Ionic Equilibria III: The Solubility Product Principle

1

20Ionic Equilibria III:

The Solubility Product Principle

2

Chapter Goals

1. Solubility Product Constants2. Determination of Solubility Product

Constants3. Uses of Solubility Product Constants4. Fractional Precipitation5. Simultaneous Equilibria Involving Slightly

Soluble Compounds6. Dissolving Precipitates

3

Solubility Product Constants

• Silver chloride, AgCl,is rather insoluble in water.• Careful experiments show that if solid AgCl is

placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water.

aqaq ClAgClAg

4


• Silver chloride, AgCl,is rather insoluble in water.• Careful experiments show that if solid AgCl is

placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water.

5


• The equilibrium constant expression for this dissolution is called a solubility product constant.– Ksp = solubility product constant

-10-sp 101.8]][Cl[AgK

6


• The solubility product constant, Ksp, for a compound is the product of the concentrations of the constituent ions, each raised to the power that corresponds to the number of ions in one formula unit of the compound.

• Consider the dissolution of silver sulfide in water.

Ag S 2 Ag + S2+ 2-

H O+ 2-2 100%

7


• The solubility product expression for Ag2S is:

K Ag Ssp

2 2 10 10 49.

8


• The dissolution of solid calcium phosphate in water is represented as:

Ca PO 3 Ca 2 PO3

2+43

2 s

H O2

43

100%

2

The solubility product constant expression is:

You do it!

K sp Ca2 3PO4

3 21.0 10 25

9


• In general, the dissolution of a slightly soluble compound and its solubility product expression are represented as:

M Y r M s Y

K M Y

r s s

H Os r

100%

sps r r s

2

10


• The same rules apply for compounds that have more than two kinds of ions.

• One example of a compound that has more than two kinds of ions is calcium ammonium phosphate.

344

2sp

3aq4

1aq4

2aqs44

PO NH CaK

PONHCa POCaNH

11

Determination of Solubility Product Constants• Example 20-1: One liter of saturated silver chloride

solution contains 0.00192 g of dissolved AgCl at 25oC. Calculate the molar solubility of, and Ksp for, AgCl.

• The molar solubility can be easily calculated from the data:

L

AgCl mol 1034.1

AgCl g 143

AgCl mol 1

L

AgCl g 00192.0

L

AgCl mol ?

5

12

Determination of Solubility Product Constants• The equation for the dissociation of silver

chloride, the appropriate molar concentrations, and the solubility product expression are:

AgCl Ag Cl

1.34 10 1.34 10

K Ag Cl

s

-5 -5

sp

M M

13

Determination of Solubility Product Constants• Substitution of the molar concentrations into the

solubility product expression gives:

K Ag Clsp

134 10 134 10

18 10

5 5

10

. .

.

14

Determination of Solubility Product Constants• Example 20-2: One liter of saturated calcium

fluoride solution contains 0.0167 gram of CaF2 at 25oC. Calculate the molar solubility of, and Ksp for, CaF2.

1. Calculate the molar solubility of CaF2.

L

CaF mol 1014.2

g 78.1

mol 1

L 1.0

CaF g 0167.0

L

CaF mol ?

24

22

15

11

244

22sp

444

-1aq

2aq2

10 92.3

1028.41014.2

F CaK

)1014.22( 1014.2 1014.2

F 2 Ca CaF

MMM

Determination of Solubility Product Constants• From the molar solubility, we can find the ion

concentrations in saturated CaF2. Then use those values to calculate the Ksp.– Note: You are most likely to leave out the factor of 2

for the concentration of the fluoride ion!

)1014.22( 1014.2 1014.2

F 2 + Ca CaF 444

-+22

MMM

16

Uses of Solubility Product Constants• The solubility product constant can be used to calculate the

solubility of a compound at 25oC.• Example 20-3: Calculate the molar solubility of barium sulfate,

BaSO4, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate at 25oC. For barium sulfate, Ksp= 1.1 x 10-10.

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Uses of Solubility Product Constants

BaSO Ba SO

Ba SO

4 s aq2+

4 aq

242

2

1011 10

xM xM xM

Ksp .

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Uses of Solubility Product Constants• Make the algebraic substitution of x’s into

solubility product expression and solve for x, giving the ion concentrations.

M

Mx

xx

100.1SOBa

100.1

101.1

524

2

5

10

19

Uses of Solubility Product Constants• Finally, to calculate the mass of BaSO4 in 1.00 L

of saturated solution, use the definition of molarity.

