20 mL 0.1 N acetic acid is mixed with 10 mL 0.1 N solution of NaOH. The pH of the resulting solution is (pK_a of acetic acid is 4.74)

$$\begin{gathered} {\mathrm{CH}}_3\mathrm{COOH}+\mathrm{NaOH}⟶{\mathrm{CH}}_3\mathrm{COONa}+{\mathrm{H}}_2 \\ \mathrm{Initially} 20\times 0.1 10\times 0.1 \\ = 2 \mathrm{mmol} = 1 \mathrm{mmol} \\ \mathrm{At} \mathrm{time} \mathrm{t} (2- 1) (1-1) 1 \mathrm{mmol} \\ = 1 \mathrm{mmol} = 0 \mathrm{mmol} \end{gathered}$$
From Henderson's equation,
$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[{\mathrm{CH}}_3\mathrm{COONa}\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}=4.74+\log \frac{1}{1}=4.74$