ch18_solubility.ppt

8/10/2019 CH18_Solubility.ppt

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Chapter 18:Solubility and

Simultaneous Equilibria

Chemistry: The Molecular Nature

of Matter, 6E

Jespersen/Brady/Hyslop


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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E

Solubility of Salts

Precipitation reactions (CH 5)

Exchange reactions in which one product is waterinsoluble compound

CaCl2 (l)+ Na2CO3 (aq)CaCO3 (s)+ 2 NaCl (aq)

Insoluble compound Compound having water solubility of less than 0.01

mole of dissolved material per liter of solution

S > 0.01 M

2


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Solubility of Salts Ch 5 Solubility Rules

Guidelines for what is insoluble Doesnt mean compound wont dissolve at all

Just not very much

Now want to Quantitate solubilities

Explore conditions under which some compoundsprecipitate and others dont

Applications in separation of ions

Especially toxic metal ions such as Hg2+, Tl3+, U3+, etc.

3


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Why Study Solubility? Tooth decay

Acids from foods dissolve enamel, [Ca5(PO4)3OH] =hydroxyapatite

Reduced by fluoride which replaces OHto formfluorapatite = [Ca5(PO4)3F] and CaF

Lower solubility means it doesnt dissolve as readilyin acid

Upper and Lower GI

X-ray of upper and lower gastrointestinal tract Clarified by barium sulfatevery insoluble

BaSO4toxic, but safe, as it doesnt dissolve

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Solubility Equilibria Solids in equilibrium with ions in solution

When ionic salt dissolves in water Assume dissociates into separate hydrated ions

Initially, no ions in solution

CaF2(s) Ca2+(aq) + 2 F

(aq) As dissolution occurs, ions build up and collide

Ca2+(aq) + 2 F(aq) CaF2(s)

At Equilibrium CaF2(s) Ca

2+(aq) + 2 F(aq)

Now have saturated solution

5


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Solubility and Solubility ProductSolubility

Amount of salt that dissolves in given amount ofsolvent to give saturated solution

Concentration Infinite number of values

Solubility product Product of molar concentrations of ions in saturated

solution raised to appropriate powers Equilibrium constant Only one value for given solid at given temperature

Temperature dependence Solubilities and thus Kspchange with T

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7/59Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E

Solubility of Salts Consider AgCl in water

Only a very small amount dissolves Equilibrium exists when solution is saturated

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

Equilibrium law:Ksp= [Ag

+][Cl]

Ksp= solubility product constant

Solubility equilibrium Reflects solubility of compound

Product of ion concentrations

7



Solubility and Solubility Product Solubility

Amount of salt that dissolves in given amount ofsolvent to give saturated solution

Solubility product Product of molar concentrations of ions in

saturated solution raised to appropriate powers Temperature dependence

Solubilites and thus Kspchange with temperature

Table 18.1 Solubility product constants at 25 C

More in Table C.6 (Appendix p. A45)

8



Ion Product vs. Solubility Product

For: AxBy(s) x Ay+(aq)+ y Bx(aq)

Ion Product Qsp= [Ay+]x[Bx]y Like solubility product, except initial concentrations

are used

Any dilution of salt that results in an unsaturatedsolution

Varies with concentration

Solubility Product Ksp

= [Ay+]x[Bx]y

Ion product value for saturated solution becomesconstant

Uses equilibrium concentration

9



Writing KspEquilibrium Laws

For: AxBy(s) xAy+ (aq)+ y Bx(aq)

Ksp= [Ay+]x[Bx]y

Ex.

BaSO4(s)Ba2+(aq)+ SO4

2(aq) Ksp= [Ba2+][SO4

2]

CaF2(s)Ca2+(aq)+ 2F(aq) Ksp= [Ca2+][F]2

Ag2CrO4(s)2Ag+(aq)+ CrO4

2(aq) Ksp= [Ag+]2[CrO4

2]

AuCl3(s)Au3+(aq)+ 3Cl(aq) Ksp= [Au

3+][Cl]3

10



Calculation using Kspand MolarSolubilities

Molar solubility Moles of salt dissolved in one liter of saturated

solution

Assume what little dissolved, dissociates 100%

Assumes there is somesolid Quantity is not important

Solid is notincluded in mass action expression

A. Given Solubilites, Calculate KspB. Given Ksp, Calculate Solubility

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A. Given Solubilites , Calculate Ksp

