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: 67a


The compound in the blank is $CN^-$. 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 $HS{O_4}^-$; 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 $H{SO_4}^-$ is $S{O_4}^{2-}$, therefore, the conjugate pair of the blank compound should be $HCN$. 5. Since the compound that is missing is a base, $HCN$ should be its conjugate acid. 6. Determine the conjugate base of $HCN$. - If we remove a proton from $HCN$, it turns to $CN^-$, therefore, this is the compound in the blank. ------ 1. To determine which side the reaction favors, identify the side with the weaker compounds. Acids: $HCN$ and $HS{O_4}^-$ ** You can find the constants values on table 14.2, on page 624. - $HCN$ is weaker; Bases: $S{O_4}^{2-}$ and $CN^-$ - $S{O_4}^{2-}$ is weaker. Therefore, this reaction is product-favored.
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