# Chapter 14 - Acids and Bases - Questions for Review and Thought - Topical Questions - Page 652d: 66a

The compound in the blank is "$N{H_4}^+$" This reaction is reactant-favored, because the weaker acid and base are on the products side.

#### Work Step by Step

1. Identify the other compound that is next to the blank, on the same side: - It is $Br^-$; Is it acting as an acid or as a base? - As a base, because it is receiving one proton. 3. If the other compound is a base, the blank should be an acid. 4. The conjugate pair of $Br^-$ is $HBr$, therefore, the conjugate pair of the blank compound should be $NH_3$. 5. Since the compound that is missing is an acid, $NH_3$ should be its conjugate base. 6. Determine the conjugate acid of $NH_3$. - If we add a proton to $NH_3$, it turns to $N{H_4}^+$, therefore, this is the compound in the blank. ------ 1. To determine which side the reaction favors, identify the side with the weaker compounds. Acids: $N{H_4}^+$ and $HBr$ - $N{H_4}^+$ is weaker; Bases: $NH_3$ and $Br^-$ - $Br^-$ is weaker. Therefore, the reactants side is favored.

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