1. Homework problems added to syllabus:
Friday 9/7 Ch. 8, # 52, 56, 60
Monday 9/10 Ch. 8, # 62, 66

2. Buffer Solutions

3. Buffer Solutions
Buffer solutions aid in maintaining
a consistent pH range for a reaction.

4. The buffer acts as a sponge for
excess acid or base.

5. Buffer Solutions
Changes in reaction pH can be
caused by:

6. Buffer Solutions
Changes in reaction pH can be
caused by:
Acidic or basic product

7. A + B HC + D

8. A + B HC + D
HC(aq) + H2O H3O+(aq) + C-(aq)

9. A + B HC + D
HC(aq) + H2O H3O+(aq) + C-(aq)
[HC]
[H3O+]

10. Buffer Solutions
Changes in reaction pH can be
caused by:
Acidic or basic product
Acidic or basic reactant

11. A + NH3(aq) + H2O(l)
A NH3(aq) + NH4+(aq) + OH-(aq)

12. A + NH3(aq) + H2O(l)
A NH3(aq) + NH4+(aq) + OH-(aq)
[NH3]

13. A + NH3(aq) + H2O(l)
A NH3(aq) + NH4+(aq) + OH-(aq)
[NH3]
[OH-]

14. A buffer consists of a weak acid and
its conjugate base or a weak base
and its conjugate acid.

15. A buffer consists of a weak acid and
its conjugate base or a weak base
and its conjugate acid.
A conjugate base or conjugate acid
does not necessarily have to come from
solution of an acid or base.

16. Weak acid
HA(aq) + H2O(aq) H3O+(aq) + A-(aq)

17. Weak acid
HA(aq) + H2O(aq) H3O+(aq) + A-(aq)
[H3O+][A-]
Ka =
[HA]

18. Weak acid
HA(aq) + H2O(aq) H3O+(aq) + A-(aq)
[H3O+][A-]
Ka =
[HA]
[HA]
[H3O+] = Ka
[A-]

19. Weak acid
HA(aq) + H2O(aq) H3O+(aq) + A-(aq)
[HA]
[H3O+] = Ka
[A-]
pH = -log10[H3O+]

20. Weak acid
HA(aq) + H2O(aq) H3O+(aq) + A-(aq)
[HA]
[H3O+] = Ka
[A-]
pH = -log10[H3O+]
[HA]
pH 
[A-]

21. Weak acid
HA(aq) + H2O(aq) H3O+(aq) + A-(aq)
[HA]
[H3O+] = Ka
[A-]
If [HA] and [A-] are large (  1 M ),
the ratio will change slowly if dilute
(0.1 M or less) strong acid is added.

22. CH3COOH M
NaCH3COO 0.5 M

23. CH3COOH M
NaCH3COO 0.5 M
CH3COOH(aq) + H2O(l)
H3O+(aq) + CH3COO-(aq)

24. CH3COOH M
NaCH3COO 0.5 M
CH3COOH(aq) + H2O(l)
H3O+(aq) + CH3COO-(aq)
CH3COOH(aq) + H2O(l) + NaCH3COO (aq)
Na+(aq) + H3O+(aq) + 2 CH3COO-(aq)

25. CH3COOH M
NaCH3COO 0.5 M
CH3COOH(aq) + H2O(l) + NaCH3COO (aq)
Na+(aq) + H3O+(aq) + 2 CH3COO-(aq)

26. CH3COOH M
Na CH3COO 0.5 M
CH3COOH(aq) + H2O(l) + NaCH3COO (aq)
Na+(aq) + H3O+(aq) + 2 CH3COO-(aq)

27. CH3COOH M
NaCH3COO 0.5 M
CH3COOH(aq) + H2O(l) + NaCH3COO (aq)
Na+(aq) + H3O+(aq) + 2 CH3COO-(aq)
[CH3COOH] [H3O+] [CH3COO-]
Start 

