Q.The pKa of acetic acid and pKb of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the pH of ammonium acetate solution.

The key idea is that ammonium acetate is a salt of a weak acid and a weak base. For such a salt, the pH is independent of concentration and given by:

Reasoning:

1. Ammonium acetate hydrolyses in water: the cation $\text{NH}_4^+$ acts as a weak acid, and the anion $\text{CH}_3\text{COO}^-$ acts as a weak base.
2. The resulting $\text{H}^+$ concentration depends on both $K_a$ and $K_b$ of the conjugate acid–base pair. The exact derivation yields the formula above. …

Why This Approach Works

Ammonium acetate ($\mathrm{CH_3COONH_4}$) is a special case: it's a salt formed from a weak acid (acetic acid, $\mathrm{CH_3COOH}$) and a weak base (ammonium hydroxide, $\mathrm{NH_4OH}$). When dissolved in water, both the cation ($\mathrm{NH_4^+}$) and the anion ($\mathrm{CH_3COO^-}$) hydrolyze — that is, they react with water.

The key insight: the pH of such a solution depends on the relative strengths of the parent acid and base. If the acid is stronger ($\mathrm{p}K_a$ smaller), the solution is slightly acidic. If the base is stronger ($\mathrm{p}K_b$ smaller), it's slightly basic. If they are nearly equal, the solution is nearly neutral.

Here, $\mathrm{p}K_a = 4.76$ and $\mathrm{p}K_b = 4.75$ — they are almost identical. So we expect a pH very close to 7. Let's calculate exactly.

Step-by-Step Derivation

1. Write the hydrolysis reactions.

The ammonium ion hydrolyses as a weak acid:

The acetate ion hydrolyses as a weak base:

2. Set up the equilibrium expressions.

For the ammonium ion:

But note: $K_a(\mathrm{NH_4^+})$ is related to $K_b(\mathrm{NH_3})$ by $K_a \times K_b = K_w$. Since we are given $\mathrm{p}K_b$ of ammonium hydroxide ($\mathrm{NH_4OH}$), which is the same as $\mathrm{p}K_b$ of $\mathrm{NH_3}$, we have:

For the acetate ion:

And $K_b$ is related to the given $K_a$ of acetic acid:

3. Derive the pH formula for a salt of weak acid + weak base.

Let the initial concentration of the salt be $c$ mol/L. At equilibrium, let $x$ be the concentration of $\mathrm{H_3O^+}$ from the first hydrolysis, and $y$ be the concentration of $\mathrm{OH^-}$ from the second hydrolysis. But here's the clever part: the two hydrolyses are coupled — the $\mathrm{H_3O^+}$ and $\mathrm{OH^-}$ produced will partially neutralize each other.

A cleaner approach: use the exact expression derived from simultaneous equilibria. For a salt $\mathrm{BA}$ where $\mathrm{B^+}$ is the conjugate acid of a weak base ($\mathrm{BOH}$) and $\mathrm{A^-}$ is the conjugate base of a weak acid ($\mathrm{HA}$), the $\mathrm{[H_3O^+]}$ is given by: