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Question
Answer the following :
A solution of CH3COONH4 is neutral. why?
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Solution
i. CH3COONH4 is a salt of a weak acid, CH3COOH (Ka = 1.8 × 10–5), and weak base, NH4OH (Kb = 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 Ka = Kb, 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.
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