Answer:
[tex]\boxed{\text{H$_{3}$O$^{+}$ + OH$^{-} \longrightarrow$ 2H$_{2}$O}}[/tex]
Explanation:
1. "Molecular" equation
[tex]\rm Ca(OH)$_{2}$ + 2H$_{3}$O$^{+} \longrightarrow \,$ Ca$^{2+}$ + 4H$_{2}$O[/tex]
2. Ionic equation
[tex]\rm \textbf{Ca}$^{2+}$ + 2OH$^{-}$ + 2H$_{3}$O$^{+}$ \longrightarrow \textbf{Ca}$^{2+}$ + 4H$_{2}$O[/tex]
3. Net ionic equation
Cancel all ions that appear on both sides of the reaction arrow (in boldface).
[tex]\rm 2OH$^{-}$ + 2H$_{3}$O$^{+} \longrightarrow$ 4H$_{2}$O[/tex]
Divide every coefficient by 2.
[tex]\rm OH$^{-}$ + H$_{3}$O$^{+} \longrightarrow$ 2H$_{2}$O[/tex]
