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Determine the molar solubility of Y for XY2 if Ksp is 2.2x10^-4.

The 2 in Xy2 is a subscript.

I found the Ksp equation which is
ksp=[x^2+][Y^-]^2

Then I did the ICE table and got x and 2x for the equilibrium concentrations. Then I did:
2.2x10^-4=(x)(2x)^2
I solved for x and got 0.0318=x

Is that correct so far? How would I find the molar solubility of Y?

1 answer

I obtained 0.038 for x.
So Y is 2x isn't it?
...........XY2 ==> X^2+ + 2Y^-
I........solid.....0.......0
C........solid.....z......2z
E........solid.....z......2z

Note that Y is just twice X.
I tried, with different manipulations, to obtain 0.0318 but couldn't do it; I don't know how you ended up with 0.0318.