when the initial [Al+3] = 0.05 M
and according to the reaction equation:
by using the ICE table:
Al(OH)3(s) → Al3+ + 3OH-
initial 0.05 0
change +X +3X
Equ 0.05+X 3X
when Ksp = [Al3+] [OH-]^3
and Ksp = 1.3 x 10^-33
and [Al3+] = (0.05+X)
[OH-]^3 = 3X
when we assume that X = the molar solubility of Al(OH)3
by substitution:
1.3 x 10^-33 = (0.05 + X) (3X)^3 by solving for X
∴ X = 8.6 X 10^-12 M
∴ the molar solubility of Al(OH)3 = 8.6 x 10^-12 M
