Chapter 8 Review, pages 576–581
Knowledge
1. (b)
2. (b)
3. (a)
4. (b)
5. (d)
6. (c)
7. (c)
8. (d)
9. (c)
10. (b)
11. (d)
12. (b)
13. (b)
14. False. A strong acid completely ionizes in solution.
15. True
16. False. The hydrogen ion concentration of a solution with a pH of 3.0 is 1 × 10−3
mol/L.
17. False. A weak acid exists primarily in the form of molecules in solution.
18. True
19. False. The reaction of a strong acid with a strong base produces a neutral salt.
20. True
21. False. The endpoint of a strong acid–strong base titration occurs at a pH near 7.
22. True
23. False. A buffer solution can absorb a small amount of acid or base without a
significant change in pH.
24. True
25. (a) (vi)
(b) (vii)
(c) (iii)
(d) (xi)
(e) (iv)
(f) (ix)
(g) (x)
(h) (viii)
(i) (i)
(j) (ii)
(k) (v)
26. Arrhenius acids form hydrogen ions in aqueous solutions and bases form hydroxide
ions in aqueous solution.
27. (a) Water behaves as a base.
(b) Water behaves as an acid.
(c) Water behaves as a base.
(d) Water behaves as an acid.
Copyright © 2012 Nelson Education Ltd.
Chapter 8: Acid-Base Equilibrium
8-2
28. (a) acid: H3PO4(aq); base: H2O(l); conjugate acid: H3O+(aq); conjugate base:
H2PO4−(aq)
(b) acid: H2O(l); base: NH3(aq); conjugate acid: NH4+(aq); conjugate base: OH−(aq)
(c) acid: HBr(aq); base: H2O(l); conjugate acid: H3O+(aq); conjugate base: Br−(aq)
(d) acid: H2O(l); base: CN−(aq); conjugate acid: HCN(aq); conjugate base: OH−(aq)
29. (a) The conjugate base of HF is fluoride ion; F–.
(b) The conjugate acid of H2O is hydronium ion; H3O+.
(c) The conjugate base of H2O is hydroxide ion; OH–.
(d) The conjugate acid of NH3 is ammonium ion; NH4+.
30. HSO4–(aq) + H2O(l) ⇌ H3O+(aq) + SO42–(aq)
HSO4–(aq) +HNO3(aq) ⇌ H2SO4(aq) + NO3–(aq)
31. pH = −log[H+]
pOH= −log[OH–]
pH + pOH = 14
(a) [H+(aq)] = 0.004 mol/L
pH = 2.4
(b) The added hydroxide ions are significantly less than the [OH] due to autoionization of
water; therefore, pH = 7
(c) [H+(aq)] = 7.3 × 10−6 mol/L
pH = 5.14
(d) [H+(aq)] = 0.25 mol/L
pH = 0.60
(e) [OH–(aq)] = 0.45 mol/L
pOH = 0.35
pH = 13.65
mol
(f) [OH ! (aq)] = 2 " 6.1 " 10!5
L
mol
[OH ! (aq)] = 1.22 " 10!4
L
!
pOH= ! log [OH (aq)]
= !log [1.22 " 10!4 ]
pOH = 3.91
pH = 14.00-3.91
=10.09
pH = 10.09
(g) [OH–(aq)] = 0.006 mol/L
pOH = 2.2
pH = 11.8
32. pOH = 14.00 – pH
+
–
!!!
"
33. H2O(l) #
!!
! H (aq) + OH (aq)
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Chapter 8: Acid-Base Equilibrium
8-3
34. Given: pH = 8.0 Required: [OH–(aq)]
Solution: [H+(aq)] = 10−pH
[H+(aq)] = 10−8.0
= 1.0 × 10−8 mol/L
1.0 " 10!14
!
[OH (aq)] =
1.0 " 10!8
–
[OH (aq)] = 1.0 × 10−6 mol/L
35. A strong acid dissociates completely in water, but a weak acid exists with molecular
and ionic species in equilibrium.
36. Hydrochloric acid is a strong acid; it will have a large Ka since the ionic products are
present in much higher concentration than the molecular reactant.
[H + (aq)][A − (aq)]
37. K a =
[HA(aq)]
(1.34 ×10−3 )(1.34 ×10−3 )
Ka =
0.1000 − 1.34 ×10−3
K a = 1.80 ×10−5
38.
Solution
pH
pOH
[H+(aq)]
[OH–]
Acidic,
basic, or
neutral?
A
6.88
7.12
1.3 × 10−7 mol/L
7.7 × 10−8 mol/L
acidic
B
0.92
13.08
0.12 mol/L
8.4 × 10−14 mol/L
acidic
C
10.89
3.11
1.3 × 10−11 mol/L
7.8 × 10−4 mol/L
basic
D
7.00
7.00
1.0 × 10−7 mol/L
1.0 × 10−7 mol/L
neutral
39. pH = 14.0 − pOH
pH = 6.4
1.0 " 10!14
40. [OH ! (aq)] =
[H + (aq)]
(a) [OH–(aq)] = 1.0 × 10–7 mol/L; neutral
(b) [OH–(aq)] = 1.2 × 10–9 mol/L; acidic
(c) [OH–(aq)] = 1.0 × 10–2 mol/L; basic
(d) [OH–(aq)] = 1.9 × 10–10 mol/L; acidic
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Chapter 8: Acid-Base Equilibrium
8-4
41. Given: pH = 4.0 Required: [OH–(aq)]
Solution: [H+(aq)] = 10−pH
[H+(aq)] = 10−4.0
= 1.0 × 10−4 mol/L
1.0 ×10−14
−
[OH (aq)] =
1.0 ×10−4
–
[OH (aq)] = 1 × 10−10 mol/L
42. (a) The ionization of H2SO4 favours the products.
