Question

Calculate the solubility of silver chromate, Ag2CrO4- in 0.005M Na22Cr)4- Ksp=2.6X10-12

HELP PLEASE......

Answer 1

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.