Question

calculate the solunility of silver chromate, Ag2CrO4, in 0.005 M NA2CrO4- Ksp=2.6x10-12

Please help.....

Answer 1

Well, at least this is not a buffer problem. This is a problem that illustrates the common ion effect on solubility. The effect is to make the solubility of a slightly soluble salt even less soluble by using a common ion. In this case the common ion is CrO4^2- from the Na2CrO4.
Let x = solubility of Ag2CrO4.
Ag2CrO4(s) ==> 2Ag^+ + CrO4^2-
...x............2x.......x

...........Na2CrO4 ==> 2Na^+ + CrO4^2-
initial....0.005M.......0........0
change....-0.005........0.005...0.005
equil.......0..........0.005......0.005

Ksp Ag2CrO4 = (Ag^+)^2(CrO4^2-)
(Ag^+) = 2x from the Ag2CrO4
(CrO4^2-) = x from Ag2CrO4 and 0.005 from Na2CrO4.
Substitute and solve for x which is the solubility of Ag2CrO4 in moles/L = M.
Post your work if you get stuck.

Answer 2

f(x)= 6/x