(b) The ionization of HCO3– favours the reactants.
(c) The ionization of HF favours the reactants.
43. (a) The autoionization of water is endothermic because the equilibrium shifts toward
ionization as the temperature is increased.
(b) Kw = [H+(aq)] [OH–(aq)]
[H+(aq)] = [OH–(aq]
5.47 ! 10"14 = [H + (aq)]2
[H + (aq)] = 5.47 ! 10"14
[H+(aq)] = 2.34 × 10–7 mol/L
[OH–(aq)] = 2.34 × 10–7 mol/L
44. (a) The solution would be a moderately weak base because one mole of hydroxide
ions are produced per mole of conjugate base; HCO3–(aq)
(b) The solution would be a strong acid that produces two moles of hydrogen ions per
mole of acid; H2SO4 (aq)
45. Given: [HF(aq)] = 0.050 mol/L; Ka = 6.6 × 10−4
Required: percent ionization
Analysis:
I
C
E
HF(aq) ⇌
0.050
−x
0.050 – x
Ka =
H+(aq) +
0
+x
x
F−(aq)
0
+x
x
[H + (aq)][F− (aq)]
[HF ( aq )]
percentage ionization [H + (aq)]
=
100
[HF ( aq )]
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Chapter 8: Acid-Base Equilibrium
8-5
(x )( x )
(0.050 – x )
(x)( x)
6.6 × 10−4 ≈
0.050
2
x ≈ 3.3 × 10−5
x = [H+(aq)]
≈ 5.7 × 10−3 mol/L
percentage ionization 0.0057
=
100
0.050
percentage ionization = 11 %
46. Given: [H2CO3(aq)] = 0.050 mol/L; Ka1 = 4.4 × 10−7; Ka2 = 4.7 × 10−11
Required: [HCO3–(aq)], [CO32–(aq)], [H+ (aq) ]
Analysis:
Solution:
I
C
E
6.6 × 10−4 =
H2CO3(aq)
0.050
−x
0.050 – x
⇌
H+(aq) +
0
+x
x
HCO3−(aq)
0
+x
x
[H + (aq)][HCO3− (aq)]
[H 2CO3 (aq)]
(x )( x )
Solution: 4.4 × 10−7 =
0.050 – x
(
x )( x )
4.4 × 10−7 ≈
0.050
x2 ≈ 2.2 × 10−8
x = [H+(aq)]
= [HCO3−(aq)]
≈ 1.5 × 10−4 mol/L
Ka =
HCO3− (aq) ⇌
H+(aq) +
I
0.00015
0.00015
C
−x
+x
E
0.00015 – x
0.00015 + x
+
2−
[H (aq)][CO3 (aq)]
Ka =
[HCO3− (aq)]
(0.00015 + x )( x )
4.7 × 10−11 =
0.00015 – x
(0.00015
)( x )
4.7 × 10−11 ≈
0.00015
x = [HCO3−(aq)]
≈ 4.7 × 10−11 mol/L
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CO32−(aq)
0
+x
x
Chapter 8: Acid-Base Equilibrium
8-6
Statement: [HCO3–(aq)] = 1.5 × 10–4 mol/L; [CO32–(aq)] = 4.7 × 10–11 mol/L;
[H+(aq)] = 1.5 × 10–4 mol/L
47. Given: [HCO2H(aq)] = 0.100 mol/L; [H+(aq)] = 1.34 × 10–3 mol/L
Required: Ka
Solution:
HCO2H(aq) ⇌ H+(aq) + HCO2−(aq)
[H + (aq)][HCO 2 − (aq)]
Ka =
[HCO 2 H ( aq ) (aq)]
(1.34 × 10–3 )(1.34 × 10–3 )
(0.100)
Ka = 1.80 × 10−5
48. Given: [HC2H3O2 (aq)] = 0.10 mol/L; Ka = 1.8 × 10−5
Required: pH
Analysis:
Ka =
HC2H3O2 (aq)
0.10
−x
0.10 – x
I
C
E
Ka =
⇌
H+(aq) +
0
+x
x
C2H3O2−(aq)
0
+x
x
[H + (aq)][C2 H3O2 − ( aq )]
[HC2 H3O2 ( aq )]
(x )( x )
0.10 – x
(x )( x )
1.8 × 10−5 ≈
0.10
2
x ≈ 1.8 × 10−6
x = [H+(aq)]
≈ 1.34 × 10−3 mol/L
−3
pH = −log(1.34 × 10 )
pH = 2.87
49. (a) An aqueous solution of sodium nitrate will be neutral.
(b) An aqueous solution of ammonium bromide will be acidic.
(c) An aqueous solution of sodium fluoride will be basic.
50. The sample is a weak acid and the titrant is a strong base
Solution:
1.8 × 10−5 =
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Chapter 8: Acid-Base Equilibrium
8-7
51.
