Calculate the molar solubility of Al(OH)3

Question

Calculate the molar solubility of Al(OH)_3. The K_sp = 2.0 x 10^-32

I can’t seem to find the right answer no matter how I try it. When I work this out I get:

Al(OH)₃ —> Al + 3 OH

K_sp = [Al] [OH]³

[Al] = x

[OH] = 3x

2 x 10^-32 = (x)(3x)³

2 x 10^-32 = 27x⁴

x⁴ = 7.41 x 10^-34

x = 5.22 x 10^-9

but that’s not the answer...

the correct answer is: 2 x 10^-11

please list correct steps so I can follow along. Thank you!

Answer 1

Janel:

ive tried working backwards from the “correct” answer. In short, I agree with your answer. The answer you quote would give a K_sp of about 10^-44. The K_sp I found online for Al(OH)3 is at least an order of magnitude lower than the one you gave as a given.

Answer 2

Because the Ksp is so very small, and we are dealing with an insoluble hydroxide, set the [OH-] to 1x10^-7 to take into account the auto ionization of water.

Ksp = [Al³⁺][OH^-]³ = 2x10-32

2x10^-32 = (x)(1x10^-7)³ = 1x10^-21 x

x = 2x10^-11 M