Ex 1.At 25 C, the solubility of AgCl is 1.34 x

105 M. Calculate the solubility product for AgCl.AgCl (s) Ag+(aq) + Cl(aq)

Ksp= [Ag+][Cl]

Ksp= (1.34 x 105)(1.34 x 105)

Ksp=1.80 x 1010

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

I 0.00 0. 00

C

E

1.34 x 105 M 1.34 x 105 M

1.34 x 105 M 1.34 x 105 M

12



Your Turn!The solubility of a salt, A2B3, is found to be

3.0105

M. What is the value ofKsp?A. 2.6 x 10-21

B. 5.4 x 10-9

C. 2.4 x 10-23D. 1.7 x 10-21

A2B3 2A3+ + 3B2-

[A] = 2(3.0 x 10-5) [B] = 3(3.0 x 10-5)

K = [A]2[B]3= (6.0 x 10-5)2(9.0 x 10-5)3

K = 2.6 x 10-21

13



B. Given Ksp, Calculate Solubility

Ex. 2 What is the molar solubility of CuI in

water? What are the equilibriumconcentrations of Cu+and I?

Step 1. Write balanced equation for

dissociation of saltCuI(s) Cu+(aq)+ I(aq)

Step 2. Write equilibrium law

Ksp= [Cu+][I]

Step 3. Kspfor salt

Ksp= 8.0 108

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Ex. 2Molar Solubilities from Ksp

Step 4. Concentration Table

Step 5. Plug into and solve Kspexpression

Ksp= 8.0 108= (x)(x)

x2= 8.0 108

x = 2.8 104 M= calculated molar solubilityof CuI = [Cu+] = [I]

Conc (M) CuI(s) Cu+ (aq) + I(aq)

Initial 0.00 0. 00

Change

Equilm

+x+x

xx

x

15



Molar Solubility and KspProblems

Strategy for solving:

1. Write balanced equation for dissociation of salt

2. Write equilibrium law

3. Kspfor salt (from table)

4. Concentration table

5a.Solve for x = solubility

Or

5b. Given solubilities, calculate Ksp

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Ex. 3 Calculate Kspfor Bi2S3 given solubility is

1.0 x 1015

M at 25 C.Step 1.Write balanced equation for

dissociation of salt

Bi2S

3(s) 2 Bi3+(aq) + 3 S2(aq)

Step 2.Write equilibrium law

Ksp= [Bi3+]2[S2]3

Step 3.Use Concentration Table todetermine concentrations of each ion

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Bi2S3(s) 2 Bi2+(aq) 3 S2(aq)

I (No entries 0.00 0.00

C in this

E column)

+3(1.0

105)+2(1.0 1015)

3.0 10152.0 1015

Ksp= (2.0 x 1015)2(3.0 x 1015)3

Ksp= (4.0 x 1030

)(27 x 1045

)

Ksp= 1.1 x 1073

Step 4.Solve for Ksp

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B.Given Ksp, Calculate Solubilities

Ex. 4 Calculate the solubility of CaF2in water

at 25 C, if Ksp= 4.0 x 1011.CaF2(s) Ca

2+ (aq)+ 2F(aq)

Step 1:Write Equilibrium Law

Ksp= [Ca2+][F]2

Step 2: Concentration Table

Conc (M) CaF2

(s) Ca2+ (aq) 2F(aq)

Initial (No entries 0.00 0.00

Change in this

Equilm column)

+2x+x

2xx

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Ex. 4Molar Solubilities from Ksp

Step 3. Plug into and solve Kspexpression

Ksp= [Ca2+][F]2= (x) (2x)2

4.01011= 4x3

4

100.4

x

113

3 11100.1

x

X = 2.15 104M = molar solubility of CaF2

[Ca2+] = X = 2.15 104M

[F

] = 2x = 2(2.15

104

M) = 4.3

104

M 20



Your Turn!Given Ksp= 1.4 x 10

7for Cu(IO3)2, calculate

the solubility of this salt.

