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: M m X n (s) ⇄ mM n+ (aq) + nX m- (aq) Solubility product, K sp = [M n+ ] m [X m- ] n

3. Solubility and Solubility Products Examples: AgCl (s) ⇌ Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] = 1.6 x 10 -10 If s is the solubility of AgCl, then: [Ag + ] = s and [Cl - ] = s K sp = (s)(s) = s 2 = 1.6 x 10 -10 s = 1.3 x 10 -5 mol/L

4. Solubility and Solubility Products Ag 2 CrO 4 (s) ⇌ 2Ag + (aq) + CrO 4 2- (aq) K sp = [Ag + ] 2 [CrO 4 2- ] = 9.0 x 10 -12 If s is the solubility of Ag 2 CrO 4, then: [Ag + ] = 2s and [CrO 4 2- ] = s K sp = (2s) 2 (s) = 4s 3 = 9.0 x 10 -12 s = 1.3 x 10 -4 mol/L

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

6. Solubility and Solubility Products Mg(OH) 2 (s) ⇌ Mg 2+ (aq) + 2 OH - (aq) K sp = [Mg 2+ ][OH - ] 2 = 8.9 x 10 -12 If the solubility of Mg(OH) 2 is s mol/L, then: [Mg 2+ ] = s mol/L and [OH - ] = 2s mol/L, K sp = (s)(2s) 2 = 4s 3 = 8.9 x 10 -12 s = 1.3 x 10 -4 mol/L

7. Solubility and Solubility Products More Examples: Ag 3 PO 4 (s) ⇌ 3Ag + (aq) + PO 4 3- (aq) K sp = [Ag + ] 3 [PO 4 3- ] = 1.8 x 10 -18 If the solubility of Ag 3 PO 4 is s mol/L, then: K sp = (3s) 3 (s) = 27s 4 = 1.8 x 10 -18 s = 1.6 x 10 -5 mol/L

8. Solubility and Solubility Products Cr(OH) 3 (s) ⇌ Cr 3+ (aq) + 3 OH - (aq) K sp = [Cr 3+ ][OH - ] 3 = 6.7 x 10 -31 If the solubility is s mol/L, then: K sp = [Cr 3+ ][OH - ] 3 = (s)(3s) 3 = 27s 4 = 6.7 x 10 -31 s = 1.3 x 10 -8 mol/L

9. Solubility and Solubility Products More Examples: Ca 3 (PO 4 ) 2 ( s ) ⇌ 3Ca 2+ (aq) + 2PO 4 3- (aq) K sp = [Ca 2+ ] 3 [PO 4 3- ] 2 = 1.3 x 10 -32 If the solubility is s mol/L, then: [Ca 2+ ] = 3s, and [PO 4 3- ] = 2s K sp = (3s) 3 (2s) 2 = 108s 5 = 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) ; K sp = 1.6 x 10 -10 ; The solubility of AgCl is 1.3 x 10 -5 mol/L at 25 o C. 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: Ag 3 PO 4 (s) ⇌ 3Ag + (aq) + PO 4 3- (aq) ; If HNO 3 is added, the following reaction occurs: H 3 O + (aq) + PO 4 3- (aq) ⇌ HPO 4 2- (aq) + H 2 O This reaction reduces PO 4 3- in solution, causing more solid Ag 3 PO 4 to dissolve. In general, the solubility of compounds such as Ag 3 PO 4, 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) ⇌ Mg 2+ (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 H 2 O, NH 3 and CO, or anions, such as F -, CN - and S 2 O 3 2-. 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) ; K sp = 1.6 x 10 -10 Ag + (aq) + 2NH 3 (aq) ⇌ Ag(NH 3 ) 2 + ( aq) ; K f = 1.7 x 10 7 Combining the two equations yields: AgCl (s) + 2NH 3 (aq) ⇌ Ag(NH 3 ) 2 + ( aq) + Cl - (aq) ; K net = K sp x K f = (1.6 x 10 -10 ) x (1.7 x 10 7 ) = 2.7 x 10 -3 K net > K sp implies that AgCl is more soluble in aqueous NH 3 than in water.

16. Solubility Exercise #1 Calculate the solubility of AgCl in water and in 1.0 M NH 3 solution at 25 o C. Solutions: Solubility in water =  (K sp ) =  (1.6 x 10 -10 ) = 1.3 x 10 -5 mol/L

17. Solubility Exercise #1 Solubility of AgCl in 1.0 NH 3 : AgCl (s) + 2NH 3 (aq) ⇌ Ag(NH 3 ) 2 + (aq) + Cl - (aq)  [Initial], M - 1.0 0.0 0.0 [Change] - -2S +S +S [Equilm.] - (1 – 2S) S S 

18. Solubility Exercise #1 Solubility of AgCl in 1.0 NH 3 (continued): S = 0.052 – 0.104S; S = 0.052/1.104 = 0.047 mol/L AgCl is much more soluble in NH 3 solution than in water.