52. (a) In a solution with a pH of 4.8, phenolphthalein will be colourless.
(b) In a solution with a pH of 4.8, bromocresol green will be green.
(c) In a solution with a pH of 4.8, phenol red will be yellow.
53. Given: [HC3H5O3(aq)] = 0.75 mol/L; [NaC3H5O3 (aq)] = 0.25 mol/L; Ka = 1.4 × 10−4
Required: pH
Analysis:
HC3H5O3(aq)
0.75
−x
0.75 – x
⇌
C3H5O3−(aq)
0.25
+x
0.25 + x
+
H+(aq)
0
+x
x
[C3H 5O3− ( aq )][H + (aq )]
Ka =
[HC3H 5O3 ( aq ) ]
(x )(0.25 + x )
Solution: 1.4 × 10−4 =
(0.75 – x )
(x )(0.25)
1.4 × 10−4 ≈
0.75
x ≈ 4.2 × 10−4
x = [H+(aq)]
≈ 4.2 × 10−4 mol/L
pH = −log(4.2 × 10−4)
pH = 3.38
Copyright © 2012 Nelson Education Ltd.
Chapter 8: Acid-Base Equilibrium
8-8
54. Given: [H2CO2(aq)] = 0.10 mol/L; VH2CO2 = 1.00 L ; [NaHCO2(aq)] = 0.10 mol/L
pH = 3.5; Ka = 3.75
Required: VNaHCO2 (aq)
Solution: pH =3.5
= –log[H+(aq)]
[H+(aq)] = 10−3.5
[H+(aq)] = 3.16 × 10−4 mol/L
[HCO2 − ( aq ) ][H + ( aq )]
Ka =
[H 2CO2 ( aq ) ]
[HCO2 − ( aq ) ](3.16 × 10−4 )
3.75 =
0.10
−4
(3.16 × 10 )
[HCO2 − ( aq ) ] =
(3.75)(0.10)
[HCO2 − ( aq ) ] = 8.4 × 10−4 mol/L
For 1.00 L, 8.4 × 10−4 mol must be added.
nNaHCO (aq ) = VNaHCO (aq ) ! cNaHCO (aq )
2
2
nNaHCO
2 (aq )
2
=
(aq )
VNaHCO
2
8.4 ! 10"4 mol
=
0.10 mol / L
= 8.4 ! 10"3 L
(aq )
VNaHCO
2 (aq )
VNaHCO
cNaHCO
2
2 (aq )
= 8.4 mL
55. (a) NH3(aq) + H2O(l) ⇌ NH4+ (aq) + OH−(aq)
(b) The equilibrium shifts toward the right because the acid reduces [OH−(aq)].
56. The conjugate base of citric acid is citrate ion.
57. Answers may vary. Sample answer: Citric acid is used for flavouring foods,
stabilizing pH, and deterring the growth of bacteria.
Understanding
58. Nitric acid produces hydrogen ions in aqueous solution.
59. Ammonia is a Brønsted–Lowry base because it accepts protons, but it is not an
Arrhenius base because it does not ionize to form hydroxide ions.
60. The concentration of liquid water does not change; so it is part of the constant, K.
61. The acid with the smallest Ka ionizes the least and therefore adds the least amount of
hydrogen ion to the solution. Therefore, the acid with a Ka of 1 × 10−6 has the higher pH.
62. When the concentration of a hydrochloric acid solution is doubled, the pH of the
solution does not decrease by a factor of two because pH is a logarithmic scale, not a
linear scale.
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Chapter 8: Acid-Base Equilibrium
8-9
pH = −log[H+(aq)]
log(1.0 × 10−7) = −7
−log(1.0 × 10−7) = 7
64. A weak acid is an acid that has a low dissociation. Concentration refers to the amount
of material in solution, not dissociation; therefore, an acid can be both weak and
concentrated.
65. According to Le Châtelier’s principle, increased pressure shifts the equilibrium
toward fewer molecules of gas. In this synthesis, there is a shift toward the product.
66. Hydrofluoric acid has the higher percentage ionization because its Ka value is larger.
67. The order is Ka1 > Ka2 > Ka3 because it is harder to remove additional protons as the
negative charge on the ion increases.
68. Water does not act as a diprotic acid because the hydroxide ion is such a weak acid
that it does not donate a proton in aqueous solutions.
69. (a) The conjugate base is a stronger base than water.
(b) [H+(aq)] > [HA(aq)]
70. The pH of the aqueous salt depends on the ability of the ions to hydrolyze and change
the pH of the solution.
71. Answers may vary. Flow charts or concept maps should include the following
information:
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
63.
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Chapter 8: Acid-Base Equilibrium
8-10
72. Given: VHC2H3O2(aq) = 100.00 mL; [HC2H3O2(aq)] = 0.200 mol/L
(a) Required: pH before titration begins
Solution:
HC2H3O2 (aq)
0.200
–x
0.200 – x
I
C
E
Ka =
⇌
H+(aq)
0
+x
+x
–
+ C2H3O2 (aq)
0
+x
+x
[H + (aq)][C 2 H 3O 2 – ( aq )]
[HC2 H 3O 2 (aq)]
[H + (aq)][C 2 H 3O 2 – ( aq )]
1.8 × 10 =
[HC2 H 3O 2 (aq)]
−5
x2
0.200 − x
x2
1.8 × 10−5 ≈
0.200
2
x ≈ 3.6 ×10−5
1.8 × 10−5 =
x ≈ 6.0 ×10−3
pH = –log[H+(aq)]
= –log [6.0 × 10–3]
pH = 2.22
Statement: The pH before titration begins is 2.22.
