Hydronium Ion Concentration at Equivalence Point of Weak Base Titration

To determine the concentration of $\text{H}^+$ at the equivalence point, we proceed with the following steps:

1. Millimoles of the weak base and strong acid:

• Millimoles of weak monoacidic base ($B$):
  $\text{mmol of } B = 2.5\text{ mL} \times \frac{2}{5}\text{ M} = 1\text{ mmol}$

At the equivalence point, the millimoles of $\text{HCl}$ required are equal to the millimoles of the weak base:
$\text{mmol of } \text{HCl} = 1\text{ mmol}$

• Volume of $\text{HCl}$ solution required:
  $V_{\text{HCl}} = \frac{1\text{ mmol}}{\frac{2}{15}\text{ M}} = \frac{15}{2}\text{ mL} = 7.5\text{ mL}$

• Total volume of the solution at the equivalence point:
  $V_{\text{total}} = 2.5\text{ mL} + 7.5\text{ mL} = 10.0\text{ mL}$

2. Concentration of the salt ($BH^+\text{Cl}^-$) formed:
$C = \frac{1\text{ mmol}}{10.0\text{ mL}} = 0.1\text{ M}$

3. Hydrolysis of the conjugate acid ($BH^+$):
The cation of the weak base undergoes hydrolysis:
$BH^+ + \text{H}_2\text{O} \rightleftharpoons B + \text{H}_3\text{O}^+$

The hydrolysis constant ($K_h$) is given by:
$K_h = \frac{K_w}{K_b} = \frac{1 \times 10^{-14}}{1 \times 10^{-12}} = 10^{-2}$

Let $h$ be the degree of hydrolysis of $BH^+$. At equilibrium:
$[BH^+] = C(1-h)$
$[B] = [\text{H}^+] = Ch$

Therefore:
$K_h = \frac{C h^2}{1 - h}$

Substituting the known values:
$10^{-2} = \frac{0.1 \cdot h^2}{1 - h}$
$0.1 = \frac{h^2}{1 - h}$
$h^2 + 0.1h - 0.1 = 0$

Solving the quadratic equation for $h$:
$h = \frac{-0.1 + \sqrt{(0.1)^2 - 4(1)(-0.1)}}{2} = \frac{-0.1 + \sqrt{0.01 + 0.4}}{2} = \frac{-0.1 + \sqrt{0.41}}{2}$
$h \approx \frac{-0.1 + 0.6403}{2} \approx 0.270$

4. Concentration of $\text{H}^+$:
$[\text{H}^+] = Ch = 0.1\text{ M} \times 0.270 \approx 2.7 \times 10^{-2}\text{ M}$

Therefore, the correct option is D.