The salt $Al {(OH)}_{3}$ involved in the following two equilibria,
$\begin{array}{l} Al (OH)_{3} (s) ⇌ {Al}^{3 +} (a q) + 3 {OH}^{-} (a q), K_{s p} \\ Al (OH)_{3} (s) + {OH}^{-} (a q) ⇌ Al {(OH)_{4}}^{-} (a q), K_{c} \end{array}$
Which of the following relationship is CORRECT at which solubility is minimum?

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$[{OH}^{-}] = {(\frac{K_{s p}}{K_{c}})}^{1 / 3}$

$[{OH}^{-}] = {(\frac{K_{c}}{K_{s p}})}^{1 / 4}$

$[{OH}^{-}] = {(\frac{3 K_{s p}}{K_{c}})}^{1 / 4}$

$[{OH}^{-}] = \sqrt{{(\frac{K_{s p}}{K_{c}})}^{1 / 4}}$

answer is D.

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Detailed Solution

$\begin{aligned} [Al (OH)_{3}] = [{Al}^{3 +}] + [Al (OH)_{4}^{-}] \\ S = \frac{K_{s p}}{{[{OH}^{-}]}^{3}} + K_{C} [{OH}^{-}] \\ \frac{d S}{d [{OH}^{-}]} = \frac{- 3 K_{s p}}{{[{OH}^{-}]}^{4}} + K_{C} = 0 \\ \Rightarrow [{OH}^{-}] = \sqrt[4]{\frac{3 K_{s p}}{K_{C}}} \end{aligned}$