(b) Required: pH after 50.0 mL of 0.100 mol/L KOH is added
Solution: Before titration begins:
n HC2H3O2(aq) = [HC2H3O2(aq)] × VHC2H3O2(aq)
= (0.200 mmol/mL)(100.00 mL)
n HC2H3O2(aq) = 20.00 mmol
Amount of KOH(aq) added:
nKOH(aq) =[KOH(aq)] × VKOH(aq)
= (0.100 mmol/mL)(50.00 mL)
nKOH(aq) = 5.00 mmol
Since KOH(aq) reacts completely with ethanoic acid:
Unreacted ethanoic acid = n HC2H3O2(aq) – nKOH(aq)
= 20.00 mmol – 5.00 mmol
Unreacted ethanoic acid = 15.0 mmol
Since 50.00 mL of KOH(aq) was added to 100.00 mL of ethanoic acid solution, the
total volume is now 150.00 mL.
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Chapter 8: Acid-Base Equilibrium
8-11
The concentration of unreacted ethanoic acid, [HC2H3O2(aq)], can be determined:
nHC2H3O2 (aq )
[HC2 H 3O 2 (aq)] =
VHC2H3O2 (aq )
(15.00 mmol)
(150.00 mL)
[HC2 H 3O 2 (aq)] = 0.100 mol/L
=
–
Similarly, the concentration of the conjugate base of ethanoic acid, [C2H3O2 (aq)], is
determined:
nC H O – (aq )
[C2 H3O2 – ( aq ) ] = 2 3 2
VC H O – (aq )
2
=
3
2
(5.00 mmol)
(150.00 mL)
[C2 H3O2 – ( aq ) ] = 0.033 mol/L
For the ethanoic acid/ethanoate solution:
HC2H3O2 (aq)
0.100
–x
0.100 – x
I
C
E
⇌
H+(aq)
0
+x
+x
–
+ C2H3O2 (aq)
0.033
+x
0.033 + x
The values can now be substituted into the equilibrium equation for the ionization of
a weak acid, where Ka = 1.8 × 10–5.
[H + (aq)][C2 H 3O 2 – ( aq )]
Ka =
[HC2 H 3O 2 (aq)]
( x)(0.033 + x)
(0.200 − x)
Using simplifying assumptions, 0.033 + x ≈ 0.033 and 0.100 – x ≈ 0.100 (valid by
the hundred rule),
(0.033)
1.8 ×10 −5 ≈ ( x)
(0.100)
1.8 ×10−5 =
x≈
(1.8 × 10 −5 )(0.100)
(0.033)
x ≈ 5.45 ×10 −5
Therefore [H+(aq)] = x ≈ 5.45 × 10–5 and the pH can be determined:
pH = –log[H+(aq)]
≈ –log(5.45 × 10–5)
pH ≈ 4.26
Statement: The pH of the solution is 4.26.
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Chapter 8: Acid-Base Equilibrium
8-12
(c) Required: pH after 100.0 mL of 0.100 mol/L KOH is added
Solution: Before titration begins:
n HC2H3O2(aq) = [HC2H3O2(aq)] × VHC2H3O2(aq)
= (0.200 mmol/mL)(100.00 mL)
n HC2H3O2(aq) = 20.00 mmol
Amount of KOH(aq) added:
nKOH(aq) =[KOH(aq)] × VKOH(aq)
= (0.100 mmol/mL)(100.00 mL)
nKOH(aq) = 10.00 mmol
Since KOH(aq) reacts completely with ethanoic acid:
Unreacted ethanoic acid = n HC2H3O2(aq) – nKOH(aq)
= 20.00 mmol – 10.00 mmol
Unreacted ethanoic acid = 10.0 mmol
Since 100.00 mL of KOH(aq) was added to 100.00 mL of ethanoic acid solution, the
total volume is now 200.00 mL.
The concentration of unreacted ethanoic acid, [HC2H3O2(aq)], can be determined:
nHC2H3O2 (aq )
[HC2 H 3O 2 (aq)] =
VHC2H3O2 (aq )
(10.00 mmol)
(200.00 mL)
[HC2 H 3O 2 (aq)] = 0.050 mol/L
=
–
Similarly, the concentration of the conjugate base of ethanoic acid, [C2H3O2 (aq)], is
determined:
nC H O – (aq )
[C2 H 3O 2 – ( aq ) ] = 2 3 2
VC H O – (aq )
2
=
3
2
(10.00 mmol)
(200.00 mL)
[C2 H 3O 2 – ( aq ) ] = 0.050 mol/L
For the ethanoic acid/ethanoate solution:
I
C
E
HC2H3O2 (aq)
0.050
–x
0.050 – x
Copyright © 2012 Nelson Education Ltd.
⇌ H+(aq)
0
+x
+x
–
+ C2H3O2 (aq)
0.050
+x
0.050 + x
Chapter 8: Acid-Base Equilibrium
8-13
The values can now be substituted into the equilibrium equation for the ionization of
a weak acid, where Ka = 1.8 × 10–5.
