz). is defined as half the peak-to-trough height; as the amplitude of a with a given frequency increases, so does its energy. As you can see in Figure $2$, two waves can have the same amplitude but different wavelengths and vice versa. The distance traveled by a per unit time is its speed ($v$), which is typically measured in meters per second (m/s). The speed of a is equal to the product of its wavelength and frequency: }} \right )\left ( \dfrac{\cancel{\text{}}}{\text{second}} \right ) &=\dfrac{\text{meters}}{\text{second}} \label{6.1.1b} \end{align} \] (the water). In contrast, energy that is transmitted, or radiated, through space in the form of periodic oscillations of electric and magnetic fields is known as . (Figure $3$). Some forms of electromagnetic radiation are shown in Figure $4$. In a vacuum, all forms of electromagnetic radiation—whether microwaves, visible light, or gamma rays—travel at the speed of light (), which turns out to be a physical constant with a value of 2.99792458 × 10 m/s (about 3.00 ×10 m/s or 1.86 × 10 mi/s). This is about a million times faster than the speed of sound. , with wavelengths of ≤ 400 nm, has enough energy to cause severe damage to our skin in the form of sunburn. Because the of the atmosphere absorbs sunlight with wavelengths less than 350 nm, it protects us from the damaging effects of highly energetic ultraviolet radiation. requires an understanding of the properties of waves and electromagnetic radiation. A is a periodic oscillation by which energy is transmitted through space. All waves are , repeating regularly in both space and time. Waves are characterized by several interrelated properties: ($λ$), the distance between successive waves; ($u$), the number of waves that pass a fixed point per unit time; ($v$), the rate at which the propagates through space; and , the magnitude of the oscillation about the mean position. The speed of a is equal to the product of its wavelength and frequency. consists of two perpendicular waves, one electric and one magnetic, propagating at the ($c$). Electromagnetic radiation is radiant energy that includes radio waves, microwaves, visible light, x-rays, and gamma rays, which differ in their frequencies and wavelengths. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/07%3A_Quantum_Theory_and_Atomic_Structure/7.01%3A_The_Nature_of_Light.txt |
Learning Objectives
• To know the relationship between atomic spectra and the electronic structure of atoms.
The concept of the photon emerged from experimentation with thermal radiation, electromagnetic radiation emitted as the result of a source’s temperature, which produces a continuous spectrum of energies.The photoelectric effect provided indisputable evidence for the existence of the photon and thus the particle-like behavior of electromagnetic radiation. However, more direct evidence was needed to verify the quantized nature of energy in all matter. In this section, we describe how observation of the interaction of atoms with visible light provided this evidence.
Line Spectra
Although objects at high temperature emit a continuous spectrum of electromagnetic radiation, a different kind of spectrum is observed when pure samples of individual elements are heated. For example, when a high-voltage electrical discharge is passed through a sample of hydrogen gas at low pressure, the resulting individual isolated hydrogen atoms caused by the dissociation of H2 emit a red light. Unlike blackbody radiation, the color of the light emitted by the hydrogen atoms does not depend greatly on the temperature of the gas in the tube. When the emitted light is passed through a prism, only a few narrow lines of particular wavelengths, called a line spectrum, are observed rather than a continuous range of wavelengths (Figure $1$). The light emitted by hydrogen atoms is red because, of its four characteristic lines, the most intense line in its spectrum is in the red portion of the visible spectrum, at 656 nm. With sodium, however, we observe a yellow color because the most intense lines in its spectrum are in the yellow portion of the spectrum, at about 589 nm.
Such emission spectra were observed for many other elements in the late 19th century, which presented a major challenge because classical physics was unable to explain them. Part of the explanation is provided by Planck’s equation: the observation of only a few values of λ (or $u$) in the line spectrum meant that only a few values of E were possible. Thus the energy levels of a hydrogen atom had to be quantized; in other words, only states that had certain values of energy were possible, or allowed. If a hydrogen atom could have any value of energy, then a continuous spectrum would have been observed, similar to blackbody radiation.
In 1885, a Swiss mathematics teacher, Johann Balmer (1825–1898), showed that the frequencies of the lines observed in the visible region of the spectrum of hydrogen fit a simple equation that can be expressed as follows:
$u=constant\; \left ( \dfrac{1}{2^{2}}-\dfrac{1}{n^{^{2}}} \right ) \label{6.3.1}$
where n = 3, 4, 5, 6. As a result, these lines are known as the Balmer series. The Swedish physicist Johannes Rydberg (1854–1919) subsequently restated and expanded Balmer’s result in the Rydberg equation:
$\dfrac{1}{\lambda }=\Re\; \left ( \dfrac{1}{n^{2}_{1}}-\dfrac{1}{n^{2}_{2}} \right ) \label{6.3.2}$
where $n_1$ and $n_2$ are positive integers, $n_2 > n_1$, and $\Re$ the Rydberg constant, has a value of 1.09737 × 107 m−1.
Johann Balmer (1825–1898)
A mathematics teacher at a secondary school for girls in Switzerland, Balmer was 60 years old when he wrote the paper on the spectral lines of hydrogen that made him famous.
Balmer published only one other paper on the topic, which appeared when he was 72 years old.
Like Balmer’s equation, Rydberg’s simple equation described the wavelengths of the visible lines in the emission spectrum of hydrogen (with n1 = 2, n2 = 3, 4, 5,…). More important, Rydberg’s equation also predicted the wavelengths of other series of lines that would be observed in the emission spectrum of hydrogen: one in the ultraviolet (n1 = 1, n2 = 2, 3, 4,…) and one in the infrared (n1 = 3, n2 = 4, 5, 6). Unfortunately, scientists had not yet developed any theoretical justification for an equation of this form.
Bohr's Model
In 1913, a Danish physicist, Niels Bohr (1885–1962; Nobel Prize in Physics, 1922), proposed a theoretical model for the hydrogen atom that explained its emission spectrum. Bohr’s model required only one assumption: The electron moves around the nucleus in circular orbits that can have only certain allowed radii. Rutherford’s earlier model of the atom had also assumed that electrons moved in circular orbits around the nucleus and that the atom was held together by the electrostatic attraction between the positively charged nucleus and the negatively charged electron. Although we now know that the assumption of circular orbits was incorrect, Bohr’s insight was to propose that the electron could occupy only certain regions of space.
Using classical physics, Niels Bohr showed that the energy of an electron in a particular orbit is given by
$E_{n}=\dfrac{-\Re hc}{n^{2}} \label{6.3.3}$
where $\Re$ is the Rydberg constant, h is Planck’s constant, c is the speed of light, and n is a positive integer corresponding to the number assigned to the orbit, with n = 1 corresponding to the orbit closest to the nucleus. In this model n = ∞ corresponds to the level where the energy holding the electron and the nucleus together is zero. In that level, the electron is unbound from the nucleus and the atom has been separated into a negatively charged (the electron) and a positively charged (the nucleus) ion. In this state the radius of the orbit is also infinite. The atom has been ionized.
Niels Bohr (1885–1962)
During the Nazi occupation of Denmark in World War II, Bohr escaped to the United States, where he became associated with the Atomic Energy Project.
In his final years, he devoted himself to the peaceful application of atomic physics and to resolving political problems arising from the development of atomic weapons.
As n decreases, the energy holding the electron and the nucleus together becomes increasingly negative, the radius of the orbit shrinks and more energy is needed to ionize the atom. The orbit with n = 1 is the lowest lying and most tightly bound. The negative sign in Equation $\ref{6.3.3}$ indicates that the electron-nucleus pair is more tightly bound (i.e. at a lower potential energy) when they are near each other than when they are far apart. Because a hydrogen atom with its one electron in this orbit has the lowest possible energy, this is the ground state (the most stable arrangement of electrons for an element or a compound) for a hydrogen atom. As n increases, the radius of the orbit increases; the electron is farther from the proton, which results in a less stable arrangement with higher potential energy (Figure $\PageIndex{2a}$). A hydrogen atom with an electron in an orbit with n > 1 is therefore in an excited state, defined as any arrangement of electrons that is higher in energy than the ground state. When an atom in an excited state undergoes a transition to the ground state in a process called decay, it loses energy by emitting a photon whose energy corresponds to the difference in energy between the two states (Figure $1$).
