Unit 4 Review, pages 586–593
Knowledge
1. (d)
2. (c)
3. (a)
4. (b)
5. (b)
6. (d)
7. (b)
8. (b)
9. (c)
10. (d)
11. (a)
12. (c)
13. (c)
14. (c)
15. (c)
16. (b)
17. (a)
18. (a)
19. (d)
20. (b)
21. (a)
22. (d)
23. False. In a closed system, reactants and products are separated from their
surroundings by a physical boundary.
24. True
25. False. When a chemical reaction reaches equilibrium, the concentrations of the
reactants and products are constant.
26. True
27. False. The rate of a chemical reaction is not proportional to the volumes of the
reacting substances.
28. False. The value of the equilibrium constant, K, changes when the temperature of a
reaction is changed.
29. True
30. True
31. False. Fritz Haber devised a method for the large-scale production of ammonia.
32. True
33. False. A shift toward a new equilibrium will occur whenever forward and reverse
reaction rates become unbalanced.
34. False. Given a balanced chemical equation, a set of instantaneous concentrations for
reactants, and a value for K, you cannot calculate whether the system is at equilibrium.
35. True
36. True
37. False. Hydrogen chloride is classified as an Arrhenius acid because it ionizes to
produce hydrogen ions in aqueous solution.
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-8
38. False. According to the Brønsted-Lowry theory, a base is a hydrogen ion acceptor.
39. True
40. False. The concentration of H+(aq) in pure water at 25 °C is 1.0 × 10−7 mol/L
41. False. A solution with a [H+(aq)] of 1.0 × 10−14 mol/L is basic.
42. False. The conjugate base of a weak acid is a much stronger base than H2O(l).
43. True
44. True
45. True
46. False. A weak acid is one for which the equilibrium lies far to the left.
47. False. When a salt is added to water, the resulting solution may be acidic or basic, or
neutral.
48. False. A strong acid reacts with a weak base to produce a basic solution.
49. True
50. True
51. False. The concentrations of buffer components in a solution should be much greater
than the concentrations of acid or base added.
52. False. Buffering capacity indicates the ability of a buffer to maintain its pH.
53. True
54. (a) (i)
(b) (iv)
(c) (iii)
(d) (ii)
55. The double arrow indicates a reversible reaction.
56. The forward and reverse reactions are proceeding at the same rate.
57. At dynamic equilibrium, both reactions are occurring at the same rate.
58. The amounts of reactants and products do not need to be equal
at equilibrium. The ratio of reactants to products depends on conditions, but the ratio is
constant because both reactants and products are being formed at the same rate.
[C]c [D]d
59. K =
[A]a [B]b
[NO(g)]2
60. (a) K =
[N 2 (g)][O 2 (g)]
(b) K =
[NO 2 (g)]2
[N 2O 4 (g)]
(c) K =
[SiCl 4 (g)][H 2 (g)]2
[SiH 4 (g)][Cl 2 (g)]2
(d) K =
[PCl3 (g)]2 [Br2 (g)]3
[PBr3 (g)]2 [Cl 2 (g)]3
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-9
[H 2O(g)][CO(g)]
=2.
[H 2 (g)][CO 2 (g)]
[H 2 (g)][CO2 (g)] 5 × 5
If 1 H2O molecules reacts,
, or about 0.5.
=
[H 2O(g)][CO(g)] 7 × 7
[H 2 (g)][CO 2 (g)] 6 × 6
If 2 H2O molecules react,
, or 1.
=
[H 2O(g)][CO(g)] 6 × 6
[H 2 (g)][CO2 (g)] 8 × 8
If 3 H2O molecules react,
, or 4.
=
[H 2O(g)][CO(g)] 4 × 4
[H 2 (g)][CO 2 (g)] 6 × 6
If 4 H2O molecules react,
. If 4 H2O molecules react, K = 2.
=
[H 2O(g)][CO(g)] 6 × 6
Therefore, at equilibrium there will be 4 molecules of H2O(g), 4 molecules of CO(g),
8 molecules of H2(g),, and 8 molecules of CO2(g),
62. (a) When water vapour is added the equilibrium will shift to the left.
(b) When the temperature is lowered, the equilibrium will shift to the left.
(c) When the pressure is decreased, the equilibrium will shift to the right.
(d) When a catalyst is added, there will be no change.
63. An enzyme is a protein catalyst that operates in a biological system.
64. The system has not reached equilibrium, and the equilibrium will shift to the right.
65. (a) The acid in this reaction is H2SO4(aq).
(b) The base in this reaction is H2O(l).
(c) The conjugate acid in this reaction is H3O+(aq).
(d) The conjugate base in this reaction is HSO4–(aq).
66. (a) The ionizing reaction for HBr(aq) is
.
HBr ( aq ) ! H + ( aq ) + Br ! ( aq )
61. Since K = 2, at equilibrium
(b) The ionizing reaction for HC2H3O2(aq) is
HC2 H 3O 2 ( aq ) ! H + ( aq ) + C2 H 3O !2 ( aq )
.
