Write the net ionic equation including the phases?

Perchloric acid is a strong acid and sodium hydroxide is a strong base, so right from the start, you should be able to say that the net ionic equation that describes this neutralization reaction is

`"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))`

Perchloric acid will ionize completely in aqueous solution to produce hydronium cations and perchlorate anions.

`"HClO"_ (4(aq)) + color(blue)("H"_ 2"O"_ ((l)))-> "H"_ 3"O"_ ((aq))^(+) + "ClO"_ (4(aq))^(-)`

Sodium hydroxide will also ionize completely in aqueous solution to produce sodium cations and hydroxide anions

`"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)`

When you mix these two solutions, you will end up with

`"HClO"_ (4(aq)) + color(blue)("H"_ 2"O"_ ((l))) + "NaOH"_ ((aq)) -> "H"_ 2"O"_ ((l)) + "NaClO"_ (4(aq)) + color(blue)("H"_ 2"O"_ ((l)))`

The complete ionic equation looks like this--keep in mind that sodium perchlorate is soluble in aqueous solution!

`"H"_ 3"O"_ ((aq))^(+) + "ClO"_ (4(aq))^(-) + "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l)) + "Na"_ ((aq))^(+) + "ClO"_ (4(aq))^(-)`

Eliminate the spectator ions

`"H"_ 3"O"_ ((aq))^(+) + color(red)(cancel(color(black)("ClO"_ (4(aq))^(-)))) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l)) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)("ClO"_ (4(aq))^(-))))`

to get the net ionic equation

`"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_ ((l))`