Question

20.0 mLs of 0.10 M NaI and 20.0 mLs of 0.10 M Pb(NO3)2 are mixed together. Calculate the final molar concentrations of Pb2 and I–. Ksp(PbI2) = 7.9x10–9.

Answer 1

20.0 mL of 0.10 M NaI and 20.0 mL of 0.10 M Pb(NO3)2 are mixed together.

0.02 L*0.10 M of NaI = 0.002 mol I- per 0.04 L

0.02 L*0.10 M of Pb(NO3)2 = 0.002 mol Pb^2+ per 0.04 L

Initial : PbI2 = Pb^2+ + 2I-

CONC.N 0.002---0.002 M

Q= [Pb^2+][I-]^2

Q= (0.002)(0.002)^2 = 8.0*10^-9

Ksp(PbI2) = 7.9x10–9. since Q = Ksp solution is saturated

Thus concentrations of Pb^2+ and I- are 0.002 M