So the difference in energy ($ΔE$) between any two orbits or energy levels is given by $\Delta E=E_{n_{1}}-E_{n_{2}}$ where n1 is the final orbit and n2 the initial orbit. Substituting from Bohr’s equation (Equation \ref{6.3.3}) for each energy value gives
\begin{align*} \Delta E &=E_{final}-E_{initial} \[4pt] &=-\dfrac{\Re hc}{n_{2}^{2}}-\left ( -\dfrac{\Re hc}{n_{1}^{2}} \right ) \[4pt] &=-\Re hc\left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \label{6.3.4} \end{align*}
If $n_2 > n_1$, the transition is from a higher energy state (larger-radius orbit) to a lower energy state (smaller-radius orbit), as shown by the dashed arrow in part (a) in Figure $3$. Substituting $hc/λ$ for $ΔE$ gives
$\Delta E = \dfrac{hc}{\lambda }=-\Re hc\left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \label{6.3.5}$
Canceling $hc$ on both sides gives
$\dfrac{1}{\lambda }=-\Re \left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \label{6.3.6}$
Except for the negative sign, this is the same equation that Rydberg obtained experimentally. The negative sign in Equations $\ref{6.3.5}$ and $\ref{6.3.6}$ indicates that energy is released as the electron moves from orbit $n_2$ to orbit $n_1$ because orbit $n_2$ is at a higher energy than orbit $n_1$. Bohr calculated the value of $\Re$ from fundamental constants such as the charge and mass of the electron and Planck's constant and obtained a value of 1.0974 × 107 m−1, the same number Rydberg had obtained by analyzing the emission spectra.
We can now understand the physical basis for the Balmer series of lines in the emission spectrum of hydrogen ($\PageIndex{3b}$); the lines in this series correspond to transitions from higher-energy orbits (n > 2) to the second orbit (n = 2). Thus the hydrogen atoms in the sample have absorbed energy from the electrical discharge and decayed from a higher-energy excited state (n > 2) to a lower-energy state (n = 2) by emitting a photon of electromagnetic radiation whose energy corresponds exactly to the difference in energy between the two states (Figure $\PageIndex{3a}$). The n = 3 to n = 2 transition gives rise to the line at 656 nm (red), the n = 4 to n = 2 transition to the line at 486 nm (green), the n = 5 to n = 2 transition to the line at 434 nm (blue), and the n = 6 to n = 2 transition to the line at 410 nm (violet). Because a sample of hydrogen contains a large number of atoms, the intensity of the various lines in a line spectrum depends on the number of atoms in each excited state. At the temperature in the gas discharge tube, more atoms are in the n = 3 than the n ≥ 4 levels. Consequently, the n = 3 to n = 2 transition is the most intense line, producing the characteristic red color of a hydrogen discharge (Figure $\PageIndex{1a}$). Other families of lines are produced by transitions from excited states with n > 1 to the orbit with n = 1 or to orbits with n ≥ 3. These transitions are shown schematically in Figure $4$
The Bohr Atom:
Using Atoms to Time
In contemporary applications, electron transitions are used in timekeeping that needs to be exact. Telecommunications systems, such as cell phones, depend on timing signals that are accurate to within a millionth of a second per day, as are the devices that control the US power grid. Global positioning system (GPS) signals must be accurate to within a billionth of a second per day, which is equivalent to gaining or losing no more than one second in 1,400,000 years. Quantifying time requires finding an event with an interval that repeats on a regular basis.
To achieve the accuracy required for modern purposes, physicists have turned to the atom. The current standard used to calibrate clocks is the cesium atom. Supercooled cesium atoms are placed in a vacuum chamber and bombarded with microwaves whose frequencies are carefully controlled. When the frequency is exactly right, the atoms absorb enough energy to undergo an electronic transition to a higher-energy state. Decay to a lower-energy state emits radiation. The microwave frequency is continually adjusted, serving as the clock’s pendulum.
In 1967, the second was defined as the duration of 9,192,631,770 oscillations of the resonant frequency of a cesium atom, called the cesium clock. Research is currently under way to develop the next generation of atomic clocks that promise to be even more accurate. Such devices would allow scientists to monitor vanishingly faint electromagnetic signals produced by nerve pathways in the brain and geologists to measure variations in gravitational fields, which cause fluctuations in time, that would aid in the discovery of oil or minerals.
Example $1$: The Lyman Series
The so-called Lyman series of lines in the emission spectrum of hydrogen corresponds to transitions from various excited states to the n = 1 orbit. Calculate the wavelength of the lowest-energy line in the Lyman series to three significant figures. In what region of the electromagnetic spectrum does it occur?
Given: lowest-energy orbit in the Lyman series
Asked for: wavelength of the lowest-energy Lyman line and corresponding region of the spectrum
Strategy:
1. Substitute the appropriate values into Equation \ref{6.3.2} (the Rydberg equation) and solve for $\lambda$.
2. Use Figure 2.2.1 to locate the region of the electromagnetic spectrum corresponding to the calculated wavelength.
Solution:
We can use the Rydberg equation to calculate the wavelength:
$\dfrac{1}{\lambda }=-\Re \left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \nonumber$
A For the Lyman series, n1 = 1. The lowest-energy line is due to a transition from the n = 2 to n = 1 orbit because they are the closest in energy.
$\dfrac{1}{\lambda }=-\Re \left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right )=1.097\times m^{-1}\left ( \dfrac{1}{1}-\dfrac{1}{4} \right )=8.228 \times 10^{6}\; m^{-1} \nonumber$
It turns out that spectroscopists (the people who study spectroscopy) use cm-1 rather than m-1 as a common unit. Wavelength is inversely proportional to energy but frequency is directly proportional as shown by Planck's formula, $E=h u$.
Spectroscopists often talk about energy and frequency as equivalent. The cm-1 unit is particularly convenient. The infrared range is roughly 200 - 5,000 cm-1, the visible from 11,000 to 25.000 cm-1 and the UV between 25,000 and 100,000 cm-1. The units of cm-1 are called wavenumbers, although people often verbalize it as inverse centimeters. We can convert the answer in part A to cm-1.
$\widetilde{ u} =\dfrac{1}{\lambda }=8.228\times 10^{6}\cancel{m^{-1}}\left (\dfrac{\cancel{m}}{100\;cm} \right )=82,280\: cm^{-1} \nonumber$
and
$\lambda = 1.215 \times 10^{−7}\; m = 122\; nm \nonumber$
This emission line is called Lyman alpha. It is the strongest atomic emission line from the sun and drives the chemistry of the upper atmosphere of all the planets, producing ions by stripping electrons from atoms and molecules. It is completely absorbed by oxygen in the upper stratosphere, dissociating O2 molecules to O atoms which react with other O2 molecules to form stratospheric ozone.
B This wavelength is in the ultraviolet region of the spectrum.
Exercise $1$: The Pfund Series
The Pfund series of lines in the emission spectrum of hydrogen corresponds to transitions from higher excited states to the n = 5 orbit. Calculate the wavelength of the second line in the Pfund series to three significant figures. In which region of the spectrum does it lie?
Answer
4.65 × 103 nm; infrared
Bohr’s model of the hydrogen atom gave an exact explanation for its observed emission spectrum. The following are his key contributions to our understanding of atomic structure:
Unfortunately, Bohr could not explain why the electron should be restricted to particular orbits. Also, despite a great deal of tinkering, such as assuming that orbits could be ellipses rather than circles, his model could not quantitatively explain the emission spectra of any element other than hydrogen (Figure $5$). In fact, Bohr’s model worked only for species that contained just one electron: H, He+, Li2+, and so forth. Scientists needed a fundamental change in their way of thinking about the electronic structure of atoms to advance beyond the Bohr model.
The Energy States of the Hydrogen Atom
Thus far we have explicitly considered only the emission of light by atoms in excited states, which produces an emission spectrum (a spectrum produced by the emission of light by atoms in excited states). The converse, absorption of light by ground-state atoms to produce an excited state, can also occur, producing an absorption spectrum (a spectrum produced by the absorption of light by ground-state atoms).
When an atom emits light, it decays to a lower energy state; when an atom absorbs light, it is excited to a higher energy state.
If white light is passed through a sample of hydrogen, hydrogen atoms absorb energy as an electron is excited to higher energy levels (orbits with n ≥ 2). If the light that emerges is passed through a prism, it forms a continuous spectrum with black lines (corresponding to no light passing through the sample) at 656, 468, 434, and 410 nm. These wavelengths correspond to the n = 2 to n = 3, n = 2 to n = 4, n = 2 to n = 5, and n = 2 to n = 6 transitions. Any given element therefore has both a characteristic emission spectrum and a characteristic absorption spectrum, which are essentially complementary images.
Emission and absorption spectra form the basis of spectroscopy, which uses spectra to provide information about the structure and the composition of a substance or an object. In particular, astronomers use emission and absorption spectra to determine the composition of stars and interstellar matter. As an example, consider the spectrum of sunlight shown in Figure $7$ Because the sun is very hot, the light it emits is in the form of a continuous emission spectrum. Superimposed on it, however, is a series of dark lines due primarily to the absorption of specific frequencies of light by cooler atoms in the outer atmosphere of the sun. By comparing these lines with the spectra of elements measured on Earth, we now know that the sun contains large amounts of hydrogen, iron, and carbon, along with smaller amounts of other elements. During the solar eclipse of 1868, the French astronomer Pierre Janssen (1824–1907) observed a set of lines that did not match those of any known element. He suggested that they were due to the presence of a new element, which he named helium, from the Greek helios, meaning “sun.” Helium was finally discovered in uranium ores on Earth in 1895. Alpha particles are helium nuclei. Alpha particles emitted by the radioactive uranium pick up electrons from the rocks to form helium atoms.
