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

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e CHEM1002 [Part 2]

Dr Michela SimoneWeeks 8 – 13

Office Hours: Monday 3-5, Friday 4-5Room: 412A (or 416)

Phone: 93512830e-mail: michela.simone@sydney.edu.au

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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 (Ksp) is the equilibrium constant for this equilibrium situation

• The solubility product (Ksp) gives the maximum concentrations of the ions in the solution and the maximum amount of solid that will dissolve

• The ionic product (Qsp) has the same form has the solubility product and is used to test whether more solid will dissolve or precipitation will occur If Qsp < Ksp, more ions can enter solution and more solid

can dissolve If Qsp > Ksp, precipitation must occur

Summary of Last Lecture

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e

Solubility Equilibria

Lecture 10:• Solubility• Blackman Chapter 10, Sections 10.4

Lecture 11:• Common ion effect• Blackman Chapter 10, Sections 10.4

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e

PbCl2(s) Pb2+(aq) + 2Cl-(aq)

Reminder: Solubility Equilibria

Ksp = [Pb2+(aq)][Cl-(aq)]2

solubility product constant

• If Pb2+(aq) and Cl-(aq) are present in the same solution at the same time, their concentrations can never be so large that [Pb2+(aq)][Cl-(aq)]2 is bigger than Ksp

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x

PbCl2(s) Pb2+(aq) + 2Cl-(aq)

Ksp = [Pb2+(aq)][Cl-(aq)]2

solubility product constant

• If x mol of PbCl2(s) dissolves in 1 L of solution, then [Pb2+(aq)] = x mol L-1 and [Cl-] = 2x mol L-1

Ksp = [Pb2+(aq)][Cl-(aq)]2 = (x)(2x)2 = 4x3

x = (Ksp/4)1/3 = solubility

• As Ksp = 1.6 x 10-5, solubility = (1.6 x 10-5/4)1/3 = 0.016 mol L-1

Solubility

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e

PbCl2(s) Pb2+(aq) + 2Cl-(aq)

Ksp = [Pb2+(aq)][Cl-(aq)]2

solubility product constant

• If NaCl is added to the solution, [Cl-(aq)] increases so equilibrium shifts to the left: less PbCl2 dissolves [Pb2+(aq)][Cl-(aq)]2+ must remain constant so [Pb2+(aq)]

decreases

The Common Ion Effect and Solubility

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x

PbCl2(s) Pb2+(aq) + 2Cl-(aq)

Ksp = [Pb2+(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 PbCl2(s) dissolve in 1 L of solution, then [Pb2+(aq)] = x mol L-1 and [Cl-] = 0.2 mol L-1

Ksp = [Pb2+(aq)][Cl-(aq)]2 = (x)(0.2)2 = 0.04x

• As Ksp = 1.6 x 10-5, solubility = 1.6 x 10-5/0.04 = 0.0004 mol L-1

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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) Mg2+(aq) + 2OH-(aq)

• In this case solubility will be higher at low pH values.

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x

• Fe3+ and Zn2+ 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

• Ksp {Fe(OH)3} = 1.0 x 10-38 = [Fe3+][OH-]3

pOH = 14.00 - 4.76 so [OH-] = 10-9.24 M

[Fe3+] = 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.

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x Separation of Cations

• Ksp {Zn(OH)2} = 1.0 x 10-15 = [Zn2+][OH-]2

[OH-] = 10-9.24 M

[Zn2+] = 1.0 x 10-15/(10-9.24)2 M = 3.0 x 103 M

Zn(OH)2 is highly soluble at this pH

Fe(OH)3 is highly insoluble at this pH.

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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

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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 MgCl2 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?