Write the net ionic equation including the phases?
Based off of this reaction -
"HClO"_4(aq) + "NaOH" (aq) -> "H"_2"O"(l) + "NaClO"_4 (aq)HClO4(aq)+NaOH(aq)→H2O(l)+NaClO4(aq)
Based off of this reaction -
1 Answer
Explanation:
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))H3O+(aq)+OH−(aq)→2H2O(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))^(-)HClO4(aq)+H2O(l)→H3O+(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))^(-)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)))HClO4(aq)+H2O(l)+NaOH(aq)→H2O(l)+NaClO4(aq)+H2O(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))^(-)H3O+(aq)+ClO−4(aq)+Na+(aq)+OH−(aq)→2H2O(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))
