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 652d: 66c


The compound on the blank is $N{H_2}^-$. This reaction is product-favored because this side has the weaker acid and base.

Work Step by Step

1. Identify the other compound that is next to the blank, on the same side: - It is $H_2O$; 2. Is it acting as an acid or as a base? - As an acid, because it is donating one proton. 3. If the other compound is an acid, the blank should be a base. 4. The conjugate pair of $H_2O$ is $OH^-$, therefore, the conjugate pair of the blank compound should be $NH_3$. 5. Since the compound that is missing is a base, $NH_3$ should be its conjugate acid. 6. Determine the conjugate base of $NH_3$. - If we remove a proton from $NH_3$, it turns to $N{H_2}^-$, therefore, this is the compound in the blank. ------ 1. To determine which side the reaction favors, identify the side with the weaker compounds. ** You can find the constants values on table 14.2, on page 624. Acids: $H_2O$ and $NH_3$ - $NH_3$ is weaker; Bases: $OH^-$ and $N{H_2}^-$ - $OH^-$ is weaker. Therefore, this reaction is product-favored.
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