Answer
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.
