Step-by-Step Solution:

1. Understand the Titration Reaction

The titration involves a weak monoacidic base (let's denote it as B) and a strong acid (HCl). The neutralization reaction is: $B + HCl \rightarrow BH^+Cl^-$ At the equivalence point, all the weak base has been converted into its conjugate acid, $BH^+$.

2. Calculate Moles of the Weak Base

First, we calculate the initial number of millimoles (mmol) of the weak base.

• Molarity of base, $M_b = \frac{2}{5} \text{ M} = 0.4 \text{ M}$
• Volume of base, $V_b = 2.5 \text{ mL}$

Moles of base = $M_b \times V_b = 0.4 \text{ mol/L} \times 2.5 \times 10^{-3} \text{ L} = 1.0 \times 10^{-3} \text{ mol} = 1.0 \text{ mmol}$

3. Determine the Volume of HCl at Equivalence Point

At the equivalence point, the moles of acid added are equal to the initial moles of the base.

• Molarity of acid, $M_a = \frac{2}{15} \text{ M}$
• Let $V_a$ be the volume of HCl required.

Moles of acid = Moles of base $M_a \times V_a = 1.0 \text{ mmol}$ $\frac{2}{15} \frac{\text{mmol}}{\text{mL}} \times V_a (\text{mL}) = 1.0 \text{ mmol}$ $V_a = \frac{1.0 \times 15}{2} = 7.5 \text{ mL}$

4. Calculate the Concentration of the Conjugate Acid at Equivalence Point

At the equivalence point, the solution contains the salt $BH^+Cl^-$. The total volume of the solution is the sum of the initial volume of the base and the volume of acid added.

• Total volume, $V_{total} = V_b + V_a = 2.5 \text{ mL} + 7.5 \text{ mL} = 10.0 \text{ mL}$
• The moles of the conjugate acid, $BH^+$, formed are equal to the initial moles of the base, which is $1.0 \text{ mmol}$.

The concentration of the conjugate acid, $[BH^+]$, is: $[BH^+] = \frac{\text{moles of } BH^+}{V_{total}} = \frac{1.0 \text{ mmol}}{10.0 \text{ mL}} = 0.1 \text{ M}$

5. Analyze the Hydrolysis of the Conjugate Acid

The conjugate acid $BH^+$ will hydrolyze in water, producing $H_3O^+$ (or $H^+$) ions, making the solution acidic. $BH^+ (aq) + H_2O (l) \rightleftharpoons B (aq) + H_3O^+ (aq)$

To find the concentration of $H^+$, we need the acid dissociation constant, $K_a$, for the conjugate acid $BH^+$. We can find it using the relationship $K_a \times K_b = K_w$.

• $K_b = 1 \times 10^{-12}$
• $K_w = 1 \times 10^{-14}$

$K_a = \frac{K_w}{K_b} = \frac{1 \times 10^{-14}}{1 \times 10^{-12}} = 1 \times 10^{-2}$

6. Calculate the H+ Concentration

We set up an ICE table for the hydrolysis of $BH^+$:

The expression for $K_a$ is: $K_a = \frac{[B][H_3O^+]}{[BH^+]} = \frac{(x)(x)}{0.1 - x}$ $1 \times 10^{-2} = \frac{x^2}{0.1 - x}$

Since the value of $K_a$ is relatively large compared to the initial concentration ($K_a/C = 10^{-2}/10^{-1} = 0.1$), the approximation $0.1 - x \approx 0.1$ is not valid. We must solve the quadratic equation.

$10^{-2} (0.1 - x) = x^2$ $10^{-3} - 10^{-2}x = x^2$ $x^2 + 10^{-2}x - 10^{-3} = 0$ $x^2 + 0.01x - 0.001 = 0$

Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-0.01 \pm \sqrt{(0.01)^2 - 4(1)(-0.001)}}{2(1)}$ $x = \frac{-0.01 \pm \sqrt{0.0001 + 0.004}}{2}$ $x = \frac{-0.01 \pm \sqrt{0.0041}}{2}$ $x = \frac{-0.01 \pm 0.06403}{2}$

Since concentration (x) must be positive, we take the positive root: $x = \frac{-0.01 + 0.06403}{2} = \frac{0.05403}{2} = 0.027015 \text{ M}$

Therefore, the concentration of $H^+$ at the equivalence point is: $[H^+] = x \approx 2.7 \times 10^{-2} \text{ M}$

This matches option D.