Calculate the pH at the equivalence point during the titration of 0.1 M, 25 mL CH₃COOH with 0.05 M NaOH solution.[K_a (CH₃COOH) = 1.8x 10^-5]

Since, at equivalence point (for acid), ${\mathrm{N}}_1{\mathrm{V}}_1={\mathrm{N}}_2{\mathrm{V}}_2$ (for base)
$\therefore \text{ Volume of NaOH required to reach equivalence point }$
$$\begin{gathered} =\frac{0.1\times 25}{0.05}=50\mathrm{mL} \\ \therefore \text{ Concentration of salt formed } \\ =\frac{\text{ millimoles of acid }}{\text{ total volume }\left(\text{ in }\mathrm{mL}\right)} \\ =\frac{25\times 0.1}{75}=\frac{0.1}{3} \\ \left[{\mathrm{H}}^{+}\right]=\sqrt{\frac{{\mathrm{K}}_{\mathrm{w}}\times {\mathrm{K}}_{\mathrm{a}}}{\mathrm{C}}} \\ =\sqrt{\frac{{10}^{-14}\times 1.8\times {10}^{-5}\times 3}{0.1}} \\ \therefore \mathrm{pH}=8.63 \end{gathered}$$