Chem 2
posted by DebC on .
Calculate the solubility of silver chromate, Ag2CrO4 in 0.005M Na22Cr)4 Ksp=2.6X1012
HELP PLEASE......

Let x = solubility Ag2CrO4 at equilibrium.
.........Ag2CrO4(s) > 2Ag^+ + CrO4^2
equil.......x............2x.......x
..........Na2CrO4==> 2Na^+ + CrO4^2
initial.....0.005M.....0.......0
change.....0.005...2*0.005...0.005
equil.......0........2*0.005..0.005
Ksp = (Ag^+)^2(CrO4^2)
Substitute from the ICE charts above.
(Ag^+) = 2x from Ag2CrO4.
(CrO4^2) = x from Ag2CrO4 + 0.005 from Na2CrO4.
Solve for x = solubility Ag2CrO4.