Q. \[ \mathrm{HNO}_3 + \mathrm{H}_2\mathrm{O} \]

Step 1. Write the molecular equation for mixing nitric acid with water. The acid molecule and the water molecule react to produce the hydronium ion and the nitrate ion. In display form, the reaction is

\[ \mathrm{HNO}_3 + \mathrm{H_2O} = \mathrm{H_3O}^+ + \mathrm{NO}_3^- \]

Step 2. Identify the acid and the base. Nitric acid, \( \mathrm{HNO}_3 \), is a strong acid. Water, \( \mathrm{H_2O} \), acts as a base by accepting a proton. The conjugate base of nitric acid is the nitrate ion, \( \mathrm{NO}_3^- \), and the proton transferred to water produces the hydronium ion, \( \mathrm{H_3O}^+ \).

Step 3. Show the dissociation of the acid into ions, and the protonation of water. Writing the same processes as separated ionic steps gives

\[ \mathrm{HNO}_3 = \mathrm{H}^+ + \mathrm{NO}_3^- \]

\[ \mathrm{H}^+ + \mathrm{H_2O} = \mathrm{H_3O}^+ \]

Step 4. Combine the ionic steps to recover the molecular reaction. Substituting the proton transfer into the dissociation yields the overall aqueous equation already given in Step 1,

\[ \mathrm{HNO}_3 + \mathrm{H_2O} = \mathrm{H_3O}^+ + \mathrm{NO}_3^- .\]

Step 5. Check mass and charge balance. Count atoms on each side: one nitrogen, one hydrogen from the acid plus two hydrogens from water give three hydrogens total, three oxygens, so atoms are balanced. Net charge on the left is 0. Net charge on the right is \(+1\) from \( \mathrm{H_3O}^+ \) and \(-1\) from \( \mathrm{NO}_3^- \), giving total 0. Charges are balanced.

Step 6. Interpret the result. Because \( \mathrm{HNO}_3 \) is a strong acid in water, it effectively dissociates completely to give \( \mathrm{H_3O}^+ \) and \( \mathrm{NO}_3^- \). The molecular equation in aqueous solution is the correct representation of this proton transfer.