Question

Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:

(a) Ag2SO4

(b) PbBr2

(c) AgI

(d) CaC2O4∙H2O

Answer 1

1.Molar solubility of Ag₂SO₄ -

Ag₂SO₄ ---> 2Ag⁺ + SO₄^2-

... x ............ 2x ......... x ......

Ksp of Ag₂SO₄ is1.2*10^-5

Here x is the molar solubility

Ksp = [Ag+]^2*[SO4 2-] = (2x)^2*x = 4x^3

1.2*10^-5 = 4x^3

x = 1.44 x10^-2M

Molar solubility = 1.44 x 10^-2 M

Ag₂SO₄ [molar mass = 312g/mol]

Molar solubility of Ag₂SO₄ in g/L =1.44 x 10^-2 M x 312g/mol

                                                          = 4.49g/L

2.Molar solubility of PbBr₂

The Ksp of PbBr₂ is 6.60× 10^–6

PbBr2---> Pb²⁺ + 2Br ^–

x            x          2x

Ksp = [Pb²⁺] [Br^-]^2

6.60× 10^–6=[x][2x]^2

6.60× 10^–6 =4x^3

So, x = 1.16 x 10^-2M

Molar solubility = 1.16 x 10^-2 M

Molar mass of PbBr₂ = 367g/mol

Molar solubility of PbBr2 in g/L =1.16 x 10^-2 M x 367g/mol

                                                          = 4.26g/L

3.Molar solubility of AgI

Ksp of AgI is 1.5 x 10-16.

AgI ---------> Ag⁺ + I^-

x                       x       x

Ksp = [Ag+][I^-]

1.5 x 10^-16 = [x] [x]

1.5 x 10^-16 = x^2

x = 1.22 x 10^-8

Molar solubility = 1.22 x 10^-8 M

Molar mass of AgI = 234.77g/mol

Molar solubility of AgI in g/L = 1.22 x 10^-8 M x 234.77g/mol = 2.86 x 10^-6 g/L