Question

Write chemical equations and corresponding equilibrium expressions for each of the two ionization steps of carbonic acid. Part A Write chemical equations for first ionization step of carbonic acid. Express your answer as a chemical equation. Identify all of the phases in your answer.

Answer 1

Solution :

$H_2CO_3$ is considered a diprotic acid.

Sp it can dissociate in solution by giving two protons.

Chemical equations for the first step of carbonic acid is :

First ionization

$H_2CO_3(aq) + H_0(1) \rightleftharpoons H_.O^+(aq) + HCO_3^-(aq)$

Equilibrium constant expression is

$K_{a}_{1}=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}$

Second ionization -

$HCO_3^-(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + CO_3^{2-}(aq)$

Equilibrium constant expression is

$K_{a2}=\frac{[H_3O^+][CO_3^{2-}]}{[HCO_3^-]}$