Question #7f381

2 Answers
Jun 23, 2017


#"HNO"_ (3(aq)) + "NaOH"_ ((aq)) -> "H"_ 2"O"_ ((l)) + "NaNO"_ (3(aq))#


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, #"NaNO"_3#.

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

#"HNO"_ (3(aq)) + "NaOH"_ ((aq)) -> "H"_ 2"O"_ ((l)) + "NaNO"_ (3(aq))#

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

#"HNO"_ (3(aq)) -> "H"_ ((aq))^(+) + "NO"_ (3(aq))^(-)#

#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#

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

#"H"_ ((aq))^(+) + "NO"_ (3(aq))^(-) + "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O" _ ((l)) + "Na"_ ((aq))^(+) + "NO"_ (3(aq))^(-)#

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

#"H"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O"_ ((l))#

Jun 23, 2017


#"HNO"_3(aq) + "NaOH"(aq) rarr "NaNO"_3 (aq) + "H"_2"O"(l)#


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

#"HNO"_3 + "NaOH" rarr "NaNO"_3 + "H"_2"O"color(green)((l))#

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 #"Na"^+# (under "alkali metal cations") and/or #"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?