ionizes as:<br><br><br><br><br>The correct information(s) related with solution is/are — Ionic Equilibrium Chemistry Question
Question
$\text{H}_2\text{CO}_3$ ionizes as:<br><br>$\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-; K_1 = 4.0 \times 10^{-6}$<br>$\text{HCO}_3^- \rightleftharpoons \text{H}^+ + \text{CO}_3^{2-}; K_2 = 5.0 \times 10^{-11}$<br><br>The correct information(s) related with $0.5\text{ M} – \text{Na}_2\text{CO}_3$ solution is/are
💡 Solution & Explanation
The first anionic hydrolysis is $\text{CO}_3^{2-} + \text{H}_2\text{O} \rightleftharpoons \text{HCO}_3^- + \text{OH}^-$ with $K_{h1} = \frac{K_w}{K_2} = \frac{10^{-14}}{5.0 \times 10^{-11}} = 2.0 \times 10^{-4}$. Degree of hydrolysis $h = \sqrt{K_{h1}/C} = \sqrt{2.0 \times 10^{-4}/0.5} = 0.02$ (Option A). $[\text{OH}^-] = C \times h = 0.5 \times 0.02 = 0.01\text{ M}$, so $\text{pOH} = 2.0$ (Option C) and $\text{pH} = 12.0$ (Option B is false). For the second hydrolysis, $K_{h2} = \frac{K_w}{K_1} = 2.5 \times 10^{-9}$. Since $[\text{HCO}_3^-] \approx [\text{OH}^-]$, $K_{h2} = \frac{[\text{H}_2\text{CO}_3][\text{OH}^-]}{[\text{HCO}_3^-]} \approx [\text{H}_2\text{CO}_3]$. Thus $[\text{H}_2\text{CO}_3] = 2.5 \times 10^{-9}\text{ M}$ (Option D). Therefore, correct answer is A,C,D.