L

BaSO g103.2

mol

g 234

L

mol101.0

L

BaSO g ?

43

54

20

Uses of Solubility Product Constants• Example 20-4: The solubility product

constant for magnesium hydroxide, Mg(OH)2, is 1.5 x 10-11. Calculate the molar solubility of magnesium hydroxide and the pH of a saturated magnesium hydroxide solution at 25oC.

You do it!

21

Uses of Solubility Product Constants• Be careful, do not forget the stoichiometric

coefficient of 2!

Mg(OH) Mg 2 OH

2

K Mg OH

2 aq2

aq

sp2 2

xM xM xM

22

Uses of Solubility Product Constants• Substitute the algebraic expressions into the

solubility product expression.

x x

x

x

x

2 15 10

4 15 10

3 75 10

16 10

2 11

3 11

3 12

4

.

.

.

. molar solubility

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Uses of Solubility Product Constants• Solve for the pOH and pH.

x x

x

x

x

M M

2 15 10

4 15 10

3 75 10

16 10

2 3 2 10

2 11

3 11

3 12

4

4

.

.

.

.

.

molar solubility

OH

pOH 3.49 pH 10.51

-

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The Common Ion Effect in Solubility Calculations• Example 20-5: Calculate the molar

solubility of barium sulfate, BaSO4, in 0.010 M sodium sulfate, Na2SO4, solution at 25oC. Compare this to the solubility of BaSO4 in pure water. (Example 20-3). (What is the common ion? How was a common ion problem solved in Chapter 19?)

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The Common Ion Effect in Solubility Calculations1. Write equations to represent the equilibria.

MxMxMx

MMM

SO Ba BaSO

)010.0( )010.0(2 010.0

SO Na 2 SONa

2aq4

2aqs4

24

+%10042

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The Common Ion Effect in Solubility Calculations2. Substitute the algebraic representations of the

concentrations into the Ksp expression and solve for x.

48

10-

1024

2sp

BaSO of solubilitymolar 101.1

101.1= 0.010

0.0100.010

applied. becan assumption gsimplifyin The

010.0

101.1SOBaK

x

x

x

xx

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The Common Ion Effect in Solubility Calculations• The molar solubility of BaSO4 in 0.010 M Na2SO4 solution is 1.1 x 10-8 M.

• The molar solubility of BaSO4 in pure water is 1.0 x 10-5 M.

– BaSO4 is 900 times more soluble in pure water than in 0.010 M sodium sulfate!

– Adding sodium sulfate to a solution is a fantastic method to remove Ba2+ ions from solution!

• If your drinking water were suspected to have lead ions in it, suggest a method to prove or disprove this suspicion.

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The Reaction Quotient in Precipitation Reactions• The reaction quotient, Q, and the Ksp of a compound are

used to calculate the concentration of ions in a solution and whether or not a precipitate will form.

• Example 20-6: We mix 100 mL of 0.010 M potassium sulfate, K2SO4, and 100 mL of 0.10 M lead (II) nitrate, Pb(NO3)2 solutions. Will a precipitate form?

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The Reaction Quotient in Precipitation Reactions1. Write out the solubility expressions.

K SO 2 K SO

Pb NO Pb NO

Will PbSO precipitate?

2 4H O 100% +

42

3 2H O 100% 2+

3-

4

2

2

2

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MM

M

MM

M

Pb2+

SO 42-

2+

42-

mL 0.10200 mL

Pb

mL 0.010200 mL

SO

1000 050

1000 0050

.

.

The Reaction Quotient in Precipitation Reactions

• Calculate the Qsp for PbSO4.– Assume that the solution volumes are additive.– Concentrations of the important ions are:

MM

MPb

2+2+

mL 0.10200 mL

Pb

100

0 050.

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The Reaction Quotient in Precipitation Reactions• Finally, calculate Qsp for PbSO4 and compare it

to the Ksp.

Q Pb SO

K for PbSO

Q K therefore solid forms

sp2

42

sp 4

sp sp

0 050 0 0050

2 5 10

18 10

4

8

. .

.

.

32

The Reaction Quotient in Precipitation Reactions• Example 20-7: Suppose we wish to remove mercury from an

aqueous solution that contains a soluble mercury compound such as Hg(NO3)2. We can do this by precipitating mercury (II) ions as the insoluble compound HgS. What concentration of sulfide ions, from a soluble compound such as Na2S, is required to reduce the Hg2+ concentration to 1.0 x 10-8 M? For HgS, Ksp=3.0 x 10-53.

