Answer
$(CH_3)_2NH(aq) + H_2O(l) \lt -- \gt (CH_3)_2N{H_2}^+(aq) + OH^-(aq)$
Work Step by Step
$(CH_3)_2NH$ is a base, therefore, it will receive one proton from the other reactant, in this case, the water molecules.
- When $(CH_3)_2NH$ receives one proton, it produces $(CH_3)_2N{H_2}^+$, and when $H_2O$ donates one proton, it will produce $OH^-$.
