1. It is Wednesday.
The first homework is due.
Place it on the bench in front.

2. Acids and bases of varying strengths.

3. Acids and bases of varying strengths.
Strong acid = 100% ionization
Strong acid = 100% donation of acidic
proton.

4. HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq)

5. HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq)
[H3O+][Cl-]
K =
[HCl]

6. HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq)
[H3O+][Cl-]
large
K = =
[HCl]

7. Generic acid
HA(aq) + H2O(l) H3O+(aq) + A-(aq)

8. Generic acid
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
[H3O+][A-]
= acidity
constant
= Ka
[HA]

9. Generic acid
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
[H3O+][A-]
= acidity
constant
= Ka
[HA]
-log10 Ka = pKa

10. Acid Ka pKa
HI  -11
HCl   -7
H2SO   -2
CH3COOH x
Table page 332

11. Base strength

12. Base strength
Inversely related to strength of
conjugate acid.

13. Base strength
Inversely related to strength of
conjugate acid.
H2O(l) + B(aq) HB+(aq) + OH-(aq)
conjugate acid

14. H2O(l) + B(aq) HB+(aq) + OH-(aq)
[HB+][OH-]
= Kb
= basicity
constant
[B]

15. [HB+][OH-]
= Kb
= basicity
constant
[B]
[H3O+][B]
= acidity
constant
= Ka
[HB+]
[H3O+][OH-] = Kw

16. [HB+][OH-]
= Kb
= basicity
constant
[B]
[H3O+][B]
= acidity
constant
= Ka
[HB+]
Conjugate acid
[H3O+][OH-] = Kw
KbKa = Kw

17. [HB+][OH-]
= Kb
= basicity
constant
[B]
[H3O+][B]
= acidity
constant
= Ka
[HB+]
[H3O+][OH-] = Kw
KbKa = Kw
pKb + pKa = pKw

18. KbKa = Kw
pKb + pKa = pKw
Expressions can be used for any
conjugate acid-base pair in water.

19. Indicators :
Usually a weak organic acid that has a
color different from its conjugate base.

20. Indicators :
Usually a weak organic acid that has a
color different from its conjugate base.
HA + H2O H3O+ + A-

21. phenolphthalein

22. Indicators :
Usually a weak organic acid that has a
color different from its conjugate base.
+ H2O H3O+ + A-

23. phenolphthalein

24. Equilibria with weak acids
and weak bases.

25. Equilibria with weak acids
and weak bases.
Weak acid:

26. Equilibria with weak acids
and weak bases.
Weak acid: Ka < 1

27. Equilibria with weak acids
and weak bases.
Weak acid: Ka < 1
[H3O+][A-]
Ka =
[HA]

28. Equilibria with weak acids
and weak bases.
Weak acid: Ka < 1
Ka (H3O+) = 1

29. Equilibria with weak acids
and weak bases.
Weak acid: Ka < 1
Ka (H3O+) = 1
Why is a weak acid weak?

30. HCl Ka  strong
HF Ka = 6.6 x weak

31. HCl Ka  strong
HF Ka = 6.6 x weak
HF more ionic than HCl

32. HCl Ka  strong
HF Ka = 6.6 x weak
HF more ionic than HCl
Relative Kas show that HF holds
proton more strongly than HCl

33. HCl Ka  strong
HF Ka = 6.6 x weak
HF more ionic than HCl
Relative Kas show that HF holds
proton more strongly than HCl
Electrostatic attraction for H+ stronger
for F- than for Cl-.

34. Organic acids Ka
CH3COOH x 10-5
CH3CH2COOH x 10-5
CCl3COOH x 10-1

35. Organic acids Ka
CH3COOH x 10-5
CH3CH2COOH x 10-5
CCl3COOH x 10-1

36. Ka
CH3COOH x 10-5
CH3CH2COOH x 10-5
CCl3COOH x 10-1
The more stable the conjugate base,
the stronger the acid.

37. Ka
CH3COOH x 10-5
CH3CH2COOH x 10-5
CCl3COOH x 10-1

38. Ka
CH3COOH x 10-5
CH3CH2COOH x 10-5
CCl3COOH x 10-1

39. Ka
CH3COOH x 10-5
CH3CH2COOH x 10-5
CCl3COOH x 10-1

40. Ka
CH3COOH x 10-5
CH3CH2COOH x 10-5
CCl3COOH x 10-1

41. Ka
CH3COOH x 10-5
CH3CH2COOH x 10-5
CCl3COOH x 10-1

42. Equilibria with weak acids
and weak bases.
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
[H3O+][A-]
= Ka
[HA]