The familiar red color of neon signs used in advertising is due to the emission spectrum of neon shown in part (b) in Figure $5$. Similarly, the blue and yellow colors of certain street lights are caused, respectively, by mercury and sodium discharges. In all these cases, an electrical discharge excites neutral atoms to a higher energy state, and light is emitted when the atoms decay to the ground state. In the case of mercury, most of the emission lines are below 450 nm, which produces a blue light (part (c) in Figure $5$). In the case of sodium, the most intense emission lines are at 589 nm, which produces an intense yellow light.
Summary
There is an intimate connection between the atomic structure of an atom and its spectral characteristics. Atoms of individual elements emit light at only specific wavelengths, producing a line spectrum rather than the continuous spectrum of all wavelengths produced by a hot object. Niels Bohr explained the line spectrum of the hydrogen atom by assuming that the electron moved in circular orbits and that orbits with only certain radii were allowed. Lines in the spectrum were due to transitions in which an electron moved from a higher-energy orbit with a larger radius to a lower-energy orbit with smaller radius. The orbit closest to the nucleus represented the ground state of the atom and was most stable; orbits farther away were higher-energy excited states. Transitions from an excited state to a lower-energy state resulted in the emission of light with only a limited number of wavelengths. Atoms can also absorb light of certain energies, resulting in a transition from the ground state or a lower-energy excited state to a higher-energy excited state. This produces an absorption spectrum, which has dark lines in the same position as the bright lines in the emission spectrum of an element. Bohr’s model revolutionized the understanding of the atom but could not explain the spectra of atoms heavier than hydrogen.
Key Concepts
• Electrons can occupy only certain regions of space, called orbits.
• Orbits closer to the nucleus are lower in energy.
• Electrons can move from one orbit to another by absorbing or emitting energy, giving rise to characteristic spectra. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/07%3A_Quantum_Theory_and_Atomic_Structure/7.02%3A_Atomic_Spectra.txt |
Learning Objectives
• To understand the wave–particle duality of matter.
Einstein’s photons of light were individual packets of energy having many of the characteristics of particles. Recall that the collision of an electron (a particle) with a sufficiently energetic photon can eject a photoelectron from the surface of a metal. Any excess energy is transferred to the electron and is converted to the kinetic energy of the ejected electron. Einstein’s hypothesis that energy is concentrated in localized bundles, however, was in sharp contrast to the classical notion that energy is spread out uniformly in a wave. We now describe Einstein’s theory of the relationship between energy and mass, a theory that others built on to develop our current model of the atom.
The Wave Character of Matter
Einstein initially assumed that photons had zero mass, which made them a peculiar sort of particle indeed. In 1905, however, he published his special theory of relativity, which related energy and mass according to the famous equation:
$E=h u=h\dfrac{c}{\lambda }=mc^{2} \label{6.4.1}$
According to this theory, a photon of wavelength $λ$ and frequency $u$ has a nonzero mass, which is given as follows:
$m=\dfrac{E}{c^{2}}=\dfrac{h u }{c^{2}}=\dfrac{h}{\lambda c} \label{6.4.2}$
That is, light, which had always been regarded as a wave, also has properties typical of particles, a condition known as wave–particle duality (a principle that matter and energy have properties typical of both waves and particles). Depending on conditions, light could be viewed as either a wave or a particle.
In 1922, the American physicist Arthur Compton (1892–1962) reported the results of experiments involving the collision of x-rays and electrons that supported the particle nature of light. At about the same time, a young French physics student, Louis de Broglie (1892–1972), began to wonder whether the converse was true: Could particles exhibit the properties of waves? In his PhD dissertation submitted to the Sorbonne in 1924, de Broglie proposed that a particle such as an electron could be described by a wave whose wavelength is given by
$\lambda =\dfrac{h}{mv} \label{6.4.3}$
where
• $h$ is Planck’s constant,
• $m$ is the mass of the particle, and
• $v$ is the velocity of the particle.
This revolutionary idea was quickly confirmed by American physicists Clinton Davisson (1881–1958) and Lester Germer (1896–1971), who showed that beams of electrons, regarded as particles, were diffracted by a sodium chloride crystal in the same manner as x-rays, which were regarded as waves. It was proven experimentally that electrons do exhibit the properties of waves. For his work, de Broglie received the Nobel Prize in Physics in 1929.
If particles exhibit the properties of waves, why had no one observed them before? The answer lies in the numerator of de Broglie’s equation, which is an extremely small number. As you will calculate in Example $1$, Planck’s constant (6.63 × 10−34 J•s) is so small that the wavelength of a particle with a large mass is too short (less than the diameter of an atomic nucleus) to be noticeable.
The de Broglie Equation:
Example $1$: Wavelength of a Baseball in Motion
Calculate the wavelength of a baseball, which has a mass of 149 g and a speed of 100 mi/h.
Given: mass and speed of object
Asked for: wavelength
Strategy:
1. Convert the speed of the baseball to the appropriate SI units: meters per second.
2. Substitute values into Equation $\ref{6.4.3}$ and solve for the wavelength.
Solution:
The wavelength of a particle is given by $λ = h/mv$. We know that m = 0.149 kg, so all we need to find is the speed of the baseball:
$v=\left ( \dfrac{100\; \cancel{mi}}{\cancel{h}} \right )\left ( \dfrac{1\; \cancel{h}}{60\; \cancel{min}} \right )\left ( \dfrac{1.609\; \cancel{km}}{\cancel{mi}} \right )\left ( \dfrac{1000\; m}{\cancel{km}} \right )$
B Recall that the joule is a derived unit, whose units are (kg•m2)/s2. Thus the wavelength of the baseball is
$\lambda =\dfrac{6.626\times 10^{-34}\; J\cdot s}{\left ( 0.149\; kg \right )\left ( 44.69\; m\cdot s \right )}= \dfrac{6.626\times 10^{-34}\; \cancel{kg}\cdot m{^\cancel{2}\cdot \cancel{s}{\cancel{^{-2}}\cdot \cancel{s}}}}{\left ( 0.149\; \cancel{kg} \right )\left ( 44.69\; \cancel{m}\cdot \cancel{s^{-1}} \right )}=9.95\times 10^{-35}\; m \nonumber$
(You should verify that the units cancel to give the wavelength in meters.) Given that the diameter of the nucleus of an atom is approximately 10−14 m, the wavelength of the baseball is almost unimaginably small.
Exercise $1$: Wavelength of a Neutron in Motion
Calculate the wavelength of a neutron that is moving at 3.00 × 103 m/s.
Answer
1.32 Å, or 132 pm
As you calculated in Example $1$, objects such as a baseball or a neutron have such short wavelengths that they are best regarded primarily as particles. In contrast, objects with very small masses (such as photons) have large wavelengths and can be viewed primarily as waves. Objects with intermediate masses, however, such as electrons, exhibit the properties of both particles and waves. Although we still usually think of electrons as particles, the wave nature of electrons is employed in an electron microscope, which has revealed most of what we know about the microscopic structure of living organisms and materials. Because the wavelength of an electron beam is much shorter than the wavelength of a beam of visible light, this instrument can resolve smaller details than a light microscope can (Figure $1$).
An Important Wave Property: Phase And Interference
A wave is a disturbance that travels in space. The magnitude of the wave at any point in space and time varies sinusoidally. While the absolute value of the magnitude of one wave at any point is not very important, the relative displacement of two waves, called the phase difference, is vitally important because it determines whether the waves reinforce or interfere with each other. Figure $\PageIndex{2A}$ shows an arbitrary phase difference between two wave and Figure $\PageIndex{2B}$ shows what happens when the two waves are 180 degrees out of phase. The green line is their sum. Figure $\PageIndex{2C}$ shows what happens when the two lines are in phase, exactly superimposed on each other. Again, the green line is the sum of the intensities. A pattern of constructive and destructive interference is obtained when two (or more) diffracting waves interact with each other. This principle of diffraction and interference was used to prove the wave properties of electrons and is the basis for how electron microscopes work.
Photograph of an interference pattern produced by circular water waves in a ripple tank.
For a mathematical analysis of phase aspects in sinusoids, check the math Libretexts library.
Standing Waves
De Broglie also investigated why only certain orbits were allowed in Bohr’s model of the hydrogen atom. He hypothesized that the electron behaves like a standing wave (a wave that does not travel in space). An example of a standing wave is the motion of a string of a violin or guitar. When the string is plucked, it vibrates at certain fixed frequencies because it is fastened at both ends (Figure $3$). If the length of the string is $L$, then the lowest-energy vibration (the fundamental) has wavelength
\begin{align} \dfrac{\lambda }{2} & =L \nonumber \ \lambda &= 2L \nonumber \end{align} \label{6.4.4}
Higher-energy vibrations are called overtones (the vibration of a standing wave that is higher in energy than the fundamental vibration) and are produced when the string is plucked more strongly; they have wavelengths given by
$\lambda=\dfrac{2L}{n} \label{6.4.5}$
where n has any integral value. When plucked, all other frequencies die out immediately. Only the resonant frequencies survive and are heard. Thus, we can think of the resonant frequencies of the string as being quantized. Notice in Figure $3$ that all overtones have one or more nodes, points where the string does not move. The amplitude of the wave at a node is zero.
