posted by Maggie .
Hi, I have a few chemistry problems that I'm having trouble solving. I put my attempted solutions after the problems in *'s. There was one problem I had no idea how to solve.
What concentration of silver chromate (Ksp = 9.0 x 10-12) will dissolve to make a saturated solution in water?
*For this one the formula is Ag2CrO4, which is [Ag]^2[CrO4]. I plugged x in for CrO4, so x^3 is equal to Ksp. I then got that x = 2.0e-4. However, that x is for CrO4, not for silver chromate. Is the answer for silver chromate just the Ksp value?*
What concentration of silver chromate (Ksp = 9.0 x 10-12) will dissolve in 500. mL of 0.01 M aqueous silver nitrate?
A saturated solution of CaSO4(aq) is made in a beaker until there is excess calcium sulfate resting at the bottom. Then solid potassium sulfate is added. Which of the following is true? (The Ksp for potassium sulfate is larger than the Ksp for calcium sulfate)
a. More calcium sulfate will start to precipitate out of solution
b. More calcium sulfate will start to dissolve in solution
*I know that when a smaller Ksp is added to a bigger Ksp it precipitates out of solution. Is that the same in the reverse?*
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!