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The Reaction Quotient in Precipitation Reactions

e.precipitat llmercury wi the100.3 exceedslightly

just toadded is S,Na of form in the ,Senough If

100.3100.1

100.3

Hg

KS

Sfor solve

100.3S HgK

HgSSHg

45

2-2

458

53

2

sp2

2

5322sp

22

M

M

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The Reaction Quotient in Precipitation Reactions• Example 20-8: Refer to example 20-7. What

volume of the solution (1.0 x 10-8 M Hg2+ ) contains 1.0 g of mercury?

gal 125,000L 100.5

Hg mol 101.0

L 1.0

Hg g 201

Hg mol 1Hg g 1.0 L ?

5

282

22

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Fractional Precipitation

• The method of precipitating some ions from solution while leaving others in solution is called fractional precipitation.– If a solution contains Cu+, Ag+, and Au+, each ion can

be precipitated as chlorides.

13

sp

10sp

7sp

100.2Cl AuKCl AuAuCl

108.1Cl AgKCl AgAgCl

109.1Cl CuKClCuCuCl

36


• Example 20-9: If solid sodium chloride is slowly added to a solution that is 0.010 M each in Cu+, Ag+, and Au+ ions, which compound precipitates first? Calculate the concentration of Cl- required to initiate precipitation of each of these metal chlorides.

AuCl. eprecipitat torequired Cl ofion concentrat theCalculate 1.

last. eprecipitat willCuCl reasoning, same By the

first. esprecipitatit so Ksmallest thehas AuCl

-

sp

37


M11

1313

sp13

-

sp

100.2010.0

100.2

Au

100.2Cl

K100.2ClAu

AuCl. eprecipitat torequired Cl ofion concentrat theCalculate 1.

last. eprecipitat willCuCl reasoning, same By the

first. esprecipitatit so Ksmallest thehas AuCl

38


• Repeat the calculation for silver chloride.

M8

1010

10sp

108.1

010.0

108.1

Ag

108.1Cl

108.1Cl AgK

39


• Finally, for copper (I) chloride to precipitate.

M5

7

+

7

7+sp

109.1

010.0

109.1

Cu

109.1Cl

109.1ClCuK

40


• These three calculations give the [Cl-] required to precipitate AuCl ([Cl-] >2.0 x 10-11 M), to precipitate AgCl ([Cl-] >1.8 x 10-8 M), and to precipitate CuCl ([Cl-] >1.9 x 10-5 M).

• It is also possible to calculate the amount of Au+ precipitated before the Ag+ begins to precipitate, as well as the amounts of Au+ and Ag+ precipitated before the Cu+ begins to precipitate.

41


• Example 20-10: Calculate the percentage of Au+ ions that precipitate before AgCl begins to precipitate.– Use the [Cl-] from Example 20-9 to determine the [Au+]

remaining in solution just before AgCl begins to precipitate.

Au Cl

AuCl

Au Au unprecipitated

2 0 10

2 0 10 2 0 10

18 10

11 10

13

13 13

8

5

.

. .

.

.

42


• The percent of Au+ ions unprecipitated just before AgCl precipitates is:

• Therefore, 99.9% of the Au+ ions precipitates before AgCl begins to precipitate.

atedunprecipit

5

original

atedunprecipit+atedunprecipit

%1.0100010.0

101.1

%100Au

AuAu %

43


• A similar calculation for the concentration of Ag+ ions unprecipitated before CuCl begins to precipitate is:

atedunprecipit Ag105.9Ag

109.1

108.1

Cl

108.1Ag

108.1Cl Ag

6

5

1010

10

44


• The percent of Ag+ ions unprecipitated just before AgCl precipitates is:

• Thus, 99.905% of the Ag+ ions precipitates before CuCl begins to precipitate.

atedunprecipit

6

original

atedunprecipit+atedunprecipit

%095.0100010.0

105.9

%100Ag

AgAg %

45

Simultaneous Equilibria Involving Slightly Soluble Compounds

• Equlibria that simultaneously involve two or more different equilibrium constant expressions are simultaneous equilibria.

46


• Example 20-12: If 0.10 mole of ammonia and 0.010 mole of magnesium nitrate, Mg(NO3)2, are added to enough water to make one liter of solution, will magnesium hydroxide precipitate from the solution?

– For Mg(OH)2, Ksp = 1.5 x 10-11. Kb for NH3 = 1.8 x 10-5.

