Question

Answer the following :

A solution of CH₃COONH₄ is neutral. why?

Answer 1

i. CH₃COONH₄ is a salt of a weak acid, CH₃COOH (K_a = 1.8 × 10^–5), and weak base, NH₄OH (K_b = 1.8 × 10^–5). When the salt CH₃COONH₄ is dissolved in water, it undergoes hydrolysis:

\[\ce{CH3COO^-_{ (aq)} + NH^+_{ 4(aq)} + H2O_{(l)} ⇌ \underset{\text{(weak acid)}}{CH_3COOH_{(aq)}} + \underset{\text{(weak base)}}{NH4OH_{(aq)}}}\]

ii. The ions of the salt react with water as

\[\ce{CH3COO^-_{ (aq)} + H2O_{(l)} ⇌ CH3COOH_{(aq)} + OH^-_{ (aq)}}\]  ...(1)

\[\ce{NH^4+_{ (aq)} + 2H2O_{(l)} ⇌ NH4OH_{(aq)} + H3O^+_{ (aq)}}\] ...(2)

iii. As K_a = K_b, the relative strength of acid and base produced in hydrolysis is the same.

iv. Therefore, the solution is neutral. Hydrolysis of `" NH"_4^+` produces as many H₃O⁺ ions as that of CH3COO^– produces OH⁻ ions.