Al(OH)3 \(\leftrightharpoons\) Al+3(aq)+ + 3OH-(aq)
using law of mass action
K = \(\frac{[Al^{+3}][OH^-]^3}{[Al(OH^-)_3]}\)
⇒ K = equilibrium constant
\(\because\) [al(OH)_3] = constant
\(\therefore\) Rearranging above equation.
K x [Al(OH)3] = [Al+3] [OH-]3
⇒ Ksp = [Al+3] [OH-]3
where Ksp = [S] [3S]3
= s x 27S3
Ksp = 27s4
We have given,
Ksp = 1 x 10-36
\(\therefore\) S4 = \(\frac1{27}\times10^{-36}\)
= 3.7 x 10-38
S4 = 370 x 10-40
S = 4.39 x 10-10 g/L
Hence, solubility of Al(OH)3 will be 4.39 x 10-10 g/L