Answer
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.
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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.
