# Hydrochloric acid, or HCI, reacts with solid NaOH. What are the products of this chemical reaction?

Apr 2, 2017

Aqueous sodium chloride and water.

#### Explanation:

Hydrochloric acid, $\text{HCl}$, is a strong acid that ionizes completely in aqueous solution to produce hydrogen cations and chloride anions

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

Sodium hydroxide is a strong base when dissolved in water, which implies that it too ionizes completely to produce sodium cations and hydroxide anions.

When you add the sodium hydroxide pellets to the hydrochloric acid solution

${\text{NaOH"_ ((s)) -> "NaOH}}_{\left(a q\right)}$

the sodium hydroxide will ionize completely, which means that you will have

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

The hydroxide anions and the hydrogen cations will neutralize each other to produce water

overbrace("H"_ ((aq))^(+))^(color(blue)("coming from HCl")) + overbrace("OH"_ ((aq))^(-))^(color(purple)("coming from NaOH")) = "H"_ 2"O"_ ((l))

The other product of the reaction is sodium chloride, $\text{NaCl}$, an ionic compound that is soluble in aqueous solution. This implies that it will exist as ions in the resulting solution.

The balanced chemical equation that describes this neutralization reaction looks like this

${\text{NaOH"_ ((s)) + "HCl"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O}}_{\left(l\right)}$

The complete ionic equation will be

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

The net ionic equation, which you get by eliminating the spectator ions, i.e. the ions that are present on both sides of the equation

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

will be

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