Question

Find moles of $N{H}_{4}Cl$ required to prevent $Mg(OH{)}_2$ from precipitating in a litre of solution which contains 0.02 mole of $N{H}_3$ and 0.001 mole of $M{g}^{2+}$ ions.
Given : $K_b\left(N{H}_3\right)={10}^{-5};K_{sp}\left[Mg\right(OH{)}_2]={10}^{-11}$

• A. ${10}^{-4}$
• B. $2\times {10}^{-3}$
• C. $0.02$
• D. $0.1$

Answer 1

The correct option is B $2\times {10}^{-3}$

$\left[O{H}^{-}\right]=\sqrt{\frac{K_{sp}}{\left[M{g}^{2+}\right]}}=\sqrt{\frac{{10}^{-11}}{{10}^{-3}}}={10}^{-4}$
$pOH=p{K}_b+\log \frac{\left[N{H}_{4}Cl\right]}{\left[N{H}_3\right]}$
$\Rightarrow 4=5+\log \frac{\left[N{H}_{4}Cl\right]}{\left[N{H}_3\right]}$
$\Rightarrow 0.10=\frac{\left[N{H}_{4}Cl\right]}{\left[N{H}_3\right]}$
moles of $N{H}_{4}Cl$ required $=0.1\times 0.02$
$=2\times {10}^{-3}$
Below this concentration, $Mg(OH{)}_2$ will precipiate.