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]

A/Prof Adam Bridgeman (Series 1)Dr Feike Dijkstra (Series 2)

Weeks 8 – 13

Office Hours: Monday 2-3, Friday 1-2Room: 543ae-mail: adam.bridgeman@sydney.edu.aue-mail: feike.dijkstra@sydney.edu.au

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Crystal structures II

• Face centred cubic : close packing with 74% efficiency, coordination number of 12 and 4 atoms per unit cell

• Hexagonal close packing : close packing with 74% packing efficiency and coordination number of 12

• Body centred cubic : not closed packed with 68% packing efficiency, coordination number of 8 and 2 atoms per unit cell

• Simple cubic : not close packed with 52% packing efficiency and coordination number of 6

• Even close packed structures have ‘holes’ (interstitial sites) and ionic structures can be considered to be based on close packed anions with cations in some or all of these sites

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 What Determines Solubility?

• Dissolution break cation-anion ionic bonds vs make new ion – water bonds

Solvent

Solute

Dissolution

Precipitation

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e Dissolving Ionic Compounds in Water

MX(s) Mn+(aq) + Xn-(aq)

Mn+(g) + Xn-(g)lattice enthalpy hydration enthalpy

enthalpy of solution

• To dissolve a salt: need to break up lattice: harder

for small ions and for those with higher charges makes new ion-water bonds: more beneficial for

small ions and for those with higher charges

both increase with ion charge and

decrease with ion size!

• Very difficult to guess what will be insoluble and what wiil be soluble!• See E2 experiment

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e Solubility and Saturated Solutions

• When dissolving a solid in solution, equilibrium means that as much solid has dissolved as possible the solution is said to be saturated the solubility is the maximum amount that will dissolve

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

• Solubility of magnesium(II) hydroxide is 9.63 mg L-1 or 1.65 × 10-4 M

• By convention, if solubility > 0.1 M then compound is soluble if solubility < 1.0 × 10-3 M then compound is insoluble intermediate cases are said to be moderately soluble

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

• Solubility equilibria are always presented as dissolution processes as in this example:

Definition: Solubility Equilibria

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

solubility product constant

Note that a high value of Ksp corresponds to a high solubility!

• Since the concentration of a pure solid is taken to be constant, the equilibrium constant is:

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x

• Imagine that someone dumps a truckload of lead(II) chloride into a lake. What is the concentration of Pb2+(aq) ions in the lake (assuming that a saturated solution results).

Ksp = [Pb2+(aq)][Cl-(aq)]2 = 1.6 x 10-5

• From chemical equation, [Cl-(aq)] = 2 x [Pb2+(aq)]

Ksp = [Pb2+(aq)](2 x [Pb2+(aq)])2 = 4 x [Pb2+(aq)]3

4[Pb2+(aq)]3 = 1.6 x 10-5 M3

[Pb2+(aq)] = 0.016 M

Determine Solubility from Ksp (I)

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

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• Important: Ksp is defined for a saturated solution in equilibrium with some solid

• When the system is NOT at equilibrium

The Ionic Product

Define: Q = reaction quotient = ionic product

For PbCl2, Q = [Pb2+(aq)][Cl-(aq)]2

• Note: Q, the ionic product is a non-equilibrium parameter! Q = Ksp only at equilibrium.

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e Ionic Product (Q) vs Solubility Product (Ksp)

For ionic compounds:

(a) when Q < Ksp, further dissolution can occur.

(b) when Q > Ksp, the solution is supersaturated, precipitation will occur.

(c) when Q = Ksp, the solution is saturated - the system is at equilibrium.

Exactly the same as Q vs K inequilibrium discussion in CHEM1001

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e Supersaturated Solutions

Sodium acetate

A) Superaturated solution.B) After seed crystal is added crystallisation starts.C) Most of the precipitate has formed.

a b c

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• What happens if 1.0 tonne of PbCl2 is dumped into a lake containing 50 cubic metres of water. Would all the PbCl2 dissolve?

Determine Solubility from Ksp (II)

• 1 tonne = (1 x106 g)/(278.11 g mol-1) = 3.6 x 103 mol• volume = 50 m3 = 50 x 103 L• [Pb2+(aq) = (3.6 x 103 mol)/(50 x 103 L) = 7.2 x 10-2 mol L-1

Q = [Pb2+(aq)][Cl-(aq)]2 = [Pb2+(aq)](2 x [Pb2+(aq)])2

= 4 x (7.2 x 10-2)3 mol3 L-3

= 1.5 x 10-3 mol3 L-3

PbCl2(s) Pb2+(aq) + 2Cl-(aq) Ksp = 1.6 x 10-5

Q > Ksp so it would not all dissolve.

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e Summary: Solubility Equilibria I

Learning Outcomes - you should now be able to:

• Complete the worksheet• Apply solubility equilibria (qualitative and

quantitative)• Use ionic product to determine solubility• Answer review problems 10.49 - 10.75 in Blackman

Next lecture:

• The common ion effect

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2. The Ksp for Cd(OH)2 is 5.9 × 10–15 at 25 oC. What is the solubility of Cd(OH)2?

(a) 7.7 × 10–8 mol L–1

(b) 5.4 × 10–8 mol L–1

(c) 1.1 × 10–5 mol L–1

(d) 5.9 × 10–15 mol L–1

(e) 1.8 × 10–5 mol L–1

Practice Examples

1.What is the solubility product constant expression for Ag3PO4?(a) Ksp = [Ag+][PO4

3–](b) Ksp = [Ag+][PO4

3–]3

(c) Ksp = [Ag+]3[PO43–]

(d) Ksp = 1/[Ag+][PO43–]

(e) Ksp = 3[Ag+][PO43–]

Cd(OH)2(s) Cd2+(aq) + 2OH–(aq)