The Ksp of chromium (III) iodate in water is 5.0 x 10-6.
Estimate the molar solubility of the compound.
Cr(IO3)3(s)  Cr3+(aq) + 3 IO3-(aq)
1 mole Cr(IO3)3 produces 1 mole Cr3+ and 3 moles IO3Ksp = [Cr3+(aq)] [IO3-(aq)3 = (s) (3s)3 = 5.0 x 10-6
where s is the solubility of Cr(IO3)3
s = 0.021
Molar solubility Cr(IO3)3 of 0.021 M
Precipitation from Solution
If equal volumes of aqueous solutions of 0.2 M Pb(NO3)2 and
KI are mixed will PbI2(s) precipitate out? Ksp of PbI2 is
1.4 x 10-8
Use the reaction quotient, Q, to predict whether precipitation
will occur
Pb(NO3)2 (aq) +2 KI(aq) -> PbI2 (s) + 2 KNO3 (aq)
Net ionic equation: Pb2+ (aq) + 2I- (aq) -> PbI2 (s)
The reverse of this reaction defines Ksp
PbI2 (s)  Pb2+ (aq) + 2I- (aq)
Ksp = [Pb2+ (aq)] [I- (aq)]2
If Q > Ksp precipitation; if Q < Ksp no precipitation
Equal volumes of Pb(NO3)2 and KI are mixed
On mixing, volume of mixed solution is twice initial volume
[Pb2+ (aq)] = 0.2M / 2 = 0.1 M
[I- (aq)] = 0.1 M
Q = [Pb2+(aq)] [I- (aq)]2 = (0.1)(0.1)2 = 0.001 M
Q > Ksp; PbI2(s) precipitates
Common Ion Effect
Adding NaCl to a saturated solution of AgCl lowers the
solubility of AgCl, reducing the amount of Ag+(aq) and
Cl- (aq)
AgCl(s)  Ag+(aq) + Cl-(aq)
The common ion effect is the reduction in the solubility of a
sparingly soluble salt by the addition of a soluble salt that
has an ion in common with it.
Example of LeChatelier’s principle.
AgCl(s)  Ag+(aq) + Cl- (aq) Ksp = 1.6 x 10-10 at 25oC
[Ag+ (aq)] = [Cl- (aq)] = 1.3 x 10-5 M
concentration of dissolved AgCl = 1.3 x 10-5 M
Dissolve AgCl in a solution of 0.10 M NaCl . What is the
solubility of AgCl in the NaCl solution?
[Cl-(aq)] = 0.10 M
Since Ksp at 25oC is a constant,
[Ag+(aq)] = Ksp / [Cl- (aq)] = 1.6 x 10-9 M
Concentration of dissolved AgCl = 1.6 x 10-9 M
Selective Precipitation
A mixture of cations in solution can be separated by adding
anions with which they form salts with different
solubilities.
A few drops of a solution
of Pb(NO3)2(aq) is
added to a solution of
KI(aq), yellow PbI2(s)
is formed
Ca(OH)2(s) (Ksp = 5.5 x 10-6 ) and Mg(OH)2(s) (Ksp = 1.1 x 10-11 ).
A sample of sea water contains, among other solutes, the
following concentrations of soluble cations: 0.050 M
Mg2+(aq) and 0.010 M Ca2+(aq). Determine the order in
which each ion precipitates as solid NaOH is added, and
give the concentration of OH- when precipitation of each
begins. Assume no volume change on addition of solid
NaOH.
M(OH)2(s)  M2+(aq) + 2 OH-(aq)
(M = Ca or Mg)
[OH- (aq)] = (Ksp / [Ca2+(aq)])0.5 = 0.023 M
[OH- (aq)] = (Ksp / [Mg2+(aq)])0.5 = 1.5 x 10-5 M
Mg(OH)2(s) will precipitate at a [OH- (aq)] ~ 1.5 x 10-5 M
Ca(OH)2(s) will precipitate at a [OH- (aq)] ~ 0.023 M
Mg(OH)2(s) precipitates first
Dissolving Precipitates
The solubility of insoluble compounds can often be increased
by addition of acids.