43. HA(aq) + H2O(l) H3O+(aq) + A-(aq)
= Ka
[HA]

44. HA(aq) + H2O(l) H3O+(aq) + A-(aq)
= Ka
[HA]
CH3COOH(aq) + H2O(l)
H3O+(aq) + CH3COO-(aq)

45. HA(aq) + H2O(l) H3O+(aq) + A-(aq)
= Ka
[HA]
CH3COOH(aq) + H2O(l)
H3O+(aq) + CH3COO-(aq)
Ka = 1.8 x 10-5

46. CH3COOH(aq) + H2O(l)
H3O+(aq) + CH3COO-(aq)
Ka = 1.8 x 10-5
1 mol CH3COOH
1 L water solution

47. CH3COOH(aq) + H2O(l)
H3O+(aq) + CH3COO-(aq)
Ka = 1.8 x 10-5
1 mol CH3COOH
Calculate pH
1 L water solution

48. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
Eq.

49. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.

50. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
1.0 -y y y

51. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
1.0 -y y y
[H3O+][A-]
= Ka
[HA]

52. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
1.0 -y y y
[H3O+][A-]
[H3O+][CH3COO-]
= Ka
[HA]
[CH3COOH]

53. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
1.0 -y y y
[H3O+][CH3COO-]
= 1.8 x 10-5
[CH3COOH]

54. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
1.0 -y y y
(y)(y)
[H3O+][CH3COO-]
= 1.8 x 10-5 =
(1-y)
[CH3COOH]

55. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
1.0 -y y y
(y)(y)
(y2)
1.8 x 10-5 = =
(1-y)
(1-y)

56. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
1.0 -y y y
(y)(y)
(y2)
1.8 x 10-5 = =
(1-y)
(1-y)
y << 1

57. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
1.0 -y y y
(y)(y)
(y2)
1.8 x 10-5 = =
(1-y)
(1-y)
y << 1
1-y  1

58. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
1.0 -y y y
(y)(y)
(y2)
1.8 x 10-5 = =
= y2
(1-y)
(1-y)
y << 1
1-y  1

59. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
1.0 -y y y
1.8 x 10-5 = y2
y = 4.24 x 10-3

60. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
1.0 -y y y
1.8 x 10-5 = y2
y = 4.24 x 10-3
-log x 10-3 =

61. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
1.0 -y y y
1.8 x 10-5 = y2
y = 4.24 x 10-3
-log x 10-3 = 2.37 = pH

62. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
1.0 -y y y
y = 4.24 x 10-3
What is the fraction of acid ionized?

63. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
1.0 -y y y
y = 4.24 x 10-3
What is the fraction of acid ionized?
[A-]
[HA]

64. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
1.0 -y y y
y = 4.24 x 10-3
What is the fraction of acid ionized?
[A-]
4.24 x 10-3 =
[HA]
1.0

65. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
1.0 -y y y
y = 4.24 x 10-3
What is the percent of acid ionized?
[A-]
4.24 x 10-3
X 100% = 0.42% =
[HA]
1.0

66. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
y y y
Calculate % ionization for M

67. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
y y y
(y2)
(y)(y) =
1.8 x 10-5 =
( y)
( y)
[A-]
[HA]

68. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
y y y
(y2)
(y)(y) =
1.8 x 10-5 =
(0.0001)
(0.0001)
[A-]
[HA]

69. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
y y y
(y2)
(y)(y) =
1.8 x 10-5 =
(0.0001)
(0.0001)
[A-]
y2 = 1.8 x 10-9
[HA]

70. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
y y y
(y2)
(y)(y) =
1.8 x 10-5 =
(0.0001)
(0.0001)
[A-]
y2 = 1.8 x 10-9
[HA]
y = 4.2 x 10-5

71. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
y y y
(y2)
(y)(y) =
1.8 x 10-5 =
(0.0001)
(0.0001)
[A-]
4.2 x 10-5
y = 4.2 x 10-5 =
1 x 10-4
42%
[HA]

72. 1 x 10-4 and 4.2 x 10-5
do not differ enough to
ignore y.

