Name______ANSWER KEY______Date______
Chapter 14/15 Review
Fill in the blanks:
Arrhenius defined acids as producing 1 in aqueous solution and bases as producing
2 in aqueous solution. The 3 definitions are more general: a(n) 4 is a proton donor
and a(n) 5 is a proton acceptor. Water is said to be 6 because it can act as either acid or
base. For strong acids and bases, the equilibrium lies far to the 7 . For weak acids and bases,
the K value is 8 . An acid containing more than one acidic proton is called 9 . The
concentrations of 10 and/or 11 ions in solution determines the pH, which is calculated by
12 .
13 can be acidic, basic, or neutral when dissolved in water. 14 is neutral when
dissolved in water. 15 is acidic when dissolved in water, and 16 is basic. When
determining pH of these solutions, a new K value must be calculated using the formula 17 .
The value of Kw is 18 .
When a weak acid and its conjugate base are dissolved in the same solution, a 19__
exists. This solution resists pH changes because the ratio of 20 is almost constant.
1. H+ / hydronium 11. OH–
2. OH– / hydroxide 12. pH = – log [H+]
3. Bronsted-Lowry 13. Salts
4. acid 14. NaCl (various answers)
5. base 15. NH4Cl (various answers)
6. amphoteric 16. NaCH3COO (various answers)
7. right 17. Ka x Kb = Kw
8. small 18. 1.0 x 10–14
9. polyprotic 19. buffer
10. H+ 20. [A–] : [HA]
Acid / Ka / Base / KbHClO2 / 1.2 x 10–2 / NH3 / 1.8 x 10–5
HF / 7.2 x 10–4 / CH3NH2 / 4.38 x 10–4
HNO2 / 4.0 x 10–4 / C6H5NH2 / 3.8 x 10–10
CH3COOH / 1.8 x 10–5 / C5H5N / 1.7 x 10–9
HOCl / 3.5 x 10–8 / (C2H5)2NH / 1.3 x 10–3
HCN / 6.2 x 10–10
1. Calculate the pH for 0.25 M solutions of the following:
a. Ba(OH)2
Ba(OH)2 → Ba2+ + 2 OH– [OH–] = 2 x 0.25 M = 0.50 M
pH = 13.70
b. LiCN
CN– + H2O D HCN + OH–
0.25 – x x x Kb = Kw = 1.0 x 10–14 = x2
Ka 6.2 x 10-10 0.25
x = [OH–] = 2.0 x 10–3 M pH = 11.30
c. C5H5N
C5H5N + H2O D C5H5NH+ + OH–
0.25 – x x x Kb = 1.7 x 10–9 = x2
0.25
x = [OH–] = 2.1 x 10–5 M pH = 9.31
d. NH4NO3
NH4+ D NH3 + H+
0.25 - x x x Ka = Kw = 1.0 x 10–14 = x2
Kb 1.8 x 10–5 0.25
x = [H+] = 1.2 x 10-5M pH = 4.93
2. Calculate the percent dissociation for a 0.35 M solution of hydrofluoric acid.
HF D H+ + F–
0.35 – x x x 7.2 x 10–4 = x2 x = [H+] = 1.6 x 10–2 M
0.35
1.6 x 10–2 M x 100 = 4.5%
0.35 M
3. The pH of a 0.063 M solution of HOBr is 4.95. Calculate Ka.
[H+] = 10–4.95 = 1.1 x 10–5 M
HOBr D H+ + OBr–
0.063 – 1.1 x 10–5 1.1 x 10–5 1.1 x 10–5 Ka = (1.1 x 10–5)2 = 2.0 x 10–9
0.063 – 1.1 x 10–5
4. Are solutions of the following salts acidic, basic, or neutral? For those which are not neutral,
write balanced equations for the reactions causing the solution to be acidic or basic.
a. KCl neutral d. KF basic
F– + H2O D HF + OH–
b. NaCH3COO basic e. NH4I acidic
CH3COO– + H2O D CH3COOH + OH– NH4+ D NH3 + H+
c. CH3NH3Cl acidic f. Mg(CN)2 basic
CH3NH3+ D CH3NH2 + H+ CN– + H2O D HCN + OH–
5. a. Calculate the pH of a solution which is 0.50 M CH3NH2 and 0.70 M CH3NH3Cl.
CH3NH2 + H2O D CH3NH3+ + OH–
0.50 – x 0.70 – x x
4.38 x 10–4 = 0.70 x x = [OH–] = 3.1 x 10–4 M pH = 10.50
0.50
b. Calculate the pH after 0.10 mol of NaOH is added to one liter of the above solution.
OH– + CH3NH3+ → CH3NH2 + H2O
0.10 0.70 0.50
-0.10 -0.10 +0.10
0 0.60 0.60 x = [OH–] = 4.38 x 10–4 M pH = 10.64
c. Calculate the pH after 0.10 mol of HCl is added to one liter of the solution in part a.
