2.5 ml of (2/5) M weak monoacidic base $({\mathrm{K}}_{\mathrm{b}}=1\times {10}^{-12}$at 25°) (2/15) M HCl in water at 25°C. The concentration of H⁺ at equivalence point is $({\mathrm{K}}_{\mathrm{w}}=1\times {10}^{-14}\text{ at }{25}^{\circ }\mathrm{C})$

Let the weak monoacidic base be BOH, then the reaction that occurs during titration is 

$$\begin{gathered} \mathrm{BOH}+\mathrm{HCl}\to \mathrm{BCl}+{\mathrm{H}}_2\mathrm{O} \\ \text{ Equilibrium: }\underset{\mathrm{c}(1-\mathrm{h})}{{\mathrm{B}}^{+}}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons \underset{\mathrm{ch}}{\mathrm{BOH}}+{\mathrm{H}}^{+} \end{gathered}$$
Using the normality equation, $\begin{array}{r}{\mathrm{N}}_1{\mathrm{V}}_1=\\ \text{ (acid) }\end{array}\begin{array}{r}{\mathrm{N}}_2{\mathrm{V}}_2\\ \text{ (base) }\end{array}$
Substituting various given values, we get 
$$\begin{gathered} \frac{2}{15}\cdot {\mathrm{V}}_1=2.5\times \frac{2}{15} \\ \text{ or }{\mathrm{V}}_1=2.5\times \frac{2}{5}\times \frac{15}{2}=2.5\times 3=7.5\mathrm{ml} \end{gathered}$$

Then the concentration of BCl in resulting solution is given by 
$$\begin{gathered} \left[\mathrm{BCl}\right]=\frac{\frac{2}{5}\times 2.5}{10}=\frac{1}{10}\text{ or }0.1\mathrm{M} \\ [\text{ Total volume }=2.5+7.5=10\mathrm{ml}] \\ \text{ Since }{\mathrm{K}}_{\mathrm{h}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{b}}} \\ \mathrm{\setminus }{\mathrm{K}}_{\mathrm{h}}=\frac{1\times {10}^{-14}}{1\times {10}^{-12}}={10}^{-2} \\ \text{ Thus }{\mathrm{K}}_{\mathrm{h}}=\frac{0.1{\mathrm{h}}^2}{(1-\mathrm{h})}\text{ or }{10}^{-2}=\frac{0.1{\mathrm{h}}^2}{(1-\mathrm{h})} \\ \text{ or }{10}^{-2}-{10}^{-2}\mathrm{h}=0.1{\mathrm{h}}^2 \\ \text{ or }0.1{\mathrm{h}}^2+{10}^{-2}\mathrm{h}-{10}^{-2}=0 \end{gathered}$$
(Solving this quadratic equation for h, we get) 
$$\begin{gathered} \text{ Using }\left[\mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}\right] \\ \mathrm{h}=\frac{-{10}^{-2}\pm \sqrt{{\left({10}^{-2}\right)}^2+4\times {10}^{-1}\times {10}^{-2}}}{2\times 0.1} \\ =\frac{-{10}^{-2}\pm \sqrt{{10}^{-4}+4^{-1}{10}^{-3}}}{2^{-}0.1} \\ =\frac{-0.01\pm \sqrt{.0001+0.004}}{0.2} \\ =\frac{-0.01\pm \sqrt{0.0041}}{0.2} \\ =\frac{-0.01\pm 0.064}{0.2} \\ =\frac{0.54}{0.2} \\ =0.27 \\ \therefore \left[{\mathrm{H}}^{+}\right]=\mathrm{c}.\mathrm{h}=0.1\times 0.27=2.7\times {10}^{-2}\mathrm{M} \\ \mathrm{Thus} \mathrm{the} \mathrm{correct} \mathrm{answer} \mathrm{is} \left[\mathrm{d}\right]. \end{gathered}$$