The solubility products of $Al (OH)_{3}$ and $Zn (OH)_{2}$ , $8.5 \times 10^{- 23}$ and $1.8 \times 10^{- 14}$ respectively. If ${NH}_{4} OH$ is added to a solution containing 0.01 M concentration of ${AI}^{3 +}$ and ${Zn}^{2 +}$ ions: then substance precipitated first is :

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none at all

$Zn (OH)_{2}$

both together

$Al (OH)_{3}$

answer is A.

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Detailed Solution

$\begin{aligned} {[{OH}^{-}]}_{1} = \sqrt[3]{\frac{8.5 \times 10^{- 23}}{0.01}} = 2.04 \times 10^{- 7} \\ {[{OH}^{-}]}_{2} = \sqrt{\frac{1.8 \times 10^{- 14}}{0.01}} = 1.34 \times 10^{- 6} \end{aligned}$

Hence, $Al (OH)_{3}$ will start precipitating first

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