50 ml? — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Adding $50\text{ ml}$ of $1.0\text{ M HCl}$ ($50\text{ mmol}$) completely neutralizes the initial $50\text{ mmol}$ of $\text{Na}_2\text{CO}_3$, bringing the titration to the first equivalence point. The solution now exclusively contains the amphiprotic salt $\text{NaHCO}_3$. The pH of an amphiprotic salt solution is $\text{pH} = \frac{pK_{a1} + pK_{a2}}{2}$. With $pK_{a1} = -\log(4.0 \times 10^{-6}) = 6 - 0.6 = 5.4$ and $pK_{a2} = 10.6$, the $\text{pH} = \frac{5.4 + 10.6}{2} = \frac{16.0}{2} = 8.0$. Therefore, correct answer is C.