H+ + CH3NH2 → CH3NH3+
0.10 0.50 0.70 4.38 x 10–4 = 0.80 x x = [OH–]= 2.2 x 10–4 M
-0.10 -0.10 +0.10 0.40
0 0.40 0.80 pH = 10.34
6. The molar solubility of silver carbonate is 0.032 M. Calculate Ksp.
Ag2CO3(s) D 2 Ag+(aq) + CO32–(aq)
0 0
+ 2x + x x = 0.032 M
0.064 0.032
Ksp = (0.064)2 (0.032) = 1.3 x 10–4
7. Calculate the molar solubility of Ag2SO4 (Ksp = 1.5 x 10–5) in the following solutions:
(a) pure water
(b) 0.15 M Al2(SO4)3
(a) Ag2SO4(s) D 2 Ag+(aq) + SO42–(aq)
0 0
+2x +x
2x x (2x)2 x = 1.5 x 10–5
x = mol sol = 0.016 M
(b) Ag2SO4(s) D 2 Ag+(aq) + SO42–(aq)
0 0.45
+ 2x + x
2x 0.45 + x (2x)2 (0.45) = 1.5 x 10–5
x = 2.9 x 10–3 M
8. A solution of hydrochloric acid of unknown concentration was titrated with 0.10 M NaOH. If
a 100.-mL sample of the HCl solution required exactly 35.0 mL of the NaOH solution to reach
the equivalence point, what was the pH of the HCl solution?
(M of HCl) (100. mL) = (0.10 M) (35.0 mL)
M of HCl = 0.035 M pH = 1.46
9. A 400.0 mL sample of 0.100 M acetic acid is titrated with 0.200 M NaOH. Calculate the pH
after the addition of the following volumes of base.
a. 30.0 mL c. 200.0 mL
b. 100.0 mL d. 300.0 mL
a. [CH3COOH] = (0.100 M)(400.0 mL) / (430.0 mL) = 0.0930 M
[OH–] = (0.200 M)(30.0 mL) / (430.0 mL) = 0.0140 M
CH3COOH + OH– → CH3COO– + H2O
0.0930 0.0140 0
– 0.0140 –0.0140 + 0.0140
0.0790 0 0.0140
CH3COOH D CH3COO– + H+
0.0790 0.0140 x 0.0140 x = 1.8 x 10–5
0.0790
x = [H+] = 1.0 x 10–4 M pH = 3.99
b. [CH3COOH] = (0.100 M)(400.0 mL) / (500.0 mL) = 0.0800 M
[OH–] = (0.200 M)(30.0 mL) / (500.0 mL) = 0.0400 M
half-equivalence point: pH = pKa = 4.74
c. [CH3COOH] = (0.100 M)(400.0 mL) / (600.0 mL) = 0.0667 M
[OH–] = (0.200 M)(30.0 mL) / (600.0 mL) = 0.0667 M
CH3COOH + OH– → CH3COO– + H2O
0.0667 0.0667 0
– 0.0667 –0.0667 + 0.0667
0 0 0.0667
CH3COO– + H2O D CH3COOH + OH–
0.0667 x x x2 = 1.0 x 10–14
0.0667 1.8 x 10–5
x = [OH–] = 6.1 x 10–6M pH = 8.78
d. [CH3COOH] = (0.100 M)(400.0 mL) / (700.0 mL) = 0.0571 M
[OH–] = (0.200 M)(30.0 mL) / (700.0 mL) = 0.0857 M
CH3COOH + OH– → CH3COO– + H2O
0.0571 0.0857 0
– 0.0571 –0.0571 + 0.0571
0 0.0286 0.0571 [OH–] = 0.0286 M pH = 12.457
10. Consider the titration of 100.0 mL of 0.100 M HCl by 0.200 M NaOH. Calculate the pH of
the resulting solution after each of the following volumes of NaOH has been added.
a. 0.0 mL c. 50.0 mL
b. 30.0 mL d. 75.0 mL
a. 0.100 M H+ pH = 1.000
b. [H+] = (0.100 M)(100.0 mL) / (130.0 mL) = 0.0769 M
[OH–] = (0.200 M)(30.0 mL) / (130.0 mL) = 0.0462 M
H+ + OH– → H2O
0.0769 0.0462
–0.0462 –0.0462
0.0307 0 [H+] = 0.0307 M pH = 1.513
c. [H+] = (0.100 M)(100.0 mL) / (150.0 mL) = 0.0667 M equivalence point
[OH–] = (0.200 M)(30.0 mL) / (150.0 mL) = 0.0667 M pH = 7.000
d. [H+] = (0.100 M)(100.0 mL) / (175.0 mL) = 0.0571 M
[OH–] = (0.200 M)(30.0 mL) / (175.0 mL) = 0.0857 M
H+ + OH– → H2O
0.0571 0.0857
–0.0571 –0.0571
0 0.0286 [OH–] = 0.0286 M pH = 12.457