100 ml? — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Adding $100\text{ ml}$ of $1.0\text{ M HCl}$ ($100\text{ mmol}$) brings the titration to the second equivalence point, completely converting the original salt into $50\text{ mmol}$ of $\text{H}_2\text{CO}_3$. The total volume is $50 + 100 = 150\text{ ml}$, making the concentration $C = 50\text{ mmol} / 150\text{ ml} = \frac{1}{3}\text{ M}$. The solution is a weak diprotic acid, where $[\text{H}^+] \approx \sqrt{K_{a1} \cdot C} = \sqrt{4.0 \times 10^{-6} \times \frac{1}{3}} = \sqrt{1.33 \times 10^{-6}} \approx 1.15 \times 10^{-3}\text{ M}$. The $\text{pH} = -\log(1.15 \times 10^{-3}) = 3 - \log 1.15 \approx 2.94$. Therefore, correct answer is A.