chemistry acid-base question?
for a 0.1mol/l of monoprotic weak acid (ha) at a certain temperature, the ratio between the unionized and ionized molecules is 50:2, find the value of ionization constant (ka) for the acid, ha.
use the ka equation: 
take the acid as ha
this will dissociate : 
ha ↔ h+ + a- 
ka = [h+] [a-] /[ha] 
because dissociation is very small we can use 0.1m for [ha] 
[h+] = 2/50*0.1 = 0.0004m 
[a-] = 0.0004m 
 ka 0.0004 * 0.0004 / 0.1 
ka = 1.6*10^-5 * 0.1 
ka = 1.6*10^-4
since the solution is a buffer, we can use the henderson-hasselbalch equation to calculate the ratio of [so32-]/[hso3-] to have a ph of 7.39:ph=pka+logconjugate baseweak acidin the solution, identify the weak acid and the conjugate base. recall from the bronsted-lowry definition that an acid is a proton (h+) donor and the base is a proton (h+) acceptor.• nahso3 is a salt that will give hso3- when it dissociatesnahso3 → na+ + hso3-• na2so3 is a salt that will give so32- when it dissociatesna2so3 → 2 na+ + so32-•hso3- is a weak acid.          - concentration of hso3- = concentration of nahso3• when hso3- loses h+ to give a conjugate base, so32- is formedhso32- - h+ → so32-- concentration of so32- = concentration of na2so3      hso3-    +   h2o     →        so32-        +       h3o+(weak acid)    (base)      (conjugate base)  (conjugate acid)we can rewrite the henderson-hasselbalch equation as:ph=pka+logso32-hso3-calculate [so32-]/[hso3-] ratio:• ph = 7.39• pka = 7.19ph=pka+logso32-hso3-7.39=7.19+logso32-hso3-logso32-hso3-=7.39-7.19logso32-hso3-log=0.2loglog has a base of 10 and when you divide a number by log you will take 10 to the power of the numberso32-hso3-=100.2[so32-]/[hso3-] = 1.58the ratio [so32-]/[hso3-] is 1.58 we’re being asked to calculate the ph of 0.03m hc9h7o4.remember that weak acids partially dissociate in water and that acids donate h+ to the base (water in this case). the dissociation of hc9h7o4 is as follows:hc9h7o4 (aq) + h2o(l) ⇌ h3o+(aq) + c9h7o4– (aq)from this, we can construct an ice table. remember that liquids are ignored in the ice table and ka expression.the ka expression for hc9h7o4  is:ka=productsreactants=[h3o+][a-][hc9h7o4]note that each concentration is raised by the stoichiometric coefficient: [hc9h7o4], [h3o+] and [c9h7o4–] are raised to 1.plugin the value of ka and the terms from the ice chart: ka=[h3o+][a-][hc9h7o4]3.6 × 10-4=[x][x][0.03 -x]check if x is negligible in the denominator: if hc9h7o4ka>500 then x is negligible in the denominatorhc9h7o4ka= 0.03 3.6 ×10-4 = 83.3 < 500since the ratio is less than 500, we can’t remove (–x) from the equation. we first need to manipulate the equation to get a quadratic equation:3.6×10-4 = xx0.03-x0.03-x3.6×10-4   =x21.08 ×10-5-3.6×10-4 x = x2x2 +1.08 ×10-5-3.6×10-4 x =0we will then use the quadratic formula to solve for x. the quadratic formula is:x = -b±b2-4ac2afrom the equation, we have a = 1, b = 1.08×10–5, and c = –3.6 × 10–4. solving for x, the equilibrium concentration of [h3o+] and [c9h7o4–]:x = -b±b2-4ac2ax = -1.08×10-3±1.08×10-32-41-3.6 ×10-421x = -1.08×10-3±0.037962the calculation then splits into two ways:x = -1.08x10-3+0.037962x = 0.01844                          x = -1.08x10-3-0.037962x = -0.03904we take the positive value of x since the concentration of h3o+ and a– must increase. this means [h3o+] = [a–] = x = 0.01844 mnow calculate the ph using the formula below:ph = -log[h3o+]ph = -log[0.01844]ph = 1.73the ph of 0.030 m of aspirin is 1.73. we’re being asked to determine the conjugate acid of o2-.recall that the conjugate acid of a base is simply the compound formed when a base gains a proton (h+). it's considered as an acid because it can lose h+ to reform the base.