[H + (aq)][C2 H 3O 2 – ( aq )]
Ka =
[HC2 H 3O 2 (aq)]
( x)(0.050 + x)
(0.050 − x)
Using simplifying assumptions, 0.050 + x ≈ 0.050 and 0.050 – x ≈ 0.050 (valid by
the hundred rule),
(0.050)
1.8 ×10−5 ≈ ( x)
(0.050)
1.8 ×10−5 =
x ≈ 1.8 ×10−5
Therefore [H+(aq)] = x ≈ 1.8 × 10–5 and the pH can be determined:
pH = –log[H+(aq)]
≈ –log(1.8 × 10–5)
pH ≈ 4.74
Statement: The pH of the solution is 4.74
(d) Required: pH after 150.0 mL of 0.100 mol/L KOH(aq) is added
Solution: Before titration begins:
n HC2H3O2(aq) = [HC2H3O2(aq)] × VHC2H3O2(aq)
= (0.200 mmol/mL)(100.00 mL)
n HC2H3O2(aq) = 20.00 mmol
Amount of KOH(aq) added:
nKOH(aq) =[KOH(aq)] × VKOH(aq)
= (0.100 mmol/mL)(150.00 mL)
nKOH(aq) = 15.00 mmol
Since KOH(aq) reacts completely with ethanoic acid:
Unreacted ethanoic acid = n HC2H3O2(aq) – nKOH(aq)
= 20.00 mmol – 15.00 mmol
Unreacted ethanoic acid = 5.0 mmol
Since 150.00 mL of KOH(aq) was added to 100.00 mL of ethanoic acid solution, the
total volume is now 250.00 mL.
The concentration of unreacted ethanoic acid, [HC2H3O2(aq)], can be determined:
nHC2 H3O2 (aq)
[HC2 H 3O 2 (aq)] =
VHC2 H3O2 (aq)
(5.00 mmol)
(250.00 mL)
[HC2 H 3O 2 (aq)] = 0.020 mol/L
=
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Chapter 8: Acid-Base Equilibrium
8-14
–
Similarly, the concentration of the conjugate base of ethanoic acid, [C2H3O2 (aq)], is
determined:
nC H O – (aq )
[C2 H 3O 2 – ( aq ) ] = 2 3 2
VC H O – (aq )
2
=
3
2
(15.00 mmol)
(250.00 mL)
[C2 H 3O 2 – ( aq ) ] = 0.060 mol/L
For the ethanoic acid/ethanoate solution:
HC2H3O2 (aq)
0.020
–x
0.020 – x
I
C
E
⇌
H+(aq)
0
+x
+x
–
+ C2H3O2 (aq)
0.060
+x
0.060 + x
The values can now be substituted into the equilibrium equation for the ionization of
a weak acid, where Ka = 1.8 × 10–5.
[H + (aq)][C2 H 3O 2 – ( aq )]
Ka =
[HC2 H 3O 2 (aq)]
( x)(0.060 + x)
(0.020 − x)
Using simplifying assumptions, 0.020 + x ≈ 0.020 and 0.060 – x ≈ 0.060 (valid by
the hundred rule),
( x)(0.060)
1.8 ×10 −5 ≈
(0.020)
1.8 ×10−5 =
x≈
(1.8 × 10 −5 )(0.020)
(0.060)
x ≈ 6.000 ×10 −6
Therefore [H+(aq)] = x ≈ 6.000 × 10–6 and the pH can be determined:
pH = –log[H+(aq)]
≈ –log(6.000 x 10–6)
pH ≈ 5.22
Statement: The pH of the solution is 5.22
(e) Required: pH after 200.0 mL of 0.100 mol/L KOH is added
Solution: Before titration begins:
n HC2H3O2(aq) = [HC2H3O2(aq)] × VHC2H3O2(aq)
= (0.200 mmol/mL)(100.00 mL)
n HC2H3O2(aq) = 20.00 mmol
Amount of KOH(aq) added:
nKOH =[KOH(aq)] × VKOH(aq)
= (0.100 mmol/mL)(200.00 mL)
nKOH = 20.00 mmol
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Chapter 8: Acid-Base Equilibrium
8-15
Since KOH(aq) reacts completely with ethanoic acid:
Unreacted ethanoic acid = n HC2H3O2 – nKOH
= 20.00 mmol – 20.00 mmol
Unreacted ethanoic acid = 0.00 mmol
Since 200.00 mL of KOH(aq) was added to 100.00 mL of ethanoic acid solution, the
total volume is now 300.00 mL. The pH of the resulting solution is determined by
the hydrolysis of the ethanoate ion.
nC H O − (aq ) = nHC2H3O2 (aq )
2
3
initial
2
nC H O − (aq ) = 20.00 mmol
2
3
2
cC H O − (aq ) =
2
3
2
nC H O − (aq )
2
3
2
VC H O − (aq )
2
3
2
20.00 mmol
300.0 mL
cC H O − (aq ) = 6.667 ×10−2 mol/L
=
2
3
2
I
C
E
Kb =
C2H3O2– (aq) + H2O(l) ⇌
6.667 × 10–2
−
−x
−
–2
6.667 × 10 – x −
HC2H3O2 (aq) + OH−(aq)
0
0
+x
+x
x
x
[HC2 H 3O 2 ( aq ) ][OH − (aq)]
[C 2 H 3O 2 − ( aq )]
1.0 ×10−14
=
1.8 ×10−5
= 5.556 ×10−10 (2 extra digits carried)
(x)(x)
5.556 ! 10"10 =
(6.667 ! 10"2 – x)
(x)(x)
6.667 ! 10"2
x2 ≈ 3.706 × 10−11
x = [OH−(aq)]
≈ 6.087 × 10−6 mol/L
pOH = −log(6.087 × 10−6)
pOH = 5.22
pH = 14.00 – 5.22
pH = 8.78
The pH of the resulting solution is 8.78.
