An aqueous solution is prepared by dissolving in sufficient water to get solution at 25°C. For , and — Ionic Equilibrium Chemistry Question
Question
An aqueous solution is prepared by dissolving $0.1\text{ mole } \text{H}_2\text{CO}_3$ in sufficient water to get $100\text{ ml}$ solution at 25°C. For $\text{H}_2\text{CO}_3$, $K_{a1} = 4.0 \times 10^{-6}$ and $K_{a2} = 5.0 \times 10^{-11}$. The only incorrect equilibrium concentration is
💡 Solution & Explanation
The initial concentration $C = 0.1\text{ mol} / 0.1\text{ L} = 1.0\text{ M}$. The hydrogen ion concentration is governed by the first dissociation: $[\text{H}^+] \approx \sqrt{K_{a1}C} = \sqrt{4.0 \times 10^{-6} \times 1.0} = 2.0 \times 10^{-3}\text{ M}$. Option (a) incorrectly claims $[\text{H}^+] = 6.32 \times 10^{-4}\text{ M}$. The other options correctly approximate $[\text{HCO}_3^-] \approx [\text{H}^+] = 2 \times 10^{-3}\text{ M}$, $[\text{CO}_3^{2-}] \approx K_{a2} = 5 \times 10^{-11}\text{ M}$, and $[\text{OH}^-] = 10^{-14}/[\text{H}^+] = 5 \times 10^{-12}\text{ M}$.