Exactly of a weak monoacidic base ( at ) is completely titrated with a aqueous solution. What is the — Ionic Equilibrium Chemistry Question
Question
Exactly $2.5 \text{ mL}$ of a $\frac{2}{5} \text{ M}$ weak monoacidic base ($K_b = 1.0 \times 10^{-12}$ at $25^\circ C$) is completely titrated with a $\frac{2}{15} \text{ M }$ aqueous $HCl$ solution. What is the exact concentration of $H^+$ ions in the resulting mixture strictly at the equivalence point? (Given $K_w = 1 \times 10^{-14}$)
💡 Solution & Explanation
Millimoles of base = $2.5 \times (2/5) = 1.0 \text{ mmol}$. At equivalence, millimoles of $HCl$ required = $1.0 \text{ mmol}$. Volume of $HCl = 1.0 / (2/15) = 7.5 \text{ mL}$. Total volume = $2.5 + 7.5 = 10 \text{ mL}$. The salt concentration $C = 1.0 \text{ mmol} / 10 \text{ mL} = 0.1 \text{ M}$. The hydrolysis constant $K_h = K_w / K_b = 10^{-14} / 10^{-12} = 10^{-2}$. Using the unapproximated formula $K_h = \frac{Ch^2}{1-h}$, we have $10^{-2} = \frac{0.1 h^2}{1-h} \Rightarrow 10h^2 + h - 1 = 0$. Solving the quadratic yields $h \approx 0.27$. For cationic hydrolysis, $[H^+] = Ch = 0.1 \times 0.27 = 2.7 \times 10^{-2} \text{ M}$.