Question

The Ksp for PbCl2 = 1.7x10^-5

a. Calculate the molar solubility of PbCl2 in pure water.

b. Calculate the molar solubility of PbCl2 in a solution that is 0.1M of the soluble salt MgCl2.

Answer 1

a)

At equilibrium:

PbCl2 <----> Pb2+ + 2 Cl-

   s 2s

Ksp = [Pb2+][Cl-]^2

1.7*10^-5=(s)*(2s)^2

1.7*10^-5= 4(s)^3

s = 1.62*10^-2 M

Answer: 1.6*10^-2 M

b)

MgCl2 here is Strong electrolyte

It will dissociate completely to give [Cl-] = 0.2 M

At equilibrium:

PbCl2 <----> Pb2+ + 2 Cl-

   s 0.2 + 2s

Ksp = [Pb2+][Cl-]^2

1.7*10^-5=(s)*(0.2+ 2s)^2

Since Ksp is small, s can be ignored as compared to 0.2

Above expression thus becomes:

1.7*10^-5=(s)*(0.2)^2

1.7*10^-5= (s) * 4*10^-2

s = 4.25*10^-4 M

Answer: 4.3*10^-4 M