Ag2CrO4 ==> 2Ag + CrO4^-2
Ksp = (Ag^+)^2(CrO4^-2)
What you do is let S = solubility of Ag2CrO4 in molarity. Then S is concn CrO4 and 2S = concn
(2S)^2(S) = Ksp
4S^3 = Ksp
Solve for S. 2S will be Ag^+, S will be Ag2CrO4 and S will be CrO4^-2.
#2. Here you have two equilibria.
Ag2CrO4 ==> 2Ag^+ + CrO4^-2
Ksp = (Ag^+)^2(CrO4^-2)
And AgNO3 ==> Ag^+ + NO3^-
AgNO3 is completely soluble BUT it contributes Ag^+. So if
S = solubility Ag2CrO4, then S = CrO4^-2, 2S+0.01 = Ag^+. Plug these into Ksp for Ag2CrO4. (This is called a common ion problem and the common ion in this problem is the Ag^+. It's another example of Le Chatelier's Principle; adding AgNO3 forces the solubility to the left in the reaction and makes it more insoluble. In most cases, 2S + 0.01 = 0.01; i.e., 2S is small and contributes only slightly to the total Ag^+ and it makes it easier to solve the equation. So you end up with
9.0 x 10^-12 = (0.01)^2(S) and solve for S which is CrO4 and Ag2CrO4. You will see that the solubility is much less in this problem than in the first problem solved.
#3. I really hope no one is trying to tell you that K2SO4 has a solubility product. It doesn't. Not even close! This is a case of common ion, also. Here is how it works.
CaSO4(s) ==> Ca^+2 + SO4^-2
Ksp = (Ca^+2)(SO4^-2).
K2SO4 ==> 2K^+ + SO4 (completely soluble).
The SO4^-2 is the common ion. By Le Chatelier's principle adding sulfate (by way of K2SO4) forces the CaSO4 equilibrium to the left which means that CaSO4 is LESS soluble with K2SO4 added than when it isn't added. Therefore, some of the CaSO4 in solution will ppt. Talk of Ksp for K2SO4 has nothing to do with it.
thank you so much!! I got a 100 on the test I was studying for!
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