Answer
$$NH{_4}^+(aq) + S{O_4}^{2-}(aq) \rightleftharpoons NH_3(aq) + HS{O_4}^-(aq) $$
Since the weaker base and weaker acid are on the reactants side, the equilibrium mixture will contain mostly reactants.
Work Step by Step
1. Identify the reactants, which one is the acid and which one is the base.
Ammonium ion: $N{H_4}^+$
Sulfate ion: $S{O_4}^{2-}$
According to table 11.3, $N{H_4}^+$ is a weak acid, and $S{O_4}^{2-}$ a weak base.
2. Therefore, $N{H_4}^+$ will donate one $H^+$ to $S{O_4}^{2-}$, we just need to draw the products.
Conjugate base for $NH_4^+$: $NH_3$
Conjugate acid for $S{O_4}^{2-}$: $HS{O_4}^-$
3. Write the full equation.
$$NH{_4}^+(aq) + S{O_4}^{2-}(aq) \rightleftharpoons NH_3(aq) + HS{O_4}^-(aq) $$
According to the same table, $NH_3$ is stronger than $S{O_4}^{2-}$, and $HS{O_4}^-$ is stronger than $N{H_4}^+$. Since the weaker base and weaker acid are on the reactants side, the equilibrium mixture will contain mostly reactants.
