Answer
$CH_3NH_2(aq) + H_2O(l) \lt -- \gt CH_3N{H_3}^+(aq) + OH^-(aq)$
Work Step by Step
$CH_3NH_2$ is a base, therefore, it will receive one proton from the other reactant, in this case, the water molecule.
When $CH_3NH_2$ receives a proton, it produces $ CH_3N{H_3}^+$, and when $H_2O$ donates this proton, it will produce $OH^-$.
