CHM 235 Quantitative Analysis
Spring 2007
Dr. S.A. Skrabal
SOLUTIONS TO PROBLEM SET 9
Review of acid-base chemistry and monoprotic weak acids and bases
3 April 2007
Please refer to Appendix G in your textbook for required values of Ka.
Note: I have expressed values of pH and pOH to two decimal places for consistency.
1. What are the pH and pOH of solutions that have the following concentrations of H3O+: (A) 9.16 x 10-4
M; (B) 4.15 x 10-10M; (C) 1.023 x 10-7 M?
pH = -log [H3O+]; pH + pOH = 14.00
(A) pH= 3.04, pOH = 10.96
(B) pH =9.38, pOH = 4.62
(C) pH = 6.99, pOH = 7.01
2. What are the pH and pOH of solutions that have the following concentrations of OH-:
(A) 4.45 x 10-4 M; (B) 7.76 x 10-12 M?
pOH = -log [OH-]
(A) pOH = 3.35, pH = 10.65
(B) pOH = 11.11, pH =2.89
3. What is the H3O+ concentration of a solution with (A) pH = 5.36; (B) pH = 2.07; (C) pOH = 2.09; (D)
pOH = 9.76?
(A) [H3O+] = 10-5.36 M = 4.37 x 10-6 M
(B) [H3O+] = 10-2.07 M =8.51 x 10-3 M
(C) pH = 14.00 – 2.09 = 11.91; [H3O+] = 10-11.91 M = 1.23 x 10-12 M
(D) pH = 14.00 – 9.76 = 4.24; [H3O+] = 10-4.24 M = 5.75 x 10-5 M
4. Referring to Table 6.1 in your textbook for the proper values of Kw, find the pH of pure H2O at the
following temperatures: (A) 10 C; (B) 35 C.
(A) Kw = 2.92 x 10-15 = [H3O+] [OH-]; [H3O+] = 5.40 x 10-8 M; pH = 7.27
(B) Kw = 2.09 x 10-14 = [H3O+] [OH-]; [H3O+] = 1.45 x 10-7 M; pH = 6.84
5. Calculate the pH of the following solutions: (A) 9.25 x 10-2 M HCl; (B) 7.3 mM KOH;
(C) 3.52 mM HNO3; (D) 2.05 x 10-2 M NaOH.
All of these substances are either strong acids (HCl, HNO3) or strong bases (KOH, NaOH).
(A) [H3O+] = 9.25 x 10-2 M; pH = 1.03
(B) [OH-] = 7.3 x 10-3 M; pOH = 2.14; pH = 11.86
(C) [H3O+] = 3.52 x 10-3 M; pH = 2.45
(D) [OH-] = 2.05 x 10-2 M; pOH = 1.69; pH = 12.31
6. What is the value of Kb for a weak acid whose pKa is 5.77? What is Ka for this weak acid? What is
its pKb?
pKa = -log Ka; Ka = 10-5.77 = 1.70 x 10-6. Ka·Kb = Kw; Kb = Kw/Ka = 1.00 x 10-14/1.70 x 10-6 =
5.88 x 10-9. pKb = -log Kb = -log(5.88 x 10-9) = 8.23
7. Suppose a 0.010 M solution of a weak base has a pH of 8.77. (A) What is the value of Kb for the
base? (B) What is the fraction of association () of the base?
(A) Kb = x2/(F-x); x =[OH-].
pOH = 14.00 – 8.77 = 5.23; [OH-] = 10-5.23 M = 5.89 x 10-6 M
Kb = (5.89 x 10-6)2/(0.010 – 5.89 x 10-6) = 3.47 x 10-9
(B)  = x/F = 5.89 x 10-6/0.010 = 5.9 x 10-4 or 0.059%
8. Suppose a 0.0125 M solution of a weak acid has a pH of 3.64. (A) What is the value of Ka for the
acid? (B) What is the fraction of dissociation () of the acid?
(A) Ka = x2/(F-x); x = [H3O+]; [H3O+] = 10-3.64 M = 2.29 x 10-4 M
Ka = (2.29 x 10-4)2/(0.0125 – 2.29 x 10-4) = 4.27 x 10-6
(B)  = x/F = 2.29 x 10-4/0.0125= 1.83 x 10-2 or 1.83%
9. Calculate the pH of the following solutions: (A) 2.50 x 10-3 M 2,2’-bipyridine; (B) 0.075 M NH3; (C)
1.78 x 10-2 M 4-methoxyaniline; (D) 3.98 x 10-3 M potassium hypochlorite (KOCl).