(c) The ionizing reaction for HCHO2(aq) is HCHO 2 ( aq ) ! H + ( aq ) + CHO 2– ( aq ) .
67. The conjugate acid-base pairs are as follows:
(a) CH3NH3+(aq) and CH3NH2(aq); ClO–(aq) and HClO(aq)
(b) HSO4–(aq) and SO42–(aq); HC2O4–(aq) and C2O42–(aq)
68. The conjugate acid of aniline is C6H5NH3+(aq).
69. The ionization of ammonia in water is represented by this equation:
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq)
70. The mathematical relationship between pH and pOH is: pH + pOH = 14.00.
71. The hydrogen ion concentration for the solution that is pH 3 is 100 times greater than
that of the solution that is pH 5.
72. pH = −log[H+]
(a) pH = 5.62
(b) pH = 6.00
(c) pH = 6.636
(d) pH = 8.04
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-10
73. [H+(aq)] = 10−pH; [OH–(aq)] = 10−pOH;
(a) For soap with a pH of 6.10: [H+(aq)] = 7.9 × 10–7 mol/L;
[OH–(aq)] = 1.3 × 10–8 mol/L
(b) For borax with a pH of 10.01: [H+(aq)] = 9.8 × 10–11 mol/L;
[OH–(aq)] = 1.0 × 10–4 mol/L
74. [H+(aq)][OH–(aq)] = 1.0 × 10−14
(a) [H+(aq)] = 1.0 × 10–7 mol/L; neutral
(b) [H+(aq)] = 1.5 × 10–6 mol/L; acidic
(c) [OH–(aq)] = 1.6 × 10–3 mol/L; basic
75. (a) pH = −log[H+(aq)]
= –log 5.6 × 10–13
= 12.25
(b) pOH = −log[OH–(aq)]
= –log 3.9 × 10–14
= 13.41
pH = 14.00 – pOH
= 14.00 – 13.41
= 0.59
(c) pH = 14.00 – pOH
= 14.00 – 10.00
= 4.00
1.0 × 10−14
−
76. [OH ( aq ) ] =
[H + ( aq ) ]
1.0 × 10−14
2.0 × 10−3
[OH − ( aq ) ] = 5.0 × 10 −12 mol/L
77. Given: [NH3(aq)] = 0.25 mol/L; Kb = 1.8 × 10−5
Required: pH
Analysis:
=
I
C
E
Kb =
+
!!
NH3(aq) + H2O(l) #
OH−(aq)
!"
! NH4 (aq) +
0.25
−
0
0
−x
−
+x
+x
0.25 – x
−
x
x
[NH 4 + (aq)][OH − (aq)]
[NH 3 (aq)]
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-11
Solution:
(x )( x )
(0.25 – x )
(x )( x )
1.8 × 10−5 ≈
0.25
2
x ≈ 4.5 × 10−6
x = [OH–(aq)] ≈ 2.1 × 10−3 mol/L
pOH = −log(2.12 × 10−3)
pOH = 2.67
pH = 14.0 – 2.67
pH = 11.33
1.0 ! 10"14
78. (a) [H + ( aq )] =
[OH " ( aq )]
1.8 × 10−5 =
1.0 ! 10"14
4.0 ! 10"12
[H + ( aq )] = 2.5 ! 10"3 mol/L
=
(b) The solution is an acid.
79. 1.0 × 10–7 mol/L
80. H2SO4(aq) →H+(aq) + HSO4–(aq)
HSO4–(aq) ⇌ H+(aq) + SO42–(aq)
81. When dissolved in water, potassium oxide, K2O(s) will produce a basic solution.
K2O(s) + H2O(l) ⇌ 2 K+(aq) + 2 OH−(aq)
82. (a) NaNO2(aq) is basic.
(b) NaHSO4(aq) is acidic.
(c) KCl(aq) is neutral.
(d) NH4Cl(aq) is acidic.
83. The products contain buffer solutions.
84. Buffers consist of a weak acid or base and its salt in solution. The molecular and ionic
forms are at equilibrium, along with hydrogen and hydroxide ions. When an acid or base
is added to the buffer, a new equilibrium forms, but if the buffer components are present
in high concentration relative to the added acid, the pH does not change very much.
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-12
Understanding
85. Answers may vary. Sample answer:
86. Answers may vary. Sample answer: A system in static equilibrium is the number of
students in each grade this year. There are a certain number of students in each grade at
my school. Every day there is the same number of students in each grade, and students do
not move from one grade to another. A system in dynamic equilibrium is the water in the
atmosphere, in lakes and oceans, and in living organisms. There is always about the same
proportion of water in these states, but individual water molecules are constantly moving
from one to another.
87. (a) K = [CO2(g)]
(b) Concentration does not change for solid phase reactants and products; they are
considered part of the constant.
88. (a) N2(g) + 3 H2(g) ⇌ 2 NH3(g)
(b) Because there are more gas molecules on the left than on the right of the equation, an
increase in pressure shifts the equilibrium toward the product.
89. (a) The system is likely endothermic since bonds in carbon dioxide must be broken in
order to produce carbon monoxide and oxygen. Therefore, an increase in temperature
shifts the equilibrium toward the right.
