Question

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

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

Answer 1

Answer: (D)

None of these

${Al{\left(OH\right)}_3}_{\left(s\right)}\rightleftharpoons {{Al}^{3+}}_{\left(aq\right)}+3{{OH}^{-}}_{\left(aq\right)};K_{sp}$

${Al{\left(OH\right)}_3}_{\left(s\right)}+{{OH}^{-}}_{\left(aq\right)}\rightleftharpoons {{Al{\left(OH\right)}_4}^{-}}_{\left(aq\right)};K_c$

Dissolved $Al{\left(OH\right)}_3$ is present in form of ${Al}^{+3}$ and ${\left[Al{\left(OH\right)}_4\right]}^{-}$

$\therefore$ Solubility $\left(s\right)={Al}^{+3}+{\left[Al{\left(OH\right)}_4\right]}^{-}$

$\Rightarrow s=\frac{K_{sp}}{{\left[{OH}^{-}\right]}^3}+K_c\left[{OH}^{-}\right]$

For minimum solubility, $\frac{ds}{d\left[{OH}^{-}\right]}=0$

$\therefore \frac{-3K_{sp}}{{\left[{OH}^{-}\right]}^4}+K_c=0$

$\Rightarrow \frac{-3K_{sp}}{{\left[{OH}^{-}\right]}^4}=-K_c$

$\Rightarrow {\left[{OH}^{-}\right]}^4=\frac{3K_{sp}}{K_c}$
$\Rightarrow \left[{OH}^{-}\right]={\left(\frac{3K_{sp}}{K_c}\right)}^{\frac{1}{4}}$