Question

What is the solubility of lead iodide in a solution of 0.14 M lead nitrate, Pb(NO3)2? Ksp of PbI2 = 9.8x10-9

Answer 1

Pb(NO3)2 here is Strong electrolyte
It will dissociate completely to give [Pb2+] = 0.14 M
At equilibrium:
PbI2      <---->     Pb2+     +         2 I-    
                     0.14 +s            2s      
Ksp = [Pb2+][I-]^2
9.8*10^-9=(0.14 + s)*(2s)^2
Since Ksp is small, s can be ignored as compared to 0.14
Above expression thus becomes:
9.8*10^-9=(0.14)*(2s)^2
9.8*10^-9= 0.14 * 4(s)^2
s = 1.3*10^-4 M
Answer: s = 1.3*10^-4 M