Question

The salt $Al(OH{)}_3$ is involved in the following two equilibria,
$Al\left(OH{)}_3\right(s)\rightleftharpoons A{l}^{3+}(aq.)+3O{H}^{-}(aq.);K_{sp}$
$Al\left(OH{)}_3\right(s)+O{H}^{-}(aq)\rightleftharpoons Al(OH{)}_4^{-}(aq.);K_c$
Which of the following relationships is correct when the solubility is minimum?

• A. $\left[O{H}^{-}\right]={\left(\frac{3K_{sp}}{K_c}\right)}^{\frac{1}{4}}$
• B. $\left[O{H}^{-}\right]={\left(\frac{K_c}{K_{sp}}\right)}^{\frac{1}{4}}$
• C. $\left[O{H}^{-}\right]=\sqrt{{\left(\frac{K_{sp}}{K_c}\right)}^{\frac{1}{4}}}$
• D. None of these

Answer 1

The correct option is A $\left[O{H}^{-}\right]={\left(\frac{3K_{sp}}{K_c}\right)}^{\frac{1}{4}}$

Dissolved $Al(OH{)}_3$ present in solution as $A{l}^{3+}$ and $Al(OH{)}_4^{-}$, lets say solubility of $Al(OH{)}_3$ is $S$ then
$S=\left[A{l}^{3+}\right]+\left[Al\right(OH{)}_4^{-}]$
$=\frac{K_{sp}}{[O{H}^{-}{]}^3}+K_c\left[O{H}^{-}\right]$
for minimum solubility $\frac{dS}{d\left[O{H}^{-}\right]}=0$ that is
$\frac{-3K_{sp}}{[O{H}^{-}{]}^4}+K_c=0$ hence ,
$\left[O{H}^{-}\right]={\left(\frac{3K_{sp}}{K_c}\right)}^{\frac{1}{4}}$