Question

A $1$ litre solution containing $N{H}_{4}Cl$ and $N{H}_{4}OH$ has hydroxide ion concentration of ${10}^{-6}$ mol/litre. Which of the following hydroxides could be precipitated when the solution is added to $1$ litre solution of $0.1M$ metal ions?
$\left(I\right)Ba\left(OH{)}_2\right(K_{sp}=5\times {10}^{-3})$
$\left(II\right)Ni\left(OH{)}_2\right(K_{sp}=1.6\times {10}^{-16})$
$\left(III\right)Mn\left(OH{)}_2\right(K_{sp}=2\times {10}^{-13})$
$\left(IV\right)Fe\left(OH{)}_2\right(K_{sp}=8\times {10}^{-16})$

• A. $I,II,IV$
• B. $IV$
• C. $II$ and $IV$
• D. $II,III,IV$

Answer 1

The correct option is D $II,III,IV$

Since 1 L of $N{H}_{4}Cl+N{H}_{4}OH$ mixture is added to 1 L of 0.1 M metal ion solution, the concentration of each species will be reduced to half.
$\left[O{H}^{-}\right]=\frac{{10}^{-6}M}{2}={10}^{-3}M$
$M^{2+}=\frac{0.1M}{2}=0.05M$
For $Ba(OH{)}_2$, $Q=\left[B{a}^{2+}\right]\times [O{H}^{-}{]}^2$
$Q=0.05\times [{10}^{-3}{]}^2$
$Q=5\times {10}^{-8}$ which is less than $K_{sp}(5\times {10}^{-3})$. Hence, no precipitation will occur.
For $Ni(OH{)}_2$, $Q=\left[N{i}^{2+}\right]\times [O{H}^{-}{]}^2$
$Q=0.05\times [{10}^{-3}{]}^2$
$Q=5\times {10}^{-8}$ which is more than $K_{sp}(1.6\times {10}^{-16})$. Hence, precipitation will occur.
For $Mn(OH{)}_2$, $Q=\left[M{n}^{2+}\right]\times [O{H}^{-}{]}^2$
$Q=0.05\times [{10}^{-3}{]}^2$
$Q=5\times {10}^{-8}$ which is more than $K_{sp}(2\times {10}^{-13})$. Hence, precipitation will occur.
For $Fe(OH{)}_2$, $Q=\left[F{e}^{2+}\right]\times [O{H}^{-}{]}^2$
$Q=0.05\times [{10}^{-3}{]}^2$
$Q=5\times {10}^{-8}$ which is more than $K_{sp}(8\times {10}^{-16})$. Hence, precipitation will occur.