A. 5.2 x 10-3

B. 3.3 x 10-3C. 2.6 x 10-4

D. 3.7 x 10-4

Cu(IO3)2(s) Cu2+ + 2IO3-Ksp= 1.4 x 10

-7= s (2s)2

s = 3.3 x 10

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Your Turn!What is the molar solubility of Ag3PO4in water? What

is the molar concentration of each ion in solution?Ksp= 8.9 1017

A. 5.4 x 10-9 [Ag+] = 5.4 x 10-9 [PO43-] = 5.4 x 10-9

B. 4.3 x 10-5 [Ag+] = 1.3 x 10-4 [PO43-] = 4.3 x 10-5

C. 5.4 x 10-9 [Ag+] = 1.6 x 10-8 [PO43-] = 5.4 x 10-9D. 9.7 x 10-7 [Ag+] = 2.7 x 10-4 [PO4

3-] = 9.7 x 10-7

Ag3PO4 (s) 3Ag+ + PO4

3-

Ksp= 8.9 x 10-17= [Ag][PO] = (3s)3(s)

s = 4.3 x 10

[Ag+] = 3s = 1.3 x 10-4M

[PO43-] = s = 4.3 x 10-5M

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Your Turn!What is the solubility of PbCl2in grams per 100.0 mL

at 25

o

? Ksp= 1.7 x 10

-5

.A. 0.56 g

B. 0.72 g

C. 0.45 g

D. 0.39 g

PbCl2(s ) Pb2+ + 2Cl-

Ksp= 1.7 x 10-5= [Pb2+][Cl-]2= s (2s)2

s = 1.62 x 10-5 M

(1.62 x 10-5mol/L) x (278.11 g/mol) x 0.1 L = 0.45 gin 100 mL

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Relative Solubilities Kspgives information about solubility of salts

Must be careful when comparing relativesolubilities

Two possible cases when comparing:

1. Must compare salts that contain the same numberof ions

AgI(s) Ksp= 1.5 x 1016

CuI(s) Ksp= 5.0 x 1012

CaSO4(s) Ksp= 6.1 x 105

Each salt dissolves to produce 2 ions

Salt cation + anion

Ksp

= [cation][anion]

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Relative Solubilities1. Must compare salts that contain the same number

of ions If solubility = x

Then [cation] = [anion] = x

Ksp= x2

So can compare solubilities by comparing Ksps

CaSO4> CuI > AgI

most soluble least soluble

largest Ksp smallest Ksp

solubilityKx sp

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Relative Solubilities2. Compare salts with different number of ions

Each produces different number of ions

Each uses different Kspexpression

No way to predict relative solubility based on Ksp

Bi2S3> Ag2S > CuSMost soluble least soluble

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Salt Ksp #ions CalcdSolubility (M)

CuS 8.5 x 1045 2 9.2 x 1023

Ag2S 1.6 x 1049 3 3.4 x 1017

Bi2S3 1.1 x 1073 5 1.0 x 1015


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Common Ion Effect Up until now all calculations in pure water

What happens if I add another salt to a solutioncontaining one of the ions in our insoluble salt?

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

Saturated solution of PbI2in water

Filter

Add KI

PbI2(yellow solid) precipitates out

Why? Le Chateliers Priniciple

Add product I

Equilibrium moves to left and solid PbI2forms

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Common Ion Effect Common ion

Ion in solution that is supplied by more than onesolute

Common Ion Effect

Lowering of solubility of ionic compound byaddition of common ion

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Ex. 5 Common Ion Effect

A. What is the molar solubility of Ag2CrO4in

0.10M AgNO3solution? Ksp= 9.0 1012.B. What is the molar solubility of

Ag2CrO4in pure water?

C. What is the molar solubility ofAg2CrO4in 0.100 M Na2CrO4?

Ag2CrO4(s) 2Ag+(aq)+ CrO4

2(aq)

Ksp= [Ag+]2[CrO4

2] = 9.0 1012

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Ex. 5 Common Ion EffectA. Molar solubility of Ag2CrO4in 0.10M AgNO3

solution? Ksp

= 9.0 1012

Ksp= 9.0 1012= (0.10M)2[x]

x = Solubility of Ag2CrO4= 9.0 1010M

[Ag+] = 0.10 M

[CrO4

2] = 9.0 1010M

)010.0(

100.9x

12

Ag2CrO4(s) 2Ag+(aq) + CrO4

2(aq)

I (No entries 0.10 M 0.00

C in thisE column)

+x

+2xx0.10

30

ff


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Ex. 5 Common Ion EffectB.Solubility of Ag2CrO4in pure water

Ag2CrO4(s) 2Ag+(aq) + CrO42(aq)I (No entries 0.00 M 0.00 M

C in this

E column)

+x+2x

x2x

Ksp= [Ag+]2[CrO4

2] = (2x)2(x) = 9.0 1012= 4x3

3 12312

1025.24

100.9

x

X = Solubility of Ag2CrO4= 1.31 104M

[CrO42] = x = 1.31 104 M

[Ag+

] = 2x = 2.62 104

M 31

C ff


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Ex. 5 Common Ion EffectC. Solubility of Ag2CrO4in 0.100 M Na2CrO4?

