When 0.2 M solution of acetic acid is neutralised with 0.2 M NaOH in 500 mL of water, the pH of the resulting solution will be: [pK_a of acetic acid = 4.74]

${\mathrm{CH}}_3\mathrm{COOH}+\mathrm{NaOH}⟶{\mathrm{CH}}_3\mathrm{COONa}+{\mathrm{H}}_2\mathrm{O}$

$\Rightarrow$At equivalence point, a salt of weak acid, strong base is formed.

$\Rightarrow \mathrm{pH}=7+\frac{1}{2}\left({\mathrm{pK}}_{\mathrm{a}}+\log \mathrm{C}\right)$

$\text{ Here }\mathrm{C}=\text{ concentration of salt }=\frac{0.2\times 500}{500+500}=0.1\mathrm{M}$

[Check that 500 mL of CH₃COOH is also required] 

$\Rightarrow \mathrm{pH}=7+\frac{1}{2}(4.74+\log 0.1)=8.87$