1. Calculate Qsp for Mg(OH)2 and compare it to Ksp.

– Mg(NO3)2 is a soluble ionic compound so [Mg2+] = 0.010 M.

– Aqueous ammonia is a weak base that we can calculate [OH-].

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xMxMMx 10.0

OHNH OH +NH -423

OH101.3= 108.1

10.0

108.1NH

OHNHK

10.0

OHNHOHNH

3-62

5

3

4b

-423

Mxx

x

xx

xMxMMx


xxx

xMxMMx

10.0

108.1NH

OHNHK

10.0

OHNH OHNH

5

3

4b

-423

48


• Once the concentrations of both the magnesium and hydroxide ions are determined, the Qsp can be calculated and compared to the Ksp.

e.precipitat willMg(OH) thus,KQ

107.1

103.1010.0

OH MgQ

2spsp

8

23

22sp

49


• Example 20-13: How many moles of solid ammonium chloride, NH4Cl, must be used to prevent precipitation of Mg(OH)2 in one liter of solution that is 0.10 M in aqueous ammonia and 0.010 M in magnesium nitrate, Mg(NO3)2 ? (Note the similarity between this problem and Example 20-12.)

You do it!• Calculate the maximum [OH-] that can exist in a

solution that is 0.010 M in Mg2+.

50


K Mg OH

OHMg

OH

sp2

2

2 11

211 11

9

5

15 10

15 10 15 100 010

15 10

3 9 10

.

. ..

.

. M

51


• Using the maximum [OH-] that can exist in solution, determine the number of moles of NH4Cl required to buffer 0.10 M aqueous ammonia so that the [OH-] does not exceed 3.9 x 10-5 M.

NH Cl NH Cl

NH + H O NH + OH

10 10 10

4100%

4

3 2 4+ -

-5 -5 -5

xM xM

M M M010 3 9 3 9 3 9. . . .

52



109.310.0

109.3109.3

108.1NH

OHNH=K

109.3 109.3 109.310.0

OH NH OH NH

ClNHClNH

5-

5-5-

5

3

+4

b

5-5-5-

-423

4100%

4

x

MMM

xMxM

53


Cl.NH mol 0.046 are there

solution, of L 1.0 is thereBecause

ClNH046.0

108.110.0

109.3

4

4

5-5

Mx

x

54


• Check these values by calculating Qsp for Mg(OH)2.

m!equilibriuat is system thisThus

KQ

105.1

109.3010.0

OH Mg=Q

spsp

11

25

2+2sp

55


• Use the ion product for water to calculate the [H+] and the pH of the solution.

H OH

HOH

H

pH

+ -14

+-14

+-14

10 10

10 10

10 10

3 9 102 6 10

9 59

510

.

.

.

..

.

M

56

Dissolving Precipitates

• For an insoluble solid, if the ion concentrations (of either the cation or anion) are decreased, a solid precipitate can be dissolved.– The trick is to make Qsp < Ksp.

• One method is to convert the ions into weak electrolytes.– Make these ions more water soluble.

• If insoluble metal hydroxides are dissolved in strong acids, they form soluble salts and water.

57


• For example, look at the dissolution of Mg(OH)2 in HCl.

.MgCl soluble more the toconverted

is Mg(OH) insoluble that theNotice

OH 2MgH 2Mg(OH)

or

OH 2MgClHCl 2Mg(OH)

2

2

2+2aq

+aq2(s)

2aq2aq2(s)

58


• A second method is to dissolve insoluble metal carbonates in strong acids.– The carbonates will form soluble salts, carbon

dioxide, and water.

dissolve. carbonate themakes and ions removes

dioxidecarbon offormation that theNotice

OHCOCa2HCaCO

or

OHCOCaCl2HClCaCO

2g2+2aq

+aqs3

2g2aq2aqs3

59


• A third method is to convert an ion to another species by an oxidation-reduction reaction.

• For example, the dissolution of insoluble metal sulfides in hot nitric acid causes the sulfide ions to be oxidized to elemental sulfur.

.K<Q themaking S toS oxidizes NO The

K108.4S Pb

OH 4NO 2S 3Pb 3NO 2H 8PbS 3

spsp02

3

sp2822

2g0s

23s

60


• A fourth method is complex ion formation.• The cations in many slightly soluble compounds will

form complex ions.– This is the method used to dissolve unreacted AgBr and

AgCl on photographic film.• Photographic “hypo” is Na2S2O3.