Quantized vibrations and overtones containing nodes are not restricted to one-dimensional systems, such as strings. A two-dimensional surface, such as a drumhead, also has quantized vibrations. Similarly, when the ends of a string are joined to form a circle, the only allowed vibrations are those with wavelength
$2πr = nλ \label{6.4.6}$
where $r$ is the radius of the circle. De Broglie argued that Bohr’s allowed orbits could be understood if the electron behaved like a standing circular wave (Figure $4$). The standing wave could exist only if the circumference of the circle was an integral multiple of the wavelength such that the propagated waves were all in phase, thereby increasing the net amplitudes and causing constructive interference. Otherwise, the propagated waves would be out of phase, resulting in a net decrease in amplitude and causing destructive interference. The nonresonant waves interfere with themselves! De Broglie’s idea explained Bohr’s allowed orbits and energy levels nicely: in the lowest energy level, corresponding to $n = 1$ in Equation $\ref{6.4.6}$, one complete wavelength would close the circle. Higher energy levels would have successively higher values of n with a corresponding number of nodes.
Like all analogies, although the standing wave model helps us understand much about why Bohr's theory worked, it also, if pushed too far, can mislead. As you will see, some of de Broglie’s ideas are retained in the modern theory of the electronic structure of the atom: the wave behavior of the electron and the presence of nodes that increase in number as the energy level increases. Unfortunately, his (and Bohr's) explanation also contains one major feature that we now know to be incorrect: in the currently accepted model, the electron in a given orbit is not always at the same distance from the nucleus.
The Heisenberg Uncertainty Principle
Because a wave is a disturbance that travels in space, it has no fixed position. One might therefore expect that it would also be hard to specify the exact position of a particle that exhibits wavelike behavior. A characteristic of light is that is can be bent or spread out by passing through a narrow slit. You can literally see this by half closing your eyes and looking through your eye lashes. This reduces the brightness of what you are seeing and somewhat fuzzes out the image, but the light bends around your lashes to provide a complete image rather than a bunch of bars across the image. This is called diffraction.
This behavior of waves is captured in Maxwell's equations (1870 or so) for electromagnetic waves and was and is well understood. An "uncertainty principle" for light is, if you will, merely a conclusion about the nature of electromagnetic waves and nothing new. De Broglie's idea of wave-particle duality means that particles such as electrons which exhibit wavelike characteristics will also undergo diffraction from slits whose size is on the order of the electron wavelength.
This situation was described mathematically by the German physicist Werner Heisenberg (1901–1976; Nobel Prize in Physics, 1932), who related the position of a particle to its momentum. Referring to the electron, Heisenberg stated that “at every moment the electron has only an inaccurate position and an inaccurate velocity, and between these two inaccuracies there is this uncertainty relation.” Mathematically, the Heisenberg uncertainty principle states that the uncertainty in the position of a particle (Δx) multiplied by the uncertainty in its momentum [Δ(mv)] is greater than or equal to Planck’s constant divided by 4π:
$\left ( \Delta x \right )\left ( \Delta \left [ mv \right ] \right )\ge \dfrac{h}{4\pi } \label{6.4.7}$
Because Planck’s constant is a very small number, the Heisenberg uncertainty principle is important only for particles such as electrons that have very low masses. These are the same particles predicted by de Broglie’s equation to have measurable wavelengths.
If the precise position $x$ of a particle is known absolutely (Δx = 0), then the uncertainty in its momentum must be infinite:
$\left ( \Delta \left [ mv \right ] \right )= \dfrac{h}{4\pi \left ( \Delta x \right ) }=\dfrac{h}{4\pi \left ( 0 \right ) }=\infty \label{6.4.8}$
Because the mass of the electron at rest ($m$) is both constant and accurately known, the uncertainty in $Δ(mv)$ must be due to the $Δv$ term, which would have to be infinitely large for $Δ(mv)$ to equal infinity. That is, according to Equation $\ref{6.4.8}$, the more accurately we know the exact position of the electron (as $Δx → 0$), the less accurately we know the speed and the kinetic energy of the electron (1/2 mv2) because $Δ(mv) → ∞$. Conversely, the more accurately we know the precise momentum (and the energy) of the electron [as $Δ(mv) → 0$], then $Δx → ∞$ and we have no idea where the electron is.
Bohr’s model of the hydrogen atom violated the Heisenberg uncertainty principle by trying to specify simultaneously both the position (an orbit of a particular radius) and the energy (a quantity related to the momentum) of the electron. Moreover, given its mass and wavelike nature, the electron in the hydrogen atom could not possibly orbit the nucleus in a well-defined circular path as predicted by Bohr. You will see, however, that the most probable radius of the electron in the hydrogen atom is exactly the one predicted by Bohr’s model.
Example $1$: Quantum Nature of Baseballs
Calculate the minimum uncertainty in the position of the pitched baseball from Example $\ref{6.4.1}$ that has a mass of exactly 149 g and a speed of 100 ± 1 mi/h.
Given: mass and speed of object
Asked for: minimum uncertainty in its position
Strategy:
1. Rearrange the inequality that describes the Heisenberg uncertainty principle (Equation $\ref{6.4.7}$) to solve for the minimum uncertainty in the position of an object (Δx).
2. Find Δv by converting the velocity of the baseball to the appropriate SI units: meters per second.
3. Substitute the appropriate values into the expression for the inequality and solve for Δx.
Solution:
A The Heisenberg uncertainty principle (Equation \ref{6.4.7}) tells us that $(Δx)(Δ(mv)) = h/4π \nonumber$. Rearranging the inequality gives
$\Delta x \ge \left( {\dfrac{h}{4\pi }} \right)\left( {\dfrac{1}{\Delta (mv)}} \right)$
B We know that h = 6.626 × 10−34 J•s and m = 0.149 kg. Because there is no uncertainty in the mass of the baseball, Δ(mv) = mΔv and Δv = ±1 mi/h. We have
$\Delta u =\left ( \dfrac{1\; \cancel{mi}}{\cancel{h}} \right )\left ( \dfrac{1\; \cancel{h}}{60\; \cancel{min}} \right )\left ( \dfrac{1\; \cancel{min}}{60\; s} \right )\left ( \dfrac{1.609\; \cancel{km}}{\cancel{mi}} \right )\left ( \dfrac{1000\; m}{\cancel{km}} \right )=0.4469\; m/s \nonumber$
C Therefore,
$\Delta x \ge \left ( \dfrac{6.626\times 10^{-34}\; J\cdot s}{4\left ( 3.1416 \right )} \right ) \left ( \dfrac{1}{\left ( 0.149\; kg \right )\left ( 0.4469\; m\cdot s^{-1} \right )} \right ) \nonumber$
Inserting the definition of a joule (1 J = 1 kg•m2/s2) gives
$\Delta x \ge \left ( \dfrac{6.626\times 10^{-34}\; \cancel{kg} \cdot m^{\cancel{2}} \cdot s}{4\left ( 3.1416 \right )\left ( \cancel{s^{2}} \right )} \right ) \left ( \dfrac{1\; \cancel{s}}{\left ( 0.149\; \cancel{kg} \right )\left ( 0.4469\; \cancel{m} \right )} \right ) \nonumber$
$\Delta x \ge 7.92 \pm \times 10^{-34}\; m \nonumber$
This is equal to $3.12 \times 10^{−32}$ inches. We can safely say that if a batter misjudges the speed of a fastball by 1 mi/h (about 1%), he will not be able to blame Heisenberg’s uncertainty principle for striking out.
Exercise $2$
Calculate the minimum uncertainty in the position of an electron traveling at one-third the speed of light, if the uncertainty in its speed is ±0.1%. Assume its mass to be equal to its mass at rest.
Answer
6 × 10−10 m, or 0.6 nm (about the diameter of a benzene molecule)
Summary
An electron possesses both particle and wave properties. The modern model for the electronic structure of the atom is based on recognizing that an electron possesses particle and wave properties, the so-called wave–particle duality. Louis de Broglie showed that the wavelength of a particle is equal to Planck’s constant divided by the mass times the velocity of the particle.
$\lambda =\dfrac{h}{mv} \nonumber$
The electron in Bohr’s circular orbits could thus be described as a standing wave, one that does not move through space. Standing waves are familiar from music: the lowest-energy standing wave is the fundamental vibration, and higher-energy vibrations are overtones and have successively more nodes, points where the amplitude of the wave is always zero. Werner Heisenberg’s uncertainty principle states that it is impossible to precisely describe both the location and the speed of particles that exhibit wavelike behavior.
