1. Chapter 16 Precipitation Equilibria

2. Outline
1. Precipitate formation: the solubility product constant (Ksp)
2. Dissolving precipitates

3. Revisiting 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

4. 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)

5. Precipitate Formation: 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)

6. Figure 16.1 – Precipitation of SrCrO4

7. 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

8. 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 °C
The product of the two concentrations at equilibrium must have this value regardless of the direction from which equilibrium is approached

9. Example 16.1

10. 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

11. Example 16.2

12. Example 16.1, (Cont’d)

13. Table 16.1

14. 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

15. 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

16. Figure 16.2

17. Example 16.3

18. Example 16.3, (Cont’d)

19. Example 16.3, (Cont’d)

20. 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:

21. Precipitation Visualized

22. Example 16.4

23. Example 16.4, (Cont'd)

24. Example 16.4, (Cont'd)

25. 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

26. Example 16.5

27. Example 16.5, (Cont'd)

28. 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 0.10 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

29. Visualizing the Common Ion Effect

30. Example 16.6

31. Example 16.6, (Cont'd)

32. 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

33. 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

34. Figure 16.3

35. Figure 16.4 – Selective Precipitation

36. Example 16.7

37. Example 16.7, (Cont'd)

38. 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

39. 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
Zn(OH)2 (s) + 2H+ (aq) ⇌ Zn2+ (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

40. Example 16.8

41. Example 16.8, (Cont'd)

42. Example 16.8, (Cont'd)

43. Insoluble Compounds that Dissolve in Strong Acid
Virtually all carbonates
The product if 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)

44. Visualizing Selective Dissolving of Precipitates

45. Example 16.9

46. 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

47. Table 16.2

48. Visualizing Dissolving by Complex Formation

49. Example 16.10

50. Example 16.10, (Cont'd)

51. Example 16.11

52. Key Concepts 1. Write the Ksp expression for an 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

53. 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