1 PRECIPITATION REACTIONS Solubility of Salts Section 18.4.

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PRECIPITATION REACTIONSSolubility of

SaltsSection 18.4

PRECIPITATION REACTIONSSolubility of

SaltsSection 18.4

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Metal Chloride SaltsMetal 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

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

3Analysis Analysis of Silver of Silver GroupGroup

Analysis 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

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

4Analysis Analysis of Silver of Silver GroupGroup

Analysis Analysis of Silver of Silver GroupGroup


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

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

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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) Chloride

PbCl2(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

Solubility of Lead(II) Iodide

Consider PbI2 dissolving in water

PbI2(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) IodideSolubility of Lead(II) IodideSolubility of Lead(II) Iodide

= 4 (solubility)3= 4 (solubility)3

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

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Precipitating an Insoluble Precipitating an Insoluble SaltSalt

Precipitating 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 Hg22+ without

forming Hg2Cl2?

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Precipitating an Insoluble SaltPrecipitating an Insoluble SaltPrecipitating an Insoluble SaltPrecipitating an Insoluble Salt

Hg2Cl2(s) = Hg22+(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 ] = Ksp

0.010 = 1.1 x 10-8 M[Cl ] =

Ksp

0.010 = 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

[Hg22+] = 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 Effect

PbCl2(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


Solution

Solubility = [Ba2+] = [SO42-] = x

Ksp = [Ba2+] [SO42-] = x2

x = (Ksp)1/2 = 1.1 x 10-5 M

Solubility in pure water = 1.1 x 10-5 M

BaSO4 in pure water

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Solution

Solubility 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.


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Solution

[Ba2+] [SO42-]

initial

change

equilib.


+ y

0.010 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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SUMMARY

Solubility in pure water = x = 1.1 x 10-5 M

Solubility in presence of added Ba2+ = 1.1 x 10-8 M

Le Chatelier’s Principle is followed!Add to the right: equilibrium goes to the left



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1 PRECIPITATION REACTIONS Solubility of Salts Section 18.4.

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