the solubility product principle. 2 silver chloride, agcl,is rather insoluble in water. careful...

The Solubility Product Principle

2

• 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.

Solubility Product Constants

AgCl(s) ⇌ Ag+(aq) + Cl-(aq)

3

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

• Ksp = solubility product constant

Solubility Product Constants

Ksp =[Ag+ ][Cl- ] =1.8 ×10 -10

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4

• 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.

Solubility Product Constants

Ag2S(s) ⇌ 2Ag+(aq) + S2-(aq) Ksp = [Ag+]2[S-2]

5

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

The solubility product constant expression is:

Solubility Product Constants

Ca3(PO4)(s) ⇌ 3Ca+2(aq) + 2PO43-(aq)

Ksp = [Ca2+]3[PO4-3]2

6

• 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.

CaNH4PO4(s) ⇌ Ca+2(aq) + NH4+ aq) + PO4

3-(aq)

Solubility Product Constants

K sp = Ca+2⎡⎣ ⎤⎦ NH4+⎡⎣ ⎤⎦ PO4

3−⎡⎣ ⎤⎦

7

• All Ksp values are small.

• All salts are only slightly soluble

Ksp Table

8

• One liter of saturated silver chloride solution contains 0.00192 g of dissolved AgCl at 25°C. Calculate the molar solubility of, and Ksp for, AgCl.

• Molar solubility can be calculated from the data:

Determination of Solubility Product Constants

? mol AgCl

L=0.00192 g AgCl

L×1 mol AgCl143 g AgCl

=1.34 ×10−5 mol AgCl

L

9

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

Determination of Solubility Product Constants

AgCl(s) Ä Ag(aq)+ + Cl(aq)

-

1.34x10-5M 1.34x10-5M

Ksp= Ag+⎡⎣ ⎤⎦ Cl-⎡⎣ ⎤⎦

10

• Substitution into the solubility product expression gives

Determination of Solubility Product Constants

Ksp = Ag+⎡⎣ ⎤⎦ Cl−⎡⎣ ⎤⎦

= 1.34 ×10−5⎡⎣ ⎤⎦ 1.34 ×10−5⎡⎣ ⎤⎦

=1.8 ×10−10

11

• One liter of saturated calcium fluoride solution contains 0.0167 gram of CaF2. Calculate the molar solubility of, and Ksp for, CaF2.

• Calculate the molar solubility of CaF2.

Determination of Solubility Product Constants

? mol CaF2

L=0.0167 g CaF2

1.0 L×1 mol78.1 g

=2.14 ×10−4 mol CaF2

L

12

• From the molar solubility, we can find the ion concentrations in saturated CaF2. Then use those values to calculate the Ksp.

Determination of Solubility Product Constants

CaF2 Ä Ca2+ + 2F-

2.14×10−4M 2(2.14 ×10−4M )

Ksp = Ca2+⎡⎣ ⎤⎦ F−⎡⎣ ⎤⎦2

= 2.14 ×10−4( ) 4.28 ×10−4( )2

=3.92 ×10−11

13

• Calculate the molar solubility of barium sulfate, BaSO4, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate. Ksp= 1.1 x 10-10.

Uses of Solubility Product Constants

x( ) x( ) =1.1×10−10

x =1.0 ×10−5M

Ba2+⎡⎣ ⎤⎦= SO42−⎡⎣ ⎤⎦=1.0 ×10

−5 M

BaSO4 Ä Ba2+ + SO42−

xM xM

K sp = Ba2+⎡⎣ ⎤⎦ SO42−⎡⎣ ⎤⎦

2=1.1 ×10−10

14

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

Uses of Solubility Product Constants

? g BaSO4

L=1.0 ×10−5mol

L×234 gmol

=2.3 ×10−3 g BaSO4

L

15

• The Ksp for magnesium hydroxide, Mg(OH)2, is 1.5 x 10-11. Calculate the molar solubility of magnesium hydroxide and the pH of saturated magnesium hydroxide solution.

Uses of Solubility Product Constants

Ksp = [Mg2+][OH-]2

Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)

16

• Substitute into the solubility product expression.

