Question

what is the solublity of lead (II) iodide (PbI2) in a solution containing 0.050 M lead (II) nitrate ((Pb(NO3)2)? the ksp value for lead (II) iodide is 9.8x10^-9

Answer 1

Pb(NO3)2 here is Strong electrolyte
It will dissociate completely to give [Pb2+] = 0.050 M
At equilibrium:
PbI2 <===>     Pb2+     +      2 I-    
0.050 +s            2s      

Ksp = [Pb2+][I-]^2
9.8*10^-9 = (0.050 + s)*(2s)^2

Since Ksp is small, s can be ignored as compared to 0.050
Above expression thus becomes:

9.8*10^-9 = (0.050)*(2s)^2
9.8*10^-9 = 0.050 * 4(s)^2
s^2 = 9.8*10^-9/0.2

s = 2.21*10^-4 M

Answer: s = 2.21*10^-4 M