PRECIPITATION REACTIONS Chapter 19 Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part

PRECIPITATION REACTIONSPRECIPITATION REACTIONSChapter 19Chapter 19

PRECIPITATION REACTIONSPRECIPITATION REACTIONSChapter 19Chapter 19

Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved

Analysis of Silver GroupAnalysis of

Silver GroupAll 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

33



Silver Group

Although all salts formed in this experiment Although all salts formed in this experiment are said to be insoluble, they do dissolve are said to be insoluble, they do dissolve to some SLIGHT extent.to some SLIGHT extent.

AgCl(s) AgCl(s) AgAg++(aq) + Cl(aq) + Cl--(aq)(aq)

When equilibrium has been established, no When equilibrium has been established, no more AgCl dissolves and the solution is more AgCl dissolves and the solution is SATURATEDSATURATED..

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

44



Silver Group

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

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

This is equivalent to the This is equivalent to the SOLUBILITYSOLUBILITY of AgCl.of AgCl.

What is [ClWhat is [Cl--]? ]?

This is also equivalent to the AgCl solubility.This is also equivalent to the AgCl solubility.

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

55



Silver Group

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

Saturated solution has Saturated solution has [Ag[Ag++] = [Cl] = [Cl--] = 1.67 x 10] = 1.67 x 10-5-5 M M

Use this to calculate KUse this to calculate Kcc

KKcc = [Ag = [Ag++] [Cl] [Cl--]]

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

= 2.79 x 10= 2.79 x 10-10-10

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

66



Silver Group

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

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

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

KKspsp = solubility product constant = solubility product constant

See Table 19.2 and Appendix JSee Table 19.2 and Appendix J

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

Ag+ Pb2+ Hg22+

AgCl PbCl2 Hg2Cl2

77


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

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

88


Consider PbIConsider PbI22 dissolving in water dissolving in water

PbIPbI22(s) (s) Pb Pb2+2+(aq) + 2 I(aq) + 2 I--(aq)(aq)

Calculate KCalculate Kspsp if solubility = if solubility =

0.00130 M0.00130 M

SolutionSolution

Solubility = [PbSolubility = [Pb2+2+] = 1.30 x 10] = 1.30 x 10-3-3 M M

[I[I--] = _______ ?] = _______ ?

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

99




Calculate KCalculate Kspsp if solubility = 0.00130 M if solubility = 0.00130 M

SolutionSolution

1.1. Solubility = [PbSolubility = [Pb2+2+] ] = 1.30 x 10= 1.30 x 10-3-3 M M

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


1010





SolutionSolution

1.1. Solubility = [PbSolubility = [Pb2+2+] = 1.30 x 10] = 1.30 x 10-3-3 M M

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

2.2. KKspsp = [Pb = [Pb2+2+] [I] [I--]]22


1111





SolutionSolution

1.1. Solubility = [PbSolubility = [Pb2+2+] = 1.30 x 10] = 1.30 x 10-3-3 M M

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

2.2. KKspsp = [Pb = [Pb2+2+] [I] [I--]]22

= [Pb= [Pb2+2+] {2 • [Pb] {2 • [Pb2+2+]}]}22

= 4 [Pb= 4 [Pb2+2+]]33


1212





SolutionSolution

2.2. KKspsp = 4 [Pb = 4 [Pb2+2+]]33 = 4 (solubility) = 4 (solubility)33

KKspsp = 4 (1.30 x 10 = 4 (1.30 x 10-3-3))33 = = 8.8 x 108.8 x 10-9-9


1313


Precipitating an Insoluble SaltPrecipitating an Insoluble Salt

HgHg22ClCl22(s) (s) Hg Hg222+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)

KKspsp = 1.1 x 10 = 1.1 x 10-18-18 = [Hg = [Hg222+2+] [Cl] [Cl--]]22

If [HgIf [Hg222+2+] = 0.010 M, what [Cl] = 0.010 M, what [Cl--] is req’d to just ] is req’d to just

begin the precipitation of Hgbegin the precipitation of Hg22ClCl22??

That is, what is the maximum [ClThat is, what is the maximum [Cl--] that can be ] that can be

in solution with 0.010 M Hgin solution with 0.010 M Hg222+2+ without without

forming Hgforming Hg22ClCl22??

1414





Recognize thatRecognize that

KKspsp = product of = product of

maximum ion concs.maximum ion concs.

Precip. begins when product of Precip. begins when product of

ion concs. EXCEEDS the Kion concs. EXCEEDS the Kspsp..

