⇒Ionic Equilibrium

${\mathrm{Na}}_2{\mathrm{SO}}_4$ is a salt of strong base $(\mathrm{NaOH})$ and strong acid $\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$.

(A) ${\mathrm{Na}}_2{\mathrm{CO}}_3$ : Salt of weak acid and strong base: Solution is basic.

(B) ${\mathrm{NH}}_4{\mathrm{NO}}_3$ : Salt of strong acid and weak base: Solution is acidic.

(C) ${\mathrm{NH}}_4\mathrm{CN}$ : Salt of weak acid and weak base for which ${\mathrm{K}}_{\mathrm{a}}<{\mathrm{K}}_{\mathrm{b}}$ : Solution is basic.

(D) $\mathrm{KCN}$ : Salt of weak acid and strong base: Solution is basic.

${\mathrm{NH}}_4\mathrm{CN}$ is the salt of weak acid $\mathrm{HCN}({\mathrm{K}}_{\mathrm{a}}=4.0\times {10}^{-10})$ and weak base ${\mathrm{NH}}_4\mathrm{OH}({\mathrm{K}}_{\mathrm{b}}=1.8\times {10}^{-5})$ showing that ${\mathrm{K}}_{\mathrm{a}}<{\mathrm{K}}_{\mathrm{b}}$. If ${\mathrm{K}}_{\mathrm{a}}<{\mathrm{K}}_{\mathrm{b}}$, the aqueous solution of the salt will be basic. Thus, the solution of ${\mathrm{NH}}_4\mathrm{CN}$ is basic and turns red litmus blue.

Using Henderson's equation,

$\begin{array}{rl}& \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+{\log }_{10}\frac{\text{ [Salt }]}{[\text{ Acid }]}\\ & \mathrm{pH}=4.7+{\log }_{10}\frac{0.2}{0.1}\\ & \mathrm{pH}=4.7+{\log }_{10}2=4.7+0.3010=5.001\end{array}$

For acidic buffer, if [Salt] > [Acid], then $\mathrm{pH}>{\mathrm{pK}}_{\mathrm{a}}$ of acid. Hence, only option (C) is valid.

$\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-12}\mathrm{mol}{\mathrm{dm}}^{-3},\left[{\mathrm{H}}^{+}\right]=?$

Product of hydrogen ion concentration and hydroxide ion concentration is equal to $1.0\times {10}^{-14}$.

$\begin{array}{rl}\left[{\mathrm{H}}^{+}\right]\times \left[{\mathrm{OH}}^{-}\right]& =1\times {10}^{-14}\\ \therefore \left[{\mathrm{H}}^{+}\right]& =\frac{1\times {10}^{-14}}{1\times {10}^{-12}}=1\times {10}^{-2}\\ & =0.01\mathrm{mol}{\mathrm{dm}}^{-3}\end{array}$