(b) A decrease in pressure shifts the equilibrium toward the right where there are more
molecules.
(c) An increase in volume decreases the pressure and shifts the equilibrium toward the
right.
(d) The addition of neon gas does not affect the equilibrium.
90. The added base reacts with the hydrogen ions and shifts the equilibrium toward the
right, producing more chromate, which is yellow
91. A catalyst does not affect chemical equilibrium because it affects the rate of reaction
in both directions.
92. Sodium hydroxide is an Arrhenius base because it dissociates to release hydroxide
ions into the solution.
93. (a) All Arrhenius acids are also Brønsted–Lowry acids because they donate protons.
Some Brønsted–Lowry acids are not Arrhenius acids because they donate protons but do
not form hydrogen ions.
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-13
(b) Brønsted–Lowry theory explains the chemical properties of bases that do not form
hydroxide ions as well as the properties of the Arrhenius bases.
94. (a)(i) SO(OH)2(aq) is sulphurous acid; H2SO3(aq).
(ii) ClO3(OH)(aq) is perchloric acid; HClO4(aq).
(iii) HPO2(OH)2(aq) is phosphoric acid; H3PO4(aq).
(b) In the acids, the bond between oxygen and hydrogen is ionic and the bond between
oxygen and the central atom is covalent. In the alkali hydroxides, the bond between
oxygen and hydrogen is covalent, while the bond between metal and hydroxide is ionic.
95. Ionization in strong acids favours the products because strong acids ionize almost
completely.
96. The hydrogen carbonate ion is classified as amphiprotic because it can donate
protons or accept protons.
NaOH(aq) + NaHCO3(aq) ⇌ Na2CO3(aq) + H2O(l)
HCl(aq) + NaHCO3(aq) ⇌ H2CO3(aq) + NaCl(aq)
97. Large amounts of sulfuric acid are used by many different industries including
agriculture, automotive, chemical, and pharmaceutical; therefore, the amount of
production of sulfuric acid represents a wide range of economic activity.
98. The buffer uses the equilibrium between ammonia and ammonium ion:
NH3(aq) + H2O(l) ⇌ NH4+ (aq)+ OH−(aq)
If a small amount of acid is added, the hydrogen ions react with hydroxide ions, but the
equilibrium between ammonia and the ammonium ion keeps the hydroxide ion
concentration constant. When a base is added, the equilibrium shifts toward reactants to
keep hydroxide ion concentration constant.
99. (a) The pH of a buffer system is mainly determined by the value of the equilibrium
constant for the equilibrium system.
(b) The capacity of a buffer is controlled by the amount of weak acid or base and its salt
that is added. Increasing concentration increases the buffer capacity.
Analysis and Application
100. (a) Homeo means “the same,” and stasis means “not changing”; so homeostasis
refers to something that does not change.
(b) Skin insulates the body from the surroundings and participates in heating and cooling
the body.
(c) Epinephrine speeds up exothermic reactions in cells.
101. (a) Lime production increases because the rotating reaction chamber constantly
removes the product, and therefore, the process continually shifts the equilibrium to the
right.
(b) The process would be run at a high temperature because the reaction is endothermic.
102. (a) The equilibrium shifts to the left because as carbon dioxide is expelled, there
would be a decrease in carbon dioxide in the blood. Since [H+(aq)] would decrease, there
would be an increase in the pH of the blood.
(b) Breathing into a paper bag causes the person to breathe air with higher CO2 levels,
and this shifts the equilibrium to the right.
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-14
103. Given: [H2(g)]initial = 0.340 mol/L; [Br2(g)]initial = 0.220 mol/L;
[IBr(g)]initial = 0.00 mol/L; K = 56.7
Required: [H2(g)]equilibrium; [Br2(g)]equilibrium; [HBr(g)]equilibrium
Analysis: Use an ICE table to find the relationship between the equilibrium
concentrations of the reactant and the products.
I
C
E
H2(g)
+
0.340
−x
0.340 – x
!!
Br2(g) #
!"
! 2 HBr(g)
0.220
0
−x
+ 2x
0.220– x
2x
Solution:
56.7 =
(2 x) 2
(0.340 − x)(0.220 − x)
56.7(0.340 − x)(0.220 − x) = 4 x 2
56.7( x 2 − 0.560 x + 0.0748) = 4 x 2
52.7 x 2 − 31.8 x + 4.24 = 0
( x − 0.199)( x − 0.404) = 0
0.404 is greater than the initial concentrations; so x = 0.199
For [H2(g)]: 0.340 – 0.199 = 0.141
For [Br2(g)]: 0.220 – 0.199 = 0.021
For [HBr(g)]: 2(0.199) = 0.398
Statement: [H2(g)] = 0.141 mol/L; [Br2(g)] = 0.021 mol/L; [HBr(g)] = 0.398 mol/L
104. Answers may vary. Sample answer: Apply to the soil an anion that forms an
insoluble salt, such as phosphate; this will bind heavy metal ions and and prevent them
from contaminating the ground water.
