1. AQUEOUS EQUILIBRIA III
Unit 5
•
•
•
•
19-1
SOLUBILITY CONSTANTS
COMPLEX IONS
NET IONIC EQUATIONS
Chapter 4.3 & 19.3-19.4 (Silberberg)

2. Chapter 19
Ionic Equilibria in Aqueous Systems
19-2

3. Ionic Equilibria in Aqueous Systems
19.3 Equilibria of Slightly Soluble Ionic Compounds
19.4 Equilibria Involving Complex Ions
19-3

4. Goals & Objectives
• See the following Learning
Objectives on page 870.
• Understand These Concepts:
• 19.9-17
• Master these Skills:
• 19.7-11, 13-16.
19-4

5. Equilibria of Slightly Soluble Ionic Compounds
Any “insoluble” ionic compound is actually slightly
soluble in aqueous solution.
We assume that the very small amount of such a compound that
dissolves will dissociate completely.
For a slightly soluble ionic compound in water, equilibrium
exists between solid solute and aqueous ions.
PbF2(s)
[Pb2+][F-]2
Qc =
[PbF2]
19-5
Pb2+(aq) + 2F-(aq)
Qsp = Qc[PbF2] = [Pb2+][F-]2

6. Solubility Product Principle
• Although all compounds dissolve in
water to some extent, some
compounds are classified as
“insoluble”. The solubility of these
insoluble compounds can be
described in terms of their solubility
product constant, Ksp.
• Memorize the Solubility
generalization handout.
19-6

7. Solubility Product Principle
•
•
•
•
•
•
19-7
Consider AgCl dissolved in water
AgCl(s) =
Ag+ + ClKsp = [Ag+][Cl-]
Consider Ag2S dissolved in water
Ag2S(s)
=
2Ag+ + S2Ksp = [Ag+]2[S2-]

8. Solubility Product Principle
• Consider Ca3(PO4)2 dissolved in
water
• Ca3(PO4)2
=
3Ca2+ + 2PO43-
• Ksp = [Ca2+]3[PO43-]2
19-8

9. Qsp and Ksp
Qsp is called the ion-product expression for a slightly
soluble ionic compound.
For any slightly soluble compound MpXq, which consists
of ions Mn+ and Xz-,
Qsp = [Mn+]p[Xz-]q
When the solution is saturated, the system is at equilibrium,
and Qsp = Ksp, the solubility product constant.
The Ksp value of a salt indicates how far the dissolution
proceeds at equilibrium (saturation).
19-9

10. Metal Sulfides
Metal sulfides behave differently from most other slightly
soluble ionic compounds, since the S2- ion is strongly
basic.
We can think of the dissolution of a metal sulfide as a
two-step process:
MnS(s)
Mn2+(aq) + S2-(aq)
S2-(aq) + H2O(l) → HS-(aq) + OH-(aq)
MnS(s) + H2O(l)
Mn2+(aq) + HS-(aq) + OH-(aq)
Ksp = [Mn2+][HS-][OH-]
19-10

11. Sample Problem 19.5
Writing Ion-Product Expressions
PROBLEM: Write the ion-product expression at equilibrium for each
compound:
(a) magnesium carbonate
(b) iron(II) hydroxide
(c) calcium phosphate
(d) silver sulfide
PLAN: We write an equation for a saturated solution of each
compound, and then write the ion-product expression at
equilibrium, Ksp. Note the sulfide in part (d).
SOLUTION:
(a)
MgCO3(s)
(b) Fe(OH)2(s)
(c)
19-11
Ca3(PO4)2(s)
Mg2+(aq) + CO32-(aq)
Fe2+(aq) + 2OH-(aq)
3Ca2+(aq) + 2PO43-(aq)
Ksp = [Mg2+][CO32-]
Ksp = [Fe2+][OH-]2
Ksp = [Ca2+]3[PO43-]2

12. Sample Problem 19.5
(d)
Ag2S(s)
2Ag+(aq) + S2-(aq)
S2-(aq) + H2O(l) → HS-(aq) + OH-(aq)
Ag2S(s) + H2O(l)
2Ag+(aq) + HS-(aq) + OH-(aq)
Ksp = [Ag+]2[HS-][OH-]
19-12

