1. Identify the type of solution

After mixing sodium acetate and acetic acid, we get a buffer solution of:

• weak acid: $\mathrm{CH_3COOH}$
• its salt: $\mathrm{CH_3COONa}$

So we use the Henderson–Hasselbalch equation:

$\mathrm{pH} = \mathrm{p}K_a + \log\frac{[\text{salt}]}{[\text{acid}]}$

2. Calculate moles before mixing

For $25\,\text{mL}$ of $0.2\,\text{M }\mathrm{CH_3COONa}$:

$n_{\text{salt}} = 0.2 \times 0.025 = 0.005\,\text{mol}$

For $25\,\text{mL}$ of $0.02\,\text{M }\mathrm{CH_3COOH}$:

$n_{\text{acid}} = 0.02 \times 0.025 = 0.0005\,\text{mol}$

3. Find ratio of salt to acid

Since both are mixed and total volume is the same for both, concentration ratio equals mole ratio:

$\frac{[\text{salt}]}{[\text{acid}]} = \frac{0.005}{0.0005} = 10$

4. Use given pH

Given:

$\mathrm{pH} = 5$

Apply Henderson equation:

$5 = \mathrm{p}K_a + \log 10$

Since $\log 10 = 1$,

$5 = \mathrm{p}K_a + 1$

$\mathrm{p}K_a = 4$

5. Calculate $K_a$

$K_a = 10^{-\mathrm{p}K_a} = 10^{-4}$

Given that dissociation constant is $x \times 10^{-5}$,

$x \times 10^{-5} = 10^{-4} = 10 \times 10^{-5}$

Hence,

$x = 10$

6. Compare with stored answer

Stored correct answer = $10$

This matches our derived answer.