$2.5 \mathrm{~mL}$ of $\frac{2}{5} \mathrm{M}$ weak monoacidic base $\left(K_b=1 \times 10^{-12}\right.$ at $\left.25^{\circ} \mathrm{C}\right)$ is titrated with $\frac{2}{15}$ $\mathrm{M} \mathrm{HCl}$ in water at $25^{\circ} \mathrm{C}$. The concentration of $\mathrm{H}^{+}$at equivalence point is $\left(K_w=1 \times 10^{-14}\right.$ at $\left.25^{\circ} \mathrm{C}\right)$

Weak monoacidic base, e.g. $\mathrm{BOH}$ is neutralised. $$ \mathrm{BOH}+\mathrm{HCl} \longrightarrow \mathrm{BCl}+\mathrm{H}_2 \mathrm{O} $$ At equivalence point all $\mathrm{BOH}$ gets converted into salt and remember! the concentration of $\mathrm{H}^{+}$(or $\mathrm{pH}$ of solution) is due to hydrolysis of resultant salt ( $\mathrm{BCl}$, cationic hydrolysis here) $$ \underset{\mathrm{C}(1-h)}{\mathrm{B}^{+}}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \underset{\mathrm{Ch}}{\rightleftharpoons} \mathrm{OH}+\underset{\mathrm{Ch}}{\mathrm{H}^{+}} $$ Volume of $\mathrm{HCl}$ used up, $$ V_a=\frac{N_b V_b}{N_a}=\frac{2.5 \times 2 \times 15}{2 \times 5}=7.5 \mathrm{~mL} $$ Concentration of salt, $$ \begin{aligned} {[\mathrm{BCl}] } & =\frac{\text { Concentration of base }}{\text { Total volume }}=\frac{2 \times 25}{5(7.5+2.5)}=\frac{1}{10}=0.1 \\ K_h & =\frac{C h^2}{1-h}=\frac{K_w}{K_b} \end{aligned} $$ ( $h$ should be estimated whether that can be neglected or not) on calculating $h=0.27$ (significant, not negligible) $$ \left[\mathrm{H}^{+}\right]=C h=0.1 \times 0.27=2.7 \times 10^{-2} \mathrm{M} $$ When $\left[\mathrm{H}^{+}\right]$is asked to calculate in connection with neutralisation, it Should be calculated: Before neutralisation Using Oslwald's dilution law During neutralisation $\quad$ Considering buffer solution Half neutralisation $\mathrm{pH}=\mathrm{p} K_a$ and $\mathrm{p} K_b$ At the end of neutralisation $\quad$ Considering hydrolysis of salt