1. Solubility Equilibrium
In saturated solutions dynamic equilibrium exists between undissolved solids and ionic species in solutions
Solids continue to dissolve and ion-pairs continue to form solids.
The rate of dissolution process is equal to the rate of precipitation.

2. Solubility Product Constant
General expression:
MmXn(s) ⇄ mMn+(aq) + nXm-(aq)
Solubility product, Ksp = [Mn+]m[Xm-]n

3. Solubility and Solubility Products
Examples:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
Ksp = [Ag+][Cl-] = 1.6 x 10-10
If s is the solubility of AgCl, then:
[Ag+] = s and [Cl-] = s
Ksp = (s)(s) = s2 = 1.6 x 10-10
s = 1.3 x 10-5 mol/L

4. Solubility and Solubility Products
Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO42-(aq)
Ksp = [Ag+]2[CrO42-] = 9.0 x 10-12
If s is the solubility of Ag2CrO4, then:
[Ag+] = 2s and [CrO42-] = s
Ksp = (2s)2(s) = 4s3 = 9.0 x 10-12
s = 1.3 x 10-4 mol/L

5. Solubility and Solubility Products
More Examples:
Ca(IO3)2(s) ⇌ Ca2+(aq) IO3-(aq)
Ksp = [Ca2+][IO3-]2 = 7.1 x 10-7
If the solubility of Ca(IO3)2(s) is s mol/L, then:
Ksp = 4s3 = 7.1 x 10-7
s = 5.6 x 10-3 mol/L

6. Solubility and Solubility Products
Mg(OH)2(s) ⇌ Mg2+(aq) OH-(aq)
Ksp = [Mg2+][OH-]2 = 8.9 x 10-12
If the solubility of Mg(OH)2 is s mol/L, then:
[Mg2+] = s mol/L and [OH-] = 2s mol/L,
Ksp = (s)(2s)2 = 4s3 = 8.9 x 10-12
s = 1.3 x 10-4 mol/L

7. Solubility and Solubility Products
More Examples:
Ag3PO4(s) ⇌ 3Ag+(aq) + PO43-(aq)
Ksp = [Ag+]3[PO43-] = 1.8 x 10-18
If the solubility of Ag3PO4 is s mol/L, then:
Ksp = (3s)3(s) = 27s4 = 1.8 x 10-18
s = 1.6 x 10-5 mol/L

8. Solubility and Solubility Products
Cr(OH)3(s) ⇌ Cr3+(aq) OH-(aq)
Ksp = [Cr3+][OH-]3 = 6.7 x 10-31
If the solubility is s mol/L, then:
Ksp = [Cr3+][OH-]3 = (s)(3s)3 = 27s4 = 6.7 x 10-31
s = 1.3 x 10-8 mol/L

9. Solubility and Solubility Products
More Examples:
Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO43-(aq)
Ksp = [Ca2+]3[PO43-]2 = 1.3 x 10-32
If the solubility is s mol/L, then:
[Ca2+] = 3s, and [PO43-] = 2s
Ksp = (3s)3(2s)2 = 108s5 = 1.3 x 10-32
s = 1.6 x 10-7 mol/L

10. Factors that affect solubility
Temperature
Solubility generally increases with temperature;
Common ion effect
Common ions reduce solubility
Salt effect
This slightly increases solubility
pH of solution
pH affects the solubility of ionic compounds in which the anions are conjugate bases of weak acids;
Formation of complex ion
The formation of complex ion increases solubility

11. AgCl(s) ⇌ Ag+(aq) + Cl-(aq); Ksp = 1.6 x 10-10;
Common Ion Effect
Consider the following solubility equilibrium:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq); Ksp = 1.6 x 10-10;
The solubility of AgCl is 1.3 x 10-5 mol/L at 25oC.
If NaCl is added, equilibrium shifts left due to increase in [Cl-] and some AgCl will precipitate out.
For example, if [Cl-] = 1.0 x 10-2 M,
Solubility of AgCl = (1.6 x 10-10)/(1.0 x 10-2)
= 1.6 x 10-8 mol/L

12. Effect of pH on Solubility
Consider the following equilibrium:
Ag3PO4(s) ⇌ 3Ag+(aq) + PO43-(aq);
If HNO3 is added, the following reaction occurs:
H3O+(aq) + PO43-(aq) ⇌ HPO42-(aq) + H2O
This reaction reduces PO43- in solution, causing more solid Ag3PO4 to dissolve.
In general, the solubility of compounds such as Ag3PO4, which anions are conjugate bases of weak acids, increases as the pH is lowered by adding nitric acid.