105. Because the pH is 2.2, this solution would be acid rain.
106. Given: [Ba(OH)2(aq)]: 3.42g/125 mL Required: pH
Analysis:
3.42 g Ba(OH)2
1 mol Ba(OH) 2
1000 mL
2 mol OH −
−
[OH (aq)] =
×
×
×
1.00 L
125 mL
171.35 g Ba(OH) 2
1 mol Ba(OH) 2
[OH − (aq)] = 0.319 mol/L
1.00 ×10−14
0.319
+
[H ( aq )] = 3.1348 ×10−14 mol/L (2 extra digits carried)
Solution: pH = −log[H+(aq)]
= −log(3.1348 × 10–14)
pH = 13.504
[H + ( aq )] =
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-15
107. Given: [HF(aq)] = 0.25 mol/L; Ka = 6.6 × 10−4
Required: pH
Analysis:
!!
HF(aq) #
H+(aq) +
!"
!
0.25
0
−x
+x
0.25 – x
x
I
C
E
F−(aq)
0
+x
x
[H + (aq)][F− (aq)]
[HF(aq)]
Solution:
(x)(x)
6.6 ! 10"4 =
(0.25 – x)
Ka =
x2
0.25
2
x # 1.65 ! 10"4
x = [H+(aq)] ≈ 1.285 × 10−2 mol/L (2 extra digits carried)
pH = −log(1.285 × 10−2)
pH = 1.89
108. Given: [HF (aq)] = 0.050 mol/L; Ka = 6.6 × 10−4
Required: pH
Analysis:
6.6 ! 10"4 #
I
C
E
!!
HF(aq) #
!"
!
0.050
−x
0.050 – x
H+(aq) +
0
+x
x
F−(aq)
0
+x
x
[H + (aq)][F− (aq)]
[HF(aq)]
Solution:
(x)(x)
6.6 ! 10"4 =
(0.050 – x)
Ka =
x2
6.6 ! 10 #
0.050
2
x # 3.30 ! 10"5
+
x = [H (aq)] ≈ 5.74 × 10−3 mol/L
pH = −log(5.74 × 10−3)
pH = 2.24
"4
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-16
109. Strontium hydroxide is a strong base because the equilibrium constant for
dissociation of the dissolved strontium hydroxide is very large.
110. Given: [HC3H5O3(aq)] = 0.85 mol/L; Ka = 7.94 × 10−5
Required: pH
Analysis:
I
C
E
HC3H5O3 (aq)
0.85
−x
0.85 – x
!!
#
!"
!
H+(aq) +
0
+x
x
C3H5O3−(aq)
0
+x
x
[H + (aq)][C3H 5O3− (aq)]
[HC3H 5O3 (aq)]
Solution:
(x)(x)
7.94 ! 10"5 =
(0.85 – x)
Ka =
x2
0.85
2
x # 6.75 ! 10"5
x = [H+(aq)] ≈ 8.22 × 10−3 mol/L
pH = −log(8.22 × 10−3)
pH = 2.09
(b) Yes, the yogurt is safe to eat because the pH is less than the dangerous range.
111. (a) Given: [H+(aq)] = 0.004 mol/L
Required: pH
Solution: pH = −log(0.004)
pH = 2.4
−
(b) Given: [OH (aq)] = 3.5 × 10−8 mol/L
Required: pH
Solution: pOH = −log(3.5 × 10−8)
pOH = 7.46
pH = 14.00 – 7.46
pH = 6.54
(c) Given: [OH−(aq)] = 1.22 × 10−4 mol/L
Required: pH
Solution: pOH = −log(1.22 × 10−4)
pOH = 3.91
pH = 14.00 – 3.91
pH = 10.09
(d) Given: [H+(aq)]= 7.3 × 10−6 mol/L
Required: pH
Solution: pH = −log(7.3 × 10−6)
pH = 5.14
7.94 ! 10"5 #
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-17
(e) Given: [H+(aq)] = 0.25 mol/L
Required: pH
Solution: pH = −log(0.25)
pH = 0.60
(f) Given: [OH−(aq)] = 0.45 mol/L
Required: pH
Solution: pOH = −log(0.45)
pOH = 0.35
pH = 14.00 – 0.35
pH = 13.65
112. Given: [chloroquinone(aq)] = 0.20 mol/L; Kb = 1 × 10−6
Required: pH
Analysis:
chloroquinone(aq)
I
C
E
0.20
−x
0.20 – x
−
+
!!
+ H2 O #
!"
! chloroquinone-H (aq) + OH (aq)
−
0
0
−
+x
+x
−
x
x
[chloroquinone − H + ( aq )]
K b = [chloroquinone − H + ( aq )][OH − ( aq )]K b =
[chloroquinone ( aq )]
Solution:
(x)(x)
1! 10"6 =
(0.20 " x)
(x)(x)
1! 10"6 #
0.20
2
x # 2 ! 10"7
x = [OH–(aq)] ≈ 4.5 × 10−4 mol/L
pOH = −log(4.5 × 10−4)
pH = 14.0 – 3.3 pOH = 3.3
pH = 10.7
113. (a) It is possible that the rainwater had a pH lower than 5.5, and lowered the pH of
the soil, making the hydrangea blooms blue.
