## Chemistry: The Central Science (13th Edition)

Products : $CH_3COOH$ and $H_2O$ The right side will be favored.
$H_3O^+$ will give one proton to $CH_3COO^-$: $CH_3COO^-(aq) + H_3O^+(aq) CH_3COOH(aq) + H_2O$ Since $CH_3COOH$$(Ka = 1.8 \times 10^{-5}) is stronger than CH_3COO^-$$(Kb = 5.6 \times 10^{-10})$, the right side will be favored.