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. nothing Request Answer Part B Complete previous part(s) Part C Write chemical equations for second ionization step of carbonic acid. Express your answer as a chemical equation. Identify all of the phases in your answer. nothing

Answer 1

Answer: The chemical equations and equilibrium constant expression for each ionization steps is written below.

Explanation:

The chemical formula of carbonic acid is $H_2CO_3$. It is a diprotic weak acid which means that it will release two hydrogen ions when dissolved in water

The chemical equation for the first dissociation of carbonic acid follows:

               $H_2CO_3(aq.)\rightleftharpoons H^+(aq.)+HCO_3^-(aq.)$

The expression of first equilibrium constant equation follows:

$Ka_1=\frac{[H^+][HCO_3^{-}]}{[H_2CO_3]}$

The chemical equation for the second dissociation of carbonic acid follows:

               $HCO_3^-(aq.)\rightarrow H^+(aq.)+CO_3^{2-}(aq.)$

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][CO_3^{2-}]}{[HCO_3^-]}$

Hence, the chemical equations and equilibrium constant expression for each ionization steps is written above.