73. [M]
[CH3COOH] [H3O+] [CH3COO-] 
initial
change
-y y y
Eq.
y y y
(y2)
(y)(y)
1.8 x 10-5 = =
( y)
( y)
34%
Full solution: pg 342

74. 1.0 M CH3COOH % ionized
M CH3COOH % ionized

75. 1.0 M CH3COOH % ionized
M CH3COOH % ionized
[A-] =
[HA]

76. 1.0 M CH3COOH % ionized
[A-] = 4.2 x 10-3 M
M CH3COOH % ionized
[A-] = 3.4 x 10-5 M
[A-] =
[HA]

77. 1.0 M CH3COOH % ionized
[A-] = 4.2 x 10-3 M
M CH3COOH % ionized
[A-] = 3.4 x 10-5 M
[A-]
[A-]  10-2 =
[HA]
[HA]  10-4

78. Problem 29:
Calculate pH of 0.35 M solution of propionic
acid.

79. Problem 29:
Calculate pH of 0.35 M solution of propionic
acid.
Ka = 1.3 x 10-5
Table 8-2

80. Problem 29:
Calculate pH of 0.35 M solution of propionic
acid.
Ka = 1.3 x 10-5
Table 8-2
HA(aq) + H2O(l) H3O+(aq) + A-(aq)

81. Problem 29:
Calculate pH of 0.35 M solution of propionic
acid.
Ka = 1.3 x 10-5
Table 8-2
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
[HA] = 0.35 M

82. Calculate pH of 0.35 M solution of propionic
acid.
Ka = 1.3 x 10-5
Table 8-2
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
[HA] = 0.35 M
[H3O+][A-]
= Ka
[HA]

83. Calculate pH of 0.35 M solution of propionic
acid.
Ka = 1.3 x 10-5
Table 8-2
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
[HA] = 0.35 M
[H3O+][A-]
(x2)
= Ka =
[HA]
[HA]

84. Calculate pH of 0.35 M solution of propionic
acid.
Ka = 1.3 x 10-5
Table 8-2
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
[HA] = 0.35 M
x2 = [HA] Ka

85. Calculate pH of 0.35 M solution of propionic
acid.
Ka = 1.3 x 10-5
Table 8-2
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
[HA] = 0.35 M
x2 = [HA] Ka
= 0.35 x 1.3 x 10-5 =

86. Calculate pH of 0.35 M solution of propionic
acid.
Ka = 1.3 x 10-5
Table 8-2
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
[HA] = 0.35 M
x2 = [HA] Ka
= 0.35 x 1.3 x 10-5 =
4.55 x 10-4

87. Calculate pH of 0.35 M solution of propionic
acid.
Ka = 1.3 x 10-5
Table 8-2
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
x2 = [HA] Ka
= 0.35 x 1.3 x 10-5 =
4.55 x 10-4
x =
4.55 x 10-4

88. Calculate pH of 0.35 M solution of propionic
acid.
Ka = 1.3 x 10-5
Table 8-2
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
x2 = [HA] Ka
= 0.35 x 1.3 x 10-5 =
4.55 x 10-4
x =
4.55 x 10-4
= 2.13 x 10-3

89. Calculate pH of 0.35 M solution of propionic
acid.
Ka = 1.3 x 10-5
Table 8-2
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
x2 = [HA] Ka
= 0.35 x 1.3 x 10-5 =
4.55 x 10-4
4.55 x 10-4
= 2.13 x 10-3
x =
-log x 10-3 =

90. Calculate pH of 0.35 M solution of propionic
acid.
Ka = 1.3 x 10-5
Table 8-2
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
x2 = [HA] Ka
= 0.35 x 1.3 x 10-5 =
4.55 x 10-4
4.55 x 10-4
= 2.13 x 10-3
x =
-log x 10-3
= 2.67

91. Equilibria involving weak bases

92. Equilibria involving weak bases
H2O(l) + B(aq) OH-(aq) + HB+(aq)

93. Equilibria involving weak bases
H2O(l) + B(aq) OH-(aq) + HB+(aq)
Acid base base acid2

94. Equilibria involving weak bases
H2O(l) + B(aq) OH-(aq) + HB+(aq)
Acid base base acid2
[OH-][HB+]
= Kb
[B]

95. Equilibria involving weak bases
H2O(l) + B(aq) OH-(aq) + HB+(aq)
Acid base base acid2
[OH-][HB+]
= Kb
[B]
Weak base: Kb < 1

