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: 67b


The compound in the blank is $HS^-$. This reaction is reactant 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 $H_3O^+$; 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 $H_3O^+$ is $H_2O$, therefore, the conjugate pair of the blank compound should be $H_2S$. 5. Since the compound that is missing is a base, $H_2S$ should be its conjugate acid. 6. Determine the conjugate base of $H_2S$. - If we remove a proton from $H_2S$, it turns to $HS^-$, therefore, this is the compound in the blank. ------ 1. To determine which side the reaction favors, identify the side with the weaker compounds. Acids: $H_2S$ and $H_3O^+$ ** You can find the constants values on table 14.2, on page 624. - $H_2S$ is weaker; Bases: $HS^-$ and $H_2O$ - $H_2O$ is weaker. Therefore, this reaction is reactant-favored.
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