For the following equilibrium, if Ksp=1.3×10−33, what is the molar solubility of aluminum hydroxide? Al(OH)3(s)↽−−⇀Al3+(aq)+3OH−(aq)

Question

For the following equilibrium, if Ksp=1.3×10−33, what is the molar solubility of aluminum hydroxide?

Al(OH)3(s)↽−−⇀Al3+(aq)+3OH−(aq)

Answer 1

Let x = the molar solubility of Al(OH)3, then:

[Al_3+][OH-]^3=K_sp --> x*(3x)^3=K_sp --> 27x^4=K_sp

Answer 2

Al(OH)₃(s) ==> Al³⁺(aq) + 3OH^-(aq)

Ksp = [Al³⁺][OH^-]³

Let x = [Al³⁺], then [OH^-] = 3x

1.3x10^-33 = (x)(3x)³

1.3x10^-33 = 27x⁴

x⁴ = 4.8x10^-35

x = 2.63x10^-9 M = molar solubility of Al(OH)₃