KSP
BaSO4 (s) ↔ Ba2+ + SO42-
K=
K[BaSO4](s) = [Ba2+][SO42-]
Ksp = [Ba2+][SO42-]
AxBy = xA+ + yBKsp = [A+]x[B-]y
Mg(OH)2 ↔ Mg2+ + 2OHKsp = [Mg2+][OH-]2
Bi2S3 ↔ 2Bi3+ + 3S2Ksp = [Bi3+]2[S2-]3
Solubility in Water
AgCl ↔ Ag+ + Cl1
1
1 (1 to 1 to 1 so concentrations are equal)
Concentration of Ag+ in terms of AgCl
Ag+ = AgCl
and the concentration of Cl- in terms of AgCl
Cl- = AgCl
KspAgCl = [Ag+][Cl-]
= [AgCl][AgCl]
Similarly for
Mg(OH)2 ↔ Mg2+ + 2OH1
1
2
Mg2+ = Mg(OH)2
OH- = 2Mg(OH)2
(Concentration of OH- 2x concentration of Mg(OH)2)
KspMg(OH)2 = [Mg2+][OH-]2
= [Mg(OH)2][2Mg(OH)2]2
Bi2S3 = 2Bi3+ + 3S21
2
3
Bi3+ = 2Bi2S3
S2- = 3Bi2S3
KspBi2S3 = [Bi3+]2[S2-]3
= [2Bi2S3]2[3Bi2S3]3
Calculate the solubility of AgCl in (a) m/l and (b) g/100 ml given the Ksp of AgCl is
1.8 x 10-10.
Ksp = [Ag+][Cl-]
1.8 x 10-10 = [AgCl][AgCl]
1.8 x 10-10 = [AgCl]2
1.3 x 10-5 m/l = [AgCl]
143 x 1.3 x 10-5 g/l
Calculate the solubility of Fe(OH)3 in (a) m/l and (b) g/100 ml given its Ksp is
1.1 x 10-36.
Fe(OH)3 → Fe3+ + 3OHFe3+ = Fe(OH)3
OH- = 3 Fe(OH)3
Ksp = [Fe3+][OH-]3
1.1 x 10-36 = [Fe(OH)3][3Fe(OH)3]3
11,000 x 10-40 = 27[Fe(OH)3]4
407 x 10-40 = [Fe(OH)3]4
4.5 x 10-10 = Fe(OH)3 m/l
Calculate the Ksp of CaCO3 given its solubility is 6.9 x 10-3 g/l.
= 6.9 x 10-5 m/l
CaCO3 → Ca2+ + CO321
1
1
Ca2+ = CaCO3
CO32- = CaCO3
Ksp = [Ca2+][ CO32-]
= [CaCO3][CaCO3]
= [6.9 x 10-5][6.9 x 10-5]
= 4.8 x 10-9
Calculate the Ksp of Ag2CrO4 given its solubility is 4.3 x 10-2 g/l.
= 1.3 x 10-4 m/l
Ag2CrO4 → 2Ag+ + CrO42Ag+ = 2Ag2CrO4
CrO42- = Ag2CrO4
Ksp = [Ag+]2[CrO42-]
= [2Ag2CrO4]2[Ag2CrO4]
= [2 x 1.3 x 10-4]2[1.3 x 10-4]
= 8.8 x 10-12
Calculate the maximum concentration of Ba2+ that can be added to a solution that is .1 M in
SO42- given the Ksp of BaSO4 is 1.1 x 10-10.
Na2SO4 → 2Na+ + SO42.1 M
.1
Ksp = [Ba2+][SO42-]
1.1 x 10-10 = [Ba2+][.1]
= [Ba2+] = 1.1 x 10-9 m/l
50 ml of .05 M Na3PO4 are added to 50 ml of .001 M BaCl2. (1) Show by calculation whether
Ba3(PO4)2 will precipitate, (2) calculate the concentration of Ba2+ in the solution and (3)
calculate the % of Ba2+ that remains in the solution.
(Note: For a precipitate to occur the product of the molar concentration of each ion raised to the
appropriate power must exceed Ksp.)
The reaction that is taking place:
Na3PO4 + BaCl2 → NaCl + Ba3(PO4)2
PO43- + Ba2+ → Ba3(PO4)2
Ba3(PO4)2 ↔ 3Ba2+ + 2PO43-
Remember mmols = (M)(ml) and M =
mmols of PO43- = 50 x .05 = 2.5
mmols of Ba2+ = 50 x .001 = .05
M PO43- =
= 2.5 x 10-2
M Ba2+ =
= 5 x 10-4
[5 x 10-4]3[2.5 x 10-2]2
[125 x 10-12][ 6.25 x 10-4]
.
