Answer
$H_3PO_4(aq) + H_2O(l) \lt -- \gt H_2P{O_4}^-(aq) + H_3O^+(aq)$
$K_a = \frac{[H_3O^+][H_2P{O_4}^-]}{[H_3PO_4]}$
Work Step by Step
1. Write the ionization chemical equation:
- Since $H_3PO_4$ is an acid, write the reaction where it donates a proton to a water molecule:
$H_3PO_4(aq) + H_2O(l) \lt -- \gt H_2P{O_4}^-(aq) + H_3O^+(aq)$
2. Now, write the $K_a$ expression:
- The $K_a$ expression is the concentrations of the products divided by the concentration of the reactants:
$K_a = \frac{[Products]}{[Reactants]}$
$K_a = \frac{[H_3O^+][H_2P{O_4}^-]}{[H_3PO_4]}$
*** We don't consider the $[H_2O]$, because it is the solvent of the solution.