$\left ( \Delta x \right )\left ( \Delta \left [ mv \right ] \right )\geqslant \dfrac{h}{4\pi } \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/07%3A_Quantum_Theory_and_Atomic_Structure/7.03%3A_The_Wave-Particle_Duality_of_Matter_and_Energy.txt |
Learning Objectives
• To apply the results of quantum mechanics to chemistry.
The paradox described by Heisenberg’s uncertainty principle and the wavelike nature of subatomic particles such as the electron made it impossible to use the equations of classical physics to describe the motion of electrons in atoms. Scientists needed a new approach that took the wave behavior of the electron into account. In 1926, an Austrian physicist, Erwin Schrödinger (1887–1961; Nobel Prize in Physics, 1933), developed wave mechanics, a mathematical technique that describes the relationship between the motion of a particle that exhibits wavelike properties (such as an electron) and its allowed energies.
Erwin Schrödinger (1887–1961)
Schrödinger’s unconventional approach to atomic theory was typical of his unconventional approach to life. He was notorious for his intense dislike of memorizing data and learning from books. When Hitler came to power in Germany, Schrödinger escaped to Italy. He then worked at Princeton University in the United States but eventually moved to the Institute for Advanced Studies in Dublin, Ireland, where he remained until his retirement in 1955.
Although quantum mechanics uses sophisticated mathematics, you do not need to understand the mathematical details to follow our discussion of its general conclusions. We focus on the properties of the wavefunctions that are the solutions of Schrödinger’s equations.
Wavefunctions
A wavefunction (Ψ) is a mathematical function that relates the location of an electron at a given point in space (identified by x, y, and z coordinates) to the amplitude of its wave, which corresponds to its energy. Thus each wavefunction is associated with a particular energy E. The properties of wavefunctions derived from quantum mechanics are summarized here:
• A wavefunction uses three variables to describe the position of an electron. A fourth variable is usually required to fully describe the location of objects in motion. Three specify the position in space (as with the Cartesian coordinates x, y, and z), and one specifies the time at which the object is at the specified location. For example, if you were the captain of a ship trying to intercept an enemy submarine, you would need to know its latitude, longitude, and depth, as well as the time at which it was going to be at this position (Figure $1$). For electrons, we can ignore the time dependence because we will be using standing waves, which by definition do not change with time, to describe the position of an electron.
• The magnitude of the wavefunction at a particular point in space is proportional to the amplitude of the wave at that point. Many wavefunctions are complex functions, which is a mathematical term indicating that they contain $\sqrt{-1}$, represented as $i$. Hence the amplitude of the wave has no real physical significance. In contrast, the sign of the wavefunction (either positive or negative) corresponds to the phase of the wave, which will be important in our discussion of chemical bonding. The sign of the wavefunction should not be confused with a positive or negative electrical charge.
• The square of the wavefunction at a given point is proportional to the probability of finding an electron at that point, which leads to a distribution of probabilities in space. The square of the wavefunction ($\Psi^2$) is always a real quantity [recall that that $\sqrt{-1}^2=-1$] and is proportional to the probability of finding an electron at a given point. More accurately, the probability is given by the product of the wavefunction Ψ and its complex conjugate Ψ*, in which all terms that contain i are replaced by $−i$. We use probabilities because, according to Heisenberg’s uncertainty principle, we cannot precisely specify the position of an electron. The probability of finding an electron at any point in space depends on several factors, including the distance from the nucleus and, in many cases, the atomic equivalent of latitude and longitude. As one way of graphically representing the probability distribution, the probability of finding an electron is indicated by the density of colored dots, as shown for the ground state of the hydrogen atom in Figure $2$.
• Describing the electron distribution as a standing wave leads to sets of quantum numbers that are characteristic of each wavefunction. From the patterns of one- and two-dimensional standing waves shown previously, you might expect (correctly) that the patterns of three-dimensional standing waves would be complex. Fortunately, however, in the 18th century, a French mathematician, Adrien Legendre (1752–1783), developed a set of equations to describe the motion of tidal waves on the surface of a flooded planet. Schrödinger incorporated Legendre’s equations into his wavefunctions. The requirement that the waves must be in phase with one another to avoid cancellation and produce a standing wave results in a limited number of solutions (wavefunctions), each of which is specified by a set of numbers called quantum numbers.
• Each wavefunction is associated with a particular energy. As in Bohr’s model, the energy of an electron in an atom is quantized; it can have only certain allowed values. The major difference between Bohr’s model and Schrödinger’s approach is that Bohr had to impose the idea of quantization arbitrarily, whereas in Schrödinger’s approach, quantization is a natural consequence of describing an electron as a standing wave.
Quantum Numbers
Schrödinger’s approach uses three quantum numbers (n, l, and ml) to specify any wavefunction. The quantum numbers provide information about the spatial distribution of an electron. Although n can be any positive integer, only certain values of l and ml are allowed for a given value of n.
The Principal Quantum Number
The principal quantum number (n) tells the average relative distance of an electron from the nucleus:
$n = 1, 2, 3, 4,… \label{6.5.1}$
As n increases for a given atom, so does the average distance of an electron from the nucleus. A negatively charged electron that is, on average, closer to the positively charged nucleus is attracted to the nucleus more strongly than an electron that is farther out in space. This means that electrons with higher values of n are easier to remove from an atom. All wavefunctions that have the same value of n are said to constitute a principal shell because those electrons have similar average distances from the nucleus. As you will see, the principal quantum number n corresponds to the n used by Bohr to describe electron orbits and by Rydberg to describe atomic energy levels.
The Azimuthal Quantum Number
The second quantum number is often called the azimuthal quantum number (l). The value of l describes the shape of the region of space occupied by the electron. The allowed values of l depend on the value of n and can range from 0 to n − 1:
$l = 0, 1, 2,…, n − 1 \label{6.5.2}$
For example, if n = 1, l can be only 0; if n = 2, l can be 0 or 1; and so forth. For a given atom, all wavefunctions that have the same values of both n and l form a subshell. The regions of space occupied by electrons in the same subshell usually have the same shape, but they are oriented differently in space.
Principal quantum number (n) & Orbital angular momentum (l): The Orbital Subshell:
The Magnetic Quantum Number
The third quantum number is the magnetic quantum number ($m_l$). The value of $m_l$ describes the orientation of the region in space occupied by an electron with respect to an applied magnetic field. The allowed values of $m_l$ depend on the value of l: ml can range from −l to l in integral steps:
$m_l = −l, −l + 1,…, 0,…, l − 1, l \label{6.5.3}$
For example, if $l = 0$, $m_l$ can be only 0; if l = 1, ml can be −1, 0, or +1; and if l = 2, ml can be −2, −1, 0, +1, or +2.
Each wavefunction with an allowed combination of n, l, and ml values describes an atomic orbital, a particular spatial distribution for an electron. For a given set of quantum numbers, each principal shell has a fixed number of subshells, and each subshell has a fixed number of orbitals.
Example$1$: n=4 Shell Structure
How many subshells and orbitals are contained within the principal shell with n = 4?
Given: value of n
Asked for: number of subshells and orbitals in the principal shell
Strategy:
1. Given n = 4, calculate the allowed values of l. From these allowed values, count the number of subshells.
2. For each allowed value of l, calculate the allowed values of ml. The sum of the number of orbitals in each subshell is the number of orbitals in the principal shell.
Solution:
A We know that l can have all integral values from 0 to n − 1. If n = 4, then l can equal 0, 1, 2, or 3. Because the shell has four values of l, it has four subshells, each of which will contain a different number of orbitals, depending on the allowed values of ml.
B For l = 0, ml can be only 0, and thus the l = 0 subshell has only one orbital. For l = 1, ml can be 0 or ±1; thus the l = 1 subshell has three orbitals. For l = 2, ml can be 0, ±1, or ±2, so there are five orbitals in the l = 2 subshell. The last allowed value of l is l = 3, for which ml can be 0, ±1, ±2, or ±3, resulting in seven orbitals in the l = 3 subshell. The total number of orbitals in the n = 4 principal shell is the sum of the number of orbitals in each subshell and is equal to n2 = 16
Exercise $1$: n=3 Shell Structure
How many subshells and orbitals are in the principal shell with n = 3?
Answer
three subshells; nine orbitals
Rather than specifying all the values of n and l every time we refer to a subshell or an orbital, chemists use an abbreviated system with lowercase letters to denote the value of l for a particular subshell or orbital:
abbreviated system with lowercase letters to denote the value of l for a particular subshell or orbital
l = 0 1 2 3
Designation s p d f
The principal quantum number is named first, followed by the letter s, p, d, or f as appropriate. (These orbital designations are derived from historical terms for corresponding spectroscopic characteristics: sharp, principle, diffuse, and fundamental.) A 1s orbital has n = 1 and l = 0; a 2p subshell has n = 2 and l = 1 (and has three 2p orbitals, corresponding to ml = −1, 0, and +1); a 3d subshell has n = 3 and l = 2 (and has five 3d orbitals, corresponding to ml = −2, −1, 0, +1, and +2); and so forth.
