Question

Calculate the molar solubility in (M) of PbCl₂ (ksp=1.2x10^-5 at 298K) in

a) pure H2O (l)

b) 0.50M NaCl (aq)

Answer 1

a)

At equilibrium:

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

   s 2s

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

1.2*10^-5=(s)*(2s)^2

1.2*10^-5= 4(s)^3

s = 1.442*10^-2 M

Answer: 1.44*10^-2 M

b)

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

1.2*10^-5=(s)*(0.5+ 2s)^2

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

Above expression thus becomes:

1.2*10^-5=(s)*(0.5)^2

1.2*10^-5= (s) * 0.25

s = 4.8*10^-5 M

Answer: 4.8*10^-5 M