Question

calculate the molar solubility of Pbl2 in 0.100 M Pb(NO3)2.
Ksp = 9.8 x 10^-9

Answer 1

..................PbI2 ==> Pb^2+ + 2I^-
I.................solid...........0............0
C................solid-x........x............2x
E.................solid...........x............2x
Ksp = (Pb^2+)(I^-)^2
........................Pb(NO3)2 ==> Pb^2+ + 2NO3^-
I.........................0.100................0.............0
C.......................-0.100...........0.100........0.200
E...........................0................0.100.........0.200
So plug into Ksp expression above the following:
(Pb^2+) = x from PbI2 and 0.100 from Pb(NO3)2 to make (0.100 + x)
(I^-) = 2x
Post your work if you get stuck.
Solve for x = solubility in mols/L