5.556 ! 10"10 #
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Chapter 8: Acid-Base Equilibrium
8-16
(f) Required: pH after 250.0 mL of 0.100 mol/L KOH is added.
Solution:
Since 200.0 mL of 0.100 mol/L KOH were required for neutralization, 50.0 mL
remain unreacted.
nKOH =[KOH(aq)] × VKOH(aq)
= (0.100 mmol/mL)(50.00 mL)
nKOH = 5.00 mmol
Total volume of solution = 350.0 mL
nKOH( aq )
cKOH(aq ) =
VKOH( aq )
5.00 mmol
350.0 mL
cKOH( aq ) = 1.4286 ×10−2 mol/L (2 extra digits carried)
=
pOH = –log[OH–(aq)]
= –log( 1.4286 ×10−2 )
= 1.85
pH = 14 – pOH
= 14 – 1.85
pH = 12.15
Statement: The pH after 250 mL of 0.100 mol/L KOH(aq) is added is 12.15.
73. Most indicators can be used for this type of titration because the pH changes from
very low to very high close to the equivalence point.
74. Methyl yellow is a better indicator because the equivalence point has a pH greater
than 7 and thymol blue changes colour at a very low pH.
75. Buffer systems can be either acidic, basic or neutral. An acidic buffer consists of a
weak acid in equilibrium with its conjugate base:
HA(aq) ⇌ H+(aq) + A−(aq)
When an acid is added to this system, the additional hydrogen ions react with the
conjugate base, A–1(aq) to form HA(aq). As a result, the system shifts to the left.
Conversely, when a base is added to this system, the hydroxide ions from the base react
with the hydrogen ions from the equilibrium, causing the system to shift to the right.
A basic buffer consists of a weak base, B(aq), in equilibrium with its conjugate acid,
BH+(aq) :
B(aq) + H2O(l) ⇌ BH+(aq) + OH−(aq)
When an acid is added to this system, the additional hydrogen ions react with the
hydroxide ions from the equilibrium. As a result, the system shifts to the left.
Conversely, when a base is added to this system, the hydroxide ions from the base react
with the conjugate acid, causing the system to shift to the right.
Copyright © 2012 Nelson Education Ltd.
Chapter 8: Acid-Base Equilibrium
8-17
76. This solution would not make a good buffer because hydrochloric acid and sodium
chloride are essentially completely ionized; no equilibrium is established to form a
buffer.
77. (a) Given: [H2NNH2(aq)] = 0.40 mol/L; [H2NNH3NO3(aq)] = 0.80 mol/L;
Kb = 1.7 × 10−6
Required: pH
Analysis:
I
C
E
H2NNH2(aq) + H2O(l)
0.40
−
−x
−
0.40 – x
−
⇌
H2NNH3+(aq) +
0.80
+x
x + 0.80
OH−(aq)
0
+x
x
[H 2 NNH 3+ ( aq ) ][OH − (aq )]
Kb =
[H 2 NNH 2 ( aq )]
(x + 0.80)( x )
Solution: 1.7 × 10−6 =
(0.40 – x )
0.80 x
1.7 × 10−6 ≈
0.40
x = [OH− (aq)]
≈ 8.5 × 10−7 mol/L
−7
pOH = −log(8.5 × 10 )
pOH = 6.07
pH = 14.00 – 6.07
pH = 7.93
Analysis and Application
78. (a) In the lungs, Hb(O2)4 is favoured; in cells, HbH44+(aq) is favoured.
(b) As the carbon dioxide level falls, the equilibrium of the carbonic acid-hydrogen
carbonate buffer shifts toward carbonic acid, which decreases pH. Breathing into a bag
increases carbon dioxide, which raises the pH and releases oxygen.
(c) Sodium hydrogen carbonate increases pH and releases oxygen in the blood.
79. (a) Given: [C6H8O6(aq)] = 2.00 g/L; Ka1 = 7.9 × 10−5; Ka2 = 1.6 × 10−12
Required: [H+], pH
Analysis:
2.00 g C6 H8O6 1.00 mol C6 H8O6
×
= 0.0114 mol/L
L
176 g C6 H8O6
I
C
E
C6H8O6(aq)
0.0114
−x
0.0114 – x
Copyright © 2012 Nelson Education Ltd.
⇌
H+(aq)
0
+x
x
+
C6H7O6−(aq)
0
+x
x
Chapter 8: Acid-Base Equilibrium
8-18
[H + (aq)][C6 H 7O6 − (aq )]
[C6 H8O6 ( aq )]
Ka1 is much greater than Ka2 ; so Ka2 has no significant contribution to
equilibrium.
