JEE Main 2024 — Ionic Equilibrium Question with Solution

To find the pH of an ammonium acetate solution, we use the fact that ammonium acetate is a salt resulting from the neutralization of a weak acid (acetic acid, $C{H}_{3}COOH$) by a weak base (ammonium hydroxide, $N{H}_{4}OH$). The respective ionization constant values for the weak acid $K_a$ and the weak base $K_b$ are given as $1.8\times 10^{-5}$.

Firstly, calculate the $p{K}_a$ and $p{K}_b$ values.

• $p{K}_a=-\log (K_a)=-\log (1.8\times 10^{-5})$
• $p{K}_b=-\log (K_b)=-\log (1.8\times 10^{-5})$

Given that the values of $K_a$ and $K_b$ are the same, their $p{K}_a$ and $p{K}_b$ values will also be the same, establishing a neutral condition where the effects of the acid and base neutralize each other.

The formula used to determine the pH of a solution of such a salt is:

$$pH=\frac{1}{2}(p{K}_w+p{K}_a-p{K}_b)$$

Where $p{K}_w$ is the ionic product of water, which is 14 at 25°C. In this specific case, since $p{K}_a=p{K}_b$, the formula simplifies to:

$$pH=\frac{1}{2}(14+p{K}_a-p{K}_a)$$

$$pH=\frac{1}{2}(14)$$

$$pH=7$$

The pH of an ammonium acetate solution in this scenario is 7, indicating a neutral solution.