Answer:
See below
Explanation:
[tex]\rm HCOOH + H_{2}O \rightleftharpoons H_{3}O^{+} + HCOO^{-}[/tex]
The general formula for an equilibrium constant expression is
[tex]K_{eq} = \dfrac{[\text{Products}]}{[\text{Reactants}]}[/tex]
For this reaction,
[tex]K_{eq} = \dfrac{[\text{HCOO}^{-}][\text{H}_{3}\text{O}^{+}]}{[\text{HCOOH}]}[/tex]