What is the molar solubility of lead(II) chromate (ksp=1.8x10^-14) in 0.13 M potassium chromate?

Answer 1

Let X = solubility Pb(CrO4)2.
....Pb(CrO4)2 ==> Pb^2+ + 2CrO4^2-
.......X...........X.......2X

K2CrO4 is 100% dissociated.
.........K2CrO4 ==> 2K^+ + CrO4^2-
initial..0.13......0.......0
change..-0.13......+0.13....0.13
equil......0.......0.13......0.13

Ksp = (Pb^2+)(CrO4^2-)
For (Pb^2+) substitute X
For (CrO4^2-) substitute 2X for that from Pb(CrO4)2 + 0.13 from K2CrO4 for total of 0.13+2X.
Solve for X.

Answer 2

4.46x10^-11 g/l