Question

To a solution of $0.1M$ $M{g}^{2+}$ and $0.8M$ $N{H}_{4}Cl$, an equal volume of $N{H}_3$ is added which just gives precipitate. Calculate $\left[N{H}_3\right]$ in solution.
$K_{sp}$ of $Mg(OH{)}_2=1.4\times {10}^{-11}$ and $K_b$ of $N{H}_{4}OH=1.8\times {10}^{-5}$

• A. $0.2345M$
• B. $0.7420M$
• C. $0.1855M$
• D. $0.3710M$

Answer 1

The correct option is D $0.3710M$

Suppose $V_{mL}$ of solution contains $0.1MM{g}^{2+}$ and $0.8M$ $N{H}_{4}Cl$.

Now $V_{mL}$ of a molarity $N{H}_3$ is added which just given precipitate of $Mg(OH{)}_2$, then

$\left[M{g}^{2+}\right][\stackrel{⊝}{O}H{]}^2=K_{sp}Mg(OH{)}_2$

$\left[\frac{0.1V}{2V}\right][\stackrel{⊝}{O}H{]}^2=1.4\times {10}^{-11}$

$\therefore [\stackrel{⊝}{O}H{]}^2=1.67\times {10}^{-5}M\left(\right[M{g}^{2+}]=\frac{Millimoles}{Totalvolume})$

Now if the $\left[\stackrel{⊝}{O}H\right]=1.67\times {10}^{-5}$, on addition of $N{H}_3$ in $N{H}_{4}Cl$, $Mg(OH{)}_2$ will precipitate.

$\therefore -log\left[\stackrel{⊝}{O}H\right]=-log{K}_b+log\frac{\left[N{H}_{4}Cl\right]}{N{H}_3}$

$-log(1.67\times {10}^{-5})=-log(1.8\times {10}^{-5})+log\frac{(0.8\times V)/2V}{(a\times V)/2V}$

$\therefore a=0.7421M$

$\therefore \left[N{H}_3\right]$ in solution $=\frac{0.7421\times V}{2V}=0.3710M$