13. Effect of pH on Solubility
Consider the following equilibrium:
Mg(OH)2(s) ⇌ Mg2+(aq) OH-(aq);
Increasing the pH means increasing [OH-] and equilibrium will shift to the left, causing some of Mg(OH)2 to precipitate out.
If the pH is lowered, [OH-] decreases and equilibrium shifts to the right, causing solid Mg(OH)2 to dissolve.
The solubility of compounds of the type M(OH)n decreases as pH is increased, and increases as pH is decreased.

14. Formation of Complex Ions on Solubility
Many transition metals ions have strong affinity for ligands to form complex ions.
Ligands are molecules, such as H2O, NH3 and CO, or anions, such as F-, CN- and S2O32-.
Complex ions are soluble – thus, the formation of complex ions increases solubility of slightly soluble ionic compounds.

15. Effect of complex ion formation on solubility
Consider the following equilibria:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq); Ksp = 1.6 x 10-10
Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2+(aq) ; Kf = 1.7 x 107
Combining the two equations yields:
AgCl(s) + 2NH3(aq) ⇌ Ag(NH3)2+(aq) + Cl-(aq);
Knet = Ksp x Kf = (1.6 x 10-10) x (1.7 x 107)
= 2.7 x 10-3
Knet > Ksp implies that AgCl is more soluble in aqueous NH3 than in water.

16. Solubility Exercise #1
Calculate the solubility of AgCl in water and in 1.0 M NH3 solution at 25oC.
Solutions:
Solubility in water = (Ksp)
= (1.6 x 10-10) = 1.3 x 10-5 mol/L

17. Solubility Exercise #1 Solubility of AgCl in 1.0 NH3:
AgCl(s) + 2NH3(aq) ⇌ Ag(NH3)2+(aq) + Cl-(aq)

[Initial], M
[Change] S S S
[Equilm.] (1 – 2S) S S

18. Solubility Exercise #1 Solubility of AgCl in 1.0 NH3 (continued):
S = – 0.104S;
S = 0.052/1.104 = mol/L
AgCl is much more soluble in NH3 solution than in water.

19. Predicting Formation of Precipitate
Qsp = Ksp  saturated solution, but no precipitate
Qsp > Ksp  saturated solution, with precipitate
Qsp < Ksp  unsaturated solution,
Qsp is ion product expressed in the same way as Ksp for a particular system.

20. Predicting Precipitation
Consider the following case:
20.0 mL of M Pb(NO3)2 is added to 30.0 mL of 0.10 M NaCl. Predict if precipitate of PbCl2 will form.
(Ksp for PbCl2 = 1.6 x 10-5)

21. Predicting Precipitation
Calculation:
[Pb2+] = (20.0 mL x M)/(50.0 mL) = M
[Cl-] = (30.0 mL x 0.10 M)/(50.0 mL) = M
Qsp = [Pb2+][Cl-]2 = (0.010 M)(0.060 M)2
= 3.6 x 10-5
Qsp > Ksp  precipitate of PbCl2 will form.

22. Practical Applications of Solubility Equilibria
Qualitative Analyses
Isolation and identification of cations and/or anions in unknown samples
Synthesis of Ionic Solids of commercial interest
Selective Precipitation based on Ksp

23. Qualitative Analysis
Separation and identification of cations, such as Ag+, Ba2+, Cr3+, Fe3+, Cu2+, etc. can be carried out based on their different solubility and their ability to form complex ions with specific reagents, such as HCl, H2SO4, NaOH, NH3, and others.
Separation and identification of anions, such as Cl-, Br-, I-, SO42-, CO32-, PO43-, etc., can be accomplished using reagents such as AgNO3, Ba(NO3)2 under neutral or acidic conditions.

