CH₃COOH + NaOH $$\to$$ CH₃COONa + H₂O

After adding 25 ml of NaOH volume of mixture = 50 + 25 = 75 ml

Initially,

Number of millimole of NaOH = 25 $$\times$$ 0.1 = 2.5 mm

Number of millimole of CH₃COOH = 50 $$\times$$ 0.1 = 5 mm

After nutrilisation,

Millimole of NaOH = 0

Millimole of CH₃COOH = 5 $$-$$ 2.5 = 2.5 mm

Millimole of CH₃COONa = 2.5

After nutrilisation,

Concentration of CH₃COOH = $$[C{H_3}COOH] = {{5 - 2.5} \over {75}} = {1 \over {30}}$$

Concentration of CH₃COONa = $$[C{H_3}COONa] = {{ 2.5} \over {75}} = {1 \over {30}}$$

$${P^H} = {P^{Ka}} + \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$$

$$ = 4.76 + \log {{{1 \over {30}}} \over {{1 \over {30}}}}$$

$$ = 4.76 + \log (1)$$

$$ = 4.76 + 0$$

$$ = 4.76$$

$$ = 4.76 \times {10^{ - 2}}$$