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Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, 2007 (John Wiley)

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Presentation on theme: "Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, 2007 (John Wiley)"— Presentation transcript:

1 Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille, Chemistry, 2007 (John Wiley) ISBN: 9 78047081 0866

2 Slide 2/13 e CHEM1002 [Part 2] Dr Michela Simone Weeks 8 – 13 Office Hours: Monday 3-5, Friday 4-5 Room: 412A (or 416) Phone: 93512830 e-mail: michela.simone@sydney.edu.au

3 Slide 3/13 e Solubility Equilibria I Saturated solutions contains solid salts in equilibrium with the maximum amount of the ions in solution that is possible The solubility product (K sp ) is the equilibrium constant for this equilibrium situation The solubility product (K sp ) gives the maximum concentrations of the ions in the solution and the maximum amount of solid that will dissolve The ionic product (Q sp ) has the same form has the solubility product and is used to test whether more solid will dissolve or precipitation will occur  If Q sp < K sp, more ions can enter solution and more solid can dissolve  If Q sp > K sp, precipitation must occur Summary of Last Lecture

4 Slide 4/13 e Solubility Equilibria Lecture 10: Solubility Blackman Chapter 10, Sections 10.4 Lecture 11: Common ion effect Blackman Chapter 10, Sections 10.4

5 Slide 5/13 e PbCl 2 (s) Pb 2+ (aq) + 2Cl - (aq) Reminder: Solubility Equilibria K sp = [Pb 2+ (aq)][Cl - (aq)] 2 solubility product constant If Pb 2+ (aq) and Cl - (aq) are present in the same solution at the same time, their concentrations can never be so large that [Pb 2+ (aq)][Cl - (aq)] 2 is bigger than K sp

6 Slide 6/13 x PbCl 2 (s) Pb 2+ (aq) + 2Cl - (aq) K sp = [Pb 2+ (aq)][Cl - (aq)] 2 solubility product constant If x mol of PbCl 2 (s) dissolves in 1 L of solution, then  [Pb 2+ (aq)] = x mol L -1 and [Cl - ] = 2x mol L -1  K sp = [Pb 2+ (aq)][Cl - (aq)] 2 = (x)(2x) 2 = 4x 3  x = (K sp /4) 1/3 = solubility As K sp = 1.6 x 10 -5,  solubility = (1.6 x 10 -5 /4) 1/3 = 0.016 mol L -1 Solubility

7 Slide 7/13 e PbCl 2 (s) Pb 2+ (aq) + 2Cl - (aq) K sp = [Pb 2+ (aq)][Cl - (aq)] 2 solubility product constant If NaCl is added to the solution, [Cl - (aq)] increases so  equilibrium shifts to the left: less PbCl 2 dissolves  [Pb 2+ (aq)][Cl - (aq)] 2+ must remain constant so [Pb 2+ (aq)] decreases The Common Ion Effect and Solubility

8 Slide 8/13 x PbCl 2 (s) Pb 2+ (aq) + 2Cl - (aq) K sp = [Pb 2+ (aq)][Cl - (aq)] 2 solubility product constant If NaCl is added to the solution, to give [Cl - (aq)] = 0.2 M The Common Ion Effect and Solubility If x mol of PbCl 2 (s) dissolve in 1 L of solution, then  [Pb 2+ (aq)] = x mol L -1 and [Cl - ] = 0.2 mol L -1  K sp = [Pb 2+ (aq)][Cl - (aq)] 2 = (x)(0.2) 2 = 0.04x As K sp = 1.6 x 10 -5,  solubility = 1.6 x 10 -5 /0.04 = 0.0004 mol L -1

9 Slide 9/13 e Solubility and pH Because of the common ion effect, solubility will be pH dependent if dissolution involves H + and OH - e.g. Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) In this case solubility will be higher at low pH values.

10 Slide 10/13 x Fe 3+ and Zn 2+ can be separated using the pH dependence of the solubilities of their salts. For example, Fe(OH) 3 and Zn(OH) 2 can be separated at a pH of 4.76 (achieved by with a buffer). Separation of Cations K sp {Fe(OH) 3 } = 1.0 x 10 -38 = [Fe 3+ ][OH - ] 3 pOH = 14.00 - 4.76 so [OH-] = 10 -9.24 M [Fe 3+ ] = 1.0 x 10 -38 /(10 -9.24 ) 3 M = 5.2 x 10 -11 M Fe(OH) 3 is highly insoluble at this pH.

11 Slide 11/13 x Separation of Cations K sp {Zn(OH) 2 } = 1.0 x 10 -15 = [Zn 2+ ][OH - ] 2 [OH - ] = 10 -9.24 M [Zn 2+ ] = 1.0 x 10 -15 /(10 -9.24 ) 2 M = 3.0 x 10 3 M Zn(OH) 2 is highly soluble at this pH Fe(OH) 3 is highly insoluble at this pH.

12 Slide 12/13 e Summary: Solubility Equilibria II Learning Outcomes - you should now be able to: Complete the worksheet Apply solubility equilibria (qualitative and quantitative) Use ionic product to determine solubility Apply the common ion effect Answer review problems 10.49 - 10.75 in Blackman Next lecture: Complexes

13 Slide 13/13 x Practice Examples 1. Identify the one correct statement concerning the solubility of Mg(OH) 2. (a) pH has no effect on the solubility of Mg(OH) 2. (b) Mg(OH) 2 is less soluble at pH 10 than at pH 7. (c) Mg(OH) 2 is more soluble in 0.1 M MgCl 2 solution than in water. (d) Mg(OH) 2 is less soluble at pH 4 than at pH 7. (e) The solubility product constant for Mg(OH) 2 is greatest at pH 7. 2. The Ksp for scandium(III) hydroxide is 2 x 10. What is the solubility of Sc(OH) 3 (in mol L-1) of a solution buffered at 6.7?


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