Chapter 18

1. Chapter 18 Solubility and Complex-Ion Equilibria

2. Overview • Solubility Equilibria • Solubility Product Constant • Solubility and Common Ion Effect • Precipitation Calculations • Effect of pH on Solubility • Complex-Ion Equilibria • Complex Ion Formation • Complex Ions and Solubility • Application of Solubility Equilibria • Qualitative analysis of metal ions

3. Solubility Equilibria • Solubility of a solid treated as with other equilibria. Solution is saturated. No more solid will dissolve since dynamic equilibrium. AgCl(s)  Ag+(aq) + Cl(aq) Ksp = [Ag+][Cl] • Solid not included in the equilibrium expression. • MyXz(s)  yM+p(aq)+zXq(aq) Ksp=[M+p]y[Xq]z where Ksp = solubility product. E.g. determine the equilibrium expression for each: PbCl2, Ag2SO4,Al(OH)3. • Ksp can be determined if the solubility is known. E.g. Determine Ksp for silver chromate (Ag2CrO4) if its solubility in water is 0.0290 g/L at 25C. • Determine molar solubility. • Determine Ksp. E.g. 2 Determine Ksp of CaF2 if its solubility is 2.20x104M.

4. SOLUBILITY FROM Ksp • Can be determined by using stoichiometry to express all quantities in terms of one variable- solubility, x. i.e. for the reaction. Use equilibrium table to write concentration of each in terms of the compound dissolving E.g. determine the solubility of PbCl2 if its Ksp = 1.2x105 PbCl2(s)  Pb2+(aq) + 2Cl(aq) E.g.1 Determine solubility of AgCl if its Ksp = 1.8x1010M2. E.g.2 Determine solubility of Ag2CO3 if its Ksp = 8.1x1012M3. E.g.3 Determine solubility of Fe(OH)3 if its Ksp = 4x1038M4

5. Factors that Affect Solubility • The common–Ion effect(Remember LeChatelier’s Principle) E.g. Determine solubility of PbCl2 (Ksp = 1.2x105)in 0.100M NaCl. • Write equilibrium table in terms of x and [Cl] a common–ion reduces the solubility of the compound. • Assume that [Cl]NaCl >>x • Solve for x. • E.g. determine the solubility of CaF2 in a solution of CaCl2. Ksp = 3.9x1011.

6. Precipitation of Ionic Compounds • Starting with two solutions, Qsp used to predict precipitation and even the extent of it. • Precipitation = reverse of dissolution • Precipitation occurs when Qsp > Ksp until Qsp = Ksp • If Qsp < Ksp, precipitation won’t occur. E.g. determine if precipitation occurs after mixing 50.00 mL 3.00x103 M BaCl2 and 50.00 mL 3.00x103 M Na2CO3. Solution: • CBaCl2 = 1.50x103 M; CNa2CO3 = 1.50x103 M • Qsp = 1.50x103 M1.50x103 M = 2.25x106 • Qsp >1.1x1010.= Ksp precipitation. E.g. 2 determine equilibrium concentration of each after precipitation occurs. Solution: • assume complete precipitation occurs; • set up equilibrium table; and solve for equilibrium concentration of barium and carbonate ion concentrations.

7. Precipitation of Ionic Compounds Eg. 3 determine the fraction of Ba2+ that has precipitated. Solution: • Use the amount remaining in solution (results of E.g. 2) divided by starting concentration to determine the fraction of barium that is left in solution. • Subtract from above. E.g.4 determine the Br concentration when AgCl starts to precipitate if the initial concentration of bromide and chloride are 0.100 M. Ksp(AgBr) = 5.0x1013; Ksp(AgCl) = 1.8x1010.

8. Factors that Affect Solubility-pH • pH of the Solution: LeChatelier’s Principle again. E.g. determine the solubility of CaF2 at a pH of 2.00. Ksp = 3.9x1011. Ka(HF) = 6.6x104. Strategy: • Determine the ratio of [F] and [HF] from the pH and Ka. • Write an expression for solubility in terms of Ka and pH and • Substitute into solubility equation to determine the solubility. Solution: • Ksp = 3.9x1011 = x[F]2 (pH changes the amount of free Fluoride.) • Let x = solubility. Then 2x = [F] + [HF] • From equilibrium equation: • 2x = [F](1+1/0.066) = 16.15*[F] or • [F] = 2*x/16.15 = 0.124*x • 3.9x1011 = x(0.0124*x)2 • x = 1.36x103 M vs. 2.13x104 M (normal solubility)

9. Separation of Ions By Selective Precipitation • Metal ions with very different Ksp can be separated. • Divalent metal ions are often separated using solubility variations for the metal sulfides. • Solution is saturated with H2S at 0.100 M; pH adjusted to keep one component soluble and the other insoluble. • H2S is diprotic acid; the overall reaction to get to sulfide is: • Combine with solubility equilibrium reaction to get the overall equilibrium expression and constant. E.g. determine the solubility of 0.00500 M Zn2+ in 0.100 M H2S at pH = 1. Ksp = 1.10x1021.

10. Complex Ions • Formation of Complex Ions (Coordination Complexation ) = an ion formed from a metal ion with a Lewis base attached to it by a coordinate covalent bond. Ag+(aq) + 2NH3(aq)  Ag(NH3)2(aq) Kf = 1.7x107 • Large equilibrium constant indicates that “free” metal is completely converted to the complex. Eg. What is the concentration of the silver amine complex above in a solution that is originally 0.100 M Ag+ and 1.00 M NH3? E.g. determine the [Ag+] (free silver concentration) in 0.100 M AgNO3 that is also 1.00 M NaCN.

11. Factors that Affect Solubility: Complexation • Free metal ion concentration in solution is reduced when complexing agent added to it; • Free metal ion concentration needed in solubility expression. E.g. determine if precipitation will occur in a solution containing 0.010 M AgNO3 and 0.0100 M Nal in 1.00 M NaCN. Recall Kf = 5.6x1018 Agl(s)Ag+ + l Ksp = 8.5.x1017 Strategy: • Determine the free metal concentration in the solution. • Use free metal concentration with iodide concentration to get Qsp • If Qsp < Ksp, no precipitation • If Qsp > Ksp, precipitation • If Qsp = Ksp, precipitation is starting.

12. Solubility with Complexing Agent E.g. Determine the solubility of AgI in 1.00 M NaCN. Recall Kf = 5.6x1018 Agl(s)Ag+ + l Ksp = 8.5.x1017 Strategy: • Combine to equilibria equations to find a single equation describing the equilibrium. the presence of a complexing agent increases the solubility • Setup equilibrium table and solve.