-aq

3

aq232-2

32s

aqaq2323aq322s

(s)

2sh

s

BrOSAgOS 2AgBr

NaBr)OAg(SNaOSNa 2AgBr

left. is AgBr film theofportion unexposed In the

BrAg 2AgBr 2

61


• Copper(II) hydroxide, which is light blue colored, dissolves in aqueous ammonia to form dark blue, [Cu(NH3)4]2+.

Cu(OH) 4 NH Cu(NH ) 2 OH

Cu(OH) Cu 2 OH

Cu 4NH Cu(NH )

2 s 3 aq 3 4 aq2

2 s aq2

aq

aq2

3 3 4 aq2

62

Complex Ion Equilibria

• A metal ion coordinated to several neutral molecules or anions forms compounds called complex ions.

• Familiar examples of complex ions include:

23223

4323

) Pt(NHCl)Co(NH

)Cu(NH )Ag(NH

63

243

43

2

d

322

43

)Cu(NH

NHCuK

NH 4Cu )Cu(NH


• The dissociation of complex ions can be represented similarly to equilibria.

• For example:

23

23

d

323

)Ag(NH

NH AgK

NH 2Ag )Ag(NH

64


• Complex ion equilibrium constants are called dissociation constants.

• Example 20-14: Calculate the concentration of silver ions in a solution that is 0.010 M in [Ag(NH3)2]+.

– Kd = 6.3 x 10-8

1. Write the dissociation reaction and equilibrium concentrations.

65


2. Substitute the algebraic quantities into the dissociation expression.

Ag(NH Ag 2 NH

2 3 3)

.2

0 010

x M xM xM

66



103.6010.0

2K

103.6)Ag(NH

NH AgK

82

d

8

23

23

d

x

xx

67


Ag104.5

106.1

103.64

4

103

103

Mx

x

x

68


• Example 20-15: How many moles of ammonia must be added to 2.00 L of water so that it will just dissolve 0.010 mole of silver chloride, AgCl?

• The reaction of interest is:

values.for these usedbeen have ratiosreaction actual The

100.5 100.5 )100.5( 2 100.5

Cl + )Ag(NH NH 2 AgCl3333

-233s

MMMM

69


• Two equilibria are involved when silver chloride dissolves in aqueous ammonia.

AgCl Ag Cl K Ag Cl

Ag(NH ) Ag 2 NH KAg NH

Ag(NH )

sp

3 2 3 d3

3 2

18 10

6 3 10

10

210

.

.

70


• The [Ag+] in the solution must satisfy both equilibrium constant expressions. Because the [Cl-] is known, the equilibrium concentration of Ag+ can be calculated from Ksp for AgCl.

Cl

108.1Ag

108.1Cl Ag10

10

71


possible. Ag maximum theisWhich

Ag 106.3

100.5

108.1

Cl

108.1Ag

108.1Cl Ag

+

+8

3

10

10

10

M

72

M094.0NH

1075.8NH

3

323


• Substitute the maximum [Ag+] into the dissociation constant expression for [Ag(NH3)2]+ and solve for the equilibrium concentration of NH3.

3

23

8

8

23

23

d

100.5

NH106.3

103.6)Ag(NH

NH AgK

73


• The amount just calculated is the equilibrium concentration of NH3 in the solution. But the total concentration of NH3 is the equilibrium amount plus the amount used in the complex formation.

amount amount

reaction mequilibriu

104.0)100.5(2094.0NH 3total3 MMM

74


• Finally, calculate the total number of moles of ammonia necessary.

? mol NH L0.104 mol NH

L33

2 0

0 21

.

. M

75

Synthesis Question

• Most kidney stones are made of calcium oxalate, Ca(O2CCO2). Patients who have their first kidney stones are given an extremely simple solution to stop further stone formation. They are told to drink six to eight glasses of water a day. How does this stop kidney stone formation?

76

Synthesis Question

stone. solid a asn rather tha

solution in ions towardsright, theto

mequilibriu theshifts water more Drinking

COOCaOHCa(COO)

is oxalate calciumfor

expressionproduct solubility The

2aq2

2aq2s2

77

Group Question

• The cavities that we get in our teeth are a result of the dissolving of the material our teeth are made of, calcium hydroxy apatite. How does using a fluoride based toothpaste decrease the occurrence of cavities?

78

20Ionic Equilibria III:

The Solubility Product Principle

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Ionic Equilibria III: The Solubility Product Principle