A volume of 2.5 ml of weak monoacidic base ( at ) is titrated with in water at . The concentration o — Ionic Equilibrium Chemistry Question
Question
A volume of 2.5 ml of $\frac{2}{5} \text{ M}$ weak monoacidic base ($K_b = 1 \times 10^{-12}$ at $25^\circ \text{C}$) is titrated with $\frac{2}{15} \text{ M} – \text{HCl}$ in water at $25^\circ \text{C}$. The concentration of $\text{H}^+$ at equivalence point is ($K_w = 1 \times 10^{-14}$ at $25^\circ \text{C}$)
💡 Solution & Explanation
Moles of base = $2.5 \text{ ml} \times 0.4 \text{ M} = 1.0 \text{ mmol}$. Volume of HCl needed = $1.0 \text{ mmol} / (2/15 \text{ M}) = 7.5 \text{ ml}$. Total volume = $2.5 + 7.5 = 10 \text{ ml}$. Concentration of salt $C = 1.0 \text{ mmol} / 10 \text{ ml} = 0.1 \text{ M}$. The hydrolysis constant $K_h = K_w / K_b = 10^{-14} / 10^{-12} = 0.01$. Because $K_h$ is quite large, the standard approximation $[\text{H}^+] = \sqrt{K_h C}$ (which gives $3.16 \times 10^{-2} \text{ M}$) fails. Using the exact quadratic: $\frac{x^2}{0.1 - x} = 0.01 \implies x^2 + 0.01x - 0.001 = 0$. Solving for $x$ (where $x = [\text{H}^+]$) yields $x = \frac{-0.01 + \sqrt{0.0001 + 0.004}}{2} = \frac{-0.01 + \sqrt{0.0041}}{2} = \frac{-0.01 + 0.064}{2} = 0.027 \text{ M} = 2.7 \times 10^{-2} \text{ M}$. Therefore, correct answer is D.