The correct option is D 8.87
0.2 M CH3COOH 0.2 M NaOH each 100 mL is taken here.
Milimoles of CH3COOH=0.2×100=20 mmol
Milimoles of CH3COOH=0.2×100=20 mmol
CH3COOH+NaOH⇌CH3COONa+H2O
20 + 20 ⇌ 0 + 0
0 + 0 ⇌ 20 + 20
So there is complete neutralisation
[salt]=20200=0.1
Using the formula,
pH=7+12log pKa+log C
pH=7+12log pKa+12log[salt]
putting the values,
pH=7+12log (4.74)+12log(0.1)
pH=7+2.37−0.5
pH=8.87