precipitation reactions solubility of salts section 18.4

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PRECIPITATION REACTIONSPRECIPITATION REACTIONSSolubility of SaltsSolubility of Salts

Section 18.4Section 18.4

PRECIPITATION REACTIONSPRECIPITATION REACTIONSSolubility of SaltsSolubility of Salts

Section 18.4Section 18.4

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Analysis of Silver Analysis of Silver GroupGroup

Analysis of Silver Analysis of Silver GroupGroup

All salts formed in All salts formed in this experiment are this experiment are said to be said to be INSOLUBLEINSOLUBLE and and form when mixing form when mixing moderately moderately concentrated concentrated solutions of the solutions of the metal ion with metal ion with chloride ions.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

Although all salts formed in this experiment

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




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




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

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2




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

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

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Lead(II) ChlorideLead(II) ChloridePbClPbCl22(s) (s) Pb Pb2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq) (aq)

KKspsp = 1.9 x 10 = 1.9 x 10-5-5

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SolutionSolution

1. 1. Solubility = [Pb2+] = 1.30 x 10-3 M

[I-] = ?

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

Solubility of Lead(II) IodideSolubility of Lead(II) IodideSolubility of Lead(II) IodideSolubility 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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SolutionSolution

2. Ksp = [Pb2+] [I-]2

= [Pb2+] [2Pb2+]2

Ksp = 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

PbI2(s) Pb2+(aq) + 2 I-(aq)

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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

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


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

Recognize that

Ksp = maximum ion concs.

Precip. begins when product of .

EXCEEDS the Ksp.

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

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

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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Precipitating an Insoluble SaltPrecipitating an Insoluble Salt


Ksp = 1.1 x 10-18

Now raise [Cl-] to 1.0 M. What is the value of [Hg2

2+]

[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.

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Common 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 in (a) pure water and (b) in 0.010 M Ba(NO3)2.

Ksp for BaSO4 = 1.1 x 10-10

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

Solution

Solubility in pure water = [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 mol/L

The Common Ion EffectThe Common Ion Effect

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SolutionSolution

Solubility in pure water = 1.1 x 10-5 mol/L.

Now dissolve BaSO4 in water already containing 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?___



LEFT

LESS

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Solution

[Ba2+] [SO42-]

initial

change

equilib.


+ y

0.010 00

+ y

0.010 + y y

Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2.

Ksp for BaSO4 = 1.1 x 10-10


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SolutionSolution

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

Because y < 1.1 x 10-5 M (= x, the solubility in pure water), this means

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



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SUMMARY

Solubility in pure water =

SO42-(aq) = x = 1.1 x 10-5 M

Solubility in presence of added Ba2+ SO4

2-(aq) = 1.1 x 10-8 M

Le Chatelier’s Principle is followed!



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Separating Metal Ions Cu2+, Ag+, Pb2++

Separating Metal Ions Cu2+, Ag+, Pb2++

Ksp Values

AgCl 1.8 x 10-10

PbCl2 1.7 x 10-5

PbCrOPbCrO4 4 1.8 x 101.8 x 10-14-14

Ksp Values

AgCl 1.8 x 10-10

PbCl2 1.7 x 10-5

PbCrOPbCrO4 4 1.8 x 101.8 x 10-14-14

24Separating Salts by Differences in Ksp

A solution contains 0.020 M Ag+ and Pb2+. Add CrO4

2-. Which precipitates first?

Ksp for Ag2CrO4 = 9.0 x 10-12

Ksp for PbCrO4 = 1.8 x 10-14

SolutionSolution

The substance whose Ksp is first exceeded precipitates first.

The ion requiring the lesser amount of CrO4

2- ppts. first.

25Separating Salts by Differences in Separating Salts by Differences in KKspsp

[CrO42-] to ppt. PbCrO4 = Ksp / [Pb2+]

= 1.8 x 10-14 / 0.020 = 9.0 x 10-13 M

Ksp for Ag2CrO4 = 9.0 x 10-12

Ksp for PbCrO4 = 1.8 x 10-1

Calculate [CrO42-] required by each ion.

[CrO42-] to ppt. Ag2CrO4 = Ksp / [Ag+]2

= 9.0 x 10-12 / (0.020)2 = 2.3 x 10-8 M

PbCrO4 precipitates first

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Ksp (Ag2CrO4)= 9.0 x 10-12

Ksp (PbCrO4) = 1.8 x 10-14

How much Pb2+ remains in solution when Ag+ begins to precipitate?

Separating Salts by Differences in Separating Salts by Differences in KKspsp

27Separating Salts by Differences in Separating Salts by Differences in KKspsp

We know that [CrO42-] = 2.3 x 10-8 M to

begin to ppt. Ag2CrO4.

What is the Pb2+ conc. at this point?

[Pb2+] = Ksp / [CrO42-] =

1.8 x 10-14 / 2.3 x 10-8 M

= 7.8 x 10-7 M

Lead ion has dropped from

0.020 M to < 10-6 M

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Separating Salts by Separating Salts by Differences in KDifferences in Kspsp


• Add CrO42- to solid PbCl2. The less

soluble salt, PbCrO4, precipitates

• PbCl2(s) + CrO42- PbCrO4 + 2 Cl-

• Salt Ksp

PbCl2 1.7 x 10-5

PbCrO4 1.8 x 10-14

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• PbCl2(s) + CrO42- PbCrO4 + 2 Cl-

Salt Ksp

PbCl2 1.7 x 10-5

PbCrO4 1.8 x 10-14

PbClPbCl22(s) (s) Pb Pb2+2+ + 2 Cl + 2 Cl-- KK11 = K = Kspsp

PbPb2+2+ + CrO + CrO442-2- PbCrO PbCrO44 KK22 = 1/K = 1/Kspsp

KKnetnet = K = K11 • K • K22 = 9.4 x 10 = 9.4 x 1088

Net reaction is product-favoredNet reaction is product-favored

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Formation of complex ions explains why you can dissolve a ppt. by forming a complex

ion.

Dissolving Precipitates Dissolving Precipitates by forming Complex Ionsby forming Complex Ions

AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-

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Examine the solubility of AgCl in ammonia.

AgCl(s) Ag+ + Cl- Ksp = 1.8 x 10-10

Ag+ + 2 NH3 Ag(NH3)2+ Kform = 1.6 x 107

-------------------------------------

AgCl(s) + 2 NH3 Ag(NH3)2+ + Cl-

Knet = Ksp • Kform = 2.9 x 10-3

By adding excess NH3, the equilibrium shifts to the right.

Dissolving Precipitates Dissolving Precipitates by forming Complex Ionsby forming Complex Ions

precipitation reactions solubility of salts section 18.4

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