Answer
$N{H_4}^+(aq) + H_2O(l) \lt -- \gt NH_3(aq) + H_3O^+(aq)$
$K_a = \frac{[H_3O^+][NH_3]}{[N{H_4}^+]}$
Work Step by Step
1. Write the ionization chemical equation:
- Since $N{H_4}^+$ is an acid, write the reaction where it donates a proton to a water molecule:
$N{H_4}^+(aq) + H_2O(l) \lt -- \gt NH_3(aq) + H_3O^+(aq)$
2. Now, write the $K_a$ expression:
- The $K_a$ expression is the concentrations of the products divided by the concentration of the reactants:
$K_a = \frac{[Products]}{[Reactants]}$
$K_a = \frac{[H_3O^+][NH_3]}{[N{H_4}^+]}$
*** We don't consider the $[H_2O]$, because it is the solvent of the solution.
