1. Equilibrium constants

2. Equilibrium constants
PCcPDd
K =
PAaPBb
[C]c[D]d
K =
[A]a[B]b

3. Equilibrium constants
PCcPDd
K =
PAaPBb
[C]c[D]d
K =
[A]a[B]b
[solid] = 1

4. Equilibrium of solids and solutions

5. Equilibrium of solids and solutions
NaCl(s) + H2O(l)

7. Equilibrium of solids and solutions
NaCl(s) + H2O(l)
Na+(aq) + Cl-(aq)
To remove an ion from the crystal lattice, the
solvating interactions must be stronger than
the lattice interactions.

11. Equilibrium of solids and solutions
NaCl(s) + H2O(l)
Na+(aq) + Cl-(aq)
Solution must be saturated for this
equilibrium to take place.

12. Dynamic equilibrium

13. Equilibrium for dissolution-precipitation
reaction:

14. Equilibrium for dissolution-precipitation
reaction:
I2(s) + CCl4(l) I2(CCl4)

15. Equilibrium for dissolution-precipitation reaction:
I2(s) + CCl4(l) I2(CCl4)
K [I2]CCl = 4

16. Equilibrium for dissolution-precipitation reaction:
I2(s) + CCl4(l) I2(CCl4)
[I2]CCl K = 4
Molarity of saturated solution = K

17. K will vary when conditions are changed
[I2]CCl K = 4
K will vary when conditions are changed
based on Le Chatelier’s Principle.

18. If solution of a material is exothermic,
[I2]CCl K = 4
If solution of a material is exothermic,
increasing the temperature will
decrease K.
A(s) A(sol)

19. If solution of a material is endothermic,
[I2]CCl K = 4
If solution of a material is endothermic,
increasing the temperature will
increase K.
A(s) A(sol)

21. Kinetics (reaction rates) must
be considered.

22. The value of K is not an indicator
of how long it takes to attain
equilibrium.

23. A K = 5 does not guarantee a 5 M
solution in a few minutes.

24. Not all solutions are ‘ideal’

25. Not all solutions are ‘ideal’
Ideal Solution:
Widely separated species
(ions or molecules) that do not
interact.

26. CsCl(s) Cs+(aq) + Cl-(aq)

27. CsCl(s) Cs+(aq) + Cl-(aq)
As the concentration of ions increases,
Cs+ to Cl- distances decrease.

28. CsCl(s) Cs+(aq) + Cl-(aq)
As the concentration of ions increases,
Cs+ to Cl- distances decrease.
Cs+…Cl-
Cs+…Cl-
Cl-…Cs+

29. CsCl(s) Cs+(aq) + Cl-(aq)
Cl-…Cs+
Ion pairing may occur
before equilibrium.

30. CsCl(s) Cs+(aq) + Cl-(aq)
Cl-…Cs+
Ion pairing may occur
before equilibrium.
These solutions are non-ideal.

31. CsCl(s) Cs+(aq) + Cl-(aq)
Salts of low solubilities allow the study
of solutions that are essentially ideal.

32. CsCl(s) Cs+(aq) + Cl-(aq)
Salts of low solubilities allow the study
of solutions that are essentially ideal.
A saturated solution of 0.1 or less is
a sign of low solubility in a salt.

33. Solubility product

34. Solubility product
AgCl(s) Ag+(aq) + Cl-(aq)

35. Solubility product
Same form as equilibrium constant
AgCl(s) Ag+(aq) + Cl-(aq)

36. Solubility product
AgCl(s) Ag+(aq) + Cl-(aq)
[Ag+][Cl-]
[AgCl]

37. Solubility product
AgCl(s) Ag+(aq) + Cl-(aq)
[Ag+][Cl-]
= Ksp 1

38. Solubility product
AgCl(s) Ag+(aq) + Cl-(aq)
[Ag+][Cl-]
= Ksp

39. Solubility product
AgCl(s) Ag+(aq) + Cl-(aq)
[Ag+][Cl-]
= Ksp
25oC Ksp = 1.6 x 10-10

40. Solubility product
AgCl(s) Ag+(aq) + Cl-(aq)
[Ag+][Cl-]
= Ksp
25oC Ksp = 1.6 x 10-10
[Ag+] = [Cl-] = y

41. Solubility product
AgCl(s) Ag+(aq) + Cl-(aq)
[Ag+][Cl-]
= Ksp
25oC Ksp = 1.6 x 10-10
[Ag+] = [Cl-] = y
y2 = Ksp = 1.6 x 10-10