Ag2CrO4(s) 2Ag+(aq) + CrO42(aq)I (No entries 0.00 M 0.10 M

C in this

E column)

+x+2x

2x 0.10

Ksp= (2x)2(0.10) = 9.0 1012= 4x2(0.10)

1112

1025.24.0

100.9

x


[CrO42] = x = 4.7 106 M

[Ag+

] = 2x = 9.5 106

M 32

E 5 C I Eff


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Ex. 5 Common Ion Effect

33

C. Solubility of Ag2CrO4in 0.100 M Na2CrO4?

Ag2CrO4(s) 2Ag+(aq) + CrO42

(aq)I (No entries 0.00 M 0.10 M

C in this

E column) 2x 0.10

Ksp= (2x)2(0.10) = 9.0 1012= 4x2(0.10)

1112

1025.24.0

100.9

x


[CrO42] = x = 4.7 106 M

[Ag+] = 2x = 9.5 106M

+x+2x

Y T !


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Your Turn!What effect would adding copper(II) nitrate

have on the solubility of CuS?A. The solubility would increase

B. The solubility would decrease

C. The solubility would not change

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Y T !


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Your Turn!The molar solubility of PbF2in 0.10 M Pb(NO3) solution

is 3.1 106M. What is Ksp

for PbF2?

A. 1.2 x 10-6

B. 3.1 x 10-7

C. 9.6 x 10-13

D. 3.8 x 10-12

PbF2(s) Pb2+ + 2F-

[Pb2+] = 0.10 M [F-] = 2(3.1 x 10-6)M

K = [Pb2+][F-]2= (0.1)(6.2 x 10-6)2

K = 3.8 x 10-12

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P di ti if P i it t ill F


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Predicting if Precipitate will Form

In making a solution containing various ions,

will the salt precipitate at the givenconcentrations?

For precipitate of salt to form, solution must besupersaturated.

Same as asking if given mixture of reactantsand products is an equilibrium mixture

Can judge by calculating ion product, Q

Comparing with solubility product, Ksp

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In making a solution containing various ions,

will the salt precipitate at the givenconcentrations?

For precipitate of salt to form, solution must besupersaturated.

Same as asking if given mixture of reactantsand products is an equilibrium mixture

Can judge by calculating ion product, Q

Comparing with solubility product, Ksp

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38

For: AxBy(s) x Ay+(aq)+ y Bx(aq)

Ion Product Qsp= [Ay+]x[Bx

]y Solubility Product Ksp= [A

y+]x[Bx]y

Precipitatewillform (untilsatisfy Ksp)

Qsp

> Ksp

supersaturated

Noprecipitatewill form

Qsp=KspQsp< Ksp

saturatedunsaturated

P di ti P i it ti


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Predicting Precipitation

Ex. 6 Does a precipitate of PbI2(Ksp=1.4x108)

form if 100.0 mL of 0.0500 M Pb(NO3)2are mixedwith 200.0mL of 0. 100 M NaI?

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

Ksp= [Pb2+][I]2= 1.4 108

Strategy for solving

1. Calculate concentrations in solution to be prepared

2. Calculate Qsp= [Pb2+][I]2

3. Compare Qspto Ksp

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E 6 P di ti P i it ti


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Ex. 6 Predicting PrecipitationStep 1. Calculate concentrations

Here diluting 100 mL to 300mL Vtotal= 100.0 mL + 200.0 mL = 300.0 mL

[Pb2+]o= 1.67 102M

[I] = 6.67 102M

mL

mLmmolmL

0.300

)/100.0)(0.200(

solutionofmL

Iofmmol]I[

mL

mLmmolmL

0.300

)/0500.0)(0.100(

solutionofmL

Pbofmmol]Pb[

22

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Ex. 6 Predicting PrecipitationStep 2. Calculate Qsp

Qsp

= [Pb2+][I]2= (1.67102M)(6.67102M)2

Qsp=7.43 105

Step 3. Compare Qspand Ksp

Qsp= 7.43 105

Ksp= 1.4 x 108

Qsp> Kspso precipitation willoccur

How much precipitate will form and what will ion

concentrations be at the end? Kspsmall

So most ions precipitate out as PbI2 Reaction essentially goes to completion

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Ex. 6 Predicting Precipitation If reaction goes essentially to completion, what

will final equilibrium concentrations be?

