The Ksp of Ag2CrO4(s) is 1.12 x 10^-12. Calculate the molar solubility of AgCrO4(s)

Question

a)in pure water
b) in a solution of 0.10 mol/L sodium chromate, Na2CrO4(s)

Answer 1

Ag2CrO4 ==> 2Ag^+ + CrO4^=
Ksp = (Ag^+)^2 (CrO4^=) = 1.12 x 10^-12
solubility Ag2CrO4 = S
(Ag^+) = 2S
(CrO4^=) = S
Plug into Ksp and solve for S.

For b.
Everything stays the same except (CrO4^=) = S + 0.1. Substitute into Ksp and solve for S. You can make the simplifying assumption that S + 0.1 = 0.1. If you do, to avoid solving a quadratic equation, then check at the end to make sure S + 0.10 = 0.1. If it isn't, then you will need to regroup and solve the quadratic or use successive approximations. I THINK S will be negligible when compared to 0.1 M.

Check my thinking. Check my work. Post your work if you get stuck.

Answer 2

yeah... its cool

Answer 3

Im not sure ...need more explanations please?