(b) CaCO3(s) and MgCO3(s) are both suitable salts to raise the pH of the soil in a garden.
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-18
114. Graphic organizers may vary. Graphic organizers should include information from
the following table:
Type of salt
Examples
Comment
pH of solution
Cation of a Group 1 KCl(aq),
Neither of the ions
neutral
or Group 2 element, NaCl(aq),
acts as an acid or a
other than Be; anion NaNO3(aq)
base
is from strong acid
Cation of a Group 1 NaC2H3O2(aq),
Anion acts as a
basic
or Group 2 element, KCN(aq),
base; cation has no
other than Be; anion NaF(aq)
effect on pH
is from weak acid
Cation is conjugate NH4Cl(aq),
Cation acts as an
acidic
acid of weak base;
NH4NO3(aq)
acid; anion has no
anion is from strong
effect on pH
acid
Cation is conjugate NH4C2H3O2(aq),
Cation acts as an
acidic if Ka > Kb
acid of weak base;
NH4CN(aq)
acid; anion acts as a basic if Kb > Ka
anion is conjugate
base
neutral if Ka = Kb
base of weak acid
Cation is highly
Al(NO3)3(aq),
Hydrated cation acts acidic
charged metal ion;
FeCl3(aq)
as an acid; anion has
anion is from strong
no effect on pH
acid
115. Given: 1.00 g of NaC2H3O2(s) dissolved in water to a final volume of 1.00 L
Required: pH
Analysis:
Amount and concentration of dissolved base (molar mass 82 g/mol):
n = (1.00 g) / (82.04 g/mol) = 0.0122 mol
c = 0.0122 mol / 1.00 L = 0.0122 mol/L
I
C
E
!!
C2H3O2−(aq) + H2O #
!"
! HC2H3O2(aq) +
0.0122
−
0
−x
−
+x
0.0122 – x
−
x
OH−(aq)
0
+x
x
Kb = [HC2H3O2(aq)][OH–(aq)] / [C2H3O2−(aq)] = 5.6 × 10–10
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-19
Solution:
5.6 ! 10"10 =
(x)(x)
(0.0122 " x)
(x)(x)
0.0122
2
x # 6.83! 10"12
x = [OH–(aq)] ≈ 2.61 × 10−6 mol/L
pOH = −log(2.61 × 10−6)
pOH = 5.58
pH = 14.00 – 5.58
pH = 8.42
116. (a) The red curve represents titration of a strong acid, since the pH at the
equivalence point is exactly 7 and no buffering occurs. The blue curve represents titration
of a weak acid, since the pH at the equivalence point is greater than 7 and a “buffer zone”
is present.
(b) Yes, methyl red could be used as an indicator, since the equivalence point occurs in
the zone in which methyl red changes colour.
117. (a) The approximate pH of the solution containing HA(aq) is 5.
(b) Given: [HA(aq)] = 1.0 mol/L
Required: Ka
Analysis:
5.6 ! 10"10 =
I
C
E
HA(aq)
1.0
−x
1.0 – x
!!
#
!"
!
H+(aq)
0
+x
x
+
A−(aq)
0
+x
x
Solution:
⎡⎣ A − ( aq )⎤⎦ ⎡⎣ H + ( aq )⎤⎦
Ka =
⎡⎣ HA ( aq )⎤⎦
( x)( x)
Ka =
(1.0 − x)
At pH = 5, [H+(aq)] = x = 1 × 10–5
(1! 10"5 )(1! 10"5 )
Ka =
1.0 " 1! 10"5 )
K a = 1! 10"10
Statement: The approximate value of Ka is 1 × 10–10.
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-20
118. Given: [HC2H3O2(aq)] = 1.0 mol/L; [NaC2H3O2(aq)] = 1.0 mol/L
Required: pH
Analysis:
I
C
E
HC2H3O2(aq)
1.0
−x
1.0 – x
!!
#
!"
!
H+(aq) +
0
+x
x
C2H3O2−(aq)
1.0
+x
1.0 + x
⎡⎣C2 H3O2 − ( aq )⎤⎦ ⎡⎣ H + ( aq )⎤⎦
Ka =
⎡⎣ HC2 H3O2 ( aq )⎤⎦
= 1.8 ×10−5
Solution:
(1.0 + x)(x)
1.8 ! 10"5 =
(1.0 " x)
(1.0)(x)
1.8 ! 10"5 #
1.0
x # 1.8 ! 10"5
+
x = [H ] ≈ 1.8 × 10−5 mol/L
pH = −log(1.8 × 10−5)
pH = 4.74
119. (a) Given: Kb = [TRISH+(aq)][OH–(aq)] / [TRIS(aq)]
= 1.19 × 10–6
Required: pH
Analysis: In an efficient buffer, [TRIS(aq)] = [TRISH+(aq)]:
Solution: [OH–(aq)] = 1.19 × 10–6
pOH = −log(1.19 × 10−6)
pOH = 5.9
pH = 14 – 5.9
pH = 8.1
Statement: TRIS buffers maintain a pH of 8.1.