1. Do stoichiometric calculations for precipitateformation

2. Then do equilibrium calculations to determine ion

concentrations in solution.Step 1. Stoichiometric Calculation

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

Beforerxn(100mL)(0.0500M)= 5.00 mmol (200mL)(0.100M)= 20.00 mmol No effect on K

sp

AfterRxn

5.005.00 =0.00 mmol

20.002(5.00)= 10 mmol

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E 6 P edicting P ecipitation


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Ex. 6 Predicting PrecipitationStep 2. System at equilibrium

Some small amount of PbI2redissolves to formequilibrium [Pb2+]

Basically a common ion problem

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

Initial concns 0.00 10.0mmol/300mL= 3.33 x 102M

Equilibrium

concns

3.33 x 102M + 2x 3.33 x 102M

+x

Ksp= 1.4 108= [Pb2+][I]2= (x)(3.33 x 102)2

]Pb[103.1

)1033.3(

104.1x 25

22

8

M

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Ex 6 Predicting Precipitation


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Ex. 6 Predicting Precipitation

3.33 x 102M >> 2x, so approximation valid

Final equilibrium concentrations of Pb2+andIin 100.0 mL of 0.0500 M Pb(NO3)2mixed

with 200.0mL of 0. 100 M NaI

[I] = 3.33 x 102M

[Pb2+] = 1.3 x 105M

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Suppose you mix 100.0 mL of 0.200 M BaCl2

with 50.0 mL of 0.0300 M Na2SO4. WillBaSO4(Ksp= 1.1 10

10) precipitate?

BaSO4(s) Ba2+(aq)+ SO4

2(aq)

Ksp= [Ba2+

][SO42

]Step 1.Calculate concentrations

[Ba2+] = 0.133 M

mL

mL

mol

molM

0.150

0.100

BaCl1

Ba1BaCl200.0]Ba[

2

2

22

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[SO42] = 0.0100 M

Step 2. Calculate Qsp

Qsp= [Ba

2+

][SO42

] = (0.133)(0.0100) Qsp= 1.33 10

3

Step 3. Compare Qspand Ksp

Ksp

= 1.1 1010

1.33 103>> 1.1 1010

Qsp>> Ksp

So BaSO4will precipitate

mL

mL

mol

molM

0.150

0.50

)(SONa1

SO1)(SONa0300.0]SO[

42

24

4224

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pH and Solubility


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pH and Solubility Mg(OH)2(s) Mg

2+(aq) + 2OH(aq)

OH

shift equilibrium to left Add H+ shift equilibrium to right

Le Chateliers Principle

Ag3PO4(s) 3Ag+

(aq) + PO43

(aq) Add H+ solubility

H+(aq) + PO43(aq) HPO4

2(aq)

AgCl(s) Ag+

(aq) + Cl

(aq) Add H+- has no effect on solubility Why?

Clis very, very weak base, so neutral anion

So adding H+doesnt effect Clconcentration47

Your Turn!


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Your Turn!What is the molar solubility of PbI2in pure

water? Ksp= 9.8 x 10-9

A. 2.1 x 10-3

B. 1.7 x 10-3

C. 4.9 x 10-5D. 1.4 x 10-3

PbI2(s) Pb2+ + 2I-

K = 9.1 x 10 = [Pb2+][I-] = s (2s)2s = 1.4 x 10-3M

48

Your Turn!


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Your Turn!What is the molar solubility of PbI2in 0.20M

NaI solution? Ksp= 7.9 109

A. 3.9 x 10-10

B. 2.4 x 10-7

C. 6.1 x 10-9D. 2.4 x 10-7

PbI2(s ) Pb2+ + 2I- [I-] = 0.20M

K = 7.9 x 10-9= [Pb2+][I-]2= [Pb2+](0.20)2[Pb2+] = 2.4 x 10-7M

Molar solubility = [Pb2+] = 2.4 x 10-7M

49

Your Turn!


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Your Turn!What is the molar solubility of PbI2in 0.20M

Pb(NO3) solution? Ksp= 7.9 109

A. 4.9 x 10-8

B. 3.7 x 10-3

C. 1.1 x 10-4D. 2.2 x 10-4

PbI2(s ) Pb2+ + 2I- [I-] = 0.20M

K = 7.9 x 10-9= [Pb2+][I-]2= (0.20) [I-]2[I-] = 2.2 x 10-4M

Molar solubility = 1/2 [I-] = 1.1 x 10-4M

50

Your Turn!