1515





SolutionSolution

[Cl[Cl--] that can exist when [Hg] that can exist when [Hg222+2+] = 0.010 M,] = 0.010 M,

1616





SolutionSolution


[Cl ] = Ksp

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

Ksp

0.010 = 1.1 x 10-8 M

1717





SolutionSolution


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

begins to precipitate.begins to precipitate.

[Cl ] = Ksp

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

Ksp

0.010 = 1.1 x 10-8 M

1818


Precipitating an Insoluble SaltPrecipitating an Insoluble SaltHgHg22ClCl22(s) (s) Hg Hg22

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

KKspsp = 1.1 x 10 = 1.1 x 10-18-18

Now raise [ClNow raise [Cl--] to 1.0 M. What is the value of ] to 1.0 M. What is the value of [Hg[Hg22

2+2+] at this point?] at this point?

SolutionSolution

[Hg[Hg222+2+] = K] = Ksp sp / [Cl/ [Cl--]]22

= K= Kspsp / (1.0) / (1.0)22 = 1.1 x 10 = 1.1 x 10-18-18 M M

The concentration of HgThe concentration of Hg222+2+ has been reduced has been reduced

by 10by 101616 ! !

1919


Separating Metal Ions Separating Metal Ions CuCu2+2+, Ag, Ag++, Pb, Pb2+2+

Separating Metal Ions Separating Metal Ions CuCu2+2+, Ag, Ag++, Pb, Pb2+2+

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

2020


Separating Salts by Differences in KspSeparating Salts by Differences in Ksp

A solution contains 0.020 M AgA solution contains 0.020 M Ag++ and Pb and Pb2+2+. Add CrO. Add CrO442-2-

to precipitate red Agto precipitate red Ag22CrOCrO44 and yellow PbCrO and yellow PbCrO44. . Which precipitates first?Which precipitates first?

KKspsp for Ag for Ag22CrOCrO44 = 9.0 x 10 = 9.0 x 10-12-12

KKsp sp for PbCrOfor PbCrO4 4 = 1.8 x 10= 1.8 x 10-14-14

SolutionSolution

The substance whose KThe substance whose Kspsp is first is first exceeded precipitates first. exceeded precipitates first.

The ion requiring the lesser amount of The ion requiring the lesser amount of CrOCrO44

2-2- ppts. first. ppts. first.

2121



A solution contains 0.020 M AgA solution contains 0.020 M Ag++ and Pb and Pb2+2+. Add . Add CrOCrO44

2-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and yellow and yellow PbCrOPbCrO44. Which precipitates first?. Which precipitates first?



SolutionSolution

Calculate [CrOCalculate [CrO442-2-] required by each ion. ] required by each ion.

2222



[CrO[CrO442-2-] to ppt. PbCrO] to ppt. PbCrO4 4 = K= Ksp sp / [Pb/ [Pb2+2+] ]

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

[CrO[CrO442-2-] to ppt. Ag] to ppt. Ag22CrOCrO4 4 = K= Ksp sp / [Ag/ [Ag++]]22

= 9.0 x 10= 9.0 x 10-12-12 / (0.020) / (0.020)22 = 2.3 x 10 = 2.3 x 10-8-8 M M

PbCrOPbCrO44 precipitates first. precipitates first.


2-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and yellow and yellow PbCrOPbCrO44. Which precipitates first?. Which precipitates first?



SolutionSolution

2323



2-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and yellow and yellow PbCrOPbCrO44. PbCrO. PbCrO44 ppts. first. ppts. first.

KKspsp (Ag (Ag22CrOCrO44)= 9.0 x 10)= 9.0 x 10-12-12

KKsp sp (PbCrO(PbCrO44) = 1.8 x 10) = 1.8 x 10-14-14

How much PbHow much Pb2+2+ remains in solution when Ag remains in solution when Ag++ begins to precipitate?begins to precipitate?


2424


A solution contains 0.020 M AgA solution contains 0.020 M Ag++ and Pb and Pb2+2+. Add CrO. Add CrO442-2- to to

precipitate red Agprecipitate red Ag22CrOCrO44 and yellow PbCrO and yellow PbCrO44. PbCrO. PbCrO44 ppts. ppts. first.first.



How much PbHow much Pb2+2+ remains in solution when Ag remains in solution when Ag++ begins to begins to precipitate?precipitate?

SolutionSolution

We know that [CrOWe know that [CrO442-2-] = 2.3 x 10] = 2.3 x 10-8-8 M to begin to ppt. M to begin to ppt.

AgAg22CrOCrO44. .

What is the PbWhat is the Pb2+2+ conc. at this point? conc. at this point?


2525



2-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and yellow and yellow PbCrOPbCrO44. .




