Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 14 - Acids and Bases - Questions for Review and Thought - Topical Questions - Page 652c: 44b


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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.