Uses of Solubility Product Constants

x( ) 2x( )2 =1.5 ×10−11

4x3 =1.5 ×10−11

x3 =3.75 ×10−12

x =1.6 ×10−4 =molar solubility

OH-⎡⎣ ⎤⎦=2 ×M =3.2 ×10−4M

pOH=3.49 pH=10.51

17

• Calculate the concentration of calcium ion in a calcium phosphate, Ca3(PO4)2, in pure water. Ksp= 2.1 x 10-33.

Ca3(PO4)2 ⇌ 3Ca2+ + 2PO43-

3x 2x2.1 x 10-33 = [3x]3[2x]2 = (27x3)(4x2)

2.1 x 10-33 = 108x5

x5 = 2.1 x 10-33/108 = 1.94 x 10-35

x = x = 1.14 x 10-7; [Ca2+] = 3x = 3.42 x 10-7

Uses of Solubility Product Constants

18

• Solubility is decreased when a common ion is added.

• This is an application of Le Châtelier’s principle:

BaSO4 ⇌ Ba2+ + SO42-

• as SO42- (from Na2SO4, say) is added, the

equilibrium shifts away from the increase.

• Therefore, BaSO4(s) is formed and precipitation occurs.

• As Na2SO4 is added to the system, the solubility of BaSO4 decreases.

Common Ion Effect

19

• Calculate the molar solubility of barium sulfate, BaSO4, in 0.010 M sodium sulfate, Na2SO4, solution. Compare this to the solubility of BaSO4 in pure water.

• Write equations to represent the equilibria.

The Common Ion Effect in Solubility

Na2SO4100%⏐ →⏐ ⏐ 2 Na++ SO4

2-

0.010M 2(0.010M) (0.010M)

BaSO4 s( ) Ä Ba2++ SO42-

x M x M x M

20

• Substitute the concentrations into the Ksp expression and solve for x.

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.900 times greater

The Common Ion Effect in Solubility

Ksp = Ba2+⎡⎣ ⎤⎦ SO42−⎡⎣ ⎤⎦=1.1×10

−10

= x( ) 0.010 + x( )The simplifying assumption can be applied.

0.010 + x( ) ≈0.010

0.010 x=1.1×10 -10

x =1.1×10−8 =molar solubility of BaSO4

21

• 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?

K2SO4 → 2K+ + SO42-

Pb(NO3)2 → Pb2+ + 2NO3-

Will PbSO4 precipitate?

The Reaction Quotient in Precipitation

22

• Calculate the Qsp for PbSO4.

• Solution volumes are additive.

• Concentrations of the important ions are:

The Reaction Quotient in Precipitation Reactions

MPb2+ =

100 mL× 0.10M200 mL

=0.050M Pb2+

MSO4

2- =100 mL× 0.010M

200 mL=0.0050M SO4

2-

23

• Finally, we calculate Qsp for PbSO4.Qsp = [Pb2+][SO4

2-] = (0.050)(0.0050) = 2.5 x 10-4

Ksp = 1.8 x 10-8

Qsp > Ksp. therefore solid forms

The Reaction Quotient in Precipitation Reactions

24

• 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 Na 2 S, is required to reduce the Hg2+ concentration to 1.0 x 10-8 M? For HgS, Ksp = 3.0 x 10-53.

The Reaction Quotient in Precipitation Reactions

25

The Reaction Quotient in Precipitation Reactions

Hg2+ +S2−→ HgS←

Ksp = Hg2+⎡⎣ ⎤⎦ S2−⎡⎣ ⎤⎦=3.0 ×10−53

solve for S2−⎡⎣ ⎤⎦

S2−⎡⎣ ⎤⎦=Ksp

Hg2+⎡⎣ ⎤⎦=3.0 ×10−53

1.0 ×10−8 =3.0 ×10−45M

If enough S2-, in the form of Na2S, is added to just

slightly exceed 3.0 ×10−45M the mercury will precipitate.

26

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

2- to precipitate red Ag2CrO4 and yellow PbCrO4. Which precipitates first?

• Ksp for Ag2CrO4 = 9.0 x 10-12

• Ksp for PbCrO4 = 1.8 x 10-14

• Solution

• The substance whose Ksp is first exceeded precipitates first.

