Answer
$HCOOH(aq) + H_2O(l) \lt -- \gt H_3O^+(aq) + HCOO^-(aq)$
$Ka = \frac{[H_3O^+][HCOO^-]}{[HCOOH]}$
Work Step by Step
The ionization equation should describe the transfer of a proton from the acid to a water molecule, therefore, the acid and the water are the reactants, and the protonated water and the conjugate base are the products.
The ionization constant expression is based on:
$K_a = \frac{[Products]}{[Reactants]}$
* The water doesn't appear, because it is the solvent, and its concentration doesn't matter.