Solubility Equilibrium

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) + 2 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) + 2 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) + 3 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. 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) + 2 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 - 1.0 0.0 0.0 • [Change] - -2S +S +S • [Equilm.] - (1 – 2S) SS 

18. Solubility Exercise #1 • Solubility of AgCl in 1.0 NH3 (continued): • S = 0.052 – 0.104S; • S = 0.052/1.104 = 0.047 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 0.025 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 0.025 M)/(50.0 mL) = 0.010 M • [Cl-] = (30.0 mL x 0.10 M)/(50.0 mL) = 0.060 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 (a) Calculate the solubility of silver chloride in 10.0 M ammonia given the following information: Ksp (AgCl) = 1.6 x 10–10 Ag+ + NH3 AgNH3+K = 2.1 x 103 AgNH3+ + NH3 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