• The ion requiring the lesser amount of CrO42-

ppts. first.

Separating Salts

27

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

2- to precipitate red Ag2CrO4 and yellow PbCrO4. Which precipitates first?

• Ksp for Ag2CrO4 = 9.0 x 10-12

• Ksp for PbCrO4 = 1.8 x 10-14

• Solution

• Calculate [CrO42-] required by each ion.

Separating Salts

28

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

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

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

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

PbCrO4 [CrO42-] = 9.0 x 10-13 M

Ag2CrO4 [CrO42-] = 2.3 x 10-8 M

PbCrO4 precipitates first.

Separating Salts

29

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

• SolutionWe know that [CrO4

2-] = 2.3 x 10-8 M to begin to ppt. Ag2CrO4.

What is the Pb2+ conc. at this point?Solution[Pb2+] = Ksp / [CrO4

2-] = 1.8 x 10-14 /2.3 x 10-8 = 7.8 x 10-7 M

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

Separating Salts

30

99.9961% of Pb2+ ions precipitates before AgCrO4 begins to precipitate.

€

% Pbunprecipitated+ =

[Pb+]unprecipitated

[Pb+]original

100x

= 7.8 10x -6

0.020 100x =0.0039 %unprecipitated

Separating Salts

31

• 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

• CuCl ⇌ Cu+ + Cl - Ksp = [Cu+][Cl -] = 1.9 x 10-7

• AgCl ⇌ Ag+ + Cl - Ksp = [Ag+][Cl -] = 1.8 x 10-10

• AuCl ⇌ Au+ + Cl- Ksp = [Au+][Cl -] = 2.0 x 10-13

Fractional Precipitation

32

• 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.

Fractional Precipitation

AuCl has the smallest Ksp so it precipitates first.

By the same reasoning, CuCl will precipitate last.

1. Calculate the concentration of Cl- required to precipitate AuCl.

Au+⎡⎣ ⎤⎦ Cl−⎡⎣ ⎤⎦=2.0 ×10−13 =Ksp

Cl−⎡⎣ ⎤⎦=2.0 ×10−13

Au+⎡⎣ ⎤⎦=2.0 ×10−13

0.010=2.0 ×10−11M

33

• Repeat the calculation for silver chloride.

Fractional Precipitation

Ksp = Ag+⎡⎣ ⎤⎦ Cl−⎡⎣ ⎤⎦=1.8 ×10−10

Cl−⎡⎣ ⎤⎦=1.8 ×10−10

Ag+⎡⎣ ⎤⎦=1.8 ×10−10

0.010

=1.8 ×10−8M

34

• For copper(I) chloride to precipitate.

Fractional Precipitation

Ksp = Cu+⎡⎣ ⎤⎦ Cl−⎡⎣ ⎤⎦=1.9 ×10−7

Cl−⎡⎣ ⎤⎦=1.9 ×10−7

Cu+⎡⎣ ⎤⎦=1.9 ×10−7

0.010

=1.9 ×10−5M

35

• We have calculated the [Cl-] required

• to precipitate AuCl, [Cl-] > 2.0 x 10-11 M

• to precipitate AgCl, [Cl-] > 1.8 x 10-8 M

• to 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.

Fractional Precipitation

36

• Calculate the percent of Au+ ions that precipitate before AgCl begins to precipitate.

• Use the [Cl-] = 1.8 x 10-8 M to determine the [Au+] remaining in solution just before AgCl begins to precipitate.

Fractional Precipitation

Au+⎡⎣ ⎤⎦ Cl−⎡⎣ ⎤⎦=2.0 ×10−13

Au+⎡⎣ ⎤⎦=2.0 ×10−13

Cl−⎡⎣ ⎤⎦=2.0 ×10−13

1.8 ×10−8

Au+⎡⎣ ⎤⎦=1.1×10−5 = Au+⎡⎣ ⎤⎦ unprecipitated

37

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

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

Fractional Precipitation

% Auunprecipitated+ =

Au+⎡⎣ ⎤⎦unprecipitated

Au+⎡⎣ ⎤⎦original

×100%

=1.1×10−5

0.010×100 =0.1%unprecipitated

38

• Similar calculations for the concentration of Ag+ ions unprecipitated before CuCl begins to precipitate gives