We can summarize the relationships between the quantum numbers and the number of subshells and orbitals as follows (Table 6.5.1):
• Each principal shell has n subshells. For n = 1, only a single subshell is possible (1s); for n = 2, there are two subshells (2s and 2p); for n = 3, there are three subshells (3s, 3p, and 3d); and so forth. Every shell has an ns subshell, any shell with n ≥ 2 also has an np subshell, and any shell with n ≥ 3 also has an nd subshell. Because a 2d subshell would require both n = 2 and l = 2, which is not an allowed value of l for n = 2, a 2d subshell does not exist.
• Each subshell has 2l + 1 orbitals. This means that all ns subshells contain a single s orbital, all np subshells contain three p orbitals, all nd subshells contain five d orbitals, and all nf subshells contain seven f orbitals.
Each principal shell has n subshells, and each subshell has 2l + 1 orbitals.
Table $1$: Values of n, l, and ml through n = 4
n l Subshell Designation $m_l$ Number of Orbitals in Subshell Number of Orbitals in Shell
1 0 1s 0 1 1
2 0 2s 0 1 4
1 2p −1, 0, 1 3
3 0 3s 0 1 9
1 3p −1, 0, 1 3
2 3d −2, −1, 0, 1, 2 5
4 0 4s 0 1 16
1 4p −1, 0, 1 3
2 4d −2, −1, 0, 1, 2 5
3 4f −3, −2, −1, 0, 1, 2, 3 7
Magnetic Quantum Number (ml) & Spin Quantum Number (ms): Magnetic Quantum Number (ml) & Spin Quantum Number (ms) YouTube(opens in new window) [youtu.be] (opens in new window)
Summary
There is a relationship between the motions of electrons in atoms and molecules and their energies that is described by quantum mechanics. Because of wave–particle duality, scientists must deal with the probability of an electron being at a particular point in space. To do so required the development of quantum mechanics, which uses wavefunctions (Ψ) to describe the mathematical relationship between the motion of electrons in atoms and molecules and their energies. Wavefunctions have five important properties:
1. the wavefunction uses three variables (Cartesian axes x, y, and z) to describe the position of an electron;
2. the magnitude of the wavefunction is proportional to the intensity of the wave;
3. the probability of finding an electron at a given point is proportional to the square of the wavefunction at that point, leading to a distribution of probabilities in space that is often portrayed as an electron density plot;
4. describing electron distributions as standing waves leads naturally to the existence of sets of quantum numbers characteristic of each wavefunction; and
5. each spatial distribution of the electron described by a wavefunction with a given set of quantum numbers has a particular energy.
Quantum numbers provide important information about the energy and spatial distribution of an electron. The principal quantum number n can be any positive integer; as n increases for an atom, the average distance of the electron from the nucleus also increases. All wavefunctions with the same value of n constitute a principal shell in which the electrons have similar average distances from the nucleus. The azimuthal quantum number l can have integral values between 0 and n − 1; it describes the shape of the electron distribution. Wavefunctions that have the same values of both n and l constitute a subshell, corresponding to electron distributions that usually differ in orientation rather than in shape or average distance from the nucleus. The magnetic quantum number ml can have 2l + 1 integral values, ranging from −l to +l, and describes the orientation of the electron distribution. Each wavefunction with a given set of values of n, l, and ml describes a particular spatial distribution of an electron in an atom, an atomic orbital. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/07%3A_Quantum_Theory_and_Atomic_Structure/7.04%3A_The_Quantum-Mechanical_Model_of_the_Atom.txt |
Learning Objectives
• To write the electron configuration of any element and relate its electron configuration to its position in the periodic table.
The quantum mechanical model allowed us to determine the energies of the hydrogen atomic orbitals; now we would like to extend this to describe the electronic structure of every element in the Periodic Table. The process of describing each atom’s electronic structure consists, essentially, of beginning with hydrogen and adding one proton and one electron at a time to create the next heavier element in the table; however, interactions between electrons make this process a bit more complicated than it sounds. All stable nuclei other than hydrogen also contain one or more neutrons. Because neutrons have no electrical charge, however, they can be ignored in the following discussion. Before demonstrating how to do this, however, we must introduce the concept of electron spin and the Pauli principle.
Orbitals and their Energies
Unlike in hydrogen-like atoms with only one electron, in multielectron atoms the values of quantum numbers n and l determine the energies of an orbital. The energies of the different orbitals for a typical multielectron atom are shown in Figure \(1\). Within a given principal shell of a multielectron atom, the orbital energies increase with increasing l. An ns orbital always lies below the corresponding np orbital, which in turn lies below the nd orbital.
These energy differences are caused by the effects of shielding and penetration, the extent to which a given orbital lies inside other filled orbitals. For example, an electron in the 2s orbital penetrates inside a filled 1s orbital more than an electron in a 2p orbital does. Since electrons, all being negatively charged, repel each other, an electron closer to the nucleus partially shields an electron farther from the nucleus from the attractive effect of the positively charged nucleus. Hence in an atom with a filled 1s orbital, the effective nuclear charge (Zeff) experienced by a 2s electron is greater than the Zeff experienced by a 2p electron. Consequently, the 2s electron is more tightly bound to the nucleus and has a lower energy, consistent with the order of energies shown in Figure \(1\).
Due to electron shielding, \(Z_{eff}\) increases more rapidly going across a row of the periodic table than going down a column.
Notice in Figure \(1\) that the difference in energies between subshells can be so large that the energies of orbitals from different principal shells can become approximately equal. For example, the energy of the 3d orbitals in most atoms is actually between the energies of the 4s and the 4p orbitals.
Electron Spin: The Fourth Quantum Number
When scientists analyzed the emission and absorption spectra of the elements more closely, they saw that for elements having more than one electron, nearly all the lines in the spectra were actually pairs of very closely spaced lines. Because each line represents an energy level available to electrons in the atom, there are twice as many energy levels available as would be predicted solely based on the quantum numbers \(n\), \(l\), and \(m_l\). Scientists also discovered that applying a magnetic field caused the lines in the pairs to split farther apart. In 1925, two graduate students in physics in the Netherlands, George Uhlenbeck (1900–1988) and Samuel Goudsmit (1902–1978), proposed that the splittings were caused by an electron spinning about its axis, much as Earth spins about its axis. When an electrically charged object spins, it produces a magnetic moment parallel to the axis of rotation, making it behave like a magnet. Although the electron cannot be viewed solely as a particle, spinning or otherwise, it is indisputable that it does have a magnetic moment. This magnetic moment is called electron spin.
In an external magnetic field, the electron has two possible orientations (Figure Figure \(2\)). These are described by a fourth quantum number (ms), which for any electron can have only two possible values, designated +½ (up) and −½ (down) to indicate that the two orientations are opposites; the subscript s is for spin. An electron behaves like a magnet that has one of two possible orientations, aligned either with the magnetic field or against it.
The Pauli Exclusion Principle
The implications of electron spin for chemistry were recognized almost immediately by an Austrian physicist, Wolfgang Pauli (1900–1958; Nobel Prize in Physics, 1945), who determined that each orbital can contain no more than two electrons. He developed the Pauli exclusion principle: No two electrons in an atom can have the same values of all four quantum numbers (n, l, ml, ms).
By giving the values of n, l, and ml, we also specify a particular orbital (e.g., 1s with n = 1, l = 0, ml = 0). Because ms has only two possible values (+½ or −½), two electrons, and only two electrons, can occupy any given orbital, one with spin up and one with spin down. With this information, we can proceed to construct the entire periodic table, which was originally based on the physical and chemical properties of the known elements.
Example \(1\)
List all the allowed combinations of the four quantum numbers (n, l, ml, ms) for electrons in a 2p orbital and predict the maximum number of electrons the 2p subshell can accommodate.
Given: orbital
Asked for: allowed quantum numbers and maximum number of electrons in orbital
Strategy:
1. List the quantum numbers (n, l, ml) that correspond to an n = 2p orbital. List all allowed combinations of (n, l, ml).
2. Build on these combinations to list all the allowed combinations of (n, l, ml, ms).
3. Add together the number of combinations to predict the maximum number of electrons the 2p subshell can accommodate.
Solution:
A For a 2p orbital, we know that n = 2, l = n − 1 = 1, and ml = −l, (−l +1),…, (l − 1), l. There are only three possible combinations of (n, l, ml): (2, 1, 1), (2, 1, 0), and (2, 1, −1).
B Because ms is independent of the other quantum numbers and can have values of only +½ and −½, there are six possible combinations of (n, l, ml, ms): (2, 1, 1, +½), (2, 1, 1, −½), (2, 1, 0, +½), (2, 1, 0, −½), (2, 1, −1, +½), and (2, 1, −1, −½).
