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.