13. Determination of Ksp
• One liter of saturated silver chloride
solution contains 0.00192g of
dissolved AgCl at 25oC. Calculate
the molar solubility of , and Ksp for,
AgCl.
• The molar solubility of CaF2 is
2.14x10-4 at 25oC. Determine Ksp for
CaF2.
19-13

14. 19-14

15. 19-15

16. Table 19.2 Solubility-Product Constants (Ksp) of Selected Ionic
Compounds at 25 C
Name, Formula
Ksp
Aluminum hydroxide, Al(OH)3
Cobalt(II) carbonate, CoCO3
1.0x10-10
Iron(II) hydroxide, Fe(OH)2
4.1x10-15
Lead(II) fluoride, PbF2
3.6x10-8
Lead(II) sulfate, PbSO4
1.6x10-8
Mercury(I) iodide, Hg2I2
4.7x10-29
Silver sulfide, Ag2S
8x10-48
Zinc iodate, Zn(IO3)2
19-16
3x10-34
3.9x10-6

17. Sample Problem 19.6
Determining Ksp from Solubility
PROBLEM: (a) Lead(II) sulfate (PbSO4) is a key component in leadacid car batteries. Its solubility in water at 25 C is
4.25x10-3 g/100 mL solution. What is the Ksp of
PbSO4?
(b) When lead(II) fluoride (PbF2) is shaken with pure
water at 25 C, the solubility is found to be 0.64 g/L.
Calculate the Ksp of PbF2.
PLAN: We write the dissolution equation and the ion-product
expression for each compound. This tells us the number of
moles of each ion formed. We use the molar mass to convert
the solubility of the compound to molar solubility (molarity),
then use it to find the molarity of each ion, which we can
substitute into the Ksp expression.
19-17

18. Sample Problem 19.6
SOLUTION:
(a)
PbSO4(s)
Pb2+(aq) + SO42-(aq)
Ksp = [Pb2+][SO42-]
Converting from g/mL to mol/L:
4.25x10-3g PbSO4 x 1000 mL x 1 mol PbSO4
100 mL soln
1L
303.3 g PbSO4
= 1.40x10-4 M PbSO4
Each mol of PbSO4 produces 1 mol of Pb2+ and 1 mol of SO42-, so
[Pb2+] = [SO42-] = 1.40x10-4 M
Ksp = [Pb2+][SO42-] = (1.40x10-4)2 = 1.96x10-8
19-18

19. Sample Problem 19.6
(b)
PbF2(s)
Pb2+(aq) + F-(aq)
Ksp = [Pb2+][F-]2
Converting from g/L to mol/L:
0.64 g PbF2 x
1 mol PbF2
1 L soln
245.2 g PbF2
= 2.6x10-3 M PbF2
Each mol of PbF2 produces 1 mol of Pb2+ and 2 mol of F-, so
[Pb2+] = 2.6x10-3 M and [F-] = 2(2.6x10-3) = 5.2x10-3 M
Ksp = [Pb2+][F-]2 = (2.6x10-3)(5.2x10-3)2
19-19
= 7.0x10-8

20. Sample Problem 19.7
Determining Solubility from Ksp
PROBLEM: Calcium hydroxide (slaked lime) is a major component of
mortar, plaster, and cement, and solutions of Ca(OH)2
are used in industry as a strong, inexpensive base.
Calculate the molar solubility of Ca(OH)2 in water if the
Ksp is 6.5x10-6.
PLAN: We write the dissolution equation and the expression for Ksp.
We know the value of Ksp, so we set up a reaction table that
expresses [Ca2+] and [OH-] in terms of S, the molar solubility.
We then substitute these expressions into the Ksp expression
and solve for S.
SOLUTION:
Ca(OH)2(s)
19-20
Ca2+(aq) + 2OH-(aq)
Ksp = [Ca2+][OH-]2 = 6.5x10-6

21. Sample Problem 19.7
Concentration (M)
Ca(OH)2(s)
Ca2+(aq) + 2OH-(aq)
Initial
-
0
0
Change
-
+S
+ 2S
Equilibrium
-
S
2S
Ksp = [Ca2+][OH-]2 = (S)(2S)2 = 4S3 = 6.5x10-6
S=
19-21
=
= 1.2x10-2 M

