2.5 ~mL of (2)/(5) M weak monoacidic base (K_b=1 × 10^ -12 . at...

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, $$\begin{aligned} {[\mathrm{BCl}]}& =\frac{\text{ Concentration\text{ of } base }}{\text{ Total volume }}=\frac{2 \times 25}{5(7.5+2.5)}=\frac{1}{10}=0.1 \\ K_h & =\frac{C h^2}{1-h}=\frac{K_w}{K_b} \end{aligned}$$(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$