Consider the following equilibrium: H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-1(aq). What is the correct equilibrium expression? Ka=[H3O+][HCO1−3][H2CO3] Ka=[H3O+][H2CO3][HCO1−3] Ka=[H2CO3][H2O][H3O+][HCO1−3] Ka=[H3O+][HCO1−3][H2CO3][H2O]

The activity of pure substance is taken as 1 because it has no effect on the reaction. Hence the equilibrium constant of the reaction is given by: Ka=[H3O+][HCO1−3]/[H2CO3]. Hope this helps :)

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RomeliaThurston RomeliaThurston · Answer 2 · 2019-04-08T07:31:11-04:00

Answer: The correct answer is \(Ka=\frac{[H_3O^+][HCO_3^{-1}]}{[H_2CO_3]}\)

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.

For the general equation:

\(aA+bB\rightarrow cC+dD\)

The expression for equilibrium constant is given as:

\(K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}\)

For the given chemical equation:

\(H_2CO_3(aq.)+H_2O(l)\rightarrow H_3O^+(aq.)+HCO_3^{-1}(aq.)\)

The expression for equilibrium constant is:

\(K_a=\frac{[H_3O^+][HCO_3^{-1}]}{[H_2CO_3]}\)

The concentration of pure solids and liquids is taken as 1, therefore \(H_2O\) is not written in the given expression.

Hence, the correct expression is given above.