Question

need help with 40 c

40. The Ksp for lead iodide (PbI2) is 1.4 X 10-8. Calculate the solu- bility of lead iodide in each of the following. ol a. water 9A 76 0oi M0 X 5 b. 0.10 M Pb(NO3)2 c. 0.010 MNal yecko

Answer 1

C)

NaI here is Strong electrolyte

It will dissociate completely to give [I-] = 0.01 M

At equilibrium:

PbI2 <----> Pb2+ + 2 I-

   s 1*10^-2 + 2s

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

1.4*10^-8=(s)*(1*10^-2+ 2s)^2

Since Ksp is small, s can be ignored as compared to 1*10^-2

Above expression thus becomes:

1.4*10^-8=(s)*(1*10^-2)^2

1.4*10^-8= (s) * 10^-4

s = 1.4*10^-4 M

Answer: 1.4*10^-4 M