# Question 7f381

Jun 23, 2017

${\text{HNO"_ (3(aq)) + "NaOH"_ ((aq)) -> "H"_ 2"O"_ ((l)) + "NaNO}}_{3 \left(a q\right)}$

#### Explanation:

Nitric acid and sodium hydroxide will neutralize each other in a $1 : 1$ mole ratio, i.e. every mole of nitric acid will consume $1$ mole of sodium hydroxide, to produce water and aqueous sodium nitrate, ${\text{NaNO}}_{3}$.

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

${\text{HNO"_ (3(aq)) + "NaOH"_ ((aq)) -> "H"_ 2"O"_ ((l)) + "NaNO}}_{3 \left(a q\right)}$

The interesting thing to notice here is that you can write a net ionic equation for this reaction by taking into account the fact that nitric acid is a strong acid and sodium hydroxide is a strong base.

You will have

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

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

Sodium nitrate is soluble in water, which implies that it exists as ions in aqueous solution.

This means that you can write a complete ionic equation for this reaction

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

Notice that some ions are present on both sides of the chemical equation--these ions are called spectator ions.

To get the net ionic equation, remove the spectator ions

"H"_ ((aq))^(+) + color(red)(cancel(color(black)("NO"_ (3(aq))^(-)))) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "OH"_ ((aq))^(-) -> "H"_ 2"O" _ ((l)) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)("NO"_ (3(aq))^(-))))

You will end up with

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

Jun 23, 2017

$\text{HNO"_3(aq) + "NaOH"(aq) rarr "NaNO"_3 (aq) + "H"_2"O} \left(l\right)$

#### Explanation:

In a neutralization reaction like this one, a (Brønsted-Lowry) acid and a (Brønsted-Lowry) base are mixed to yield a salt and water (most often in the liquid phase, so we put an color(green)((l)).

The reaction we have so far is

$\text{HNO"_3 + "NaOH" rarr "NaNO"_3 + "H"_2"O} \textcolor{g r e e n}{\left(l\right)}$

The acid and base are most generally present as ions dissolved in solution, so we write an color(red)((aq) after each:

"HNO"_3color(red)((aq)) + "NaOH"color(red)((aq)) rarr "NaNO"_3 + "H"_2"O"color(green)((l)

To find out if the salt is soluble, we refer to a solubility guidelines:

Compounds containing ${\text{Na}}^{+}$ (under "alkali metal cations") and/or ${\text{NO}}_{3}^{-}$ are seen here to be quite soluble, so it is also present as ions, so we write an color(blue)((aq) next to it:

"HNO"_3color(red)((aq)) + "NaOH"color(red)((aq)) rarr "NaNO"_3color(blue)((aq)) + "H"_2"O"color(green)((l)#

This is the molecular equation for this reaction. Would you know how to make the ionic equations for this reaction?