22. Uses of Ksp
• Determine the molar solubility of
BaSO4 in pure water at 25oC.
Ksp=1.1x10-10
• Calculate the molar solubility and pH
of a saturated solution of Mg(OH)2 at
25oC. Ksp= 1.5x10-11
• Determine the molar solubility of
BaSO4 in 0.010M sodium sulfate
solution at 25oC. Ksp= 1.1x10-10
19-22

23. 19-23

24. 19-24

25. 19-25

26. Table 19.3 Relationship Between Ksp and Solubility at 25 C
No. of Ions Formula Cation/Anion
Ksp
Solubility (M)
2
MgCO3
1/1
3.5x10-8
1.9x10-4
2
PbSO4
1/1
1.6x10-8
1.3x10-4
2
BaCrO4
1/1
2.1x10-10
1.4x10-5
3
Ca(OH)2
1/2
6.5x10-6
1.2x10-2
3
BaF2
1/2
1.5x10-6
7.2x10-3
3
CaF2
1/2
3.2x10-11
2.0x10-4
3
Ag2CrO4
2/1
2.6x10-12
8.7x10-5
The higher the Ksp value, the greater the solubility, as long as we
compare compounds that have the same total number of ions in
their formulas.
19-26

27. Figure 19.12
The effect of a common ion on solubility.
PbCrO4(s)
Pb2+(aq) + CrO42-(aq)
If Na2CrO4 solution is added to a saturated solution of PbCrO4, it
provides the common ion CrO42-, causing the equilibrium to shift to
the left. Solubility decreases and solid PbCrO4 precipitates.
19-27

28. Sample Problem 19.8
Calculating the Effect of a Common Ion
on Solubility
PROBLEM: In Sample Problem 19.7, we calculated the solubility of
Ca(OH)2 in water. What is its solubility in 0.10 M
Ca(NO3)2? Ksp of Ca(OH)2 is 6.5x10-6.
PLAN: The addition of Ca2+, an ion common to both solutions, should
lower the solubility of Ca(OH)2. We write the equation and Ksp
expression for the dissolution and set up a reaction table in
terms of S, the molar solubility of Ca(OH)2. We make the
assumption that S is small relative to [Ca2+]init because Ksp is
low. We can then solve for S and check the assumption.
SOLUTION:
Ca(OH)2(s)
19-28
Ca2+(aq) + 2OH-(aq)
Ksp = [Ca2+][OH-]2

29. Sample Problem 19.8
[Ca2+]init = 0.10 M because Ca(NO3)2 is a soluble salt, and dissociates
completely in solution.
Concentration (M)
Ca(OH)2(s)
Ca2+(aq) + 2OH-(aq)
Initial
-
0.10
0
Change
-
+S
+ 2S
Equilibrium
-
0.10 + S
2S
Ksp = [Ca2+][OH-]2 = 6.5x10-6 ≈ (0.10)(2S)2 = (0.10)(4S2)
4S2
6.5x10-6
≈
0.10
so S ≈
= 4.0x10-3 M
-3
Checking the assumption: 4.0x10 M x 100 = 4.0% < 5%
0.10 M
19-29

30. Effect of pH on Solubility
Changes in pH affects the solubility of many slightly
soluble ionic compounds.
The addition of H3O+ will increase the solubility of a salt
that contains the anion of a weak acid.
CaCO3(s)
Ca2+(aq) + CO32-(aq)
CO32-(aq) + H3O+(aq) → HCO3-(aq) + H2O(l)
HCO3-(aq) + H3O+(aq) → [H2CO3(aq)] + H2O(l) → CO2(g) + 2H2O(l)
The net effect of adding H3O+ to CaCO3 is the removal
of CO32- ions, which causes an equilibrium shift to the
right. More CaCO3 will dissolve.
19-30

31. Sample Problem 19.9
Predicting the Effect on Solubility of
Adding Strong Acid
PROBLEM: Write balanced equations to explain whether addition of
H3O+ from a strong acid affects the solubility of each ionic
compound:
(a) lead(II) bromide (b) copper(II) hydroxide
(c) iron(II) sulfide
PLAN: We write the balanced dissolution equation for each compound
and note the anion. The anion of a weak acid reacts with H3O+,
causing an increase in solubility.
SOLUTION:
(a)
PbBr2(s)
Pb2+(aq) + 2Br-(aq)
Br- is the anion of HBr, a strong acid, so it does not react with H3O+. The
addition of strong acid has no effect on its solubility.
19-31