42. Solubility product
AgCl(s) Ag+(aq) + Cl-(aq)
[Ag+][Cl-]
= Ksp
y2 = Ksp = 1.6 x 10-10
[Ag+] = [Cl-]
= 1.26 x 10-5 M

43. Solubility product
AgCl(s) Ag+(aq) + Cl-(aq)
[Ag+][Cl-]
= Ksp
y2 = Ksp = 1.6 x 10-10
1.8 x 10-3 g/L
[Ag+] = [Cl-]
= 1.26 x 10-5 M

44. BaF2(s) Ba2+(aq) + 2 F-(aq)

45. BaF2(s) Ba2+(aq) + 2 F-(aq)
[Ba2+][F-]2 = Ksp

46. Fe3+(aq) + 3 OH-(aq)
Fe(OH)3

47. Fe3+(aq) + 3 OH-(aq)
Fe(OH)3
[Fe3+][OH-]3 = 1.1 x 10-36

48. Fe3+(aq) + 3 OH-(aq)
Fe(OH)3
[Fe3+][OH-]3 = 1.1 x 10-36
For every Fe3+ that goes into solution,
3 OH- go into solution.

49. Fe3+(aq) + 3 OH-(aq)
Fe(OH)3
[Fe3+][OH-]3 = 1.1 x 10-36
[y][3y]3 = 1.1 x 10-36

50. Fe3+(aq) + 3 OH-(aq)
Fe(OH)3
[Fe3+][OH-]3 = 1.1 x 10-36
[y][3y]3 = 1.1 x 10-36
If there is another source of OH- (NaOH)
that provides a higher [OH-] then that is
the value of [OH-] to be used.

51. Fe3+(aq) + 3 OH-(aq)
Fe(OH)3
[Fe3+][OH-]3 = 1.1 x 10-36
[y][3y]3 = 1.1 x 10-36
27y4 = 1.1 x 10-36

52. Fe3+(aq) + 3 OH-(aq)
Fe(OH)3
[Fe3+][OH-]3 = 1.1 x 10-36
[y][3y]3 = 1.1 x 10-36
27y4 = 1.1 x 10-36
1.1 x 10-36
y4 =
= 4.1 x 10-38 27

53. Fe3+(aq) + 3 OH-(aq)
Fe(OH)3
[Fe3+][OH-]3 = 1.1 x 10-36
[y][3y]3 = 1.1 x 10-36
27y4 = 1.1 x 10-36
1.1 x 10-36
y4 =
= 4.1 x 10-38 27
y = 4.5 x 10-10

54. Ksp CuS = 2 x 10-47
CuS(s) Cu2+(aq) + S2-(aq)

55. Ksp CuS = 2 x 10-47
CuS(s) Cu2+(aq) + S2-(aq)
[Cu2+][S2-] = 2 x 10-47

56. Ksp CuS = 2 x 10-47
CuS(s) Cu2+(aq) + S2-(aq)
[Cu2+][S2-] = 2 x 10-47
y2 = 2 x 10-47

57. Ksp CuS = 2 x 10-47
CuS(s) Cu2+(aq) + S2-(aq)
[Cu2+][S2-] = 2 x 10-47
y2 = 2 x 10-47
y = 4.5 x = [Cu2+] = [S2-]

58. Ksp CuS = 2 x 10-47
CuS(s) Cu2+(aq) + S2-(aq)
y = 4.5 x = [Cu2+] = [S2-]
mw CuS = 95.6

59. Ksp CuS = 2 x 10-47
CuS(s) Cu2+(aq) + S2-(aq)
y = 4.5 x = [Cu2+] = [S2-]
mw CuS = 95.6
4.5 x (95.6) = 4.3 x g/L

60. Ksp CuS = 2 x 10-47
CuS(s) Cu2+(aq) + S2-(aq)
y = 4.5 x = [Cu2+] = [S2-]
4.5 x (95.6) = 4.3 x g/L
Atoms/L = moles x Ao

61. Ksp CuS = 2 x 10-47
CuS(s) Cu2+(aq) + S2-(aq)
y = 4.5 x = [Cu2+] = [S2-]
4.5 x (95.6) = 4.3 x g/L
Atoms/L = moles x Ao = (4.5 x 10-24)(6 x 1023) =

62. Ksp CuS = 2 x 10-47
CuS(s) Cu2+(aq) + S2-(aq)
y = 4.5 x = [Cu2+] = [S2-]
4.5 x (95.6) = 4.3 x g/L
Atoms/L = moles x Ao = (4.5 x 10-24)(6 x 1023) =
2.7 atoms/L

63. Example 9-3
Determine Ksp for a salt based
on solubility data.

64. Example 9-3
Determine Ksp for a salt based
on solubility data.
PbCl g/L for a saturated solution

65. Example 9-3
Determine Ksp for a salt based
on solubility data.
PbCl g/L for a saturated solution
PbCl Pb2+(aq) + 2 Cl-(aq)

