Question

The solubility product constant of lead(ii) chloride , PbCl2 is 1.7*10^-5 . calculate the molar solubility in pure water and in a 0.50 m solution of sodium chloride.

Answer 1

1)

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: 1.62*10^-2 M

2)

NaCl here is Strong electrolyte

It will dissociate completely to give [Cl-] = 0.5 M

At equilibrium:

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

   s 0.5 + 2s

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

1.7*10^-5=(s)*(0.5+ 2s)^2

Since Ksp is small, s can be ignored as compared to 0.5

Above expression thus becomes:

1.7*10^-5=(s)*(0.5)^2

1.7*10^-5= (s) * 0.25

s = 6.80*10^-5 M

Answer: 6.80*10^-5 M