24. Selective Precipitation (Mixtures of Metal Ions)
Use a reagent whose anion forms a precipitate with only one or a few of the metal ions in the mixture.
Example:
Solution contains Ba2+ and Ag+ ions.
Adding NaCl will form a precipitate with Ag+ (AgCl), while still leaving Ba2+ in solution.

25. Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S
At a low pH, [S2–] is relatively low and only the very insoluble HgS and CuS precipitate.
When OH– is added to lower [H+], the value of [S2–] increases, and MnS and NiS precipitate.

26. Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S

27. Separating the Common Cations by Selective Precipitation

28. Synthesis of Ionic Solids
Chemicals such as AgCl, AgBr, and AgI that are important in photography are prepared by precipitation method.
AgNO3(aq) + KBr(aq)  AgBr(s) + KNO3(aq)

29. Selective Precipitation
Compounds with different solubility can be selectively precipitated by adjusting the concentration of the precipitating reagents.
For example, AgCl has a much lower Ksp than PbCl2
If Ag+ and Pb2+ are present in the same solution, the Ag+ ion can be selectively precipitated as AgCl, leaving Pb2+ in solution.

30. Complex Ion Equilibria
Complex ions are ions consisting central metal ions and ligands covalently bonded to the metal ions;
Ligands can be neutral molecules such as H2O, CO, and NH3, or anions such as Cl-, F-, OH-, and CN-;
For example, in the complex ion [Cu(NH3)4]2+, four NH3 molecules are covalently bonded to Cu2+.

31. Formation of Complex Ions
In aqueous solutions, metal ions form complex ions with water molecules as ligands.
If stronger ligands are present, ligand exchanges occur and equilibrium is established.
For example:
Cu2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4]2+(aq)

32. Stepwise Formation of Complex Ion
At molecular level, ligand molecules or ions combine with metal ions in stepwise manner;
Each step has its equilibrium and equilibrium constant;
For example:
(1) Ag+(aq) + NH3(aq) ⇌ Ag(NH3)+(aq)
(2) Ag(NH3)+(aq) + NH3(aq) ⇌ Ag(NH3)2+(aq);

33. Stepwise Formation of Complex Ion
Individual equilibrium steps:
(1) Ag+(aq) + NH3(aq) ⇌ Ag(NH3)+(aq); Kf1 = 2.1 x 103
(2) Ag(NH3)+(aq) + NH3(aq) ⇌ Ag(NH3)2+(aq); Kf2 = 8.2 x 103
Combining (1) and (2) yields:
Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2+(aq);

34. Stepwise complex ion formation for Cu(NH3)42+
Individual equilibrium steps:
Cu2+(aq) + NH3(aq) ⇌ Cu(NH3)2+(aq); K1 = 1.9 x 104
Cu(NH3)2+(aq) + NH3(aq) ⇌ Cu(NH3)22+(aq); K2 = 3.9 x 103
Cu(NH3)22+(aq) + NH3(aq) ⇌ Cu(NH3)32+(aq); K3 = 1.0 x 103
Cu(NH3)32+(aq) + NH3(aq) ⇌ Cu(NH3)42+(aq); K4 = 1.5 x 102
Combining equilibrium:
Cu2+(aq) + 4NH3(aq) ⇌ Cu(NH3)42+(aq);
Kf = K1 x K2 x K3 x K4 = 1.1 x 1013

35. Complex Ions and Solubility
Two strategies for dissolving a water–insoluble ionic solid.
If the anion of the solid is a good base, the solubility is greatly increased by acidifying the solution.
In cases where the anion is not sufficiently basic, the ionic solid often can be dissolved in a solution containing a ligand that forms stable complex ions with its cation.

36. Concept Check Ksp (AgCl) = 1.6 x 10–10 Ag+ + NH3 AgNH3+ K = 2.1 x 103
(a) Calculate the solubility of silver chloride in 10.0 M ammonia given the following information:
Ksp (AgCl) = 1.6 x 10–10
Ag+ + NH AgNH K = 2.1 x 103
AgNH3+ + NH Ag(NH3)2+ K = 8.2 x 103
(b) Calculate the concentration of NH3 in the final equilibrium mixture.
Answers: (a) 0.48 M; (b) 9.0 M
a) 0.48 M
b) 9.0 M
This problem is discussed at length in the text.