Posted by Deb on .
calculate the solunility of silver chromate, Ag2CrO4, in 0.005 M NA2CrO4 Ksp=2.6x1012
Please help.....

College Chem 2 
DrBob222,
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. 
College Chem 2 
Anonymous,
f(x)= 6/x