66. Example 9-3
Determine Ksp for a salt based
on solubility data.
PbCl g/L for a saturated solution
PbCl Pb2+(aq) + 2 Cl-(aq)
[Pb2+][Cl-]2 = Ksp

67. Example 9-3
PbCl g/L for a saturated solution
PbCl Pb2+(aq) + 2 Cl-(aq)
[Pb2+][Cl-]2 = Ksp
(y)(2y)2 = Ksp

68. Example 9-3
PbCl g/L for a saturated solution
PbCl Pb2+(aq) + 2 Cl-(aq)
[Pb2+][Cl-]2 = Ksp
(y)(2y)2 = Ksp
4y2 = Ksp

69. Example 9-3
PbCl g/L for a saturated solution
PbCl Pb2+(aq) + 2 Cl-(aq)
[Pb2+][Cl-]2 = Ksp
4y2 = Ksp
8.67 g PbCl2 =
278.1 g/mol

70. Example 9-3
PbCl g/L for a saturated solution
PbCl Pb2+(aq) + 2 Cl-(aq)
[Pb2+][Cl-]2 = Ksp
4y3 = Ksp
8.67 g PbCl2
= mol
278.1 g/mol
< 0.1 mol

71. Example 9-3
PbCl g/L for a saturated solution
PbCl Pb2+(aq) + 2 Cl-(aq)
[Pb2+][Cl-]2 = Ksp
4y3 = Ksp
8.67 g PbCl2
= mol
278.1 g/mol
Ksp = 4(0.031)3

72. Example 9-3
PbCl g/L for a saturated solution
PbCl Pb2+(aq) + 2 Cl-(aq)
[Pb2+][Cl-]2 = Ksp
4y3 = Ksp
8.67 g PbCl
= mol
278.1 g/mol
Ksp = 4(0.031)3
=1.2 x 10-4

73. Example 9-3
PbCl g/L for a saturated solution
PbCl Pb2+(aq) + 2 Cl-(aq)
[Pb2+][Cl-]2 = Ksp
4y3 = Ksp
8.67 g PbCl
= mol
278.1 g/mol
Table = 1.6 x 10-5
Ksp = 4(0.031)3
=1.2 x 10-4

74. Example 9-3
PbCl g/L for a saturated solution
PbCl Pb2+(aq) + 2 Cl-(aq)
[Pb2+][Cl-]2 = Ksp
4y3 = Ksp
8.67 g PbCl
= mol
278.1 g/mol
Table = 1.6 x 10-5
=1.2 x 10-4
Ksp = 4(0.031)3
Non-ideal solution

75. Precipitation from solution

76. Precipitation from solution
Start with solution that is not saturated.

77. Precipitation from solution
Start with solution that is not saturated.
Cause solution to become saturated.

78. Precipitation from solution
Start with solution that is not saturated.
Cause solution to become saturated.
Remove solvent by evaporation.

79. Precipitation from solution
Start with solution that is not saturated.
Cause solution to become saturated.
Remove solvent by evaporation.
Change nature of solvent.

80. Change nature of solvent.
The polarity of a solvent
system can be adjusted.

81. Change nature of solvent.
The polarity of a solvent
system can be adjusted.
NaCl(s) + H2O(l) Na+(aq) + Cl-(aq)
Not saturated

82. Change nature of solvent.
The polarity of a solvent
system can be adjusted.
NaCl(s) + H2O(l) Na+(aq) + Cl-(aq)
Not saturated
Add EtOH to solution.

83. Precipitation from solution
Start with solution that is not saturated.
Cause solution to become saturated.
Remove solvent by evaporation.
Change nature of solvent.
Cool or warm system.

84. Precipitation of a product from
a reaction.
Will it precipitate?

85. Determine reaction quotient.
AgNO3 (solution) mixed with
NaCl (solution)
Will AgCl precipitate?

86. Determine reaction quotient.
AgNO3 (solution) mixed with
NaCl (solution)
Will AgCl precipitate?
Q = [Ag+][Cl-]
Q = reaction quotient

87. AgNO3 (solution) mixed with
NaCl (solution)
Q = [Ag+][Cl-]
Q = reaction quotient
Qinit = conditions just as solutions are mixed

88. AgNO3 (solution) mixed with
NaCl (solution)
Q = [Ag+][Cl-]
Q = reaction quotient
Qinit = conditions just as solutions are mixed
If Qinit < Ksp then no AgCl will
precipitate.
[Ag+][Cl-] too low

89. AgNO3 (solution) mixed with
NaCl (solution)
Q = [Ag+][Cl-]
Q = reaction quotient
Qinit = conditions just as solutions are mixed
If Qinit > Ksp then AgCl will precipitate.
Ag+ Cl- too high