[readmore]for o2–, simply add a hydronium ion from the formula. note that we also need to add the +1 charge of the proton (h+) to o2–, giving our conjugate acid a net charge = -1.o2– + h+ → oh–  –2    +   1     =       –1this means the conjugate acid of o2– is oh–. recall that at the equivalence point of a titration:[readmore]moles acid=moles basealso, recall that moles = molarity × volume. solving for molarity of naoh:0.568 g 1 mol khp204 g1 mol oh-1 mol khp1 mol naoh1 mol oh-136.78 ml1 ml1x10-3 l = 0.0757 mollor mtherefore, the molarity of naoh needed is 0.0757 m a) he helium (he) is an element by itself. he atom can only exhibit dispersion forces.b) hclh          1a                   1 × 1 e– = 1 e–cl         7a                   1 × 7 e– = 7 e–   total: 8 valence e–[readmore]cl is more electronegative than h, so hcl is a polar covalent compound. hcl exhibits dipole-dipole interaction.c) h2oh          1a                   1 × 2 e– = 2 e–o          6a                   1 × 6 e– = 6 e–   total: 8 valence e– since hydrogen is bonded directly to oxygen.h2o exhibits hydrogen bonding.d) cl2cl         7a                   1 × 7 e– = 14 valence e–composed of the same atom. there's no electronegativity difference.cl2 can only exhibit dispersion forces. we’re being asked to determine which of the given solutions is a good buffer system.recall that a buffer can be composed of the following:• weak base and its conjugate acid• weak acid and its conjugate base• weak acid and a strong base ▪ concentration of weak acid > concentration of strong base• weak base and a strong acid ▪ concentration of weak base > concentration of strong acid[readmore]a. 0.19 m potassium hydroxide + 0.27 m potassium bromide.potassium hydroxide → koh → group 1a k+ + oh- → strong basepotassium bromide → kbr breaks up in the solution: kbr → k+ + br-▪ k+ is a neutral ion → will not contribute to ph  hbr       +      h2o       →      br-      +         h3o+(weak acid)                            (conjugate base)•  strong base + weak base• not a good buffer systemb. 0.28 m hydroiodic acid + 0.18 m sodium iodide. hydroiodic acid → hi → weak acidsodium iodide → nai → breaks up in the solution: nai → na+ + i-hi       +      h2o       →      i-      +         h3o+(weak acid)                            (conjugate base)• weak acid + conjugate base = buffer• the solution is a good buffer systemc. 0.30 m ammonia + 0.37 m ammonium nitrate. ammonia → nh3 → weak baseammonium nitrate → nh4no3 → breaks up in the solution: nh4no3 → nh4+ + no3-nh3       +      h2o       →      nh4+      +         oh-(weak base)                            (conjugate acid)• weak base + conjugate acid = buffer• the solution is a good buffer systemd. 0.17 m hypochlorous acid + 0.12 m hydrobromic acid. hypochlorous acid → hclo → weak acidhydrobromic acid → hbr → weak acid•  weak acid + weak acid• not a good buffer systeme. 0.38 m calcium bromide + 0.30 m barium bromide.calcium bromide → cabr → breaks up in the solution: cabr → ca+ + br-barium bromide → babr → breaks up in the solution: babr → ba+ + br-•  conjugate base + conjugate base• not a good buffer systemanswer: the following aqueous solutions are good buffer systems b. 0.28 m hydroiodic acid + 0.18 m sodium iodide and c. 0.30 m ammonia + 0.37 m ammonium nitrate AgNO3(aq) + NaCl(s) = AgCl(s) + NaNO3  
