Answer
$$HNO_2(aq) + OH^-(aq) \rightleftharpoons N{O_2}^-(aq) + H_2O(l) $$
Since the weaker base and weaker acid are on the products side, the equilibrium mixture will contain mostly products.
Work Step by Step
1. Identify the reactants, which one is the acid and which one is the base.
Nitrous acid: $HNO_2$
Hydroxide ion: $OH^-$
According to table 11.3, $HNO_2$ is a weak acid, and $OH^-$ a strong base.
2. Therefore, $HNO_2$ will donate one $H^+$ to $OH^-$, we just need to write the products.
Conjugate base for $HNO_2$: $N{O_2}^-$
Conjugate acid for $OH^-$: $H_2O$
3. Write the full equation.
$$HNO_2(aq) + OH^-(aq) \rightleftharpoons N{O_2}^-(aq) + H_2O(l) $$
According to the same table, $OH^-$ is stronger than $N{O_2}^-$, and $HNO_2$ is stronger than $H_2O$. Since the weaker base and weaker acid are on the products side, the equilibrium mixture will contain mostly products.
