1. Chapter 15 Complex Ion and Precipitation Equilibria

2. Outline Complex ion equilibria; formation constant (Kf)
Solubility; solubility product constant (Ksp)
Precipitate formation
Dissolving precipitates

3. Complex Ions Recall (chapter 13) that metal ions are Lewis acids
Compounds with atoms containing lone pairs are Lewis bases
Metal ions can combine with molecules or ions to produce complex ions

4. Copper ion and ammonia Cu2+ (aq) + 4NH3 (aq) ⇌ Cu(NH3)42+ (aq)
The equilibrium constant for the formation of the complex ion is called the formation constant, Kf
For this complex ion, Kf = 2 X 1012, so the reaction goes essentially to completion

5. Table 15.1

6. Example 15.1

7. Example 15.1, (Cont’d)

8. Example 15.1, (Cont’d)

9. Example 15.1, (Cont’d)

10. Two coordination complexes of Co3+
On the left: [Co(NH3)63+]
On the right: [Co(NH3)5Cl2+]

11. Revisiting Solubility and Precipitation
In Chapter 4 we learned that there are compounds that do not dissolve in water
These were called insoluble
A reaction that produces an insoluble precipitate was assumed to go to completion
In reality, even insoluble compounds dissolve to some extent, usually small
An equilibrium is set up between the precipitate and its ions
Precipitates can be dissolved by forming complex ions

12. Two Types of Equilibria
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
Solid exists in equilibrium with the ions formed when a small amount of solid dissolves
AgCl (s) + 2NH3 (aq) ⇌ Ag(NH3)2+ (aq) + Cl- (aq)
Formation of a stable complex ion can cause an otherwise insoluble compound to dissolve
There are multiple equilibria at work in this example, in similar fashion to the equilibria underlying the function of a buffer (Chapter 14)

13. Solubility Product Constant, Ksp
Consider mixing two solutions:
Sr(NO3)2 (aq)
K2CrO4 (aq)
The following net ionic equation describes the reaction:
Sr2+ (aq) + CrO42- (aq)  SrCrO4 (s)

14. Figure 15.3 – A precipitate of SrCrO4

15. Ksp Expression SrCrO4 (s) ⇌ Sr2+ (aq) + CrO42- (aq)
The solid establishes an equilibrium with its ions once it forms
We can write an equilibrium expression, leaving out the term for the solid (recall that its concentration does not change as long as some is present)
Ksp is called the solubility product constant

16. Interpreting the Solubility Expression and Ksp
Ksp has a fixed value at a given temperature
For strontium chromate, Ksp = 3.6 X 10-5 at 25 ºC
The product of the two concentrations at equilibrium must have this value regardless of the direction from which equilibrium is approached

17. Example 15.2

18. Ksp and the Equilibrium Concentration of Ions
Ksp SrCrO4 = [Sr2+][CrO42-] = 3.6 X 10-5
This means that if we know one ion concentration, the other one can easily be calculated
If [Sr2+] = 1.0 X 10-4 M, then
If [CrO42-] = 2.0 X 10-3, then

19. Example 15.3

20. Example 15.3, (Cont’d)

21. Example 15.3, (Cont’d)

22. Table 15.2

23. Ksp and Water Solubility
One way to establish a solubility equilibrium
Stir a slightly soluble solid with water
An equilibrium is established between the solid and its ions
BaSO4 (s) ⇌ Ba2+ (aq) + SO42- (aq)
If we set the concentration of the ions equal to a variable, s:

24. Example 15.4

25. Example 15.4, (Cont’d)

26. Example 15.4, (Cont’d)

27. Calculating Ksp Given Solubility
Instead of calculating solubility from Ksp, it is possible to calculate Ksp from the solubility
Recall that solubility may be given in many different sets of units
Convert the solubility to moles per liter for use in the Ksp expression

28. Example 15.5

29. Ksp and the Common Ion Effect
BaSO4 (s) ⇌ Ba2+ (aq) + SO42- (aq)
How would you expect the solubility of barium sulfate in water to compare to its solubility in M Na2SO4?
Solubility must be less than it is in pure water
Recall LeChâtelier’s Principle
The presence of the common ion, SO42-, will drive the equilibrium to the left
Common ions reduce solubility