90. AgNO3 (solution) mixed with
NaCl (solution)
Q = [Ag+][Cl-]
Q = reaction quotient
Qinit = conditions just as solutions are mixed
If Qinit > Ksp then AgCl will precipitate.
As the reaction continues, precipitation will
stop when Q = Ksp

91. precipitate
No precipitate

92. Exercise page 384
TlIO ml M
NaIO ml M
Will TlIO3 precipitate at
equilibrium?

93. Exercise page 384
TlIO ml M
NaIO ml M
Will TlIO3 precipitate at
equilibrium?
Ksp = 3.1 x 10-6

94. TlIO ml M
NaIO ml M
Ksp = 3.1 x 10-6
Q = [Tl+][IO3-]
[Tl3+] = .555(0.0022)
= M
1.000

95. TlIO ml M
NaIO ml M
Ksp = 3.1 x 10-6
Q = [Tl+][IO3-]
[Tl+] = M

96. TlIO ml M
NaIO ml M
Ksp = 3.1 x 10-6
Q = [Tl+][IO3-]
.555 L x M =
moles
[Tl+] = M
[IO3-] = M M =

97. TlIO ml M
For this
solution
NaIO ml M
Ksp = 3.1 x 10-6
Q = [Tl+][IO3-] = (0.0012)(.00099) = 1.2 x 10-6
.555 L x M =
moles
[Tl+] = M
[IO3-] = M

98. TlIO ml M
For mixed
solutions
NaIO ml M
Ksp = 3.1 x 10-6
Q = [Tl+][IO3-] = (0.0012)(.0022) = 2.6 x 10-6
Q < Ksp
No precipitate

99. Common ion effect

100. Common ion effect
Adding more of a cation or anion
that is already in solution will
cause Q to change.

101. Common ion effect
Adding more of a cation or anion
that is already in solution will
cause Q to change.
Q = [A+][B-]

102. Ag+(aq) + Cl-(aq)
AgCl(s)
Q = [Ag+][Cl-]
Add more Cl- (NaCl)

103. Q > Ksp
Q < Ksp

104. Ag+(aq) + Cl-(aq)
AgCl(s)
Q = [Ag+][Cl-]
Add more Cl- (NaCl)
This raises Q above Ksp

105. Common ion effect
If a solution and a solid salt to be
dissolved in it have an ion in common,
the solubility of the salt is depressed.

106. Exercise page 387
Tl+(aq) + IO3-(aq)
TlIO3(s)
Ksp = 3.1 x 10-6
Q = [Tl+][IO3-]
0.050 M KIO3
What is the solubility of TlIO3
in this solution?

107. Exercise page 387
Tl+(aq) + IO3-(aq)
TlIO3(s)
Ksp = 3.1 x 10-6
0.050 M KIO3
Q = [Tl+][IO3-]
Ksp = [Tl+][IO3-]
[Tl+] = y

108. Exercise page 387
Tl+(aq) + IO3-(aq)
TlIO3(s)
Ksp = 3.1 x 10-6
0.050 M KIO3
Q = [Tl+][IO3-]
Ksp = [Tl+][IO3-]
[Tl+] = y
[IO3-] = y

109. Exercise page 387
Tl+(aq) + IO3-(aq)
TlIO3(s)
Ksp = 3.1 x 10-6
0.050 M KIO3
Q = [Tl+][IO3-]
Ksp = [Tl+][IO3-]
[Tl+] = y
[IO3-] = y  0.050

110. Exercise page 387
Tl+(aq) + IO3-(aq)
TlIO3(s)
Ksp = 3.1 x 10-6
0.050 M KIO3
Q = [Tl+][IO3-]
Ksp = [Tl+][IO3-]
[Tl+] = y
[IO3-] = y  0.050
3.1 x 10-6 = (y)(0.050) = 0.050y

111. Exercise page 387
Tl+(aq) + IO3-(aq)
TlIO3(s)
Ksp = 3.1 x 10-6
0.050 M KIO3
Ksp = [Tl+][IO3-]
[Tl+] = y
[IO3-] = y  0.050
3.1 x 10-6 = (y)(0.050) = 0.050y
y = 6.2 x 10-5 mol/L

112. Exercise page 387
Tl+(aq) + IO3-(aq)
TlIO3(s)
Ksp = 3.1 x 10-6
0.050 M KIO3
Ksp = [Tl+][IO3-]
[Tl+] = y
[IO3-] = y  0.050
3.1 x 10-6 = (y)(0.050) = 0.050y
No common ion
y = 6.2 x 10-5 mol/L
[Tl+] = 1.8 x 10-3