96. Equilibria involving weak bases
H2O(l) + NH3(aq) OH-(aq) + NH4+(aq)
Acid base base acid2
Kb = ?

97. Equilibria involving weak bases
H2O(l) + NH3(aq) OH-(aq) + NH4+(aq)
Acid base base acid2
Kb = ?
KaKb = Kw

98. Equilibria involving weak bases
H2O(l) + NH3(aq) OH-(aq) + NH4+(aq)
Acid base base acid2 Kw
Kb = Ka

99. Equilibria involving weak bases
H2O(l) + NH3(aq) OH-(aq) + NH4+(aq)
Acid base base acid2 Kw
Kb = Ka

100. Equilibria involving weak bases
H2O(l) + NH3(aq) OH-(aq) + NH4+(aq)
Acid base base acid2 Kw
Kb = Ka
Table page 332: Ka = 5.6 x 10-10

101. Equilibria involving weak bases
H2O(l) + NH3(aq) OH-(aq) + NH4+(aq)
Acid base base acid2
1 x 10-14 Kw
Kb = =
5.6 x 10-10 Ka

102. Equilibria involving weak bases
H2O(l) + NH3(aq) OH-(aq) + NH4+(aq)
Acid base base acid2
1 x 10-14 Kw
Kb = =
= 1.8 x 10-5
5.6 x 10-10 Ka

103. Hydrolysis

104. Hydrolysis
The reaction of water with an ion
resulting in a change in the [H3O+]-
[OH-] balance.

105. Hydrolysis
The reaction of water with an ion
resulting in a change in the [H3O+]-
[OH-] balance.
Hydrolysis of anions raises pH.

106. Hydrolysis
The reaction of water with an ion
resulting in a change in the [H3O+]-
[OH-] balance.
Hydrolysis of anions raises pH.
Hydrolysis of cations lowers pH.

107. Hydrolysis
H2O(l) + F-(aq) HF(aq) + OH-(aq)

108. Hydrolysis
H2O(l) + F-(aq) HF(aq) + OH-(aq)
Ka (HF) = 6.6 x 10-4

109. Hydrolysis
H2O(l) + F-(aq) HF(aq) + OH-(aq)
Ka (HF) = 6.6 x 10-4
[HF][OH-]
Kb =
[F-]

110. Hydrolysis
H2O(l) + F-(aq) HF(aq) + OH-(aq)
Ka (HF) = 6.6 x 10-4
[HF][OH-]
Kb =
[F-]
pH M NaF

111. Hydrolysis
H2O(l) + F-(aq) HF(aq) + OH-(aq)
Ka (HF) = 6.6 x 10-4
[HF][OH-] Kw =
Kb =
[F-] Ka
pH M NaF

112. Hydrolysis
H2O(l) + F-(aq) HF(aq) + OH-(aq)
Ka (HF) = 6.6 x 10-4
[HF][OH-] Kw
[F-]Kw =
[OH-]2 =
[F-] Ka Ka
pH M NaF

113. Hydrolysis
H2O(l) + F-(aq) HF(aq) + OH-(aq)
Ka (HF) = 6.6 x 10-4
[F-]Kw
(0.095)(1 x 10-14)
[OH-]2 = =
(6.6 x 10-4) Ka
pH M NaF

114. Hydrolysis
H2O(l) + F-(aq) HF(aq) + OH-(aq)
[F-]Kw
(0.095)(1 x 10-14) = = =
[OH-]2
(6.6 x 10-4) Ka

115. Hydrolysis
H2O(l) + F-(aq) HF(aq) + OH-(aq)
[F-]Kw
(0.095)(1 x 10-14) =
[OH-]2 = =
(6.6 x 10-4) Ka
1.44 x 10-12
[OH-]2 =

116. Hydrolysis
H2O(l) + F-(aq) HF(aq) + OH-(aq)
[F-]Kw
(0.095)(1 x 10-14) =
[OH-]2 = =
(6.6 x 10-4) Ka
1.44 x 10-12
[OH-]2 =
[OH-] = 1.2 x 10-6

117. Hydrolysis
H2O(l) + F-(aq) HF(aq) + OH-(aq)
[OH-] = 1.2 x 10-6
1 x 10-14
[H3O+] = =
1.2 x 10-6

118. Hydrolysis
H2O(l) + F-(aq) HF(aq) + OH-(aq)
[OH-] = 1.2 x 10-6
1 x 10-14
8.34 x 10-9
[H3O+] = =
1.2 x 10-6

119. Hydrolysis
H2O(l) + F-(aq) HF(aq) + OH-(aq)
[OH-] = 1.2 x 10-6
1 x 10-14
8.34 x 10-9
[H3O+] = =
1.2 x 10-6
-log10
(8.34 x 10-9) = 8.08

120. Hydrolysis
H2O(l) + F-(aq) HF(aq) + OH-(aq)
pH = 8.08

121. It is Wednesday.
The first homework is due.
Place it on the bench in front.