How do you write a total ionic equation and a net ionic equation for this reaction? HCl(aq) + NaOH (aq) -> NaCl (aq) + H_2O(l)?

Dec 6, 2016

Here's how you can do that.

Explanation:

The thing to recognize here is the fact that you're dealing with a neutralization reaction that features sodium hydroxide, $\text{NaOH}$, a strong base, and hydrochloric acid, $\text{HCl}$, a strong acid.

This tells you that the two reactants will dissociate completely in aqueous solution to produce cations and anions. More specifically, you will have

${\text{NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

${\text{HCl"_ ((aq)) -> "H"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

Now, when these two solutions are mixed, the hydroxide anions produced by the strong base and the hydrogen ions produced by the strong acid will neutralize each other to produce water.

The sodium cations and the chloride anions act as spectator ions because they are present on both sides of the chemical equation as ions.

You can tell that this is the case because sodium chloride, $\text{NaCl}$, one of the two products of the reaction, is soluble in aqueous solution.

So, the complete ionic equation will be

${\text{H"_ ((aq))^(+) + "Cl"_ ((aq))^(-) + "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "Na"_ ((aq))^(+) + "Cl" _ ((aq))^(-) + "H"_ 2"O}}_{\left(l\right)}$

To get the net ionic equation, simply remove the spectator ions

${\text{H"_ ((aq))^(+) + color(red)(cancel(color(black)("Cl"_ ((aq))^(-)))) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "OH"_ ((aq))^(-) -> color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)("Cl"_ ((aq))^(-)))) + "H"_ 2"O}}_{\left(l\right)}$

You will end up with

${\text{H"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O}}_{\left(l\right)}$

SIDE NOTE You will sometimes see the hydrogen ion being replaced by the hydronium ion, ${\text{H"_3"O}}^{+}$.

In this case, the dissociation of hydrochloric acid is represented by the chemical equation

${\text{HCl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

The net ionic equation will now take the form

${\text{H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O}}_{\left(l\right)}$