28. CH3COOH M
NaCH3COO 0.5 M
CH3COOH(aq) + H2O(l) + NaCH3COO (aq)
Na+(aq) + H3O+(aq) + 2 CH3COO-(aq)
[CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y

29. CH3COOH M
Na CH3COO 0.5 M
CH3COOH(aq) + H2O(l) + NaCH3COO (aq)
Na+(aq) + H3O+(aq) + 2 CH3COO-(aq)
[CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.0-y y y

30. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.0-y y y
[H3O+][CH3COO-]
Ka =
[CH3COOH]

31. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.0-y y y
[H3O+][CH3COO-]
(y)(0.5+y)
Ka = =
(1-y)
[CH3COOH]
Ka = 1.8 x 10-5

32. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.0-y y y
(y)(0.5+y)
1.8 x 10-5 =
(1-y)
Assume y << 1

33. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.0-y y y
(y)(0.5)
1.8 x 10-5 =
(1)
Assume y << 1

34. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.0-y y y
y = 2( 1.8 x 10-5) = 3.6 x 10-5

35. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.0-y y y
y = 2( 1.8 x 10-5) = 3.6 x 10-5
pH = -log10(3.6x10-5) = 4.44

36. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.0-y y y
y = 2( 1.8 x 10-5) = 3.6 x 10-5
pH = -log10(3.6x10-5) = 4.44
buffer
pH = 2.37 for 1 M CH3COOH

37. pH = -log10(3.6x10-5) = 4.44
buffer
pH = 2.37 for 1 M CH3COOH
[H3O+][CH3COO-]
Ka =
[CH3COOH]
CH3COOH(aq) + H2O(l)
H3O+(aq) + CH3COO-(aq)

38. pH = -log10(3.6x10-5) = 4.44
buffer
pH = 2.37 for 1 M CH3COOH
[H3O+][CH3COO-]
Ka =
[CH3COOH]
[CH3COOH]
[H3O+] = Ka
[CH3COO-]

39. Prepare buffer solution of CH3COOH and
CH3COO- and include 0.1 M HCl.

40. Prepare buffer solution of CH3COOH and
CH3COO- and include 0.1 M HCl.
HCl(aq) + H2O(aq) H3O+ + Cl-(aq)

41. Prepare buffer solution of CH3COOH and
CH3COO- and include 0.1 M HCl.
HCl(aq) + H2O(aq) H3O+ + Cl-(aq)
HCl is a strong acid % dissociation

42. Prepare buffer solution of CH3COOH and
CH3COO- and include 0.1 M HCl.
HCl(aq) + H2O(aq) H3O+ + Cl-(aq)
HCl is a strong acid % dissociation
H3O+(aq)+ CH3COO-(aq) CH3COOH(aq) + H2O(l)

43. Prepare buffer solution of CH3COOH and
CH3COO- and include 0.1 M HCl.
HCl(aq) + H2O(aq) H3O+ + Cl-(aq)
HCl is a strong acid % dissociation
H3O+(aq)+ CH3COO-(aq) CH3COOH(aq) + H2O(l)
Ka of CH3COOH favors products

44. Assume all HCl reacts with CH3COO-
to form an extra 0.1 M CH3COOH and
reduces CH3COO- by 0.1 M.

45. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.1-y y y
[CH3COOH]
[H3O+] = Ka
[CH3COO-]

46. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.1-y y y
[CH3COOH]
[H3O+] = Ka
[CH3COO-]
(1.1-y)
[H3O+] = Ka =
(0.4+y)

47. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.1-y y y
[CH3COOH]
[H3O+] = Ka
[CH3COO-]
(1.1-y)
1.1
[H3O+] = Ka
= Ka
(0.4+y)
0.4

48. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.1-y y y
[CH3COOH]
[H3O+] = Ka
[CH3COO-]
1.1
[H3O+]
= Ka
= (1.8 x 10-5)(2.75)=
0.4

49. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.1-y y y
[CH3COOH]
[H3O+] = Ka
[CH3COO-]
1.1
[H3O+]
= Ka
= (1.8 x 10-5)(2.75)=
4.95 x 10-5
0.4

50. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.1-y y y
[CH3COOH]
[H3O+] = Ka
[CH3COO-]
[H3O+]
= 4.95 x 10-5
pH = 4.31

51. pH 1.0 M CH3COOH
pH 1.0 M CH3COOH/
0.5 M CH3COO
0.5 M CH3COO-/
0.1 M HCl

52. pH 1.0 M CH3COOH
pH 1.0 M CH3COOH/
0.5 M CH3COO
0.5 M CH3COO-/
0.1 M HCl
The pH decreases due to the increase in
CH3COOH concentration.

53. pH 1.0 M CH3COOH
pH 1.0 M CH3COOH/
0.5 M CH3COO
0.5 M CH3COO-/
0.1 M HCl
pH 0.1 M HCl

54. The action of the buffer involves the
reaction of the strong acid with the
conjugate base of a weak acid.

55. The action of the buffer involves the
reaction of the strong acid with the
conjugate base of a weak acid.
This generates a weak acid of roughly
the equivalent of the added strong acid.

56. This increased amount of weak acid
decreases the pH of the solution, but
it is a much smaller decrease than would
have occurred with the addition of strong
acid only.

57. CH3COOH(aq) + H2O(l) + NaCH3COO (aq)
Na+(aq) + H3O+(aq) + 2 CH3COO-(aq)

58. Prepare same buffer with 0.5 M HCl

59. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.5-y y y
[CH3COOH]
[H3O+] = Ka
[CH3COO-]

60. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.5-y y y
[CH3COOH]
[H3O+] = Ka
[CH3COO-]
y2 = Ka[CH3COOH]

61. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.5-y y y
[CH3COOH]
[H3O+] = Ka
[CH3COO-]
y2 = Ka[CH3COOH] = (1.8 x 10-5)(1.5)

62. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.5-y y y
[CH3COOH]
[H3O+] = Ka
[CH3COO-]
y2 = Ka[CH3COOH] = (1.8 x 10-5)(1.5) =
2.7 x 10-5

63. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.5-y y y
[CH3COOH]
[H3O+] = Ka
[CH3COO-]
y2 = 2.7 x 10-5
y = 5.20 x 10-3

64. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.5-y y y
[CH3COOH]
[H3O+] = Ka
[CH3COO-]
y2 = 2.7 x 10-5
y = 5.20 x 10-3
pH = 2.28

65. pH 1.0 M CH3COOH
pH 1.0 M CH3COOH/
0.5 M CH3COO
0.5 M CH3COO-/
0.1 M HCl
pH buffer M HCl
pH 0.1 M HCl

66. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.0-y y y
[H3O+][CH3COO-]
Ka =
[CH3COOH]
Add 0.51 M HCl

67. [CH3COOH] [H3O+] [CH3COO-]
Start
Change y y y
Equilibrium 1.5-y y y
[H3O+][CH3COO-]
Ka =
[CH3COOH]
Add 0.51 M HCl

68. [CH3COOH] [H3O+] [CH3COO-]
Start
Change y y y
Equilibrium 1.5-y y y
(0.01+y)(y)
[H3O+][CH3COO-]
Ka = =
[CH3COOH]
(1.5-y)

69. [CH3COOH] [H3O+] [CH3COO-]
Start
Change y y y
Equilibrium 1.5-y y y
(0.01+y)(y)
[H3O+][CH3COO-]
Ka = = =
(1.5-y)
[CH3COOH]
(0.01y+y2) =
(1.5-y)

70. [CH3COOH] [H3O+] [CH3COO-]
Start
Change y y y
Equilibrium 1.5-y y y
(0.01+y)(y)
[H3O+][CH3COO-]
Ka = = =
(1.5-y)
[CH3COOH]
y y x 10-5 = 0

