The compound in the blank is $CH_3COO^-$. 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 $HCN$; 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 $HCN$ is $CN^-$, therefore, the conjugate pair of the blank compound should be $CH_3COOH$. 5. Since the compound that is missing is a base, $CH_3COOH$ should be its conjugate acid. 6. Determine the conjugate base of $CH_3COOH$. - If we remove a proton from $CH_3COOH$, it turns to $CH_3COO^-$, 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 $CH_3COOH$ - $HCN$ is weaker; Bases: $CH_3COO^-$ and $CN^-$ - $CH_3COO^-$ is weaker. Therefore, this reaction is product-favored.