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?
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.
Yes, you are correct so far. You have correctly set up the Ksp expression for XY2 as Ksp = [X^2+][Y^-]^2.
To find the molar solubility of Y, you need to determine the concentration of Y^- ions in solution at equilibrium. In this case, the concentration of Y^- ions is 2x, where x is the molar solubility of Y.
Substituting the value of x that you found (x = 0.0318) into the expression for the concentration of Y^- ions, you have:
[Y^-] = 2x = 2(0.0318) = 0.0636 M
Therefore, the molar solubility of Y in the solution of XY2 is 0.0636 M.
Yes, you are on the right track. Your calculation to find x, which represents the molar solubility of Y, is correct. However, there is a small mistake in your equation. The Ksp expression for XY2 should be:
Ksp = [X^2+][Y^-]^2
Based on your ICE table, you correctly identified that the concentration of XY2 is x, and the concentration of Y^- (or [Y^-]) is 2x.
So, the correct equation to solve for x is:
2.2x10^-4 = (x)(2x)^2
Now, let's solve it:
2.2x10^-4 = 8x^3
Divide both sides by 8:
2.75x10^-5 = x^3
Take the cube root of both sides to solve for x:
x ≈ 0.0312
Therefore, the molar solubility of Y (2x) is approximately 0.0624 mol/L.