The equilibrium carbonate ion concentration after equal volumes of and solutions are mixed, is ( and — Ionic Equilibrium Chemistry Question
Question
The equilibrium carbonate ion concentration after equal volumes of $0.7\text{ M} – \text{Na}_2\text{CO}_3$ and $0.7\text{ M} – \text{HCl}$ solutions are mixed, is ($K_{a1}$ and $K_{a2}$ for $\text{H}_2\text{CO}_3$ are $4.9 \times 10^{-6}$ and $4.0 \times 10^{-11}$, respectively)
💡 Solution & Explanation
Mixing equal volumes and molarities of $\text{Na}_2\text{CO}_3$ and $\text{HCl}$ fully converts them to $\text{NaHCO}_3$. The concentration halves to $0.35\text{ M}$. For the amphiprotic $\text{HCO}_3^-$ ion, $[\text{H}^+] = \sqrt{K_{a1} \times K_{a2}} = \sqrt{4.9 \times 10^{-6} \times 4.0 \times 10^{-11}} = \sqrt{19.6 \times 10^{-17}} = 1.4 \times 10^{-8}\text{ M}$. Using the second dissociation expression $K_{a2} = \frac{[\text{H}^+][\text{CO}_3^{2-}]}{[\text{HCO}_3^-]}$, we substitute the values: $4.0 \times 10^{-11} = \frac{(1.4 \times 10^{-8})[\text{CO}_3^{2-}]}{0.35}$. Solving yields $[\text{CO}_3^{2-}] = \frac{1.4 \times 10^{-11}}{1.4 \times 10^{-8}} = 10^{-3}\text{ M} = 0.001\text{ M}$. Therefore, correct answer is D.