Answer
$NH_2CH_2COO^-(aq) + H_2O(l) \lt -- \gt NH_3CH_2COO(aq) + OH^-(aq)$
$NH_3CH_2COO(aq) + H_2O(l) \lt -- \gt NH_3CH_2COOH^+(aq) + OH^-(aq)$
Work Step by Step
1. Identify if the compound is a polyprotic acid or base:
$NH_2CH_2COO^-$ is a base.
2. Write the ionization equation for the original base.
** It is important to known how to identify that, in this case, we have an amine group, and a "$-COO^-$" group, where both are capable of receiving a proton.
- Write an equation where $NH_2CH_3COO^-$ receives a proton from a water molecule.
3. Now, for a polyprotic base, the conjugate acid is capable of receiving one proton too, so write an equation where it does that.
- $NH_3CH_2COO$ can act as a base too, so write an equation where it receives a proton from a water molecule.
- Repeat "Step 3" for the next reaction, until the conjugate acid is incapable of receiving one proton.
