Name____________ANSWER KEY____________
Date____________________
Chapter 14/15 Review
Fill in the blanks:
Arrhenius defined acids as producing 1
2
in aqueous solution. The 3
and a(n) 5
in aqueous solution and bases as producing
definitions are more general: a(n) 4
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
the K value is
is a proton donor
7 . For weak acids and bases,
8 . An acid containing more than one acidic proton is called
concentrations of 10
and/or
11
9 . The
ions in solution determines the pH, which is calculated by
12 .
13
can be acidic, basic, or neutral when dissolved in water.
dissolved in water.
15
is acidic when dissolved in water, and
16
14
is neutral when
is basic. When
determining pH of these solutions, a new K value must be calculated using the formula
The value of Kw is
17 .
18 .
When a weak acid and its conjugate base are dissolved in the same solution, a
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]
19__
Acid
HClO2
HF
HNO2
CH3COOH
HOCl
HCN
Ka
1.2 x 10–2
7.2 x 10–4
4.0 x 10–4
1.8 x 10–5
3.5 x 10–8
6.2 x 10–10
Base
NH3
CH3NH2
C6H5NH2
C5H5N
(C2H5)2NH
Kb
1.8 x 10–5
4.38 x 10–4
3.8 x 10–10
1.7 x 10–9
1.3 x 10–3
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– +
0.25 – x
H2 O 
HCN +
x
x
OH–
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  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+ 
0.25 - x
NH3 + H+
x
x
Kw = 1.0 x 10–14 = x2
Kb
1.8 x 10–5
0.25
+
x = [H ] = 1.2 x 10-5 M
Ka =
pH = 4.93
2. Calculate the percent dissociation for a 0.35 M solution of hydrofluoric acid.
HF

0.35 – x
H+ + F–
x
x
7.2 x 10–4 =
x2
0.35
x = [H+] = 1.6 x 10–2 M
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

0.063 – 1.1 x 10–5
H+
+
1.1 x 10–5
OBr–
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  HF + OH–
b. NaCH3COO
basic
CH3COO– + H2O  CH3COOH + OH–
e. NH4I
acidic
NH4+  NH3 + H+
c. CH3NH3Cl
acidic
+
CH3NH3  CH3NH2 + H+
f. Mg(CN)2 basic
CN– + H2O  HCN + OH–
5. a. Calculate the pH of a solution which is 0.50 M CH3NH2 and 0.70 M CH3NH3Cl.
CH3NH2 + H2O  CH3NH3+ + OH–
0.50 – x
0.70 – x
x
4.38 x 10–4 = 0.70 x
0.50
x = [OH–] = 3.1 x 10–4 M
pH = 10.50
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+ +
0.10
-0.10
0
CH3NH2 → CH3NH3+
0.50
0.70
-0.10
+0.10
0.40
0.80
4.38 x 10–4 = 0.80 x
0.40
x = [OH–]= 2.2 x 10–4 M
pH = 10.34
6. The molar solubility of silver carbonate is 0.032 M. Calculate Ksp.
Ag2CO3(s)  2 Ag+(aq) + CO32–(aq)
0
0
+ 2x
+x
0.064
0.032
x = 0.032 M
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) 
(b) Ag2SO4(s) 
2 Ag+(aq) + SO42–(aq)
0
0
+2x
+x
2x
x
2 Ag+(aq) + SO42–(aq)
0
0.45
+ 2x
+x
2x
0.45 + x
(2x)2 x = 1.5 x 10–5
x = mol sol = 0.016 M
(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–
→
0.0930
0.0140
– 0.0140
–0.0140
0.0790
0
CH3COO– + H2O
0
+ 0.0140
0.0140
CH3COOH  CH3COO– + H+
0.0790
0.0140
x
b.
0.0140 x = 1.8 x 10–5
0.0790
x = [H+] = 1.0 x 10–4 M pH = 3.99
[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–
→
0.0667
0.0667
– 0.0667
–0.0667
0
0
CH3COO– + H2O
0
+ 0.0667
0.0667
CH3COO– + H2O  CH3COOH + OH–
0.0667
x
x
x2 = 1.0 x 10–14
0.0667 1.8 x 10–5
x = [OH–] = 6.1 x 10–6 M 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–
→
0.0571
0.0857
– 0.0571
–0.0571
0
0.0286
CH3COO– + H2O
0
+ 0.0571
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+
+
0.0769
–0.0462
0.0307
OH– → H2O
0.0462
–0.0462
0
[H+] = 0.0307 M
pH = 1.513
c. [H+] = (0.100 M)(100.0 mL) / (150.0 mL) = 0.0667 M
[OH–] = (0.200 M)(30.0 mL) / (150.0 mL) = 0.0667 M
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+
+
0.0571
–0.0571
0
OH– → H2O
0.0857
–0.0571
0.0286
[OH–] = 0.0286 M
pH = 12.457
equivalence point
pH = 7.000