32. Sample Problem 19.9
(b)
Cu(OH)2(s)
Cu2+(aq) + 2OH-(aq)
OH- is the anion of H2O, a very weak acid, and is in fact a strong base. It
will react with H3O+:
OH-(aq) + H3O+(aq) → 2H2O(l)
The addition of strong acid will cause an increase in solubility.
(c)
FeS(s)
Fe2+(aq) + S2-(aq)
S2- is the anion of HS-, a weak acid, and is a strong base. It will react
completely with water to form HS- and OH-. Both these ions will react
with added H3O+:
HS-(aq) + H3O+(aq) → H2S(aq) + H2O(l)
OH-(aq) + H3O+(aq) → 2H2O(l)
The addition of strong acid will cause an increase in solubility.
19-32

33. Figure 19.14
Limestone cave in Nerja, Málaga, Spain.
Limestone is mostly CaCO3 (Ksp = 3.3x10-9).
Ground water rich in CO2 trickles over
CaCO3, causing it to dissolve. This
gradually carves out a cave.
Water containing HCO3- and Ca2+ ions
drips from the cave ceiling. The air has
a lower PCO than the soil, causing CO2
2
to come out of solution. A shift in
equilibrium results in the precipitation
of CaCO3 to form stalagmites and
stalactites.
CO2(g)
CO2(aq)
CO2(aq) + 2H2O(l)
H3O+(aq) + HCO3-(aq)
CaCO3(s) + CO2(aq) + H2O(l)
19-33
Ca2+(aq) + 2HCO3-(aq)

34. Predicting the Formation of a Precipitate
For a saturated solution of a slightly soluble ionic salt,
Qsp = Ksp.
When two solutions containing the ions of slightly soluble
salts are mixed,
If Qsp = Ksp,
the solution is saturated and no change will occur.
If Qsp > Ksp,
a precipitate will form until the remaining solution is saturated.
If Qsp =< Ksp,
no precipitate will form because the solution is unsaturated.
19-34

35. Sample Problem 19.10
Predicting Whether a Precipitate Will
Form
PROBLEM: A common laboratory method for preparing a precipitate is
to mix solutions containing the component ions. Does a
precipitate form when 0.100 L of 0.30 M Ca(NO3)2 is mixed
with 0.200 L of 0.060 M NaF?
PLAN: First we need to decide which slightly soluble salt could form,
look up its Ksp value in Appendix C, and write the dissolution
equation and Ksp expression. We find the initial ion
concentrations from the given volumes and molarities of the
two solutions, calculate the value for Qsp and compare it to Ksp.
SOLUTION:
The ions present are Ca2+, NO3-, Na+, and F-. All Na+ and NO3- salts are
soluble, so the only possible precipitate is CaF2 (Ksp = 3.2x10-11).
CaF2(s)
19-35
Ca2+(aq) + 2F-(aq)
Ksp = [Ca2+][F-]2

36. Sample Problem 19.10
Ca(NO3)2 and NaF are soluble, and dissociate completely in solution.
We need to calculate [Ca2+] and [F-] in the final solution.
Amount (mol) of Ca2+ = 0.030 M Ca2+ x 0.100 L = 0.030 mol Ca2+.
0.030 mol Ca2+ = 0.10 M Ca2+
2+]
[Ca init =
0.100 L + 0.200 L
Amount (mol) of F- = 0.060 M F- x 0.200 L = 0.012 mol F-.
[F-]init
=
0.012 mol F= 0.040 M F0.100 L + 0.200 L
Qsp = [Ca2+]init[F-]2init = (0.10)(0.040)2 = 1.6x10-4
Since Qsp > Ksp, CaF2 will precipitate until Qsp = 3.2x10-11.
19-36

37. Uses of Ksp
• Predict whether a precipitate of
PbSO4 will form in a solution having
[Pb2+] = 0.050M and [SO42-] =
0.0050M. Ksp = 1.8x10-8
19-37

