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: 43e

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.
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