Answer
Products : $CH_3COOH$ and $H_2O$
The right side will be favored.
Work Step by Step
$H_3O^+$ 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.