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Your Turn!From the previous two Your Turn problems,

what can you conclude about the relativeeffect of added Pb2+vs. added I?

A. Adding Pb2+increases the solubility of PbI2

more than I

-

.B. Adding I-decreases the solubility of PbI2more than Pb2+.

C. Adding Pb2+

decreases the solubility ofPbI2 more than I-.

D. They both have the same effect on thesolubility of PbI

2

.51


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Solubility and Simultaneous

Equilibria

Part 2

Metal Oxides Undergo Reaction


53/59


Metal Oxides Undergo Reactionwith Water

53

Usually ignore reaction of ionic solid with water If the anion of a salt is very basic, a subsequent

reaction of the anion with water occurs

Such is the case of many metal oxides

Kbfor O2= 1 1022

So Ag2O actually dissociates to form Ag+and OH

Kspvalue listed takes this subsequent reaction into account

Ag2O(s)2Ag+(aq)+ O2(aq) KspO2(aq)+ H2O 2OH

(aq) KbAg2O(s)+ H2O 2Ag

+(aq)+ 2OH(aq) Knet

Metal Sulfides Also Undergo Reaction


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Metal Sulfides Also Undergo ReactionWith Water

54

Sulfide ion (S2

) is also very basic Doesnt exist in aqueous solution

Metal sulfides also undergo a subsequent

reaction with waterAg2S(s) 2Ag

+(aq) + S2(aq) KspS2(aq)+ H2O OH

(aq)+ HS(aq) Kb

Ag2S(s)+ H2O

2Ag+

(aq)+ OH

(aq)+ HS

(aq)KnetActual Ksp=[Ag

+]2[OH][HS]

Most Water-Insoluble Metal Oxides


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H+ reacts with O2to produce H2O

Releases metal ion from solid

Recall that oxide ion is too powerful a base to exist inaqueous solution

Ex. Fe2O3(s)+ 6H+

(aq)

2Fe3+

(aq)+ 3H2OHow do water-insoluble metal oxides form?

Usually in basic solution

Elimination of water is often involved

Metal ion must be capable of reacting with OHto extractoxide ion and leave water or H+

Ex: 2Ag+(aq)+ 2OH(aq)Ag2O(s)+ H2O

2Fe3+(aq)+ 3OH(aq)Fe2O3(s)+ 3H+(aq)

Most Water-Insoluble Metal OxidesDissolve in Acid

55

Formation of Insoluble Metal Sulfides


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Sulfur is below oxygen in Group VIA

As a result, metal sulfides are similar tometal oxides

S2(like O2) is too strong of a base to exist in

water Sulfides dissolve by reacting with water

Ex. Na2S(s)+ H2O 2Na+(aq)+ HS(aq)+ OH(aq)

Formation of Insoluble Metal Sulfides

56

Basic Salts Are More Soluble In


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Basic Salts Are More Soluble InAcids

57

Subsequent reactions assist the solubility of solids If anion of salt is basic, it will react in acidic

solution to dissolve more fully

Net reaction of such dissolutions is called Kspa

ZnS(s) Zn2+(aq) + S2(aq) KspS2(aq)+ H+(aq)HS(aq) 1/Ka2HS(aq)+ H+(aq)H2S

(g) 1/Ka1

ZnS(s)+ 2H+(aq)Zn2+(aq) + H2S(aq) Kspa

Kspa(acidic) and Ksp(basic) are listed in Table 18.2

Learning Check


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Learning Check

58

What is the molar solubility of BaCO3in 3.0 M

HCl? KspBaCO3= 5.0 x 10

9

H2CO3: Ka1 = 4.3 x 107Ka2 = 4.7 x 10

11

BaCO3(s)+ 2H+(aq) Ba2+(aq) + H2CO3(aq)

X =1.5 M

Reaction of Metal ions with HS


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Reaction of Metal ions with HS Metal sulfides can form

Some metal ions are so reactive that they reactwith H2S directly

These active ions include Cu2+, Pb2+, and Ni2+

A typical reaction is:

Cu2+(aq)+ H2S(aq) CuS(s)+ 2H+(aq)

Large value of Kindicates that

Equilibrium lies far to right

Only forward reaction important

Sulfides require closer investigation

ch18_solubility.ppt

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