chemistry
posted by Anonymous on .
What is the molar solubility of lead(II) chromate (ksp=1.8x10^14) in 0.13 M potassium chromate?

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. 
4.46x10^11 g/l