# Question #12933

May 19, 2016

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

#### Explanation:

You're dealing with a neutralization reaction in which nitric acid, ${\text{HNO}}_{3}$, reacts with lithium hydroxide, $\text{LiOH}$, to produce aqueous lithium nitrate, ${\text{LiNO}}_{3}$, and water.

The balanced chemical equation that describes this reaction looks like this

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

Nitric acid is a strong acid, which means that it ionizes completely in aqueous solution to form hydronium cations, ${\text{H"_3"O}}^{+}$, and nitrate anions, ${\text{NO}}_{3}^{-}$

${\text{HNO"_ (3(aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "NO}}_{3 \left(a q\right)}^{-}$

Lithium hydroxide is a strong base, so it too will ionize completely in aqueous solution to form lithium cations, ${\text{Li}}^{+}$, and hydroxide anions, ${\text{OH}}^{-}$

${\text{LiOH"_ ((aq)) ->."Li"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

The complete ionic equation for this reaction will look like this

${\text{H"_ 3"O"_ ((aq))^(+) + "NO"_ (3(aq))^(-) + "Li"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "Li"_ ((aq))^(+) + "NO"_ (3(aq))^(-) + 2"H"_ 2"O}}_{\left(l\right)}$

The net ionic equation, which doesn't include spectator ions, will look like this

${\text{H"_ 3"O"_ ((aq))^(+) + color(red)(cancel(color(black)("NO"_ (3(aq))^(-)))) + color(red)(cancel(color(black)("Li"_ ((aq))^(+)))) + "OH"_ ((aq))^(-) -> color(red)(cancel(color(black)("Li"_ ((aq))^(+)))) + color(red)(cancel(color(black)("NO"_ (3(aq))^(-)))) + 2"H"_ 2"O}}_{\left(l\right)}$

This is equivalent to

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

This tells you that the hydronium cations coming from the acid and the hydroxide anions coming from the base neutralize each other to form water.