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