[TRIS(aq)]
(b) Required:
ratio at pH 7 and at pH 9
[TRISH + (aq)]
Analysis: At pH 7.00, pOH = 7.00; [OH–] = 1 × 10−7
Solution:
[TRISH + (aq)][OH − (aq)]
Kb =
[TRIS(aq)]
1.19 ×10−6 =
[TRISH + (aq)](1×10 −7 )
[TRIS(aq)]
[TRIS(aq)]
= 0.084
[TRISH + (aq)]
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-21
Analysis: At pH 9.00, pOH = 5.00; [OH–] = 1 × 10−5
Solution:
[TRISH + (aq)][OH − (aq)]
Kb =
[TRIS(aq)]
1.19 ×10−6 =
[TRISH + (aq)](1×10 −5 )
[TRIS(aq)]
[TRIS(aq)]
= 8.4
[TRISH + (aq)]
Statement: The ratio is 0.084 at pH 7.00 and 8.4 at pH 9.00.
(c) Required: pH of buffer
Solution:
Initial pH:
Amount and concentration of dissolved TRIS base (molar mass 121 g/mol):
(50.0 g)
nTRIS =
(120.94 g/mol)
= 0.413 mol
(0.413 mol)
[TRIS(aq)] =
(2.00 L)
= 0.207 mol/L
Amount and concentration of dissolved TRISHCl (molar mass 157.5 g/mol):
nTRISHCl = (65.0 g) / (157.5 g/mol) = 0.413 mol
cTRISHCl = (0.413 mol) / (2.00 L) = 0.207 mol/L
+
−
!!
TRIS(aq) + H2O(l) #
!"
! TRISH (aq) + OH (aq)
0.413
−
0.413
0
−x
−
+x
+x
0.413 – x −
0.413 + x
x
+
!
[TRISH (aq)][OH (aq)]
Kb =
[TRIS(aq)]
I
C
E
1.19 " 10!6 =
(0.413+ x)(x)
(0.413! x)
1.19 " 10!6 #
(0.413)(x)
(0.413)
x # 1.19 " 10!6
x = [OH (aq)] ≈ 1.19 × 10–6 mol/L
pOH = −log(1.19 × 10–6)
pOH = 5.92
pH = 14.0 – 5.92
pH = 8.08
Required: pH after 0.500 mL of 12 mol/L HCl(aq) is added to a 200.0 mL portion of
the buffer.
–
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-22
Solution: Amount of HCl(aq) added: nHCl = (0.500 mL)(12 mmol/mL) = 6 mmol
Amounts of TRIS and TRISH+: n = (200.0 mL) (0.207 mmol/mL) = 41.4 mmol
HCl will react with TRIS to produce TRISH+:
nTRIS = 41.4 mmol – 6 mmol
= 35.4 mmol;
(35.4 mmol)
[TRIS]=
(200.0 mL)
=0.177 mol/L
nTRISH+ = 41.4 mmol + 6 mmol
= 47.4 mmol;
(47.4 mmol)
[TRISH + (aq)] =
(200.0 mL)
= 0.237 mol/L
+
−
!!
TRIS(aq) + H2O(l) #
!"
! TRISH (aq) + OH (aq)
0.177
−
0.237
0
−x
−
+x
+x
0.177 – x −
0.237 + x
x
+
!
[TRISH (aq)][OH (aq)]
Kb =
[TRIS(aq)]
I
C
E
1.19 " 10!6 =
(0.237 + x)(x)
(0.177 ! x)
1.19 " 10!6 #
(0.237)(x)
(0.177)
x # 8.89 " 10!7
x = [OH (aq)] ≈ 8.89 × 10–7 mol/L
pOH = −log(8.89 × 10–7)
pOH = 6.05
pH = 14.0 – 6.05
pH = 7.95
Statement: The pH of the buffer is 8.08. The pH after 0.500 mL of 12 mol/L HCl is
added to a 200.0 mL portion of the buffer is 7.95.
120. (a) Given: 0.100 mol ammonia; 0100 mol ammonium chloride;
final volume: 1.00L; Kb for ammonia = 1.8 × 10–5.
Required: pH of the buffer
Analysis:
Concentration of NH3(aq) and NH4+(aq) = 0.100 mol/L
–
I
C
E
+
!!
NH3(aq) + H2O(l) #
!"
! NH4 (aq) +
0.100
−
0.100
−x
−
+x
0.100 – x
−
0.100 + x
Copyright © 2012 Nelson Education Ltd.
OH−(aq)
0
+x
x
Unit 4: Chemical Systems and Equilibrium U4-23
Solution:
Kb =
[NH +4 (aq)][OH - (aq)]
[NH 3 (aq)]
1.8 ! 10"5 =
(0.100 + x)(x)
(0.100 " x)
1.8 ! 10"5 #
(0.100)(x)
(0.100)
x # 1.8 ! 10"5
x = [OH ] ≈ 1.8 × 10–5 mol/L
pOH = −log(1.8 × 10–5)
pOH = 4.74
pH = 14.00 – 4.74
pH = 9.26
Statement: The pH of the buffer is 9.26.