30. Visualizing the Common Ion Effect

31. Figure 15.2

32. Example 15.6

33. Example 15.6, (Cont’d)

34. Ksp and Precipitate Formation
Ksp values can be used to predict whether a precipitate will form when two solutions are mixed
Recall the use of Q, the reaction quotient, from Chapter 12
We can calculate Q at any time and compare it to Ksp
The relative magnitude of Q vs. Ksp will indicate whether or not a precipitate will form

35. Figure 15.3: Precipitation of strontium chromate

36. Q and Ksp
If Q > Ksp, a precipitate will form, decreasing the ion concentrations until equilibrium is established
If Q < Ksp, the solution is unsaturated; no precipitate will form
If Q = Ksp, the solution is saturated just to the point of precipitation

37. Figure 15.4

38. Example 15.7

39. Example 15.7, (Cont’d)

40. Example 15.7, (Cont’d)

41. Example 15.7, (Cont’d)

42. Precipitation Visualized

43. Selective Precipitation
Consider a solution of two cations
One way to separate the cations is to add an anion that precipitates only one of them
This approach is called selective precipitation
Related approach
Consider a solution of magnesium and barium ions

44. Selective Precipitation, (Cont'd)
Ksp BaCO3 = 2.6 X 10-9
Ksp MgCO3 = 6.8 X 10-6
Carbonate ion is added
Since BaCO3 is less soluble than MgCO3, BaCO3 precipitates first, leaving magnesium ion in solution
Differences in solubility can be used to separate cations

45. Figure 15.5 – Selective Precipitation

46. Example 15.8

47. Example 15.8, (Cont'd)

48. Example 15.8, (Cont’d)

49. Dissolving Precipitates
Bringing water-insoluble compounds into solution
Adding a strong acid to react with basic anions
Adding an agent that forms a complex ion to react with a metal cation

50. Strong Acid Zn(OH)2 (s) + 2H+ (aq)  Zn2+ (aq) + 2H2O
This reaction takes place as two equilibria:
Zn(OH)2 (s) ⇌ Zn2+ (aq) + 2OH- (aq)
2H+ (aq) + 2OH- (aq) ⇌ 2H2O
Because the equilibrium constant for the neutralization is so large, the reaction goes essentially to completion
Note that for the second equilibrium, K = (1/Kw)2 = 1 X 1028

51. Example 15.9

52. Example 15.9, (Cont'd)

53. Example 15.9, (Cont'd)

54. Insoluble Compounds that Dissolve in Strong Acid
Virtually all carbonates
The product of the reaction is H2CO3, a weak acid that decomposes to carbon dioxide
H2CO3 (aq)  H2O + CO2 (g)
Many sulfides
The product of the reaction is H2S, a gas that is also a weak acid
H2S (aq) ⇌ H+ (aq) + HS- (aq)

55. Visualizing Selective Dissolving of Precipitates

56. Example 15.10

57. Complex Formation
Ammonia and NaOH can dissolve compounds whose metal cations form complexes with NH3 and OH-
As with the addition of a strong acid, multiple equilibria are at work:
Zn(OH)2 (s) ⇌ Zn2+ (aq) + 2OH- (aq) Ksp
Zn2+ (aq) + 4NH3 (aq) ⇌ Zn(NH3)42+ (aq) Kf
Net: Zn(OH)2 (s) + 4NH3 (aq)  Zn(NH3)42+ (aq) + 2OH- (aq)
Knet = KspKf = 4 X X 3.6 X 108 = 1 X 10-8

58. Table 15.3

59. Visualizing Dissolving by Complex Formation

60. Example 15.11

61. Example 15.11, (Cont'd)

62. Example 15.11, (Cont’d)

63. Example 15.12

64. Key Concepts
Relate Kf to the ratio of concentration of complex ion to metal ion
Write the Ksp expression for any insoluble ionic solid
2. Use the value of Ksp to
A. Calculate the concentration of one ion, knowing the other
B. Determine whether a precipitate will form
C. Calculate the water solubility of a compound
D. Calculate the solubility of a compound in a solution of a common ion
E. Determine which ion will precipitate first

65. Key Concepts 3. Calculate K for
A. Dissolving a metal hydroxide in a strong acid
B. Dissolving a precipitate in a complexing agent
4. Write balanced, net ionic equations to explain why a precipitate dissolves in
A. Strong acid
B. Ammonia or hydroxide solution