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

20. Predicting Precipitation Consider the following case: 20.0 mL of 0.025 M Pb(NO 3 ) 2 is added to 30.0 mL of 0.10 M NaCl. Predict if precipitate of PbCl 2 will form. (Ksp for PbCl 2 = 1.6 x 10-5)

21. Predicting Precipitation Calculation: [Pb 2+ ] = (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 Q sp = [Pb 2+ ][Cl - ] 2 = (0.010 M)(0.060 M) 2 = 3.6 x 10-5 Q sp > K sp  precipitate of PbCl 2 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 K sp

23. Qualitative Analysis Separation and identification of cations, such as Ag +, Ba 2+, Cr 3+, Fe 3+, Cu 2+, etc. can be carried out based on their different solubility and their ability to form complex ions with specific reagents, such as HCl, H 2 SO 4, NaOH, NH 3, and others. Separation and identification of anions, such as Cl -, Br -, I -, SO 4 2-, CO 3 2-, PO 4 3-, etc., can be accomplished using reagents such as AgNO 3, Ba(NO 3 ) 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 Ba 2+ and Ag + ions.  Adding NaCl will form a precipitate with Ag + (AgCl), while still leaving Ba 2+ in solution.

25. Separation of Cu 2+ and Hg 2+ from Ni 2+ and Mn 2+ using H 2 S At a low pH, [S 2– ] is relatively low and only the very insoluble HgS and CuS precipitate. When OH – is added to lower [H + ], the value of [S 2– ] increases, and MnS and NiS precipitate.

26. Separation of Cu 2+ and Hg 2+ from Ni 2+ and Mn 2+ using H 2 S

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. AgNO 3 (aq) + KBr (aq)  AgBr (s) + KNO 3 (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 K sp than PbCl 2 If Ag + and Pb 2+ are present in the same solution, the Ag + ion can be selectively precipitated as AgCl, leaving Pb 2+ 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 H 2 O, CO, and NH 3, or anions such as Cl -, F -, OH -, and CN - ; For example, in the complex ion [Cu(NH 3 ) 4 ] 2+, four NH 3 molecules are covalently bonded to Cu 2+.

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: Cu 2+ (aq) + 4NH 3 (aq) ⇌ [Cu(NH 3 ) 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) + NH 3 (aq) ⇌ Ag(NH 3 ) + (aq) (2) Ag(NH 3 ) + (aq) + NH 3 (aq) ⇌ Ag(NH 3 ) 2 + (aq) ;

33. Stepwise Formation of Complex Ion Individual equilibrium steps: (1) Ag + (aq) + NH 3 (aq) ⇌ Ag(NH 3 ) + (aq) ; K f1 = 2.1 x 10 3 (2) Ag(NH 3 ) + (aq) + NH 3 (aq) ⇌ Ag(NH 3 ) 2 + (aq) ; K f2 = 8.2 x 10 3 Combining (1) and (2) yields: Ag + (aq) + 2NH 3 (aq) ⇌ Ag(NH 3 ) 2 + (aq) ;

34. Stepwise complex ion formation for Cu(NH 3 ) 4 2+ Individual equilibrium steps: 1.Cu 2+ (aq) + NH 3 (aq) ⇌ Cu(NH 3 ) 2+ (aq) ; K 1 = 1.9 x 10 4 2.Cu(NH 3 ) 2+ (aq) + NH 3 (aq) ⇌ Cu(NH 3 ) 2 2+ (aq) ; K 2 = 3.9 x 10 3 3.Cu(NH 3 ) 2 2+ (aq) + NH 3 (aq) ⇌ Cu(NH 3 ) 3 2+ (aq) ; K 3 = 1.0 x 10 3 4.Cu(NH 3 ) 3 2+ (aq) + NH 3 (aq) ⇌ Cu(NH 3 ) 4 2+ (aq) ; K 4 = 1.5 x 10 2 Combining equilibrium: Cu 2+ (aq) + 4NH 3 (aq) ⇌ Cu(NH 3 ) 4 2+ (aq) ; K f = K 1 x K 2 x K 3 x K 4 = 1.1 x 10 13

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: K sp (AgCl) = 1.6 x 10 –10 Ag + + NH 3 AgNH 3 + K = 2.1 x 10 3 AgNH 3 + + NH 3 Ag(NH 3 ) 2 + K = 8.2 x 10 3 (b) Calculate the concentration of NH 3 in the final equilibrium mixture. Answers: (a) 0.48 M ;(b) 9.0 M