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 bepKa of acetic acid = 4.74]

CH3COOH+NaOH⟶CH3COONa+H2O⇒At equivalence point, a salt of weak acid, strong base is formed.⇒pH=7+12pKa+log⁡C Here C= concentration of salt =0.2×500500+500=0.1M[Check that 500 mL of CH3COOH is also required]⇒pH=7+12(4.74+log⁡0.1)=8.87