C Hence the 2p subshell, which consists of three 2p orbitals (2px, 2py, and 2pz), can contain a total of six electrons, two in each orbital.
Exercise \(1\)
List all the allowed combinations of the four quantum numbers (n, l, ml, ms) for a 6s orbital, and predict the total number of electrons it can contain.
Answer
(6, 0, 0, +½), (6, 0, 0, −½); two electrons
Magnetic Quantum Number (ml) & Spin Quantum Number (ms): Magnetic Quantum Number (ml) & Spin Quantum Number (ms), YouTube(opens in new window) [youtu.be] (opens in new window)
Summary
The arrangement of atoms in the periodic table arises from the lowest energy arrangement of electrons in the valence shell. In addition to the three quantum numbers (n, l, ml) dictated by quantum mechanics, a fourth quantum number is required to explain certain properties of atoms. This is the electron spin quantum number (ms), which can have values of +½ or −½ for any electron, corresponding to the two possible orientations of an electron in a magnetic field. The concept of electron spin has important consequences for chemistry because the Pauli exclusion principle implies that no orbital can contain more than two electrons (with opposite spin). | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/08%3A_Electron_Configuration_and_Chemical_Periodicity/8.01%3A_Characteristics_of_Many-Electron_Atoms.txt |
Learning Objectives
• To understand the basics of adding electrons to atomic orbitals
• To understand the basics of the Aufbau principle
The electron configuration of an element is the arrangement of its electrons in its atomic orbitals. By knowing the electron configuration of an element, we can predict and explain a great deal of its chemistry.
The Aufbau Principle
We construct the periodic table by following the aufbau principle (from German, meaning “building up”). First we determine the number of electrons in the atom; then we add electrons one at a time to the lowest-energy orbital available without violating the Pauli principle. We use the orbital energy diagram of Figure \(1\), recognizing that each orbital can hold two electrons, one with spin up ↑, corresponding to ms = +½, which is arbitrarily written first, and one with spin down ↓, corresponding to ms = −½. A filled orbital is indicated by ↑↓, in which the electron spins are said to be paired. Here is a schematic orbital diagram for a hydrogen atom in its ground state:
From the orbital diagram, we can write the electron configuration in an abbreviated form in which the occupied orbitals are identified by their principal quantum number n and their value of l (s, p, d, or f), with the number of electrons in the subshell indicated by a superscript. For hydrogen, therefore, the single electron is placed in the 1s orbital, which is the orbital lowest in energy (Figure \(1\)), and the electron configuration is written as 1s1 and read as “one-s-one.”
A neutral helium atom, with an atomic number of 2 (Z = 2), has two electrons. We place one electron in the orbital that is lowest in energy, the 1s orbital. From the Pauli exclusion principle, we know that an orbital can contain two electrons with opposite spin, so we place the second electron in the same orbital as the first but pointing down, so that the electrons are paired. The orbital diagram for the helium atom is therefore
written as 1s2, where the superscript 2 implies the pairing of spins. Otherwise, our configuration would violate the Pauli principle.
The next element is lithium, with Z = 3 and three electrons in the neutral atom. We know that the 1s orbital can hold two of the electrons with their spins paired; the third electron must enter a higher energy orbital. Figure 6.29 tells us that the next lowest energy orbital is 2s, so the orbital diagram for lithium is
This electron configuration is written as 1s22s1.
The next element is beryllium, with Z = 4 and four electrons. We fill both the 1s and 2s orbitals to achieve a 1s22s2 electron configuration:
When we reach boron, with Z = 5 and five electrons, we must place the fifth electron in one of the 2p orbitals. Because all three 2p orbitals are degenerate, it doesn’t matter which one we select. The electron configuration of boron is 1s22s22p1:
At carbon, with Z = 6 and six electrons, we are faced with a choice. Should the sixth electron be placed in the same 2p orbital that already has an electron, or should it go in one of the empty 2p orbitals? If it goes in an empty 2p orbital, will the sixth electron have its spin aligned with or be opposite to the spin of the fifth? In short, which of the following three orbital diagrams is correct for carbon, remembering that the 2p orbitals are degenerate?
Because of electron-electron interactions, it is more favorable energetically for an electron to be in an unoccupied orbital than in one that is already occupied; hence we can eliminate choice a. Similarly, experiments have shown that choice b is slightly higher in energy (less stable) than choice c because electrons in degenerate orbitals prefer to line up with their spins parallel; thus, we can eliminate choice b. Choice c illustrates Hund’s rule (named after the German physicist Friedrich H. Hund, 1896–1997), which today says that the lowest-energy electron configuration for an atom is the one that has the maximum number of electrons with parallel spins in degenerate orbitals. By Hund’s rule, the electron configuration of carbon, which is 1s22s22p2, is understood to correspond to the orbital diagram shown in c. Experimentally, it is found that the ground state of a neutral carbon atom does indeed contain two unpaired electrons.
When we get to nitrogen (Z = 7, with seven electrons), Hund’s rule tells us that the lowest-energy arrangement is
with three unpaired electrons. The electron configuration of nitrogen is thus 1s22s22p3.
At oxygen, with Z = 8 and eight electrons, we have no choice. One electron must be paired with another in one of the 2p orbitals, which gives us two unpaired electrons and a 1s22s22p4 electron configuration. Because all the 2p orbitals are degenerate, it doesn’t matter which one has the pair of electrons.
Similarly, fluorine has the electron configuration 1s22s22p5:
When we reach neon, with Z = 10, we have filled the 2p subshell, giving a 1s22s22p6 electron configuration:
Notice that for neon, as for helium, all the orbitals through the 2p level are completely filled. This fact is very important in dictating both the chemical reactivity and the bonding of helium and neon, as you will see.
Electron Configuration of Atoms:
Valence Electrons
As we continue through the periodic table in this way, writing the electron configurations of larger and larger atoms, it becomes tedious to keep copying the configurations of the filled inner subshells. In practice, chemists simplify the notation by using a bracketed noble gas symbol to represent the configuration of the noble gas from the preceding row because all the orbitals in a noble gas are filled. For example, [Ne] represents the 1s22s22p6 electron configuration of neon (Z = 10), so the electron configuration of sodium, with Z = 11, which is 1s22s22p63s1, is written as [Ne]3s1:
Electron Configuration of Neon and Sodium
Neon Z = 10 1s22s22p6
Sodium Z = 11 1s22s22p63s1 = [Ne]3s1
Because electrons in filled inner orbitals are closer to the nucleus and more tightly bound to it, they are rarely involved in chemical reactions. This means that the chemistry of an atom depends mostly on the electrons in its outermost shell, which are called the valence electrons. The simplified notation allows us to see the valence-electron configuration more easily. Using this notation to compare the electron configurations of sodium and lithium, we have:
Electron Configuration of Sodium and Lithium
Sodium 1s22s22p63s1 = [Ne]3s1
Lithium 1s22s1 = [He]2s1
It is readily apparent that both sodium and lithium have one s electron in their valence shell. We would therefore predict that sodium and lithium have very similar chemistry, which is indeed the case.
As we continue to build the eight elements of period 3, the 3s and 3p orbitals are filled, one electron at a time. This row concludes with the noble gas argon, which has the electron configuration [Ne]3s23p6, corresponding to a filled valence shell.
Example \(1\): Electronic Configuration of Phosphorus
Draw an orbital diagram and use it to derive the electron configuration of phosphorus, Z = 15. What is its valence electron configuration?
Given: atomic number
Asked for: orbital diagram and valence electron configuration for phosphorus
Strategy:
1. Locate the nearest noble gas preceding phosphorus in the periodic table. Then subtract its number of electrons from those in phosphorus to obtain the number of valence electrons in phosphorus.
2. Referring to Figure \(1\), draw an orbital diagram to represent those valence orbitals. Following Hund’s rule, place the valence electrons in the available orbitals, beginning with the orbital that is lowest in energy. Write the electron configuration from your orbital diagram.
3. Ignore the inner orbitals (those that correspond to the electron configuration of the nearest noble gas) and write the valence electron configuration for phosphorus.
Solution:
A Because phosphorus is in the third row of the periodic table, we know that it has a [Ne] closed shell with 10 electrons. We begin by subtracting 10 electrons from the 15 in phosphorus.
B The additional five electrons are placed in the next available orbitals, which Figure \(1\) tells us are the 3s and 3p orbitals:
Because the 3s orbital is lower in energy than the 3p orbitals, we fill it first:
Hund’s rule tells us that the remaining three electrons will occupy the degenerate 3p orbitals separately but with their spins aligned:
The electron configuration is [Ne]3s23p3.
C We obtain the valence electron configuration by ignoring the inner orbitals, which for phosphorus means that we ignore the [Ne] closed shell. This gives a valence-electron configuration of 3s23p3.
Exercise \(1\)
Draw an orbital diagram and use it to derive the electron configuration of chlorine, Z = 17. What is its valence electron configuration?
