Answer
$N{H_4}^+(aq) + OH^-(aq) \lt -- \gt NH_3(aq) + H_2O(l)$
Work Step by Step
1. Write the reaction where the reactants exchange theirs cations/anions:
$N{H_4}Cl(aq) + NaOH(aq) \lt -- \gt NaCl(aq) + NH_3(aq) + H_2O(l)$
** When $N{H_4}^+$ reacts with $OH^-$, it produces a base, and $H_2O$.
2. Identify the 100% dissociated/ionizated compounds, and separate them in ions:
${NH_4}^+ (aq) + Cl^-(aq) + Na^+(aq) + OH^-(aq) \lt -- \gt Na^+(aq) + Cl^-(aq) + NH_3(aq) + H_2O(l)$
Salts: Totally dissociated.
$NaOH$: Strong base, totally ionizated.
3. Eliminate the repeated ions/molecules:
$N{H_4}^+(aq) + OH^-(aq) \lt -- \gt NH_3(aq) + H_2O(l)$
- This is the net ionic equation.
