Solubility and Complex-Ion Equilibria Chapter 18 Solubility and Complex-Ion Equilibria John A. Schreifels Chemistry 212

Overview Solubility Equilibria Complex-Ion 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 John A. Schreifels Chemistry 212

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. John A. Schreifels Chemistry 212

PbCl2(s)  Pb2+(aq) + 2Cl(aq) 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 John A. Schreifels Chemistry 212

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. John A. Schreifels Chemistry 212

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. assume complete precipitation occurs; set up equilibrium table; and solve for equilibrium concentration of barium and carbonate ion concentrations. John A. Schreifels Chemistry 212

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. John A. Schreifels Chemistry 212

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) John A. Schreifels Chemistry 212

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. John A. Schreifels Chemistry 212

Ag+(aq) + 2NH3(aq)  Ag(NH3)2(aq) Kf = 1.7x107 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. John A. Schreifels Chemistry 212

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. John A. Schreifels Chemistry 212

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. John A. Schreifels Chemistry 212