Answer
[Ne]3s23p5; 3s23p5
Definition of Valence Electrons:
The general order in which orbitals are filled is depicted in Figure \(2\). Subshells corresponding to each value of n are written from left to right on successive horizontal lines, where each row represents a row in the periodic table. The order in which the orbitals are filled is indicated by the diagonal lines running from the upper right to the lower left. Accordingly, the 4s orbital is filled prior to the 3d orbital because of shielding and penetration effects. Consequently, the electron configuration of potassium, which begins the fourth period, is [Ar]4s1, and the configuration of calcium is [Ar]4s2. Five 3d orbitals are filled by the next 10 elements, the transition metals, followed by three 4p orbitals. Notice that the last member of this row is the noble gas krypton (Z = 36), [Ar]4s23d104p6 = [Kr], which has filled 4s, 3d, and 4p orbitals. The fifth row of the periodic table is essentially the same as the fourth, except that the 5s, 4d, and 5p orbitals are filled sequentially.
The sixth row of the periodic table will be different from the preceding two because the 4f orbitals, which can hold 14 electrons, are filled between the 6s and the 5d orbitals. The elements that contain 4f orbitals in their valence shell are the lanthanides. When the 6p orbitals are finally filled, we have reached the next noble gas, radon (Z = 86), [Xe]6s24f145d106p6 = [Rn]. In the last row, the 5f orbitals are filled between the 7s and the 6d orbitals, which gives the 14 actinide elements. Because the large number of protons makes their nuclei unstable, all the actinides are radioactive.
Example \(2\): Electron Configuration of Mercury
Write the electron configuration of mercury (Z = 80), showing all the inner orbitals.
Given: atomic number
Asked for: complete electron configuration
Strategy:
Using the orbital diagram in Figure \(1\) and the periodic table as a guide, fill the orbitals until all 80 electrons have been placed.
Solution:
By placing the electrons in orbitals following the order shown in Figure \(2\) and using the periodic table as a guide, we obtain
Solution to Example 6.8.2
1s2 row 1 2 electrons
2s22p6 row 2 8 electrons
3s23p6 row 3 8 electrons
4s23d104p6 row 4 18 electrons
5s24d105p6 row 5 18 electrons
row 1–5 54 electrons
After filling the first five rows, we still have 80 − 54 = 26 more electrons to accommodate. According to Figure \(2\), we need to fill the 6s (2 electrons), 4f (14 electrons), and 5d (10 electrons) orbitals. The result is mercury’s electron configuration:
1s22s22p63s23p64s23d104p65s24d105p66s24f145d10 = Hg = [Xe]6s24f145d10
with a filled 5d subshell, a 6s24f145d10 valence shell configuration, and a total of 80 electrons. (You should always check to be sure that the total number of electrons equals the atomic number.)
Exercise \(2\): Electron Configuration of Flerovium
Although element 114 is not stable enough to occur in nature, atoms of element 114 were created for the first time in a nuclear reactor in 1998 by a team of Russian and American scientists. The element is named after the Flerov Laboratory of Nuclear Reactions of the Joint Institute for Nuclear Research in Dubna, Russia, where the element was discovered. The name of the laboratory, in turn, honors the Russian physicist Georgy Flyorov. Write the complete electron configuration for element 114.
Answer
s22s22p63s23p64s23d104p65s24d105p66s24f145d106p67s25f146d107p2
The electron configurations of the elements are presented in Figure \(2\), which lists the orbitals in the order in which they are filled. In several cases, the ground state electron configurations are different from those predicted by Figure \(1\). Some of these anomalies occur as the 3d orbitals are filled. For example, the observed ground state electron configuration of chromium is [Ar]4s13d5 rather than the predicted [Ar]4s23d4. Similarly, the observed electron configuration of copper is [Ar]4s13d10 instead of [Ar]s23d9. The actual electron configuration may be rationalized in terms of an added stability associated with a half-filled (ns1, np3, nd5, nf7) or filled (ns2, np6, nd10, nf14) subshell. (In fact, this "special stability" is really another consequence of the instability caused by pairing an electron with another in the same orbital, as illustrated by Hund's rule.) Given the small differences between higher energy levels, this added stability is enough to shift an electron from one orbital to another. In heavier elements, other more complex effects can also be important, leading to many additional anomalies. For example, cerium has an electron configuration of [Xe]6s24f15d1, which is impossible to rationalize in simple terms. In most cases, however, these apparent anomalies do not have important chemical consequences.
Additional stability is associated with half-filled or filled subshells.
Electron Configuration of Transition Metals: Electron Configuration of Transition Metals, YouTube(opens in new window) [youtu.be]
Summary
Based on the Pauli principle and a knowledge of orbital energies obtained using hydrogen-like orbitals, it is possible to construct the periodic table by filling up the available orbitals beginning with the lowest-energy orbitals (the aufbau principle), which gives rise to a particular arrangement of electrons for each element (its electron configuration). Hund’s rule says that the lowest-energy arrangement of electrons is the one that places them in degenerate orbitals with their spins parallel. For chemical purposes, the most important electrons are those in the outermost principal shell, the valence electrons.
Learning Objectives
• To correlate the arrangement of atoms in the periodic table results in blocks corresponding to filling of the ns, np, nd, and nf orbitals
As you have learned, the electron configurations of the elements explain the otherwise peculiar shape of the periodic table. Although the table was originally organized on the basis of physical and chemical similarities between the elements within groups, these similarities are ultimately attributable to orbital energy levels and the Pauli principle, which cause the individual subshells to be filled in a particular order. As a result, the periodic table can be divided into “blocks” corresponding to the type of subshell that is being filled, as illustrated in Figure \(1\). For example, the two columns on the left, known as the s block, consist of elements in which the ns orbitals are being filled. The six columns on the right, elements in which the np orbitals are being filled, constitute the p block. In between are the 10 columns of the d block, elements in which the (n − 1)d orbitals are filled. At the bottom lie the 14 columns of the f block, elements in which the (n − 2)f orbitals are filled. Because two electrons can be accommodated per orbital, the number of columns in each block is the same as the maximum electron capacity of the subshell: 2 for ns, 6 for np, 10 for (n − 1)d, and 14 for (n − 2)f. Within each column, each element has the same valence electron configuration—for example, ns1 (group 1) or ns2np1 (group 13). As you will see, this is reflected in important similarities in the chemical reactivity and the bonding for the elements in each column.
Because each orbital can have a maximum of 2 electrons, there are 2 columns in the s block, 6 columns in the p block, 10 columns in the d block, and 14 columns in the f block.
Hydrogen and helium are placed somewhat arbitrarily. Although hydrogen is not an alkali metal, its 1s1 electron configuration suggests a similarity to lithium ([He]2s1) and the other elements in the first column. Although helium, with a filled ns subshell, should be similar chemically to other elements with an ns2 electron configuration, the closed principal shell dominates its chemistry, justifying its placement above neon on the right.
Example \(1\)
Use the periodic table to predict the valence electron configuration of all the elements of group 2 (beryllium, magnesium, calcium, strontium, barium, and radium).
Given: series of elements
Asked for: valence electron configurations
Strategy:
1. Identify the block in the periodic table to which the group 2 elements belong. Locate the nearest noble gas preceding each element and identify the principal quantum number of the valence shell of each element.
2. Write the valence electron configuration of each element by first indicating the filled inner shells using the symbol for the nearest preceding noble gas and then listing the principal quantum number of its valence shell, its valence orbitals, and the number of valence electrons in each orbital as superscripts.
Solution:
A The group 2 elements are in the s block of the periodic table, and as group 2 elements, they all have two valence electrons. Beginning with beryllium, we see that its nearest preceding noble gas is helium and that the principal quantum number of its valence shell is n = 2.
B Thus beryllium has an [He]s2 electron configuration. The next element down, magnesium, is expected to have exactly the same arrangement of electrons in the n = 3 principal shell: [Ne]s2. By extrapolation, we expect all the group 2 elements to have an ns2 electron configuration.
Exercise \(1\)
Use the periodic table to predict the characteristic valence electron configuration of the halogens in group 17.
Answer
All have an ns2np5 electron configuration, one electron short of a noble gas electron configuration. (Note that the heavier halogens also have filled (n − 1)d10 subshells, as well as an (n − 2)f14 subshell for Rn; these do not, however, affect their chemistry in any significant way.
Summary
The arrangement of atoms in the periodic table results in blocks corresponding to filling of the ns, np, nd, and nf orbitals to produce the distinctive chemical properties of the elements in the s block, p block, d block, and f block, respectively. | textbooks/chem/General_Chemistry/Map%3A_Chemistry_-_The_Molecular_Nature_of_Matter_and_Change_(Silberberg)/08%3A_Electron_Configuration_and_Chemical_Periodicity/8.02%3A_The_Quantum-Mechanical_Model_and_the_Periodic_Table.txt |
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