(b) Required: pH of the buffer if 0.010 mol HCl is added.
Analysis: HCl will react with NH3(aq) to produce NH4+(aq):
nNH3 = 0.100 mol – 0.010 mol
= 0.090 mol;
[NH3(aq)] = 0.090 mol/L
nNH4+ = 0.100 mol + 0.010 mol
= 0.110 mol;
+
[NH4 (aq)] = 0.110 mol/L
–
I
C
E
NH3(aq) + H2O(l)
0.090
−
−x
−
0.090 – x
−
Solution:
Kb =
+
!!
#
!"
! NH4 (aq) +
0.110
+x
0.110 + x
OH−(aq)
0
+x
x
[NH +4 ( aq )] "#OH ! ( aq ) $%
[NH 3 ( aq )]
1.8 & 10!5 =
(0.110 + x)(x)
(0.090 ! x)
1.8 & 10!5 '
(0.110)(x)
(0.090)
x ' 1.47 & 10!5
x = [OH–(aq)] ≈ 1.47 × 10–5 mol/L
pOH = −log(1.47 × 10–5)
pOH = 4.83
pH = 14.00 – 4.83
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-24
pH = 9.17
Statement: The pH of the buffer if 0.010 mol HCl(aq) is added is 9.17.
(c) Given: 1.00 L water; 0.010 mol HCl
Required: change in pH
Solution: [H+(aq)] = 0.010 mol / L
pH = −log(0.010)
pH = 2
Statement: If the same amount of HCl had been added to 1.00 L of water the pH
would have dropped by 2.
Evaluation
121. System b produces the highest percentage of products; system a produces the lowest
percentage of products. Explanation may vary. Sample explanations: System b has the
highest equilibrium constant, K = 7, indicating that the equilibrium position is toward the
right of the reaction. Therefore the reaction favours the products and this system will
produce the highest percentage of products. System a has the lowest equilibrium constant,
K = 4.1 × 10–4, indicating the equilibrium position is toward the left of the reaction.
Therefore the reaction favours the reactants and this system will produce the lowest
percentage of products.
122. Answers may vary. Sample answer: According to the Arrhenius theory, in aqueous
solution an acid produces hydrogen ions, H+, and a base produces hydroxide ions, OH–.
According to the Brønsted–Lowry theory, an acid is a hydrogen ion (proton) donor and a
base is a hydrogen ion acceptor. The Brønsted–Lowry theory is an improvement over the
Arrhenius theory since it is more general. It applies to some acids and bases that are not
categorized by the Arrhenius definition, recognizes that some substances can act as acids
in one reaction and as bases in another, and is not limited to aqueous solutions.
123. Calculating the pH of both samples:
For the sample with a concentration of 0.1 mol/L:
I
C
E
HC2H3O2(aq)
0.1
−x
0.1 – x
!!
#
!"
!
H+(aq) +
0
+x
x
C2H3O2−(aq)
0
+x
x
"#C2 H 3O !2 ( aq ) $% "# H + ( aq ) $%
Ka =
"# HC2 H 3O 2 ( aq ) $%
1.8 & 10!5 =
(x)(x)
(0.1! x)
1.8 & 10!5 '
(x)(x)
(0.1)
x 2 ' 1.8 & 10!6
x = [H+(aq)] ≈ 1.34 × 10−3 mol/L
pH = −log(1.34 × 10−3)
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-25
pH = 2.87
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-26
For the sample with a concentration of 0.5 mol/L:
I
C
E
HC2H3O2(aq)
0.5
−x
0.5 – x
!!
#
!"
!
H+(aq) +
0
+x
x
C2H3O2−(aq)
0
+x
x
"#C2 H 3O !2 ( aq ) $% "# H + ( aq ) $%
Ka =
"# HC2 H 3O 2 ( aq ) $%
1.8 & 10!5 =
(x)(x)
(0.5 ! x)
1.8 & 10!5 '
(x)(x)
(0.5)
x ' 9.0 & 10!6
x = [H+(aq)] ≈ 3.0 × 10−3 mol/L
pH = −log(3.0 × 10−3)
pH = 2.52
The pH of the 0.1 mol/L ethanoic acid sample is 2.87 and the pH of the 0.5 mol/L
ethanoic acid sample is 2.52. The pH range of colour change for phenolphthalein is 8.0 to
9.8, where it changes from colourless acid to pink base. Therefore addition of
phenolphthalein to either sample will result in a colourless solution for both and it won’t
be possible to determine which is which. An alternative solution might be to perform a
titration with a strong base and equal volumes of each sample. Phenolphthalein could be
used as an indicator to estimate when the equivalence point is reached. (For weak acids,
the equivalence point will be at a pH greater than 7, so phenolphthalein is likely to be a
good choice of indicator.) Because one sample is less concentrated, it will require the
addition of less base to reach the equivalence point. Therefore the less concentrated
ethanoic acid sample will change colour first.
124. From the titration curve, the equivalence point occurs at approximately pH 8.
Bromocresol green changes colour between pH 3.8 and 5.4. Therefore it would change
colour well before the equivalence point is reached and is not a good indicator for
titrations of this acid.