Fractional Precipitation

Ag+⎡⎣ ⎤⎦ Cl−⎡⎣ ⎤⎦=1.8 ×10−10

Ag+⎡⎣ ⎤⎦=1.8 ×10−10

Cl−⎡⎣ ⎤⎦=1.8 ×10−10

1.9 ×10−5

Ag+⎡⎣ ⎤⎦=9.5 ×10−6 = Ag+⎡⎣ ⎤⎦ unprecipitated

39

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

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

Fractional Precipitation

% Agunprecipitated+ =

Ag+⎡⎣ ⎤⎦unprecipitated

Ag+⎡⎣ ⎤⎦original

×100%

=9.5 ×10−6

0.010×100 =0.095%unprecipitated

40

• 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.

• 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-]

Simultaneous Equilibria

41

Simultaneous Equilibria

NH3 +H2O→ NH4

+ +OH-←

0.10 −x( )M xM xM

Kb =NH4

+⎡⎣ ⎤⎦ OH−⎡⎣ ⎤⎦NH3[ ]

=1.8 ×10−5

=x( ) x( )0.10 −x( )

x2 =1.8 ×10−6 ∴x=1.3 ×10 -3M = OH−⎡⎣ ⎤⎦

42

• Since we now have the concentrations of both the magnesium and hydroxide ions, we can calculate the Qsp and compare it to the Ksp.

Simultaneous Equilibria

Qsp = Mg2+⎡⎣ ⎤⎦ OH−⎡⎣ ⎤⎦2

= 0.010( ) 1.3 ×10−3( )2

=1.7 ×10−8

Qsp >Ksp, thus Mg(OH)2 will precipitate.

43

• 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?

• Calculate the maximum [OH-] that can exist in a solution that is 0.010 M in Mg2+.

Simultaneous Equilibria

44

• Calculate the maximum [OH-] that can exist in a solution that is 0.010 M in Mg2+.

Simultaneous Equilibria

Ksp = Mg2+⎡⎣ ⎤⎦ OH−⎡⎣ ⎤⎦2=1.5 ×10−11

OH−⎡⎣ ⎤⎦2=1.5 ×10−11

Mg2+⎡⎣ ⎤⎦=1.5 ×10−11

0.010

=1.5 ×10−9

OH−⎡⎣ ⎤⎦≤3.9 ×10−5M

45

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

Simultaneous Equilibria

NH4Cl 100%⏐ →⏐ ⏐ NH4+ + Cl-

xM xM

NH3 + H2O Ä NH4+ + OH−

0.10-3.9´10 -5( )M 3.9´10 -5M 3.9´10 -5M

Kb=NH4

+⎡⎣ ⎤⎦ OH-⎡⎣ ⎤⎦NH3[ ]

=1.8´10 -5

=x+3.9´10 -5( ) 3.9´10 -5( )

0.10-3.9´10 -5( )

The simplifying assumption can be applied.

46

NH3 + H2O ⇌ NH4+ + OH-

(0.10- 3.9 x 10-5) xM + 3.9 x 10-5 3.9 x 10-5

Simultaneous Equilibria

3.9×10 -5( )x

0.10=1.8 ×10−5

x =0.046M = NH4Cl[ ]Because there is 1.0 L of solution,

there are 0.046 mol NH4Cl.

47

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

Simultaneous Equilibria

Qsp= Mg2+⎡⎣ ⎤⎦ OH−⎡⎣ ⎤⎦2

= 0.010( ) 3.9 ×10−5( )2

=1.5 ×10−11

Qsp =Ksp

Thus this system is at equilibrium!

48

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

Simultaneous Equilibria

H+⎡⎣ ⎤⎦ OH−⎡⎣ ⎤⎦=1.0 ×10-14

H+⎡⎣ ⎤⎦=1.0 ×10 -14

OH−⎡⎣ ⎤⎦

H+⎡⎣ ⎤⎦=1.0 ×10 -14

3.9 ×10−5 =2.6 ×10−10M

pH=9.59