71. [CH3COOH] [H3O+] [CH3COO-]
Start
Change y y y
Equilibrium 1.5-y y y
(0.01+y)(y)
[H3O+][CH3COO-]
Ka = = =
(1.5-y)
[CH3COOH]
y y x 10-5 = 0
y = 9.93 x 10-3

72. [CH3COOH] [H3O+] [CH3COO-]
Start
Change y y y
Equilibrium 1.5-y y y
(0.01+y)(y)
[H3O+][CH3COO-]
Ka = = =
(1.5-y)
[CH3COOH]
y y x 10-5 = 0
y = 9.93 x 10-3
pH = 2.00

73. pH 1.0 M CH3COOH
pH 1.0 M CH3COOH/
0.5 M CH3COO
0.5 M CH3COO-/
0.1 M HCl
pH buffer M HCl
pH buffer M HCl
pH 0.01 M HCl

75. Additional conjugate base will create
a solution which can react with more
acid before the pH decreases appreciably.

76. Additional conjugate base will create
a solution which can react with more
acid before the pH decreases appreciably.
This implies that the weak acid
concentration should be raised as well.

77. Determining correct acid-conjugate
base combination and concentration
for a desired buffer pH.

78. Determining correct acid-conjugate
base combination and concentration
for a desired buffer pH.
Ka > pH < 7

79. Determining correct acid-conjugate
base combination and concentration
for a desired buffer pH.
Ka > pH < 7
Ka < pH > 7

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

81. HA(aq) + H2O(aq) H3O+(aq) + A-(aq)
Ka =
[HA]
[A- ]o,[HA]o = original concentrations

82. HA(aq) + H2O(aq) H3O+(aq) + A-(aq)
Ka =
[HA]
[A- ]o,[HA]o = original concentrations
[HA]  [HA]o
[A-]  [A-]o

83. HA(aq) + H2O(aq) H3O+(aq) + A-(aq)
[H3O+][A-]o
Ka = 
[HA]
[HA]o

84. HA(aq) + H2O(aq) H3O+(aq) + A-(aq)
[H3O+][A-]o
Ka = 
[HA]
[HA]o
[HA]o
[H3O+]  Ka
[A-]o

85. HA(aq) + H2O(aq) H3O+(aq) + A-(aq)
[H3O+][A-]o
Ka = 
[HA]
[HA]o
[HA]o
[H3O+]  Ka
[A-]o
[HA]o
[H3O+]  Ka
-log10 =
[A-]o

86. HA(aq) + H2O(aq) H3O+(aq) + A-(aq)
[HA]o
[H3O+]  Ka
[A-]o
[HA]o
-log10
[H3O+]  Ka =
[A-]o
[HA]o
pH  pKa -log10
[A-]o

87. HA(aq) + H2O(aq) H3O+(aq) + A-(aq)
[HA]o
pH  pKa -log10
[A-]o
Henderson-Hasselbach equation

88. [CH3COOH] [H3O+] [CH3COO-]
Start 
Change y y y
Equilibrium 1.0-y y y
[CH3COOH]
[H3O+] = Ka
[CH3COO-]
Henderson-Hasselbach equation
pH = -log10(3.6x10-5) = 4.44

89. [HA]o
pH  pKa -log10
[A-]o
To prepare a buffer of a given pH, start
with an acid with pKa  pH.

90. [HA]o
pH  pKa -log10
[A-]o
To prepare a buffer of a given pH, start
with an acid with pKa  pH.
[HA]o
 1
[A-]o

91. [HA]o
pH  pKa -log10
[A-]o
To prepare a buffer of a given pH, start
with an acid with pKa  pH.
[HA]o
Log10(1) = 0
 1
[A-]o

92. [HA]o
pH  pKa -log10
[A-]o
Design a buffer with a pH of 9.20

93. [NH4+]o
pH  pKa -log10
[NH3]o
Design a buffer with a pH of 9.20
pKa NH4+ = 9.25

94. [NH4+]o
pH  pKa -log10
[NH3]o
[NH4+]
log10
= pKa - pH
[NH3]
Design a buffer with a pH of 9.20
pKa NH4+ = 9.25

