Question

The solubility product of $Al(OH{)}_3$ is $2.7\times {10}^{-11}$. Calculate its solubility in $g{L}^{-1}$ and also find out $pH$ of this solution. (Atomic mass of $Al=27u)$.

Answer 1

Let $S$ be the solubility of $Al(OH{)}_3$.

$Al(OH{)}_3\rightleftharpoons A{l}^{3+}+3(OH{)}^{-}$

$\therefore K_{sp}=\left[A{l}^{3+}\right][O{H}^{-}{]}^3=2.7\times {10}^{-11}$

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

$=S\times (3S{)}^3=2.7\times {10}^{-11}$

Where, $S=$ Molar solubility

$\therefore S^4=\frac{K_{sp}}{27}=\frac{2.7\times {10}^{-11}}{27}=1\times {10}^{-12}$

$\therefore S=1\times {10}^{-3}mol/L$

Solubility of $Al(OH{)}_3=1\times {10}^{-3}mol/L$

So, Solubility of $Al(OH{)}_3$ in $g/L=$ Molar mass $\times$ Solubility in $mol/L$

$=78\times {10}^{-3}g/L$

$=7.8\times {10}^{-2}g/L$

Concentration of $\left[O{H}^{-}\right]=3S$

$=3\times {10}^{-3}mol/L$ (from $Al\left(OH{)}_3\right)$

$pOH=-lo{g}_{10}\left[O{H}^{-}\right]=-lo{g}_{10}(3\times {10}^{-3})$

$=3-log3=2.522$

$\therefore pH=14-2.522$

$=11.478$

Thus, the solubility in $g/L$ is $7.8\times {10}^{-2}g/L$ and $pH=11.478$.