Answer
This reation is reactant-favored.
- $H_2O(l) + HNO_3(aq) \lt -- \gt H_3O^{+}(aq) + N{O_3}^-(aq)$
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Other reactions:
- $H_2O(l) + N{O_3}^-(aq) \lt -- \gt OH^-(aq) + HNO_3(aq)$**
** This reaction is very insignificant, because $N{O_3}^-$ is a very weak acid.
- $2H_2O(l) \lt -- \gt H_3O^+(aq) + OH^-(aq)$
Work Step by Step
1. Identify the acids and bases:
- $H_2O$ is amphiprotic.
- $HNO_3$ is an acid.
- Therefore, the only reaction that can occur is with $HNO_3$ as an acid.
2. Write the reaction
- $H_2O(l) + HNO_3(aq) \lt -- \gt H_3O^{+}(aq) + N{O_3}^-(aq)$
3. Identify the weaker acid (or base).
$HNO_3$ and $H_3O^+$: $H_3O^+$ is weaker.
Therefore, the side that has the $H_3O^+$ is favored;
Other possible reaction?
- $H_2O$ can act as an acid too:
$H_2O(l) + N{O_3}^-(aq) \lt -- \gt OH^-(aq) + HNO_3(aq)$
$HNO_3$ can act only as an acid.
$H_3O^+$ can act only as an acid.
$N{O_3}^-$ can act only as a base.
And we have the auto-ionization of water:
$2H_2O(l) \lt -- \gt H_3O^+(aq) + OH^-(aq)$
