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The solubility products of $Al(OH)_3$ and $Zn(OH)_2$ are $8.5 \times 10^{-23}$ and $1.8 \times 10^{-14}$ at room temperature. If the solution contains $Al^{3+}$ and $Zn^{2+}$ ions, the ion first precipitated by adding $NH_4OH$ is
$Al^{3+}$
$Zn^{2+}$
both
none of these
Solution
The value of solubility product for $Al ( OH )_3$ is $8.5 \times 10^{-23}$.
It is much lower than the value of the solubility product for $Zn ( OH )_2$ which is $1.8 \times 10^{-14}$
Hence, when both $Al ^{3+}$ and $Zn ^{2+}$ ions are present in the solution and as $NH _4 OH$ is added, $Al ( OH )_3$ will precipitate first as the solubility product of $Al ( OH )_3$ is quite low.
Hence, its ionic product will exceed its solubility product at a very low concentration of $OH ^{-}$ ions.
Option $A$ is correct.
