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

Products : $CH_3COO^-$ e $H_2S$ The right side (Products) will be favored.
$CH_3COOH$ will give one proton to $HS^-$. $CH_3COOH(aq) + HS^-(aq) CH_3COO^-(aq) + H_2S(aq)$ Since $CH_3COOH$$(Ka = 1.8 \times 10^{-5}) is stronger than H_2S$$(Ka = 1.1 \times 10^{-7})$, the products side will be favored.