$$20$$ mL of $$0.1$$ M NaOH is added to $$50$$ mL of $$0.1$$ M acetic acid solution. The pH of the resulting solution is _____ $$\times 10^{-2}$$. (Nearest integer) Given : pKa CH$$_3$$COOH $$= 4.76$$
log 2 = 0.30
log 3 = 0.48

20 mL of 0.1 M NaOH + 50 mL of 0.1 M CH₃COOH.

Moles of NaOH = 0.002 mol

Moles of CH₃COOH = 0.005 mol

After reaction: NaOH fully consumed, remaining CH₃COOH = 0.003 mol, CH₃COONa formed = 0.002 mol.

This is a buffer solution. Using Henderson-Hasselbalch equation:

$$pH = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]} = 4.76 + \log\frac{0.002}{0.003}$$

$$= 4.76 + \log\frac{2}{3} = 4.76 + \log 2 - \log 3$$

$$= 4.76 + 0.30 - 0.48 = 4.58$$

In the required form: $$458 \times 10^{-2}$$

This matches the answer key value of $$\mathbf{458}$$.