38. 19-38

39. 19-39

40. Selective Precipitation
Selective precipitation is used to separate a solution
containing a mixture of ions.
A precipitating ion is added to the solution until the Qsp
of the more soluble compound is almost equal to its Ksp.
The less soluble compound will precipitate in as large a
quantity as possible, leaving behind the ion of the more
soluble compound.
19-40

41. Sample Problem 19.12
Separating Ions by Selective Precipitation
PROBLEM: A solution consists of 0.20 M MgCl2 and 0.10 M CuCl2.
Calculate the [OH-] that would separate the metal ions as
their hydroxides. Ksp of Mg(OH)2= is 6.3x10-10; Ksp of
Cu(OH)2 is 2.2x10-20.
PLAN: Both compounds have 1/2 ratios of cation/anion, so we can
compare their solubilities by comparing their Ksp values.
Mg(OH)2 is 1010 times more soluble than Cu(OH)2, so
Cu(OH)2 will precipitate first. We write the dissolution
equations and Ksp expressions. Using the given cation
concentrations, we solve for the [OH-] that gives a saturated
solution of Mg(OH)2. Then we calculate the [Cu2+] remaining
to see if the separation was successful.
19-41

42. Sample Problem 19.12
SOLUTION:
Mg(OH)2(s)
Mg2+(aq) + 2OH-(aq)
Ksp = [Mg2+][OH-]2 = 6.3x10-10
Cu(OH)2(s)
Cu2+(aq) + 2OH-(aq)
Ksp = [Cu2+][OH-]2 = 2.2x10-20
[OH-] =
=
= 5.6x10-5 M
This is the maximum [OH-] that will not precipitate Mg2+ ion.
Calculating the [Cu2+] remaining in solution with this [OH-]
[Cu2+]
Ksp
2.2x10-20
=
=
-]2
[OH
(5.6x10-5)2
= 7.0x10-12 M
Since the initial [Cu2+] is 0.10 M, virtually all the Cu2+ ion is precipitated.
19-42

43. Chemical Connections
Figure B19.1
Formation of acidic precipitation.
Since pH affects the solubility of many slightly soluble ionic compounds, acid
rain has far-reaching effects on many aspects of our environment.
19-43

44. Complex Ion Equilibria
• Many complex ions are known to
exist. The majority consist of a metal
ion with several anions or molecules
coordinated to it. The class of
compounds is called coordination
compounds.
19-44

45. Figure 19.15
Cr(NH3)63+, a typical complex ion.
A complex ion consists of a central metal ion covalently bonded to
two or more anions or molecules, called ligands.
19-45

46. Figure 19.16
The stepwise exchange of NH3 for H2O in M(H2O)42+.
The overall formation constant is given by
[M(NH3)42+]
Kf =
[M(H2O)42+][NH3]4
19-46

47. Complex Ion Equilibria
• Some examples of complex ions:
• NH3 complexes
– Ag(NH3)2+
– Cu(NH3)42+
– Zn(NH3)42+
• OH- complexes
– Al(OH)4¯
– Cr(OH)4¯
– Zn(OH)4219-47

48. Table 19.4 Formation Constants (Kf) of Some Complex Ions at
25 C
19-48

49. Sample Problem 19.13
Calculating the Concentration of a
Complex Ion
PROBLEM: An industrial chemist converts Zn(H2O)42+ to the more
stable Zn(NH3)42+ by mixing 50.0 L of 0.0020 M
Zn(H2O)42+ and 25.0 L of 0.15 M NH3. What is the final
[Zn(H2O)42+] at equilibrium? Kf of Zn(NH3)42+ is 7.8x108.
PLAN:
We write the reaction equation and the Kf expression, and use
a reaction table to calculate equilibrium concentrations. To set
up the table, we must first find [Zn(H2O)42+]init and [NH3]init
using the given volumes and molarities. With a large excess
of NH3 and a high Kf, we assume that almost all the
Zn(H2O)42+ is converted to Zn(NH3)42+.
SOLUTION:
Zn(H2O)42+(aq) + 4NH3(aq)
Kf =
Zn(NH3)42+(aq) + 4H2O(l)
[Zn(NH3)42+]
[Zn(H2O)42+][NH3]4
19-49

