Answer
When $NH_3$ act as a base:
$NH_3(aq) + H_2O(l) \lt -- \gt N{H_4}^+(aq) + OH^-(aq)$
$K_b = \frac{[OH^-][N{H_4}^+]}{[NH_3]}$
When $NH_3$ acts as an acid:
$NH_3(aq) + H_2O(l) \lt -- \gt N{H_2}^-(aq) + H_3O^+(aq)$
$K_a = \frac{[H_3O^+][N{H_2}^-]}{[NH_3]}$
Work Step by Step
- $NH_3$ is amphiprotic, therefore, it can act as an acid or as a base, so, let's do both cases:
1. Write the ionization chemical equation:
- For when $NH_3$ act as a base, write the reaction where it takes a proton from a water molecule:
$NH_3(aq) + H_2O(l) \lt -- \gt N{H_4}^+(aq) + OH^-(aq)$
- For when $NH_3$ act as an acid, write the reaction where it donates a proton to a water molecule:
$NH_3(aq) + H_2O(l) \lt -- \gt N{H_2}^-(aq) + H_3O^+(aq)$
2. Now, write the $K_a$ and $K_b$ expressions:
- The $K_b$ and $K_a$ expressions are the concentrations of the products divided by the concentration of the reactants:
For when it acts as a base:
$K_b = \frac{[Products]}{[Reactants]}$
$K_b = \frac{[OH^-][N{H_4}^+]}{[NH_3]}$
For when it acts as an acid:
$K_a = \frac{[Products]}{[Reactants]}$
$K_a = \frac{[H_3O^+][N{H_2}^-]}{[NH_3]}$
*** We don't consider the $[H_2O]$, because it is the solvent of the solution.
