Question

To a solution of $0.1M$ ${Mg}^{2+}$ and $0.8M$ ${NH}_{4}Cl$, an addition of equal volume of ${NH}_3$ gives precipitate. $\left[{NH}_3\right]$ in solution is:

[$K_{{sp}_{Mg{\left(OH\right)}_2}}=1.4\times {10}^{-11}$ and $K_{b_{{NH}_{4}OH}}=1.8\times {10}^{-5}$].

• A. $0.37M$
• B. $0.40M$
• C. $0.33M$
• D. $0.30M$

Answer 1

The correct option is A $0.37M$

Suppose $VmL$ of solution containing $0.1M$ ${Mg}^{2+}$ and $0.8M$ ${NH}_{4}Cl$.

Now $VmL$ of ${NH}_3$ of a $M$ is added to it in order to have just precipitation of $Mg{\left(OH\right)}_2$, then

$\left[{Mg}^{2+}\right]{\left[{OH}^{-}\right]}^2=K_{{sp}_{Mg{\left(OH\right)}_2}}$

or $\left[\frac{0.1\times V}{2V}\right]{\left[{OH}^{-}\right]}^2=1.4\times {10}^{-11}[\therefore \left[{Mg}^{2+}\right]=\frac{millimole}{totalvolume}]$

$\therefore$ $\left[{OH}^{-}\right]=1.67\times {10}^{-5}M$

The solution must therefore, contain $\left[{OH}^{-}\right]=1.67\times {10}^{-5}M$ which is obtained by buffer solution of ${NH}_3$ and ${NH}_{4}Cl$

$\therefore$ $-\log \left[OH\right]=-\log K_b+\log \frac{\left[ConjugateAcid\right]}{\left[Base\right]}$

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

$\therefore$ $a=0.74M$

$\therefore$ $\left[{NH}_3\right]$ in solution $=\frac{0.74\times V}{2V}=0.37M$