n(AgCl) = 0.765 g / M(AgCl)  
M(AgCl) = 108 + 35.5 = 143.5 g/mol  
n(AgCl) = 0.765 g /143.5 g/mol = 0.00533 mol  
n(AgCl) = chloride ions = 0.00533 mol  
C(chloride ions) = n/V  
C(chloride ions) = 0.00533 mol/1l = 0.00533 mol/ L (a)  
m(NaCl)=n(NaCl)*M(NaCl)  
n(NaCl) = 0.00533 mol  
M(NaCl) = 23 + 35.5 = 58.5 g/mol  
m(NaCl) = 0.00533 mol * 58.5 g/mol = 0.3115 g (b)  
  
**Answer:**  
**a) 0.00533 mol/ L**  
**b) 0.3115 g** [readmore]typed answerh2 + i2 ⇌ 2hi      kc = 4.6i            0.5   0.5     0.5c          -x      -x       +2xe        0.5-x 0.5-x   0.5+2xkc=hi2hii2kc=0.5+2x20.5-x0.5-xkc=0.5+2x20.5-x24.6=0.5+2x0.5-xx=0.138hi=0.5+2x=0.776 m generaloraverage rate is the change in the concentration of a compound over a period of time.the average rate the reactants and products in a reaction are equal when the coefficient for each is 1.the rate for the reactants are negative. this is because the reactants are being consumed in the reaction. the rate for the product c is positive because it is being produced in the reaction.for a reaction where the coefficients of the compounds are different, the rate is multiplied by the reciprocal of the coefficient.example: 2a + b → 3cthe given reaction is:3o2(g) → 2o3(g)­given the rate of o2 consumption is -2.17x10-5 mol/l∙snote that value is negative to indicate that it is decreasing as the reaction proceeds.rearranging the rate equation to give the reaction rate is terms of [o3]: hi there!we are asked to determine the reaction order of a and b using the given data.recall that the rate law only focuses on the reactant concentrations and has a general form of:rate= kaxbyk = rate constanta & b = reactantsx & y = reactant ordersreactant ain order to calculate the order of the reaction with respect to a, we must look for the experimental data where the concentration of a changes while the concentration of b remains the same.trial[a] (mol/l)[b] (mol/l)initial rate (m/s)10.1000.100520.3000.1004530.1000.2001040.3000.20090the rate laws are:rate= kaxbytrial 1:5 m/s= k0.100x0.100ytrial 2:45 m/s= k0.300x0.100ysolve for x using the ratio of the two rates:***the larger rate should be on the numeratorrate 2rate 1=45 m/s5 m/s= k0.300x0.100yk0.100x0.100y• for reactions at a constant temperature, the rate constant remains the same. thus, it could be canceled out.• since the reaction order ‘y’ of b is also the same, it could also be canceled out.45 m/s5 m/s= k0.300x0.100yk0.100x0.100y9 = 3xlog 9=x·log3log 9log3=x·log3log3x=2the reaction order, x, with respect to a is 2. reactant bin order to calculate the order of the reaction with respect to a, we must look for the experimental data where the concentration of b changes while the concentration of a remains the same.trial[a] (mol/l)[b] (mol/l)initial rate (m/s)10.1000.100520.3000.1004530.1000.2001040.3000.20090the rate laws are:rate= kaxbytrial 1:5 m/s= k0.100x0.100ytrial 3:10 m/s= k0.100x0.200ysolve for x using the ratio of the two rates:***the larger rate should be on the numeratorrate 3rate 1=10 m/s5 m/s= k0.100x0.200yk0.100x0.100y• for reactions at a constant temperature, the rate constant remains the same. thus, it could be canceled out.• since the reaction order ‘x’ of a is also the same, it could also be canceled out.10 m/s5 m/s= k0.100x0.200yk0.100x0.100y2 = 2ylog 2=y·log2log 2log2=y·log2log2y=1the reaction order, y, with respect to b is 1.