What is the molar solubility of barium phosphate in a 0.282M barium acetate solution?

If I set up and ice table I would have 0.282M - X = ? Would I use the molar mass of barium acetate to complete the equation to find X?

Ba3(PO4)2 ==> 3Ba^2+ + 2PO4^3-
Ba(C2H3O2)2 ==> Ba^2+ + 2C2H3O2^-
(Ba^2+) = 3x from Ba3(PO4)2 + 0.282 from Ba(C2H3O2)
(PO4^3-) = 2x from Ba3(PO4)2
Substitute into Ksp and solve for x = molar solubility Ba3(PO4)2.


So for Ksp I would do (3x+0.282)(2x) ?

Yes. You can assume 3x + 0.282 = 0.282. Look up Ksp (or it may be in the problem) and solve for x.

Yes, you are correct! To find the molar solubility of barium phosphate (Ba3(PO4)2) in a 0.282M barium acetate (Ba(C2H3O2)2) solution, you can use the following steps:

1. Write the balanced equation for the dissociation of barium phosphate:
Ba3(PO4)2 ⇌ 3Ba^2+ + 2PO4^3-

2. Write the balanced equation for the dissociation of barium acetate:
Ba(C2H3O2)2 ⇌ Ba^2+ + 2C2H3O2^-

3. Set up an ICE table (Initial, Change, Equilibrium) for the dissociation of barium phosphate:
Initial: Ba3(PO4)2 (s) → 0 M
Change: Ba3(PO4)2 (s) → +3x M, 3Ba^2+ → +3x M, 2PO4^3- → +2x M
Equilibrium: Ba3(PO4)2 (s) → 3x M, 3Ba^2+ → 3x M, 2PO4^3- → 2x M

4. Substitute the equilibrium concentrations into the solubility product expression (Ksp) for barium phosphate:
Ksp = [Ba^2+]^3 * [PO4^3-]^2

5. Since the concentration of barium acetate is given as 0.282M, you can substitute this value into the equation:
[Ba^2+] = 3x + 0.282M
[PO4^3-] = 2x

6. Substitute these expressions into the Ksp equation:
Ksp = (3x + 0.282M)^3 * (2x)^2

7. Solve the equation for x to find the molar solubility of barium phosphate (Ba3(PO4)2).

So, the correct expression for Ksp is (3x + 0.282M)^3 * (2x)^2.