Question

The solubility products of $Al{\left(OH\right)}_3$ and $Zn{\left(OH\right)}_2$ 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 ${NH}_{4}OH$ is:

• A. ${Al}^{3+}$
• B. ${Zn}^{2+}$
• C. Both
• D. None of these

Answer 1

The correct option is B ${Zn}^{2+}$

$Al{\left(OH\right)}_3\rightleftharpoons {Al}^{3+}+3{OH}^{-}$

$K_s=\left[{Al}^{3+}\right]{\left[{OH}^{-}\right]}^3$

$=S_1\times {\left(3S_1\right)}^3$

$=27S_1^4$ but $K_s=8.5\times {10}^{-23}$

$\therefore S_1=1.32\times {10}^{-6}$.

$Zn{\left(OH\right)}_2\rightleftharpoons {Al}^{2+}+2{OH}^{-}$

$K_s=\left[{Zn}^{2+}\right]{\left[{OH}^{-}\right]}^2$

$=S_2\times {\left(2S_2\right)}^2$

$=4S_2^3$ but $K_s=1.8\times {10}^{-14}$

$S_2=1.64\times {10}^{-5}$.

So $Z{n}^{2+}$ will precipitate first.

Hence, the correct option is $\text{B}$