SolutionSolution

[Pb[Pb2+2+] = K] = Kspsp / [CrO / [CrO442-2-] = 1.8 x 10] = 1.8 x 10-14-14 / 2.3 x 10 / 2.3 x 10-8-8 M M

= 7.8 x 10= 7.8 x 10-7-7 M M


2626



2-2- to precipitate red Ag to precipitate red Ag22CrOCrO44 and yellow and yellow PbCrOPbCrO44. .




SolutionSolution

[Pb[Pb2+2+] = K] = Kspsp / [CrO / [CrO442-2-] = 1.8 x 10] = 1.8 x 10-14-14 / 2.3 x 10 / 2.3 x 10-8-8 M M

= 7.8 x 10= 7.8 x 10-7-7 M M

Lead ion has dropped from 0.020 M to < 10Lead ion has dropped from 0.020 M to < 10-6-6 M M


2727


Common Ion EffectCommon Ion EffectAdding an Ion “Common” to an Adding an Ion “Common” to an

EquilibriumEquilibrium

Common Ion EffectCommon Ion EffectAdding an Ion “Common” to an Adding an Ion “Common” to an

EquilibriumEquilibrium

2828


Calculate the solubility of BaSOCalculate the solubility of BaSO4 4 in (a) pure in (a) pure water and (b) in 0.010 M Ba(NOwater and (b) in 0.010 M Ba(NO33))22..

KKspsp for BaSO for BaSO4 4 = 1.1 x 10= 1.1 x 10-10-10

BaSOBaSO44(s) Ba(s) Ba2+2+(aq) + SO(aq) + SO442-2-(aq)(aq)

SolutionSolution

Solubility in pure water = [BaSolubility in pure water = [Ba2+2+] = [SO] = [SO442-2-] = x] = x

KKspsp = [Ba = [Ba2+2+] [SO] [SO442-2-] = x] = x22

x = (Kx = (Kspsp))1/21/2 = 1.1 x 10 = 1.1 x 10-5-5 M M

The Common Ion EffectThe Common Ion EffectThe Common Ion EffectThe Common Ion Effect

2929




BaSOBaSO44(s) (s) Ba Ba2+2+(aq) + SO(aq) + SO442-2-(aq)(aq)

SolutionSolution

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

Now dissolve BaSONow dissolve BaSO44 in water already in water already containing 0.010 M Bacontaining 0.010 M Ba2+2+. .


3030


Calculate the solubility of BaSOCalculate the solubility of BaSO4 4 in (a) pure water in (a) pure water and (b) in 0.010 M Ba(NOand (b) in 0.010 M Ba(NO33))22..



SolutionSolution

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

Now dissolve BaSONow dissolve BaSO44 in water already containing in water already containing 0.010 M Ba0.010 M Ba2+2+. .

Which way will the “common ion” shift the Which way will the “common ion” shift the equilibrium? ___ Will solubility of BaSOequilibrium? ___ Will solubility of BaSO44 be less be less than or greater than in pure water?___than or greater than in pure water?___


3131





SolutionSolution

[Ba[Ba2+2+]] [SO[SO442-2-]]

initialinitial

changechange

equilib.equilib.


3232





SolutionSolution

[Ba[Ba2+2+]] [SO[SO442-2-]]

initialinitial 0.0100.010 0 0

changechange + y+ y + y + y

equilib.equilib. 0.010 + y0.010 + y y y


3333


Calculate the solubility of BaSOCalculate the solubility of BaSO4 4 in (a) pure water and (b) in (a) pure water and (b) in 0.010 M Ba(NOin 0.010 M Ba(NO33))22..



SolutionSolution

KKspsp = [Ba = [Ba2+2+] [SO] [SO442-2-] = (0.010 + y) (y)] = (0.010 + y) (y)

Because y < 1.1 x 10Because y < 1.1 x 10-5-5 M (= x, the solubility in pure M (= x, the solubility in pure water), this means 0.010 + y is about equal to 0.010. water), this means 0.010 + y is about equal to 0.010. Therefore,Therefore,

KKspsp = 1.1 x 10 = 1.1 x 10-10-10 = (0.010)(y) = (0.010)(y)

y = 1.1 x 10y = 1.1 x 10-8-8 M = solubility in presence of added Ba M = solubility in presence of added Ba2+2+ ion.ion.


3434





SolutionSolution

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

Solubility in presence of added BaSolubility in presence of added Ba2+2+ = 1.1 x 10= 1.1 x 10-8-8 M M

Le Chatelier’s Principle is followed!Le Chatelier’s Principle is followed!


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PRECIPITATION REACTIONS Chapter 19 Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of any part