(A) Weak acid: Ka = 4.5 x 10-5 = Ka = x2/(F-x) = x2/(2.50 x 10-3 – x)
Multiply out to get: x2 + 4.5 x 10-5x – 1.125 x 10-7 = 0; x = [H3O+] = 3.14 x 10-4 M; pH = 3.50
Or assume Ka = x2/F to get x = 3.35 x 10-4 M; pH = 3.47.
(B) Weak base: Kb = Kw/Ka = 1.00 x 10-14/5.70 x 10-10 = 1.75 x 10-5
Kb = 1.75 x 10-5 = x2/(F – x) = x2/(0.075 – x). Multiply out to get: x2 + 1.75 x 10-5x –1.31 x 10-6 = 0
x = [OH-] = 1.14 x 10-3 M; pOH = 2.94; pH = 11.06
Or assume Kb = x2/F to get x = 1.15 x 10-3 M; pOH = 2.94; pH = 11.06
(C) Weak acid: Ka =4.40 x 10-6 = x2/(F – x) = x2/(1.78 x 10-2 – x)
Multiply out to get: x2 + 4.40 x 10-6x – 7.83 x 10-8 = 0
x = [H3O+] = 2.78 x 10-4 M; pH = 3.56
Or assume Ka = x2/F to get x = 2.80 x 10-4 M; pH = 3.55
(D) Weak base: Kb = Kw/Ka = 1.00 x 10-14/3.0 x 10-8 = 3.3 x 10-7
Kb = 3.3 x 10-7 = x2/(F – x) = x2/(3.98 x 10-3 – x). Multiply out to get:
x2 + 3.3 x 10-7x – 1.31 x 10-9 = 0; x = [OH-] = 3.60 x 10-5 M; pOH = 4.44; pH = 9.56.
Or assume Kb = x2/F to get same answer.
10. What is the pH of a solution prepared by dissolving 3.76 g of sodium benzoate (NaC7H5O2; FW =
144.11) in H2O to a total volume of 500 mL?
(3.76 g NaC7H5O2)(1 mol NaC7H5O2/144.11 g NaC7H5O2)(1/0.500 L) = 5.22 x 10-2 M
Weak base: Kb = Kw/Ka = 1.00 x 10-14/6.28 x 10-5 = 1.59 x 10-10 = x2/(F – x) = x2/(5.22 x 10-2 – x)
Multiply out to get: x2 + 1.59 x 10-10x – 8.30 x 10-12 = 0
x = [OH-] = 2.88 x 10-6 M; pOH = 5.54; pH = 8.46
Or assume Kb = x2/F to get same answer.
11. What is the pH of a solution prepared by dissolving 1.89 g of iodoacetic acid (ICH2CO2H; FW =
185.95) in H2O to a total volume of 250 mL?
(1.89 g ICH2CO2H)(1 mol ICH2CO2H/185.95 g ICH2CO2H)(1/0.250 L) = 4.07 x 10-2 M
Weak acid: Ka = 6.68 x 10-4 = x2/(F – x) = x2/(4.07 x 10-2 – x).
Multiply out to get: x2 + 6.68 x 10-4x – 2.72 x 10-5 = 0
x= [H3O+] = 4.89 x 10-3 M; pH = 2.31. Or assume Ka = x2/F to get x = 5.21 x 10-3 M; pH = 2.28.
12. Suppose a 0.015 M solution of a weak acid is 3.5% dissociated (i.e.,  = 0.035). Find Ka for this
acid.
 = x/F = 0.035 = x/0.015; x = [H3O+] = [A-] = 5.25 x 10-4 M
Ka = x2/(F – x) = (5.25 x 10-4)2 / (0.015 – 5.25 x 10-4) = 1.90 x 10-5
13. Calculate the value of the equilibrium constant (K) for the following reactions:
(A) HF + OH-  F- + H2O
(B) HCO2- + H3O+  HCO2H + H2O
(A) This is the reverse of the Kb reaction: F- + H2O  HF + OHF- is the conjugate base to HF (hydrofluoric acid; Ka = 6.8 x 10-4.
Then Kb = Kw/Ka = 1.00 x 10-14/6.8 x 10-4 = 1.5 x 10-11
Krxn = 1/Kb = 1/1.5 x 10-11 = 6.7 x 1010. Large K means highly product-favored reaction. Strong base
(OH-) reacts with WB (F-) completely.
(B) This the reverse of the Ka reaction: HCO2H + H2O  HCO2- + H3O+.
Ka for HCO2H (formic acid) = 1.80 x 10-4. Krxn = 1/Ka = 1/1.80 x 10-4 = 5.56 x 103.
Large K indicates strong acid (H3O+) reacts with weak base (HCO2-) completely.