50. Sample Problem 19.13
[Zn(H2O)42+]initial = 50.0 L x 0.0020 M = 1.3x10-3 M
50.0 L + 25.0 L
25.0 L x 0.15 M
= 5.0x10-2 M
[NH3]initial =
50.0 L + 25.0 L
4 mol of NH3 is needed per mol of Zn(H2O4)2+, so
[NH3]reacted = 4(1.3x10-3 M) = 5.2x10-3 M and
[Zn(NH3)42+] ≈ 1.3x10-3 M
Concentration (M)
Initial
Change
Equilibrium
19-50
Zn(H2O)42+(aq) + 4NH3(aq)
1.3x10-3
~(-1.3x10-3)
x
5.0x10-2
~(-5.2x10-3)
4.5x10-2
Zn(NH3)42+(aq) + 4H2O(l)
0
-
~(+1.3x10-3)
-
1.3x10-3
-

51. Sample Problem 19.13
Kf =
[Zn(NH3)42+]
[Zn(H2O)42+][NH3]4
=
7.8x108
x = [Zn(H2O)42+ = 4.1x10-7 M
19-51
=
(1.3x10-3)
x(4.5x10-2)4

52. Complex Ion Equilibria
• Dissociation of complex ions
• Ag(NH3)2+ =
• Kd = [Ag+][NH3]2
•
___________
•
[Ag(NH3)2+]
19-52
Ag+ + 2NH3

53. Complex Ion Equilibria
• Cr(OH)4¯ =
Cr3+ + 4OH¯
• Kd = [Cr3+][OH¯]4
_________________
•
•
[Cr(OH)4-1]
19-53

54. Complex Ion Equilibria


19-54
Determine the concentration of silver ions in
a solution that is 0.10M in [Ag(NH3)2]+. Kd =
6.3x10-8
Determine the [Ag+] in a solution that is
0.10M in Ag(NH3)2NO3 and 0.10M in NH3.

55. 19-55

56. 19-56

57. Complex Ion Equilibria
Ag+
 Cu2+
 Zn2+

Al3+
 Cr3+
 Zn2+

19-57

58. Net Ionic Equations
• Rules for writing:
– A. List predominant species
• All soluble salts, strong acids and strong bases
are written as their component ions. All others
are written as the molecule.
19-58

59. Net Ionic Equations
• Rules (continued):
– B. Combine ions of opposite charge
and look for one or more of the
following:
•
•
•
•
•
19-59
1. formation of a weak acid
2. formation of a weak base
3. formation of water
4. formation of an insoluble substance
5. formation of a complex ion

60. Net Ionic Equations
• Rules (continued):
– C. If no reaction in B. above, then look
at each of the following in the order
given as a source of secondary species:
•
•
•
•
•
19-60
1. solubility equilibria
2. complex equilibria
3. weak acid equilibria
4. weak base equilibria
5. hydrolysis equilibria (only if necessary)

61. Net Ionic Equations
• Rules (continued):
– D. If no reaction in C, then try secondary
species reacting with secondary species
again leaving hydrolysis to last.
19-61

62. Writing Net Ionic Equations
• Give net ionic equations for each of
the following reactions occurring in
aqueous solution. Also indicate the
form of K in terms of other constants
such as Ka, Kb, Kw, Ksp or Kd.
– 1. AgNO3 + NaCl
– 2. BaCl2 + Na2SO4
– 3. FeCl3 + NaOH
19-62

63. Writing Net Ionic Equations
– 4. CuS + HNO3
– 5. AlCl3 + NH3
– 6. Cr(NO3)3 + NaOH
– 7. Cr(NO3)3 + NaOH(xs)
– 8. Cr(NO3)3 + NH3(xs)
– 9. ZnCl2 + NH3(xs)
– 10. Cu(OH)2 + NH3(xs)
– 11. K2Zn(OH)4 + HNO3(xs)
19-63

64. Writing Net Ionic Equations
– 12. AgCl + NH3(xs)
– 13. Ni(CH3COO)2 + H2O
– 14. Cr(OH)3 + NaOH(xs)
19-64

65. 19-65

66. 19-66