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