## Chemistry: The Molecular Science (5th Edition)

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: $NH_3(aq) + H_2O(l) \lt -- \gt N{H_2}^-(aq) + H_3O^+(aq)$ $K_a = \frac{[H_3O^+][N{H_2}^-]}{[NH_3]}$
- $NH_3$ 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.