Question

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]

Answer 1

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 :)

Answer 2

Answer: The correct answer is
`Ka=([H_3O^+][HCO_3^(-1)])/([H_2CO_3])`

Step-by-step 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)=([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=([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.