Answer
$C_{10}H_{15}NH_2(aq) + H_2O(l) \lt -- \gt C_{10}H_{15}{NH_3}^+(aq) + OH^-(aq)$
Work Step by Step
- Amantadine is a weak base, therefore, when $C_{10}H_{15}NH_2$ reacts with water, it will take one proton from $H_2O$, and will produce $OH^-$ and $ C_{10}H_{15}{NH_3}^+$.
- Since it produces $OH^-$, the solution will be considered $basic$.
