PRECIPITATION REACTIONS Solubility of Salts Section 18.4

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PRECIPITATION REACTIONSSolubility of

SaltsSection 18.4

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Metal Chloride Salts

These products are said to be

INSOLUBLE and form when mixing

moderately concentrated

solutions of the metal ion with

chloride ions.

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

3Analysis Analysis of Silver of Silver GroupGroup

The products are said to be insoluble, they do dissolve to some SLIGHT extent.

AgCl(s) = Ag+(aq) + Cl-(aq)When equilibrium has been established, no

more AgCl dissolves and the solution is SATURATED.

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

4Analysis Analysis of Silver of Silver GroupGroup

AgCl(s) = Ag+(aq) + Cl-(aq)

When solution is SATURATED, expt. shows that [Ag+] = 1.67 x 10-5 M.

This is equivalent to the SOLUBILITY of AgCl.

What is [Cl-]? [Cl-] is equivalent to the AgCl solubility.

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

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AgCl(s) = Ag+(aq) + Cl-(aq)Saturated solution has [Ag+] = [Cl-] = 1.67 x 10-5 M

Use this to calculate Kc

Kc = [Ag+] [Cl-]

= (1.67 x 10-5)(1.67 x 10-5) = 2.79 x 10-10

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Kc = [Ag+] [Cl-] = 2.79 x 10-10

Because this is the product of “solubilities”, we call it

Ksp = solubility product constant

See Table 18.2 and Appendix J

Solubility Product Constant

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Lead(II) ChlorideLead(II) ChloridePbCl2(s) = Pb2+(aq) + 2 Cl-(aq)

Ksp = 1.9 x 10-5

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SolutionSolution

Solubility = [Pb2+] = 1.30 x 10-3 M [I-] = ?

[I-] = 2 x [Pb2+] = 2.60 x 10-3 M

Solubility of Lead(II) Iodide

Consider PbI2 dissolving in waterPbI2(s) = Pb2+(aq) + 2 I-(aq)

Calculate Ksp if solubility = 0.00130 M

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Calculate Ksp

Ksp = [Pb2+] [I-]2

= X{2 X}2

Ksp = 4 X3 = 4[Pb2+]3

Solubility of Lead(II) IodideSolubility of Lead(II) Iodide

= 4 (solubility)3

Ksp = 4 (1.30 x 10-3)3 = 8.8 x 10-9

11Precipitating an Insoluble Precipitating an Insoluble SaltSalt

Hg2Cl2(s) = Hg22+(aq) + 2 Cl-(aq)

Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2

If [Hg22+] = 0.010 M, what [Cl-] is req’d to just

begin the precipitation of Hg2Cl2?

That is, what is the maximum [Cl-] that can be in solution with 0.010 M Hg2

2+ without forming

Hg2Cl2?

12Precipitating an Insoluble SaltPrecipitating an Insoluble SaltHg2Cl2(s) = Hg2

2+(aq) + 2 Cl-(aq)

Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2

Recognize that

Ksp = product of maximum ion concs..

Precip. begins when product of ion concs. EXCEEDS the Ksp.

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Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2

Solution

[Cl-] that can exist when [Hg22+] = 0.010

M,

[Cl ] = Ksp0.010

= 1.1 x 10-8 M

If this conc. of ClIf this conc. of Cl-- is just exceeded, Hg is just exceeded, Hg22ClCl22 begins to precipitate.begins to precipitate.

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Hg2Cl2(s) = Hg22+(aq) + 2 Cl-(aq)

Ksp = 1.1 x 10-18

Now use [Cl-] = 1.0 M. What is the value of [Hg2

2+] at this point?

Solution[Hg2

2+] = Ksp / [Cl-]2

= Ksp / (1.0)2 = 1.1 x 10-18 M

The concentration of Hg22+ has been

reduced by 1016 !

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The Common Ion EffectThe Common Ion EffectAdding an ion “common” to an equilibrium causes the

equilibrium to shift back to reactant.

16Common Ion EffectCommon Ion EffectPbCl2(s) = Pb2+(aq) + 2 Cl-(aq)

Ksp = 1.9 x 10-5

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Calculate the solubility of BaSO4

BaSO4(s) = Ba2+(aq) + SO42-(aq)

(a) In pure water and

(b) in 0.010 M Ba(NO3)2.

Ksp for BaSO4 = 1.1 x 10-10

The Common Ion EffectThe Common Ion Effect

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Ksp for BaSO4 = 1.1 x 10-10

BaSO4(s) = Ba2+(aq) + SO42-(aq)

SolutionSolubility = [Ba2+] = [SO4

2-] = xKsp = [Ba2+] [SO4

2-] = x2

x = (Ksp)1/2 = 1.1 x 10-5 MSolubility in pure water = 1.1 x 10-5 M

BaSO4 in pure water

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SolutionSolubility in pure water = 1.1 x 10-5 mol/L.Now starting with 0.010 M Ba2+.

Which way will the “common ion” shift the equilibrium? ___ Will

solubility of BaSO4 be less than or greater than in pure water?___

BaSO4 in in 0.010 M Ba(NO3)2.BaSO4(s) = Ba2+(aq) + SO4

2-(aq)

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Solution [Ba2+] [SO4

2-]initialchange equilib.

The Common Ion EffectThe Common Ion Effect

+ y0.010 0

+ y

0.010 + y y

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Ksp = [Ba2+] [SO42-] = (0.010 + y) (y)

Because y < 1.1 x 10-5 M (pure), 0.010 + y is about equal to 0.010.

Therefore,Ksp = 1.1 x 10-10 = (0.010)(y)

y = 1.1 x 10-8 M = solubility in presence of added Ba2+ ion.

The Common Ion EffectThe Common Ion EffectSolutionSolution

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SUMMARYSolubility in pure water = x = 1.1 x 10-

5 MSolubility in presence of added Ba2+

= 1.1 x 10-8 MLe Chatelier’s Principle is followed!

Add to the right: equilibrium goes to the left

The Common Ion EffectThe Common Ion EffectBaSO4(s) = Ba2+(aq) + SO4

2-(aq)

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The Common Ion EffectThe Common Ion Effect

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The Common Ion EffectThe Common Ion Effect