780 x 10-16 = 7.8 x 10-14 > 10-29 so Ba3(PO4)2 will precipitate
(The concentration of SO42- is 50 times the concentration of the Ba2+. Therefore we can assume
the concentration of the PO43- to be a constant or changed very little when Ba3(PO4)2
precipitates.)
Ksp = [Ba2+]3[PO43-]2
1.3 x 10-29 = [Ba2+]3[2.5 x 10-2]2
= [Ba2+]3
2.08 x 10-26 = [Ba2+]3
2.7 x 10-9 m/l = [Ba2+]
(concentration of Ba2+ in the solution)
% Ba2+ =
=
x 100
x 100
= 2.7 x 10-4 %
A test tube contains:
.1 M Zn2+ Ksp Zn(OH)2
4.5 x 10-17 and
.1 M Mg2+ Ksp Mg(OH)2 1.5 x 10-11
(Note: Mg(OH)2 is the most soluble.)
Calculate the maximum concentration of OH- that can be added to a test tube without affecting
the precipitation of Mg(OH)2 and determine the concentration of Zn2+ that remains in the
solution.
Ksp = [Mg2+][OH-]2
1.5 x 10-11 = [.1][OH-]2
= [OH-]2
1.5 x 10-11 = [OH-]2
1.2 x 10-5 m/l = OH-
Ksp = [Zn2+][OH-]2
= [.1][1.5 x 10-10] > 10-17
Zn will precipitate while Mg(OH)2 will not.
The concentration of Zn2+ in the solution is calculated in the following way:
4.5 x 10-17 = [Zn2+][1.2 x 10-5]2
= [Zn2+]
3 x 10-7 m/l = [Zn2+]
% of Zn2+ in the solution:
% Zn2+ =
x 100
= 3 x 10-4 %
Calculate the pH needed to prevent the precipitation of Mg(OH)2 in the above problem.
H+ =
H+ = 8.2 x 10-10
pH = 9.1
(Note: If the pH of the solution is 10, both of the compounds will precipitate because increasing
the pH decreases the H+ concentration which means the OH- concentration will increase.)
Given the Ksp of ZnS is 1 x 10-27 calculate the concentration of the H+ needed to prevent the
precipitation of a solution of ZnS from a solution that contains .1 M Zn2+ and the solution is
saturated with H2S.
Zn2+ + S2- → ZnS
1 x 10-27 = [Zn2+][S2-]
1 x 10-21 = [.1][S2-]
= [S2-]
1 x 10-26 = [S2-]
(Max. concentration of S2- that can be added to .1 M Zn2+ without
affecting the precipitation of ZnS [a saturated solution is formed])
H2S ↔ H+ + HSHS- ↔ H+ + S2-
Ka2
Ka1 =
Ka2 =
Ka1Ka2 =
Ka1Ka2 =
Ka1
1 x 10-7
1.3 x 10-13
From table
(1 x 10-7)(1.3 x 10-13) =
1.3 x 10-21 = [H+]2[S2-]
(This is the important relationship.)
1.3 x 10-21 = [H+]2[1 x 10-26]
= [H+]2
1.3 x 10-5 = [H+]2
3.6 x 10-2 = [H+]
pH = -2.5
Given the Ksp of NiS is 2 x 10-27 and MnS is 4.3 x 10-22 (1) calculate the H+ concentration and
pH that will precipitate NiS without precipitating MnS and (2) calculate the % Ni2+ that stays in
the solution that contains .05 M Mn2+ and .05 M Ni2+.
MnS ↔ Mn2+ + S2-
4.3 x 10-22 = [5 x 10-2][S2-]
= [S2-]
8.6 x 10-21 = [S2-]
(Max. concentration of S2- that can be added without precipitating. MnS
the most soluble compound.)
Concentration of Ni2+ in the solution in which the S2- is 8.6 x 10-21:
2 x 10-27 = [Ni2+][8.6 x 10-21]
= [Ni2+]
2.3 x 10-7 = [Ni2+] m/l of Ni2+ in the solution
x 100 = 4.6 x 10-4 % of Ni2+ in the solution
1.3 x 10-21 = [H+]2[S2-]
= [H+]2
.15 = [H+]2
.39 = [H+]
pH = 4.1