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?

you are right, I was wrong yesterday.

solve for x.

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.

I get 3.69E-9

Yes, you have set up the solubility product expression correctly for Al(OH)3. The equation you wrote, Al^3+ + 3OH^- ⇌ Al(OH)3, represents the dissociation of Al(OH)3 into Al^3+ ions and OH^- ions in solution.

The solubility product expression, Ksp, is written as the product of the concentration of the ions raised to their stoichiometric coefficients. In this case, the Ksp expression is:

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

To calculate the molar solubility of Al(OH)3, you assume that x is the molar solubility of Al^3+ and 3x is the molar solubility of OH^-. By substituting these values into the Ksp expression, you can solve for x.

In your equation, you set up the following equation:

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

Now you can proceed with solving for x.