0 ml? — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Before any HCl is added, the solution is purely $1.0\text{ M Na}_2\text{CO}_3$. The $\text{CO}_3^{2-}$ ion undergoes anionic hydrolysis: $\text{CO}_3^{2-} + \text{H}_2\text{O} \rightleftharpoons \text{HCO}_3^- + \text{OH}^-$. The hydrolysis constant $K_h = \frac{K_w}{K_{a2}} = \frac{10^{-14}}{2.5 \times 10^{-11}} = 4.0 \times 10^{-4}$. $[\text{OH}^-] = \sqrt{K_h \cdot C} = \sqrt{4.0 \times 10^{-4} \times 1.0} = 2.0 \times 10^{-2}\text{ M}$. The $\text{pOH} = -\log(2.0 \times 10^{-2}) = 2 - 0.3 = 1.7$. The $\text{pH} = 14 - 1.7 = 12.3$. Therefore, correct answer is B.