(x )( x )
Solution: 7.9 × 10−5 =
(0.0114 – x )
(x )( x )
7.9 × 10−5 ≈
0.0114
x2 ≈ 9.006 × 10−7 (2 extra digits carried)
x = [H+(aq)] ≈ 9.4899 × 10−4 mol/L
pH = −log (9.4899 × 10–4)
pH = 3.02
Statement: [H+(aq)] ≈ 9.5 × 10−4 mol/L
(b) Given: [C6H8O6(aq)] = 0.0114 mol/L; Ka1 = 7.9 × 10−5; pH = 1.0
Required: percentage ionization
Analysis:
pH=1.0 so [H+] = 1.00 × 10−1 mol/L
[H + (aq)][C6 H 7O6 − (aq )]
Ka1 =
[C6 H8O6 ( aq )]
Solution:
[H + (aq)][C6 H 7O6 − (aq )]
Ka1 =
[C6 H8O6 ( aq )]
Ka1 =
(0.1)[C6 H 7O6 − (aq )]
7.9 × 10 =
[C6 H8O6 (aq )]
−5
percentage ionization =
[C6 H7 O6 − ( aq )]
×100 %
[C6 H8O6 ( aq )]
percentage ionization = 7.9 × 10−4 × 100 %
percentage ionization = 0.079 %
80. (a) Both solutions are acidic:
3 NO2(g) + H2O(l) ⇌ NO(g) + 2 HNO3(aq)
SO2(g) + H2O(l) ⇌ H2SO3(aq)
(b) Lakes on limestone bedrock are less affected by these acidic solutions because
calcium carbonate in the limestone reacts with H2SO3 to neutralize the acid.
81. (a) Approximately 200 mL of sodium hydroxide has been added at the second
equivalence point.
(b)(i) When 0 mL NaOH(aq) is added, the major entities present are H2A(aq) and H2O(l).
(ii) When between 0 and100 mL NaOH(aq) is added, the major entities present are
H2A(aq), HA–(aq), H2O(l), and Na+(aq).
(iii) When 100 mL NaOH(aq) is added, the major entities present are HA–(aq), H2O(l),
and Na+(aq).
Copyright © 2012 Nelson Education Ltd.
Chapter 8: Acid-Base Equilibrium
8-19
(iv) When between 100 and 200 mL NaOH(aq) is added, the major entities present are
HA–(aq), A2–(aq), H2O(l), and Na+(aq).
(v) When 200 mL NaOH(aq) is added, the major entities present are A2–(aq), H2O(l), and
Na+(aq).
(vi) When more than 200 mL NaOH is added, the major entities present are A2–(aq),
H2O(l), Na+(aq), and OH–(l).
(c) Given: pH = 4; titration is half completed
Required: Ka1
Solution: At 50 mL, 50 % of H2A(aq) has reacted so [H2A(aq)] = [HA−(aq)]
[H+(aq)] = 10–pH
[H+(aq)] = 1.0 × 10−4
[H + (aq)][HA − (aq)]
K a1 =
[H 2 A(aq)]
K a1 =
[1.0 ×10−4 ][HA − (aq)]
[HA − (aq)]
K a1 = 1.0 ×10−4
Given: pH = 8; second titration is half completed
Required: Ka2
Solution: At 50 mL, 50 % of HA− has reacted so [HA−] = [A2−]
[H+] = 10–pH
[H+] = 1.0 × 10−8
[H + ][A 2 − ]
K a2 =
[HA − ]
K a2 =
[1.0 ×10−8 ][A 2 − ]
[A 2 − ]
K a2 = 1.0 ×10−8
82. Answers may vary. Sample answer: Prepare buffer solutions with known pH across
the entire pH range. Add the indicator to each solution and record the colour.
83. Answers may vary. Sample answer: Prepare buffer solutions with known pH across
the entire pH range. Prepare a solution of a weak acid with pH = 4. Add methyl orange
indicator to make an orange-coloured solution. Prepare a solution of a weak base at the
same concentration with pH = 10. Add thymolphthalein to make a blue solution.
Combine equal amounts of the two coloured solutions to make a colourless solution.
84. Answers may vary. Sample answer: Choose an acid or base whose Ka is close to the
desired [H+]. Calculate the concentrations of the acid or base and its salt needed to
produce a buffer at the pH. Prepare the solution. Determine the pH and adjust by adding
the acid or base or its salt to get to the exact pH required.
85. Answers may vary. Sample answer: Titrate samples of the buffer with a strong acid
and with a strong base to determine the amount of hydrogen ions and the amount of
hydroxide ions that can be added without substantial change to the pH.
86. (a) To prepare a buffer with pH 3.0, I would use phosphoric acid and potassium
dihydrogen phosphate.
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Chapter 8: Acid-Base Equilibrium
8-20
(b) To prepare a buffer with pH 4.0, I would use hydrofluoric acid and potassium
fluoride.
(c) To prepare a buffer with pH 5.0, I would use ethanoic acid and sodium ethanoate.
(d) To prepare a buffer with pH 7.0, I would use benzoic acid and sodium benzoate.
87. Answers will vary. Concept maps should include Arrhenius acids and bases,
Brønsted–Lowry acids and bases, proton donors, proton acceptors, the formation of salts
and H2O, and conjugate acids and bases.
