Question

What is the molar solubility of $Al(OH{)}_3$ in $0.2MNaOH$ solution ? Given that, solubility product of $Al(OH{)}_3=2.4\times {10}^{-24}$

• A. $12\times {10}^{-23}$
• B. $12\times {10}^{-21}$
• C. $3\times {10}^{-19}$
• D. $3\times {10}^{-22}$

Answer 1

The correct option is D $3\times {10}^{-22}$

$Al(OH{)}_3\leftrightharpoons A{l}^{+3}+3O{H}^{-}$

$K_{sp}=\left[A{l}^{3+}\right]\times [O{H}^{-}{]}^3$

Now,

Let, $S^{\prime }$ be the solubility.

$\left[O{H}^{-}\right]=3\left(S^{{}^{\prime }}\right)+0.2M\simeq 0.2M$

$K_{sp}=S^{{}^{\prime }}\times (0.2{)}^3=2.4\times {10}^{-24}$

$S^{{}^{\prime }}=3\times {10}^{-22}M$