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Chapter 16

Precipitation EquilibriumSolubility Product Principle

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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

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Solubility Product Constants

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

Molar concentration of ions raised to their stoichiometric powers at equilibrium 10sp 101.8ClAgK

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Solubility Product Constants

Solubility product constant 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%

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Solubility Product Constants

Its solubility product expression is

K Ag Ssp

2 2 10 10 49.

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Solubility Product Constants

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

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Solubility Product Constants

Its solubility product constant expression is

K Ca POsp2 3

43 2

10 10 25.

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Solubility Product Constants

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

Ksp has a fixed value for a given system at a given temperature

M Y r M s Y

K M Y

r s s

H Os r

100%

sps r r s

2

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Solubility Product Constants

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

An example is calcium ammonium phosphate.

CaNH PO Ca NH PO

K Ca NH PO

4 4 s2

4 43

sp2

4 43

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Determination of Solubility Product Constants

Example: 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.

Molar solubility can be calculated from the data:

? .

.

mol AgClL

g AgClL

mol AgCl143 g AgCl

mol AgClL

0 00192 1

134 10 5

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Determination of Solubility Product Constants

The equation for the dissociation of silver chloride and its solubility product expression are

AgCl Ag Cl

1.34 10 1.34 10

K Ag Cl

s

-5 -5

sp

M M

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Determination of Solubility Product Constants

Substitution into the solubility product expression gives

K Ag Clsp

134 10 134 10

18 10

5 5

10

. .

.

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Uses of Solubility Product Constants

We can use the solubility product constant to calculate the solubility of a compound at 25oC.

Example: 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.

Ksp= 1.1 x 10-10.

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Uses of Solubility Product Constants

Example: 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. Ksp= 1.1 x 10-10.

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

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

x x

x M

11 10

10 10

10

5

.

.

Ba SO242

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Uses of Solubility Product Constants

Now we can calculate the mass of BaSO4 in 1.00 L of saturated solution.

?

.

g BaSOL

1.0 10 molL

234 gmol

g

45

BaSOL

4

2 3 10 3

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The Reaction Quotient in Precipitation Reactions

Use solubility product constants to calculate the concentration of ions in a solution and whether or not a precipitate will form.

Example: 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 Reactions

Example: 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?

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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The Reaction Quotient in Precipitation Reactions

Calculate the Qsp for PbSO4. Solution volumes are additive. Concentrations of the important ions are:

MM

M

MM

M

Pb2+

SO 42-

2+

42-

mL 0.10200 mL

Pb

mL 0.010200 mL

SO

1000 050

1000 0050

.

.

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The Reaction Quotient in Precipitation Reactions

Finally, we calculate Qsp for PbSO4.

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

. .

.

.

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Fractional Precipitation

Fractional precipitation is a method of precipitating some ions from solution while leaving others in solution. Look at a solution that contains Cu+, Ag+, and Au+ We could precipitate them as chlorides

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Fractional Precipitation

Fractional precipitation is a method of precipitating some ions from solution while leaving others in solution. Look at a solution that contains Cu+, Ag+, and Au+ We could precipitate them as chlorides

13

sp

10sp

7sp

100.2ClAuKCl AuAuCl

108.1ClAgKCl AgAgCl

109.1ClCuKClCuCuCl

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Fractional Precipitation

Example: 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 I chlorides.

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Fractional Precipitation

Example: 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 I chlorides.

AuCl. eprecipitat torequired Cl ofion concentrat theCalculate

last. eprecipitat willCuCl reasoning, same By the

first. esprecipitatit so Ksmallest thehas AuCl

-

sp

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Fractional Precipitation

Example: 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 I chlorides.

M11

1313

sp13

-

sp

100.2010.0

100.2

Au

100.2Cl

K100.2ClAu

AuCl. eprecipitat torequired Cl ofion concentrat theCalculate

last. eprecipitat willCuCl reasoning, same By the

first. esprecipitatit so Ksmallest thehas AuCl

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Fractional Precipitation

Repeat the calculation for silver chloride.

K Ag Cl

ClAg

sp

18 10

18 10 18 100 010

18 10

10

10 10

8

.

. ..

. M

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Fractional Precipitation

For copper (I) chloride to precipitate.

K Cu Cl

ClCu

sp+

+

19 10

19 10 19 100 010

19 10

7

7 7

5

.

. ..

. M

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Fractional Precipitation

We have calculated the [Cl-] required to precipitate AuCl, [Cl-] >2.0 x 10-11 Mto precipitate AgCl, [Cl-] >1.8 x 10-8 Mto precipitate CuCl, [Cl-] >1.9 x 10-5 M

We can calculate the amount of Au+ precipitated before Ag+ begins to precipitate, as well as the amounts of Au+ and Ag+ precipitated before Cu+ begins to precipitate.

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Fractional Precipitation Example: Calculate the percent of Au+ ions that

precipitate before AgCl begins to precipitate. Use the [Cl-] from before to determine the [Au+]

remaining in solution just before AgCl begins to precipitate.

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Fractional Precipitation Example: Calculate the percent of Au+ ions that

precipitate before AgCl begins to precipitate. Use the [Cl-] from before 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

.

. .

.

.

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Fractional Precipitation

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

% Au

Au

Auunprecipitated+ unprecipitated

original

unprecipitated

100%

11 100 010

100 0 00011%8.

..

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Fractional PrecipitationThe percent of Au+ ions unprecipitated just before AgCl precipitates is

Therefore, 99.99989% of the Au+ ions precipitates before AgCl begins to precipitate.

% Au

Au

Auunprecipitated+ unprecipitated

original

unprecipitated

100%

11 100 010

100 0 00011%8.

..

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Fractional PrecipitationSimilar calculations for the concentration of Ag+

ions unprecipitated before CuCl begins to precipitate gives atedunprecipit Ag105.9Ag

109.1

108.1

Cl

108.1Ag

108.1ClAg

6

5

1010

10

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Fractional Precipitation

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

atedunprecipit

6

original

atedunprecipit+atedunprecipit

%095.0100010.0

105.9

%100Ag

AgAg %

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Fractional PrecipitationThe percent of Au+ 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 %

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Factors that Affect Solubility

CaF2 (s) Ca+2 (aq)+ 2F- (aq)

Addition of a Common ion (F- from NaF) Solubility decreases Equilibrium shifts to left

Changes in pH (H+ reacts with F-) Solubility increases (with increasing pH) Equilibrium shifts to right

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Factors that Affect SolubilityAg+(aq) + 2NH3(aq) Ag(NH3)2

+ (aq)

complex ion

Formation of a complex ion Lewis Acid base chemistry

Calculate Kf Formation Constant

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Factors that Affect SolubilityAg+(aq) + 2NH3(aq) Ag(NH3)2

+ (aq)

complex ionAgCl(s) Ag+(aq) + Cl-(aq)

In formation of complex ion Removed Ag+ from the equilibrium Equilibrium shifts to right Favors dissolving AgCl

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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?

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Synthesis Question

stone. solida as than

rather solution in ions towardsright, theto

mequilibriu theshifts water more Drinking

COOCaOHCa(COO)

oxalate calciumfor

expressionproduct Solubility

2aq2

2aq2s2

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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?