$$50$$ mL of $$0.1$$ M $$CH_3COOH$$ is being titrated against $$0.1$$ M NaOH. When $$25$$ mL of NaOH has been added, the pH of the solution will be ______ $$\times 10^{-2}$$. (Nearest integer)
(Given : $$pK_a(CH_3COOH) = 4.76$$)

We need to find the pH when 25 mL of 0.1 M NaOH is added to 50 mL of 0.1 M $$CH_3COOH$$.

Calculate moles of acid and base

Moles of $$CH_3COOH$$ = 50 × 0.1 = 5 mmol

Moles of NaOH = 25 × 0.1 = 2.5 mmol

After reaction

$$CH_3COOH + NaOH \to CH_3COONa + H_2O$$

NaOH is the limiting reagent. After the reaction:

- $$CH_3COOH$$ remaining = 5 - 2.5 = 2.5 mmol

- $$CH_3COONa$$ formed = 2.5 mmol

This is a buffer solution (half-equivalence point)

Since moles of acid = moles of conjugate base (salt), this is the half-equivalence point.

Using the Henderson-Hasselbalch equation:

$$pH = pK_a + \log\frac{[CH_3COO^-]}{[CH_3COOH]}$$

$$pH = pK_a + \log\frac{2.5}{2.5}$$

$$pH = pK_a + \log(1)$$

$$pH = pK_a + 0$$

$$pH = 4.76$$

Express the answer in the required format

The pH = 4.76 = $$476 \times 10^{-2}$$

The correct answer is 476.