### CHEM1310 Lecture - Georgia Institute of Technology

```Chapter 10
Properties of Solutions
17.1 Solution Composition
17.2 The Thermodynamics of Solution Formation (skip)
17.3 Factors Affecting Solubility
17.4 The Vapor Pressures of Solutions
17.5 Boiling-Point Elevation and Freezing-Point Depression
17.6 Osmotic Pressure
17.7 Colligative Properties of Electrolyte Solutions
17.8 Colloids
GAS
SOLID
Melting
Freezing
LIQUID
Colligative Properties of Solutions
•
•
For Colligative properties, the difference
between a pure solvent and dilute solution
depends only on the number of solute
particles present and not on their
chemical identity.
Examples
–
–
–
–
Vapor Pressure Depression
Boiling Point Elevation
Melting Point Depression
Osmotic Pressure
Lowering of Vapor Pressure
– Vapor Pressure of a solvent above a dilute solution
is always less than the vapor pressure above the
pure solvent.
Elevation of Boiling Point
– The boiling point of a solution of a non-volatile
solute in a volatile solvent always exceeds the
boiling point of a pure solvent
Boiling
•
liquid in equilibrium with its vapor at the external pressure.
Boiling Point
•
Vapor press = external pressure
Normal boiling point
•
Vap press. = 1 atm
Elevation of Boiling
Point & Vapor
Pressure Depression
T  K m
b solut e
T is the boiling point elevat ion
Kb is molal boiling- point elevation constant
m
is the molalit y of t he solute in solut
solut e

Phase diagrams for pure water (red
lines)
and for an aqueous solution
containing a nonvolatile
solution (blue lines).
Solution Composition
The solute and solvent can be any
combination of solid (s), liquid (l), and
gaseous (g) phases.
Dissolution: Two (or more) substances mix at the level of
individual atoms, molecules, or ions.
Solution: A homogeneous mixture (mixed at level of atoms
molecules or ions
Solvent: The major component
Solute: The minor component
Solution Composition
Mass Fraction, Mole Fraction, Molality and
Molarity
Mass percentage (weight percentage):
mass percentage of the component =
mass of component
X 100%
total mass of mixture
Mole fraction: The amount of a given component (in moles)
divided by the total amount (in moles)
X1 = n1/(n1 + n2) for a two component system
X2 = n2/(n1 + n2) = 1 – X1 or X1+X2=1
Molality
msolute =
moles solute per kilogram solvent
= moles per kg or (mol kg-1)
Molarity (biochemists pay attention)
csolute =
moles solute per volume solution
= moles per liter of solution (mol L-1)
Factors Affecting Solubility
1. Molecular Interactions
– Review chapter 4
– Polar molecules, water soluble, hydrophilic (water
loving)
•
E.g., Vitamins B and C; water-soluble
– Non-polar molecules, soluble in non-polar
molecules, hydrophobic (water fearing)
•
E.g., Vitamins A, D, K and E; fat-soluble
Factors Affecting Solubility of Gases
1. Structure Effects
2. Pressure Effects
Henry’s Law (for dilute solutions)
The mole fraction of volatile solute is proportional
to the vapor pressure of the solute.
P = kH X
kH = Henry’s Law constant, X = mole fraction.
Increasing the partial pressure of a gas over a liquid
increases the amount of gas disolved in the liquid.
kH depends on temperature.
When the partial pressure of nitrogen over a sample of water at
19.4°C is 9.20 atm, the concentration of nitrogen in the water is
5.76 x 10-3 mol L-1. Compute Henry’s law constant for nitrogen
in water at this temperature.
Given
PN  9.20 at m
2
c N  [N2 ]  5.76x103 mol/L
2
Henry's Law
PN  k N X N
2
XN 
2
2
2
nN
nN  nH
2

2
2O
nN
nH
2
2O
When the partial pressure of nitrogen over a sample of water at
19.4°C is 9.20 atm, then the concentration of nitrogen in the
water is 5.76 x 10-3 mol L-1. Compute Henry’s law constant for
nitrogen in water at this temperature.
PN  k N X N
Given
2
PN  9.20 atm
2
c N  [N 2 ]  5.76x10 3 mol/l
PN  k N X N
2
XN 
2
2
kN 
nN

2
nN  nH
2
2
2
2
O
nN
nH
2
rearrange
2
Henry' s Law
2
PN
2
XN
Given

Find
2
2
2
O
Next assume 1 Liter
When the partial pressure of nitrogen over a sample
of water at 19.4°C is 9.20 atm, then the
concentration of nitrogen in the water is 5.76 x 10-3
mol L-1. Compute Henry’s law constant for nitrogen
PN
in water at this temperature.
Given
2
kN
XN
2

nN
nH
2
2O

2
XN

Find
2
3
5 .7 6 x 1 0
m o l/l

 1 0 0 0 g/l 

 1 8 g/m o l



 1 .0 3 7 8x 1 0 4
kN
2

PN
2
XN
2
Giv en
9 .2 0at m


4
Fin d
1 .0 3 7 8 x 1 0
 8 .8 6 x 1 04 at m
Factors Affecting Solubility
1. Structure Effects
2. Pressure Effects
3. Temperature Effects for Aqueous Solutions
The solubility of some
solids as a function of
temperature.
The aqueous solubilities of most solids
increase with increasing temperature,
some decrease with temp.
Endothermic – heat is absorbed by the
system (think evaporation of water, or
melting of ice)
Exothermic – heat is evolved by the system
(think fire, or freezing of water).
Factors Affecting Solubility
1. Structure Effects
2. Pressure Effects
3. Temperature Effects for Aqueous Solutions
The solubility of some gases in
water as a function of
temperature at a constant
pressure of 1 atm.
The greatest gas solubility for a gas in
solution is predicted under what
conditions?
1)
2)
3)
4)
5)
low T, low P
low T, high P
high T, low P
high T, high P
solubility of gases does not depend
upon temperature
The greatest gas solubility for a gas in
solution is predicted under what
conditions?
1)
2)
3)
4)
5)
low T, low P
low T, high P
high T, low P
high T, high P
solubility of gases does not depend
upon temperature
According to Henry's Law, the solubility
of a gas in a liquid
1)
2)
3)
4)
5)
depends on the polarity of the liquid
depends on the liquid's density
remains the same at all temperatures
increases as the gas pressure above the
solution increases
decreases as the gas pressure above the
solution increases
According to Henry's Law, the solubility of a
gas in a liquid
1) depends on the polarity of the liquid
2) depends on the liquid's density
3) remains the same at all temperatures
4) increases as the gas pressure above
the solution increases
5) decreases as the gas pressure above the
solution increases
The Person Behind the Science
Francois-Marie Raoult (1830-1901)
Highlights
– 1886 Raoult's law , the partial pressure
of a solvent vapor in equilibrium with a
solution is proportional to the ratio of the
number of solvent molecules to nonvolatile solute molecules.
– allows molecular weights to be
determined, and provides the
explanation for freezing point depression
and boiling point elevation.
Moments in a Life
Psoln =
XsolventP°solvent
– Raoult was a prominent member of the
group which created physical chemistry,
including Arrhenius, Nernst, van t'Hoff,
Planck.
For ideal
solutions
Raoult’s Law, non-volatile solute
• Consider a non-volatile solute (component 2)
dissolved in a volatile solvent (component 1).
• X1 = the mole fraction of solvent
Raoult’s Law
P1=X1 P°1
P°1 = the vapor
pressure of pure
component 1
Raoult’s Law, volatile solute
• Volatile solute (component 1)
• Volatile solvent (component 2)
P1 = X1 P°1
P2 = X2 P°2
Ptot = P1+ P2
Vapor pressure for a solution of two volatile liquids.
Positive deviation
= solute-solvent attractions < solvent-solvent attractions
For non-ideal Solutions
Negative deviation
= solute-solvent attractions > solvent-solvent attractions
boiling point: T  K m
b solute
freezing point: T  K m
f solute
Osmotic Pressure
PV = nRT
Fourth Colligative Property
• Important for transport of
molecules across cell
membranes, called
semipermeable membranes
• Osmotic Pressure = Π
Π = M RT
ΠV = n RT
Molarity (M) = moles/L or n/V
Osmotic Pressure
The normal flow of solvent into the solution
(osmosis) can be prevented by applying an
external pressure to the solution.
