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

Products: $NH_3$ and $H_2O$. The right side will be favored.
$NH_4^+$ will donate one proton to $OH^-$ $NH_4^+(aq) + OH^-(aq) NH_3(aq) + H_2O(l)$ Since $NH_3$$(Kb = 1.8 \times 10 ^{-5}) is stronger than NH_4^+$$(Ka = 5.6 \times 10^{-10})$, the rigth side will be favored.