ZnCO3 (s)  Zn2+ (aq) + CO32- (aq)
Adding acid like HNO3(aq)
CO32- (aq) + 2 HNO3(aq) -> CO2 (g) + H2O(l) + 2 NO3- (aq)
Addition of acids reacts with the anions in solution lowering
the concentration of the anion.
The insoluble compound then dissolves further to increase
the concentration of the anion in solution.
Another example of LeChatelier’s principle in action.
The solubility of a solid can be increased by removing an ion
from solution.
Acids can be used to dissolve hydroxides, sulfides, sulfites,
or carbonate precipitates.
Mg(OH)2(s)  Mg2+(aq) + 2 OH-(aq)
Ksp = 1.1 x 10-11
In acidic pH concentration of OH- is lowered; increases
solubility of the metal hydroxide
Estimate the solubility of Fe(OH)3 at 25oC in a solution
buffered to a pH of 2.9. Ksp (Fe(OH)3) = 1.1 x 10-36
pOH = 11.1
[OH-(aq)] = 7.94 x 10-12 M
Fe(OH)3(s)  Fe3+(aq) + 3 OH-(aq)
Ksp = 1.1 x 10-36 = [Fe3+(aq)] [OH-(aq)]3
[Fe3+(aq)] = Ksp / [OH-(aq)]3 = 1.1 x 10-36 / (7.94 x 10-12 )3
= 2.2 x 10-3 M = molar solubility of Fe(OH)3
In pure water, molar solubility of Fe(OH)3 is ~ 4.5 x 10-10 M
Complex Ion Formation
The formation of a complex can remove an ion, affecting the
solubility equilibrium.
Example: reaction between a Lewis acid such as a metal
cation and a Lewis base such as NH3.
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
If NH3 is added to a saturated solution of AgCl, the Ag+
complexes with the NH3, removing the Ag+ from solution,
increasing the solubility of AgCl
If enough NH3 is added, all the AgCl will dissolve.
Both dissolution and complex formation are at equilibrium
AgCl(s)  Ag+(aq) + Cl-(aq)
Ksp = [Ag+(aq)] [Cl-(aq)]
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
Formation constant, Kf: equilibrium constant for complex
formation
Kf = [Ag(NH3)2+(aq) ] / ([Ag+(aq) ] [NH3(aq) ]2)
= 1.6 x 107 at 25oC
Calculate the molar solubility of AgCl in 0.10 M NH3(aq) given
that Ksp = 1.6 x 10-10 for AgCl and Kf = 1.6 x 107 for
Ag(NH3)2+.
AgCl(s)  Ag+(aq) + Cl-(aq)
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
Overall: AgCl(s) + 2 NH3(aq)  Ag(NH3)2+(aq) + Cl-(aq)
K = Ksp Kf
Molar solubility of AgCl = [Cl-(aq) ]
Also, [Ag(NH3)2+(aq)] = [Cl-(aq)]
NH3(aq)
Ag(NH3)2+(aq)
Cl-(aq)
Initial
0.10
0
0
Change
-2x
x
x
Equilibrium
0.10 - 2x
x
x
K = [Ag(NH3)2+(aq)] [Cl-(aq)] / [NH3(aq)]2 = Ksp Kf = 2.6 x 10-3
x = 4.6 x 10-3
Molar solubility of AgCl is 4.6 x 10-3 M
Compare with 1.3 x 10-5 M in pure water
Qualitative Analysis
Qualitative Analysis involves the separation and identification
of ions by techniques such as complex formation, selective
precipitation, and control of the pH of a solution.
A solution of Pb2+(aq), Hg22+(aq), Ag+ (aq), Cu2+(aq), Zn2+(aq)
(1)
(2)
(3)
(1) Add HCl. Precipitate Hg2Cl2, AgCl, PbCl2
(2) Add H2S. Precipitate CuS
(3) Make solution basic (add NH3), precipitates ZnS
1) Precipitate of Hg2Cl2, AgCl, PbCl2
Rinse the precipitate in hot water; PbCl2 dissolves
Add CrO42- to precipitate Pb2+ as PbCrO4(s)
To the Hg2Cl2, AgCl precipitates add NH3 to form Ag(NH3)2+
complex which dissolves.
Ag(NH3)2+(aq) + Cl-(aq) + 2 H3O+(aq)  AgCl(s) + 2 NH4+(aq) +
2 H2O(l)