SOLUBILITY PRODUCT PRINCIPLE
Section I
Examine the ionization of the general compound AxBy.
xA+ + yB-
AxBy
 A    B y
Eq. 1: Ke =


A
x By

Eq. 2: Ksp = [A+]  [B-]
Ag+ + Cl-
Example: AgCl (s)
Ag Cl 
Eq. 3: Ke =


AgCl
Eq. 4: Ksp = [Ag+] [Cl-]
Example 1
Calculate the concentration of AgCl in m/l given the Ksp of AgCl = 1 x 10-10
Solution
The Ksp expression for AgCl is given by Equation 4.
AgCl
Ag+ = Cl-
The ionization of AgCl shows that the ratio of AgCl: Ag+ : Cl- is 1 : 1 : 1. That is, the concentration
of Ag+ = the concentration of Cl- and the concentration of Cl- = the concentration of AgCl.
Substituting AgCl for Ag+ and for Cl1 x 10-10 = [AgCl] [AgCl]
1 x 10-10 = [AgCl]2
1  10 10 = AgCl
1 x 10-5 = AgCl
therefore, the solubility of AgCl is 1 x 105 m/l.
If the solubility of AgCl is to be expressed in g/l then:
143.5 g/m x 1 x 10-5 m/l = 1.435 x 10-3 g/l
Another approach is to use ratio and proportion.
1 mole of AgCl = 143.5 g
1 x 10-5 = X
X = 1 x 10-5 x 143.5
= 1.435 x 10-3 g
If the solubility of AgCl is to be expressed in g/200 ml then:
1.435 x 10-3 g/l x
1 liter
x 200 ml = 2.87 x 10-4 g
1000 ml
Another approach is to us ratio and proportion.
1.435 x 10-3 = 1000 ml
X = 200 ml
200  1435
.
 10 3
= 2.87 x 10-4 g
1000
Example 2
The solubility of BaSO4 in water is 2.42 x 10-4 g/100 ml. Calculate its Ksp.
Solution
Obtain the solubility of BaSO4 in m/l.
2.42 x 10-4 g/100 ml x
1000 ml
= 2.42 x 10-3 g/l
liter
2.42  10 3
= 1.04 x 10-5 m/l
233 g / m
From the ionization of BaSO4 : BaSO4
concentration SO4= = [BaSO4]
Ba+2 + SO4= concentration Ba+2 = [BaSO4]
Ksp expression of BaSO4 is given by Equation 5.
Eq. 5: Ksp = [Ba+2] [SO4=]
Substitute BaSO4 for Ba+2 and for SO4=
Eq. 6: Ksp = [BaSO4] [BaSO4]
The solubility of BaSO4 was found to be 1.04 x 10-5 m/l
Ksp = [1.04 x 10-5] [1.04 x 10-5]
= 1.08 x 10-10
Example 3
The Ksp of Ni(OH)2 is 1.6 x 10-14. Calculate the solubility of Ni(OH)2 in (a) m/l; (b) g/l.
Solution
The Ksp expression for Ni (OH)2 is given by Equation 7.
Eq. 7: Ksp = [Ni+2] [OH-2]
For the ionization of Ni(OH)2
Ni+2 + 2OH-
The concentration of Ni+2 = Ni(OH)2 .
The concentration of OH- = 2Ni(OH)2.
Substitute these values in the Ksp expression given by Eq. 7.
Ksp = [Ni(OH)2 [2Ni(OH)2]2
1.6 x 10-14 = 4[Ni(OH)2]3
.4 x 10-14 = [Ni(OH)2]3
4 x 10-15 = [Ni(OH2]3
Take the cube root of both sides.
Ni(OH)2 = 1.59 x 10-5 m/l
(b)
1.59 x 10-5 m/l x 92.7 g/m = 147 x 10-5 g/l
= 1.47 x 10-3 g/l
If the [Ag+] is equal to [Cl-] and each is 1 x 10-6 m/l
[Ag+] = [Cl-] = 1 x 10-6
then the product of the [Ag+] [Cl-] = 1 x 10-12 which is less than 1 x 10-10.
Thus, AgCl will not form a precipitate. In another example the [Ba+2] is 1 x 10-3
[SO4=] is 1 x 10-4
[Ba+2] [SO4=] = 1 x 10-7
Ksp of BaSo4 is 1 x 10-10 therefore, BaSO4 will precipitate.
Example 4
A solution contains .1M AgNO3. Calculate the concentration of Cl- needed to start the
precipitation of AgCl. Given Ksp of AgCl is 1 x 10 -10.
Solution
Ksp expression for AgCl is given by Equation 4
Ksp = [Ag+] [Cl-]
The concentration of Ag+ is .1M and the Ksp of AgCl is given (1 x 10-10).
1 x 10-10 = [.1] [Cl-]
1  10 10
= [Cl-]
.1
1 x 10-9 = [Cl-]m/l
Section II
1. Calculate the [Ag+] in m/l in a saturate solution of AgBr,
Ksp of AgBr is 3.3 x 10-13.
2. The solubility of PbSO4 in water is 3.03 x 10-3 g/100 ml. Calculate the Ksp.
3. The solubility of Ag2BrO4 is 2.4 x 10-2 g/l. Calculate its Ksp.
4. What is the solubility of Pb (103)2 in
(a)
mol/l
(b)
g/l
(c)
g/200 ml
Given the Ksp of Pb (103)2 is 2.5 x 10-13
5. A solution contains .001M Ca+2 and .001M OH2. Will Ca(OH)2 precipitate? Ksp of Ca(OH)2 is 7.9
x 10-6.
6. Calculate the concentration of Cl- needed to start precipitation in a .005M solution of Ag+. Ksp of
AgCl is 1 x 10-10.
7. Calculate the concentration of Ag+ left in solution of AgCl is precipitated by adding HCl to a
solution of AgNO3 to make the final concentration of the Cl- .05 M. Ksp AgCl 1 x 10-10.
8. Calculate the concentration of the H+ required to prevent the precipitation of ZnS in a solution that
0.1 M in ZnCl2 saturated with H2S (.1 M H2S). Given: Ksp ZnS 1.1 x 10-21
[H+]2[S2] = 1 x 10-21
ANSWERS
1. 5.7 x 10-7 m/l
2. 1 x 10-8
3. 1.2 x 10-12
4. (a) 4 x 10-5 m/l
(b) 2.2 x 10-2 g/l
(c) 4.4 x 10-3 g/200 ml
5. No precipitate
Ksp = [Ca+2] [OH-2]
= [1 x 10-3] [1 x 10-3]2
= [1 x 10-9]
1 x 10-9 is smaller than Ksp; therefore, no precipitate forms.
6. 2 x 10-8 m/l
7. 2 x 10-9 m/l
8. .30 m/l