What is molar solubility of Al(OH)3 at 25 degrees C given the solubility product Ksp = 5.0e-33?

I set this up as Al^3+ + 3OH^-

Ksp=[Al^3+][OH^-]^3

5.0e-33 = (x)(3x)^3

Before I go any further did I do this correctly?

1. you are right, I was wrong yesterday.

solve for x.

posted by bobpursley
2. So it would be 27x^4, so I divide 5.0e-33 by 27 to get 1.8e-34 and then raise this to the 1/4 power which I got 4e-9.

posted by Hannah
3. I get 3.69E-9

posted by DrBob222

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