Question

The compound whose 0.1 M solution at ${25}^{\circ }C$ is basic is :

• A. $N{H}_{4}Cl$
• B. $(N{H}_4{)}_{2}S{O}_4$
• C. $C{H}_{3}COONa$
• D. All of the above.

Answer 1

The correct option is C $C{H}_{3}COONa$

Option A : Ammonium chloride $N{H}_{4}Cl$ is a salt made of strong acid $HCl$ and weak base $N{H}_{4}OH$ . Due to cationic hydrolysis, $H^{+}$ is generated which makes the solution acidic.

Option B : $(N{H}_4{)}_{2}S{O}_4$ is salt of strong acid $H_{2}S{O}_4$ and weak base $N{H}_{4}OH$. Cation $N{H}_4^{+}$ will undergo hydrolysis and the solution will be acidic in nature due to H+ formation.

Complete dissociation :

$N{H}_{4}Cl⟶N{H}_4^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$

Hydrolysis :

$N{H}_4^{+}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons N{H}_{4}OH\left(aq\right)+H^{+}\left(aq\right)$

Option C : $C{H}_{3}COONa$ is a salt of strong base $NaOH$ and weak acid $C{H}_{3}COOH$.

On complete dissociation $C{H}_{3}COONa\left(aq\right)\to C{H}_{3}CO{O}^{-}\left(aq\right)+N{a}^{+}\left(aq\right)$

On hydrolysis $C{H}_{3}CO{O}^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons C{H}_{3}COOH\left(aq\right)+O{H}^{-}\left(aq\right))$

The anion $C{H}_{3}CO{O}^{-}$ will undergo hydrolysis and the solution will be basic in nature due to $O{H}^{-}$ formation.