Answer
Reactants:
Acid: $CH_3COOH$
Base: $CN^-$
Products:
Conjugate base: $CH_3COO^-$
Conjugate acid: $HCN$
Work Step by Step
Reactants: Left side of the reaction.
Products: Right side.
1. Identify the acid and the base of the reactants:
- $CH_3COOH$ is turning into $CH_3COO^-$; therefore, it is donating one proton, so it is the acid.
- $CN^-$ is turning into $HCN$; therefore, it is receiving one proton, so it is the base.
2. Identify the conjugate acid and base:
- The conjugate base is the result of an acid losing a $H^+$. Since $CH_3COOH$ is the acid, $CH_3COO^-$ is the conjugate base.
- The conjugate acid is the result of a base gaining a $H^+$. Since $CN^-$ is the base, $HCN$ is the conjugate acid.
