what is the Ka equation for HCHO2

i thought it was K_{a} = [CHO_2][H^+]/[HCHO_2] but its not correct

To obtain the correct Ka equation for HCHO2, we need to consider the dissociation of the acid HCHO2 (formic acid) in water.

The dissociation reaction is as follows:

HCHO2 (aq) ⇌ H+ (aq) + CHO2- (aq)

To write the Ka expression, we need to determine the concentration of each species at equilibrium. Let's represent the equilibrium concentration of HCHO2 as [HCHO2], the concentration of H+ as [H+], and the concentration of CHO2- as [CHO2-].

Based on the stoichiometry of the dissociation reaction, we can conclude that the concentration of HCHO2 at equilibrium is equal to the initial concentration of HCHO2 minus the concentration of H+, since one HCHO2 molecule is converted into one H+ ion.

Therefore, [HCHO2] = [initial HCHO2] - [H+].

The Ka expression for HCHO2 can be written as:

Ka = [H+][CHO2-]/[HCHO2]

Please note that the stoichiometric coefficients of reactants and products are not included in the Ka expression, as they are not part of the equilibrium expression.