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8–1 John A. Schreifels Chemistry 212 Chapter 18-1 Chapter 18 Solubility and Complex-Ion Equilibria

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8–2 John A. Schreifels Chemistry 212 Chapter 18-2 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

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8–3 John A. Schreifels Chemistry 212 Chapter 18-3 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)K sp = [Ag + ][Cl ] Solid not included in the equilibrium expression. –M y X z (s) yM +p (aq)+zX q (aq) K sp =[M +p ] y [X q ] z where K sp = solubility product. E.g. determine the equilibrium expression for each: PbCl 2, Ag 2 SO 4,Al(OH) 3. K sp can be determined if the solubility is known. E.g. Determine K sp for silver chromate (Ag 2 CrO 4 ) if its solubility in water is 0.0290 g/L at 25 C. –Determine molar solubility. –Determine Ksp. E.g. 2 Determine K sp of CaF 2 if its solubility is 2.20x10 4 M.

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8–4 John A. Schreifels Chemistry 212 Chapter 18-4 SOLUBILITY FROM K sp 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 PbCl 2 if its K sp = 1.2x10 5 PbCl 2 (s) Pb 2+ (aq) + 2Cl (aq) E.g.1 Determine solubility of AgCl if its K sp = 1.8x10 10 M 2. E.g.2 Determine solubility of Ag 2 CO 3 if its K sp = 8.1x10 12 M 3. E.g.3 Determine solubility of Fe(OH) 3 if its K sp = 4x10 38 M 4

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8–5 John A. Schreifels Chemistry 212 Chapter 18-5 Factors that Affect Solubility The common–Ion effect (Remember LeChateliers Principle) E.g. Determine solubility of PbCl 2 (K sp = 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 [C l ] NaCl >>x –Solve for x. E.g. determine the solubility of CaF2 in a solution of CaCl2. Ksp = 3.9x10 11.

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8–6 John A. Schreifels Chemistry 212 Chapter 18-6 Precipitation of Ionic Compounds Starting with two solutions, Q sp used to predict precipitation and even the extent of it. –Precipitation = reverse of dissolution –Precipitation occurs when Q sp > K sp until Q sp = K sp –If Qsp < Ksp, precipitation wont occur. E.g. determine if precipitation occurs after mixing 50.00 mL 3.00x10 3 M BaCl 2 and 50.00 mL 3.00x10 3 M Na 2 CO 3. Solution: –C BaCl2 = 1.50x10 3 M; C Na2CO3 = 1.50x10 3 M –Qsp = 1.50x10 3 M 1.50x10 3 M = 2.25x10 6 –Q sp >1.1x10 10.= K sp 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.

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8–7 John A. Schreifels Chemistry 212 Chapter 18-7 Precipitation of Ionic Compounds Eg. 3 determine the fraction of Ba 2+ 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. K sp (AgBr) = 5.0x10 13 ; K sp (AgCl) = 1.8x10 10.

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8–8 John A. Schreifels Chemistry 212 Chapter 18-8 Factors that Affect Solubility-pH pH of the Solution: LeChateliers Principle again. E.g. determine the solubility of CaF 2 at a pH of 2.00. K sp = 3.9x10 11. K a (HF) = 6.6x10 4. Strategy: –Determine the ratio of [F ] and [HF] from the pH and K a. –Write an expression for solubility in terms of K a and pH and –Substitute into solubility equation to determine the solubility. Solution: –K sp = 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)

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8–9 John A. Schreifels Chemistry 212 Chapter 18-9 Separation of Ions By Selective Precipitation Metal ions with very different K sp can be separated. Divalent metal ions are often separated using solubility variations for the metal sulfides. Solution is saturated with H 2 S 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 Zn 2+ in 0.100 M H 2 S at pH = 1. K sp = 1.10x10 21.

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8–10 John A. Schreifels Chemistry 212 Chapter 18-10 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) + 2NH 3 (aq) Ag(NH 3 ) 2 (aq) K f = 1.7x10 7 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 NH 3 ? E.g. determine the [Ag + ] (free silver concentration) in 0.100 M AgNO 3 that is also 1.00 M NaCN.

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8–11 John A. Schreifels Chemistry 212 Chapter 18-11 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 AgNO 3 and 0.0100 M Nal in 1.00 M NaCN. Recall K f = 5.6x10 18 Agl(s) Ag + + l K sp = 8.5.x10 17 Strategy: –Determine the free metal concentration in the solution. –Use free metal concentration with iodide concentration to get Q sp If Q sp < K sp, no precipitation If Q sp > K sp, precipitation If Q sp = K sp, precipitation is starting.

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8–12 John A. Schreifels Chemistry 212 Chapter 18-12 Solubility with Complexing Agent E.g. Determine the solubility of AgI in 1.00 M NaCN. Recall K f = 5.6x10 18 Agl(s) Ag + + l K sp = 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.

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