Answer
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.
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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.
