Determine the molar solubility for al(oh)3 in pure water. ksp for al(oh)3 = 1.3 x 10-33.

When Al(OH)₃ dissolves it dissociates as follows;
Al(OH)₃ ---> Al³⁺ + 3OH⁻
molar solubility is the number of moles that can be dissolved in 1 L of solution
if the molar solubility of Al(OH)₃ is x then molar solubility of Al³⁺ is x and OH⁻ is 3x
then solubility product constant - ksp formula is as follows 
ksp = [Al³⁺][OH⁻]³
ksp = (x)(3x)³
ksp = x * 9x³
ksp = 9x⁴
ksp = 1.3 x 10⁻³³ 
9x⁴ = 1.3 x 10⁻³³ 
x = 3.4 x 10⁻⁹ M
molar solubility of Al(OH)₃ is therefore 3.4 x 10⁻⁹ M

Sorry i dont understand the question can you rephase it?