Osmotic Pressure useful for
 Determining the Molar Mass of
protein and other macromolecules
 small concentrations cause
large osmotic pressures
 Can prevent transfer of all solute
particles
 Dialysis at the wall of most
plant and animal cells
Dialysis: Representation of the functioning
of an artificial kidney
A cellophane (polymeric)
tube acts as the semipermeable membrane
 Purifies blood by
washing impurities
(solutes) into the
dialyzing solution.
A dilute aqueous solution of a non-dissociating compound
contains 1.19 g of the compound per liter of solution and
has an osmotic pressure of 0.0288 atm at a temperature of
37°C. Compute the molar mass of the compound.
Strategy
1) use   MRT t o find M (mol/L)
mass
2) Recall t hat #of moles 
mwt
g
g
1.19
L
3) Rearrange mwt 


mole mole
M
L
A dilute aqueous solution of a non-dissociating
compound contains 1.19 g of the compound per liter
of solution and has an osmotic pressure of 0.0288 atm
at a temperature of 37°C. Compute the molar mass
of the compound
Solut ion

RT

0.0288 at m
c

RT
(0.0820 L at m m1
olK 1 )(37 273.15K)
1) use   MRT or M
3
M  1.132x10
m ol/L
2) Rearrange M 
M
1.19g L
g
g
L

m ole m ole
L
3
1.132x10
m ol/L
 1.05x103 g/mol
The Person Behind the Science
J.H. van’t Hoff (1852-1901)
Highlights
– Discovery of the laws of chemical
dynamics and osmotic pressure in
solutions
– Mathematical laws that closely
resemble the laws describing the
behavior of gases.
– his work led to Arrhenius's theory
of electrolytic dissociation or
ionization
– Studies in molecular structure laid
the foundation of stereochemistry.
Moments in a Life
– 1901 awarded first Noble Prize in
Chemistry
van’t Hoff Factor (i)
moles of particles in solution
i
moles of solute dissolved
ΔT = − i m K
Colligative Properties of Electrolyte Solutions
Elevation of Boiling Point
ΔTb = m Kb
Where m = molality
(Molality is moles of solute per kilogram of solvent)
The Effect of Dissociation
ΔTb = i m Kb
i = the number of particles released into the
solution per formula unit of solute
e.g., NaCl dissociates into i = 2
e.g., Na2SO4 dissociates into i = 3
(2 Na+ + 1 SO4-2)
e.g., acetic acid (a weak acid and weak
electrolyte) does not dissociate i = 1
also
Depression of
Freezing Point
ΔTf = − m Kf
ΔTf = − i m Kf
Which aqueous solution would be expected to have
the highest boiling point?
1) 0.100 m NaCl
2) 0.100 m CaCl2
3) 0.080 m Fe(NO3)3
4) 0.080 m Fe(NO3)2
5) 0.080 m Co(SO4)
Which aqueous solution would be expected to have the
highest boiling point?
1)
0.100 m NaCl
ΔTb = (2)(0.100) Kb = 0.200 Kb
2)
0.100 m CaCl2
ΔTb = (3)(0.100) Kb = 0.300 Kb
3)
0.080 m Fe(NO3)3 ΔTb = (4)(0.080) Kb = 0.320 Kb
4)
0.080 m Fe(NO3)2
ΔTb = (3)(0.080) Kb = 0.240 Kb
5)
0.080 m Co(SO4)
ΔTb = (2)(0.080) Kb = 0.160 Kb
Elevation of Boiling Point
The Effect of Dissociation
ΔTb = i m Kb
Colloids: Colloidal Dispersions
• Colloids are large particles dispersed in
solution
– 1nm to 1000 nm in size
– E.g., Globular proteins 500 nm
• Examples
–
–
–
–
–
–
–
Opal (water in solid SiO2)
Aerosols (liquids in Gas)
Smoke (solids in Air)
Milk (fat droplets & solids in water)
Mayonnaise (water droplets in oil)
Paint (solid pigments in liquid)
Biological fluids (proteins & fats in water)
• Characteristics
– Large particle size colloids: translucent, cloudy,
milky)
– Small particle size colloids: can be clear
Colloidal Dispersions
– Tyndall Effect
• Light Scattering
```
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