Question

Solubility products of $Al(OH{)}_3$ and $Zn(OH{)}_2$ are $2.7\times {10}^{-23}and3.2\times {10}^{-14}$ respectively. If to a solution of 0.1 M each of $A{l}^{3+}andZ{n}^{2+}ions,N{H}_{4}OH$ is added in increasing amounts which of the following will be precipitated first ?

• A. $Zn(OH{)}_2$
• B. $Al(OH{)}_3$
• C. $Bothofthem$
• D. $Noneofthem$

Answer 1

The correct option is B

$Al(OH{)}_3$

$Al(OH{)}_{3\left(S\right)}\rightleftharpoons A{l}^{3+}+3O{H}^{-};K_{SP}=2.7\times {10}^{-23}=27S^4\Rightarrow S=1.0\times {10}^{-6}M$
$\left[O{H}^{-}\right]$ required for precipitation of \)Al(OH)_3 = 3S = 3.0 \times 10^{-6}mol~L^{-1}\)
$Zn(OH{)}_{2\left(S\right)}\rightleftharpoons Z{n}_{aq}^{2+}+2O{H}_{\left(aq\right)}^{-}{K}_{SP}=3.2\times {10}^{-14}=4S^3\Rightarrow S=2\times {10}^{-5}mol{L}^{-1}$
$[OH{]}^{-}$ required for precipitation of $Zn(OH{)}_2$
$=2S=4\times {10}^{-5}mol{L}^{-1}$