95. [NH4+]o
pH  pKa -log10
[NH3]o
[NH4+]
log10
= = 0.05
[NH3]
Design a buffer with a pH of 9.20
pKa NH4+ = 9.25

96. [NH4+]o
pH  pKa -log10
[NH3]o
[NH4+]
log10
= = 0.05
[NH3]
[NH4+] =
[NH3]

97. [NH4+]o
pH  pKa -log10
[NH3]o
[NH4+]
log10
= = 0.05
[NH3]
[NH4+]
= = 1.12
[NH3]

98. [NH4+]o
pH  pKa -log10
[NH3]o
[NH4+]
= = 1.12
[NH3]
[NH4Cl] = 1.12 M
[NH3] = 1.0 M

99. [NH4+]o
pH  pKa -log10
[NH3]o
[NH4+]
= = 1.12
[NH3]
[NH4Cl] = 1.12 M
[NH4Cl] = M
[NH3] = 1.0 M
[NH3] = 0.10 M

100. Buffers in biological systems

101. Buffers in biological systems
Human blood pH

102. Buffers in biological systems
Human blood pH
Buffer system
Carbonic acid - hydrogen carbonate

103. Buffer system
Carbonic acid - hydrogen carbonate
H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l)
CO2(g) + H2O(l)

104. H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l)
CO2(g) + H2O(l)
pKa1 H2CO3 = 6.1
Human blood pH

105. H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l)
CO2(g) + H2O(l)
pKa1 H2CO3 = 6.1
Human blood pH
[HCO3-]
Must significantly differ from 1
[H2CO3]

106. [HCO3-]
calculate
[H2CO3]

107. [HCO3-]
calculate
[H2CO3]
[A-]o
pH  pKa +log10
[HA]o
Henderson-Hasselbach equation

108. [HCO3-]
calculate
[H2CO3]
[A-]o
pH  pKa +log10
[HA]o
[A-]o
log10
= pH - pKa
[HA]o

109. pH = 7.4
[HCO3-]
calculate
[H2CO3]
pKa = 6.1
[A-]o
log10
= pH - pKa
[HA]o

110. pH = 7.4
[HCO3-]
calculate
[H2CO3]
pKa = 6.1
[A-]o
log10
= pH - pKa
[HA]o
[A-]o
log10
= = 1.3
[HA]o

111. pH = 7.4
[HCO3-]
calculate
[H2CO3]
pKa = 6.1
[A-]o
log10
= pH - pKa
[HA]o
[A-]o
log10
= = 1.3
[HA]o
101.3 = 19.95

112. pH = 7.4
[HCO3-]
calculate
[H2CO3]
pKa = 6.1
[A-]o
log10
= pKb - pH
[HA]o
[A-]o
log10
= = 0.4
[HA]o
[A-]o
 20
[HA]o

113. pH = 7.4
[HCO3-]
calculate
[H2CO3]
pKa = 6.1
[A-]o
log10
= pKb - pH
[HA]o
[A-]o
log10
= = 0.4
[HA]o
[HCO3-] = M
[A-]o
 20
[H2CO3] = M
[HA]o

114. [HCO3-] = M
[H2CO3] = M
High acid capacity
Low base capacity

115. H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l)
CO2(g) + H2O(l)
pH < acidosis

116. H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l)
CO2(g) + H2O(l)
pH < acidosis
Failure to eliminate CO2

117. H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l)
CO2(g) + H2O(l)
pH < acidosis
Failure to eliminate CO2
Heavy exercise

118. H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l)
CO2(g) + H2O(l)
pH > alkalosis

119. H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l)
CO2(g) + H2O(l)
pH > alkalosis
Eliminating too much CO2 -
hyperventilation

120. H3O+(aq) + HCO3-(aq) H2CO3(aq) + H2O(l)
CO2(g) + H2O(l)
pH > alkalosis
Eliminating too much CO2 -
hyperventilation
Vomiting - removes acid from stomach