Consider the reaction CO₃^2-(aq)+ H₂O(l) HCO₃^-(aq)+ OH^-(aq).K_b for CO₃^2- is 2.1 × 10^-4 at 25°C.What is K_a for the HCO₃^- ion at 25°C?

Question

Quizplus · Accepted Answer

First, we can write the balanced chemical equation for the reaction: CO₃^2-(aq) + H₂O(l) ⇌ HCO₃^-(aq) + OH^-(aq) Using the given equilibrium constant: Kc = [HCO₃^-][OH^-] / [CO₃^2-][H₂O] Kc = 2.1 × 10^-4 We want to find the equilibrium constant, K_a, for the ion HCO₃^-. This ion is produced by the dissociation of the acid HCO₃^-: HCO₃^-(aq) ⇌ H+(aq) + CO₃^2-(aq) The equilibrium constant for this reaction is the acid dissociation constant, Ka: Ka = [H+(aq)][CO₃^2-(aq)] / [HCO₃^-(aq)] We can express [H+(aq)] in terms of [HCO₃^-(aq)] using the equilibrium expression for the acid dissociation reaction and the equilibrium constant for the overall reaction: Ka = [H+(aq)][CO₃^2-(aq)] / [HCO₃^-(aq)] Kc = [HCO₃^-(aq)][OH^-(aq)] / [CO₃^2-(aq)][H₂O(l)] [H+(aq)][CO₃^2-(aq)] / [HCO₃^-(aq)] = Kc / Ka [H+(aq)] = (Kc / Ka) x [H₂O(l)] / [OH^-(aq)] Substituting this expression for [H+(aq)] into the expression for Ka, we get: Ka = Kc x [H₂O(l)] / ([OH^-(aq)] x [HCO₃^-(aq)]) We can now substitute the given values of Kc and the concentrations into this equation to solve for Ka: Ka = (2.1 × 10^-4) x (1) / ([1] x [1]) = 2.1 × 10^-4 Finally, we can express K_a in terms of Ka using the expression: K_a = Kw / Ka where Kw is the ion product constant for water, which is 1.0 x 10^-14 at 25°C: K_a = (1.0 x 10^-14) / (2.1 x 10^-4) = 4.8 x 10^-11 Therefore, the answer is choice B, 4.8 x 10^-11.

Consider the Reaction CO32-(aq)+ H2O(l) HCO3-(aq)+ OH-(aq)

Consider the Reaction CO₃^2-(aq)+ H₂O(l) HCO₃^-(aq)+ OH^-(aq)