Evaluation
88. Answers may vary. Spreadsheets should include a table with the following
comparative information:
Antacid brand Active ingredient mol of active
Cost per package Cost per mole
ingredient per
of antacid
package
89. Phenolphthalein is not an appropriate indicator for this titration because it does not
change colour at a pH near the equivalence point. Methyl red is a better indicator for this
titration.
90. A titration of a weak acid with a weak base is not feasible because there would be a
buffering effect around the endpoint, and so the curve would not have a sharp break.
Reflect on Your Learning
91. Answers will vary. Sample answer: There may be many examples of acids and bases
in foods, cleaning solutions, medicines, and other everyday substances. A pH meter or
pH indicator strips could be used to search for examples.
92. Answers will vary. Sample answer: Many materials may act as indicators if they are
acidic or basic and have a different colour than their salt. A stain remover could work by
changing the pH of the coloured substance to a form that does not have colour.
93. Answers will vary based on student experience.
Research
94. Answers will vary. Sample answer: Because acid forms as substances in the paper
break down, books that are printed on acidic paper are stored at cool temperatures and
low humidity to slow the reactions. Valuable books can often be treated with a slightly
basic buffer solution to neutralize the acid and react with the additional acid that forms.
95. Answers will vary. Sample answer: Arab scientists learned to make alkali hydroxides
at least 1200 years ago. The first alkali hydroxides to be isolated, sodium hydroxides and
potassium hydroxide, were produced from the ashes of burnt wood. The word alkali
comes from the Arab word for ashes.
Copyright © 2012 Nelson Education Ltd.
Chapter 8: Acid-Base Equilibrium
8-21
96.
97. Answers will vary. Students should trace acid rain through the pH change in lakes
and streams, and connect this information to the presence of nitrogen and sulfur
compounds in the atmosphere. By relating higher concentrations of these compounds to
power plant locations and population centres, students can point to the combustion of
fuels as the original source.
98. Answers will vary. Sample answer: The main goal of food preservation is to kill and
prevent growth of microorganisms such as bacteria, yeasts, and fungi. Most organisms
exist in a particular pH range. Acidic foods, such as tomatoes, naturally resist spoiling
when they are canned. Other foods can be preserved by pickling in vinegar, which
decreases its pH, or sometimes by treating with lye to increase the pH.
99. Answers may vary. Sample answer: Limestone is made of calcium carbonate, a weak
base, which can neutralize the acid in acid rain and act as a buffer to maintain lake water
at an optimum pH. Limestone is used for treating lakes and streams because it is a natural
material, is mined in many locations around the world, and dissolves slowly, producing a
long-lasting treatment.
100. Answers may vary. Sample answer: Svante Arrhenius was a research chemist and
professor in Sweden during the late 1800s and early 1900s. His contributions to
chemistry include the observation that pure water does not conduct electric current but a
salt solution does conduct because salts form ions in solution. He expanded on the theory
of ions to include a definition of acids and bases as substances that form hydrogen ions
and hydroxide ions in solution. In addition to his studies of ions, he developed the
concept of activation energy in chemical reactions, one of the basic concepts of collision
theory. Arrhenius was the first scientist to relate temperature on Earth to the
concentration of carbon dioxide in the atmosphere and also to predict global warming as
a result of burning fossil fuels.
101. Answers may vary. Sample answer: Acid indigestion and acid reflux occur when
stomach acids leak into the esophagus and damage tissues. The simplest medical
treatment is an antacid, which is a basic compound that reduces the acidity of the stomach
contents. Other treatments reduce the amount of acid by causing cells in the digestive
system to reduce the production of acidic compounds.
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Chapter 8: Acid-Base Equilibrium
8-22
102. Answers may vary. Sample answer: Buffers are used in medicines to control the
effect of pH changes on the pharmaceutical compound and to match the pH of the
medicine to that of the body system where it is used. Most body fluids contain buffering
compounds that control pH. Matching the medicine to the pH of the body increases the
effectiveness of delivering the medicine and protects tissues from possibly harmful pH
changes.
103. Answers may vary. Sample answer: Most toothpastes for controlling tooth decay use
fluoride salts such as stannous fluoride. The fluoride ion replaces the hydroxide ion in
hydroxyapatite, forming a surface coating that has a lower reactivity with acids.
104. Answers may vary. Sample answer: Occupations in which people would monitor pH
regularly include manufacturing chemists and technicians who use pH probes and
titrations to measure acids and bases during chemical reactions, environmental
technicians who use probes to monitor and control acidity in streams and lakes, medical
lab technicians who measure the pH of body fluids, and food chemists who use
information about pH to process foods safely.
105. Answers may vary. Sample answer: Household products use a wide variety of
acids and bases. Acids used at home include the acetic acid in vinegar, which is a
component of many cleaning compounds, and citric acid, used to preserve foods.
Household bases include ammonia (base), used in cleaning and disinfecting solutions and
in treatments for insect bites and stings, and sodium carbonate, a component of laundry
detergents.
106. Answers may vary. Sample answer: Most plants grow best within a particular pH
range. Depending on the type of plant and its specific needs for elements and compounds
in soil, the optimum pH range can vary from about 6 to 7.5. Farmers adjust soil pH by
adding mineral elements and compounds to the soil. The most common mineral soil
conditioners are lime, composed of calcium carbonate and/or magnesium carbonate,
which is used to increase soil pH, and sulfur, used to decrease soil pH.
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Chapter 8: Acid-Base Equilibrium
8-23