Question

The solubility product of PbBr2 is 8.9 10-6. Determine the molar solubility in the following.

(a) pure water

(b) 0.26 M KBr solution

(c) 0.21 M Pb(NO3)2 solution

Answer 1

a)

The salt dissolves as:

PbBr2 <----> Pb2+ + 2 Br-

   s 2s

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

8.9*10^-6=(s)*(2s)^2

8.9*10^-6= 4(s)^3

s = 1.30*10^-2 M

Answer: 1.30*10^-2 M

b)

KBr here is Strong electrolyte

It will dissociate completely to give [Br-] = 0.26 M

At equilibrium:

PbBr2 <----> Pb2+ + 2 Br-

   s 0.26 + 2s

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

8.9*10^-6=(s)*(0.26+ 2s)^2

Since Ksp is small, s can be ignored as compared to 0.26

Above expression thus becomes:

8.9*10^-6=(s)*(0.26)^2

8.9*10^-6= (s) * 6.76*10^-2

s = 1.32*10^-4 M

Answer: 1.32*10^-4 M

c)

Pb(NO3)2 here is Strong electrolyte

It will dissociate completely to give [Pb2+] = 0.21 M

At equilibrium:

PbBr2 <----> Pb2+ + 2 Br-

   0.21 +s 2s

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

8.9*10^-6=(0.21 + s)*(2s)^2

Since Ksp is small, s can be ignored as compared to 0.21

Above expression thus becomes:

8.9*10^-6=(0.21)*(2s)^2

8.9*10^-6= 0.21 * 4(s)^2

s = 3.26*10^-3 M

Answer: 3.26*10^-3 M