125. (a) Answers may vary. Sample answer: Although both solutions have equal amounts
of salt dissolved, and both compounds are strong bases, Ba(OH)2 will produce a more
concentrated base than KOH since two hydroxide ions are formed from each molecule of
the salt. Either solution would be suitable to use as the titrant, but KOH might be a better
choice since the stoichiometry of the reaction is simpler, making the concentration of the
weak acid easier to calculate.
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-27
(b) Explanations may vary. Sample answer: Since a weak acid is being titrated, the strong
base will react with the weak acid to form the conjugate base of the weak acid. Halfway
to the equivalence point, when the amounts of weak acid and its conjugate base are
approximately equal, the solution is considered a buffer. However at the equivalence
point, enough base has been added to react completely with the weak acid and the
solution is no longer a buffer. Therefore the solution will not be a buffer after the
titration.
126. Answers may vary. Sample answer: The buffer was only effective for a short time,
after which the solution became acidic; the conjugate base was depleted too quickly. This
suggests that the buffering capacity of the system was poor. To increase the amount of
time that the buffer maintains a constant pH, a more concentrated buffer should be made
so that more of it is available to react with the excess acid. The amounts of both the weak
acid and its conjugate base should be increased so that the ratio of the conjugate acidbase pair remains the same. A greater concentration of the buffering components will
increase the buffering capacity of the system.
Reflect on Your Learning
127. Answers may vary. Students’ answers should include the major concepts and terms
discussed in Chapter 7, and how they relate to the central concept of chemical
equilibrium. Concepts that could be included are reversible reaction, equilibrium
constant, Le Châtelier’s principle, reaction quotient, solubility product constant, common
ion effect, and many others. The Chapter 7 Summary Vocabulary on page 478 may be
used as a guideline.
128. Answers may vary. Students’ answers should follow the question guidelines, and the
four reactions should be related to practical applications. For example, calcium hydroxide
is a strong base:
Ca(OH)2(aq) → Ca2+(aq) + 2 OH−(aq)
In the above reaction, the equilibrium lies far to the right and complete dissociation
occurs. Calcium hydroxide treatment can be used to counteract the effects of acid
precipitation on lakes, ponds, and streams.
Research
129. Answers may vary. Students’ answers should follow the question guidelines and
include applications of both enzymes and inhibitors. Possible topics include the use of
enzymes in the food industry, and pharmaceutical drugs that act as inhibitors of certain
enzymes.
130. Answers may vary. Increasing atmospheric carbon dioxide levels leads to ocean
acidification; when carbon dioxide dissolves in water, it forms carbonic acid. Students’
answers should follow the question guidelines and include the reaction between water
and carbon dioxide that results in carbonic acid formation, as well as the dissociation of
carbonic acid, which leads to lowered pH:
H2O(l) + CO2(g) ⇌ H2CO3(aq)
H2CO3(aq)⇌ H+(aq)+ HCO3−(aq)
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-28
131. Answers may vary. An example of a treatment for acute heavy metal poisoning is
chelation therapy. Chelating agents are compounds that sequester metal ions by binding
to them. The complex of metal ion and chelator is then excreted from the body, removing
the toxic heavy metal.
132. Answers may vary. Students’ answers should include the structure of salicylic acid
(2-hydroxybenzoic acid) and an evaluation of its acidic properties.
The carboxylic acid group hydrogen is ionizable. The phenolic hydroxyl group hydrogen
is also weakly ionizable.
133. (a) When ants bite, they may spray methanoic acid (formic acid) into the wound.
Ammonia is a base and reacts with the methanoic acid:
HCO2H(aq) + NH3(aq) ⇌ NH4+(aq) + HCO2−(aq)
This reaction produces an ammonium ion, the conjugate acid of ammonia, and a
methanoate ion, the conjugate base of methanoic acid.
(b) The treatment reduces the stinging sensation because the acid is neutralized by the
application of ammonia.
134. Answers may vary. Students’ answers should include the information that oven
cleaners contain basic compounds such as sodium hydroxide (NaOH) and potassium
hydroxide (KOH). These bases react with the lipids present in grease to form soaps
(sodium or potassium salts of fatty acids), which dissolve in water. The compounds
present in oven cleaners are corrosive and care should be taken to handle them safely.
135. Answers may vary. Students’ answers should follow the question guidelines and
include at least four acidic or basic compounds. For example, ammonia is a basic
household cleaner suitable for general use.
136. Answers may vary. Students’ answers should include how oxides of sulfur and
nitrogen released into the atmosphere react to form sulfuric and nitric acid, respectively,
and how this leads to acid precipitation, which acidifies a body of water. The effects of
other factors such as the buffering capacity of the body of water may also be discussed.
137. Answers may vary. Students’ answers should include the importance of the carbonic
acid/bicarbonate buffering system in maintaining blood pH:
H2CO3(aq) ⇌ H+(aq) + HCO3−(aq)
The ability of blood hemoglobin to bind excess H+ may also be discussed.
Copyright © 2012 Nelson Education Ltd.
Unit 4: Chemical Systems and Equilibrium U4-29