To determine the total ionic equation for the reaction between nitric acid (HNO₃) and sodium hydroxide (NaOH), let's break down each step:
1. Dissociation of Reactants:
- Nitric Acid (HNO₃) dissociates in water to form hydrogen ions (H⁺) and nitrate ions (NO₃⁻):
[tex]\[
\text{HNO}_3 \rightarrow \text{H}^+ + \text{NO}_3^-
\][/tex]
- Sodium Hydroxide (NaOH) dissociates in water to form sodium ions (Na⁺) and hydroxide ions (OH⁻):
[tex]\[
\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-
\][/tex]
2. Combining all Ions:
- When these dissociated ions are in solution, we have:
[tex]\[
\text{H}^+ + \text{NO}_3^- + \text{Na}^+ + \text{OH}^-
\][/tex]
3. Reaction that Occurs:
- The H⁺ ions react with OH⁻ ions to form water (H₂O):
[tex]\[
\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}
\][/tex]
- The sodium ions (Na⁺) and nitrate ions (NO₃⁻) remain unchanged in the solution as spectator ions:
4. Total Ionic Equation:
- Combining these steps, the complete ionic equation is:
[tex]\[
\text{H}^+ + \text{NO}_3^- + \text{Na}^+ + \text{OH}^- \rightarrow \text{Na}^+ + \text{NO}_3^- + \text{H}_2\text{O}
\][/tex]
Therefore, the total ionic equation for the reaction between HNO₃ and NaOH is:
[tex]\[
\text{H}^+ + \text{NO}_3^- + \text{Na}^+ + \text{OH}^- \rightarrow \text{Na}^+ + \text{NO}_3^- + \text{H}_2\text{O}
\][/tex]
Thus, the correct answer is:
[tex]\[
\boxed{\text{H}^+ + \text{NO}_3^- + \text{Na}^+ + \text{OH}^- \rightarrow \text{Na}^+ + \text{NO}_3^- + \text{H}_2\text{O}}
\][/tex]
