Question

Question #7f381

Answer 1

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

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, `"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))`

Answer 2

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

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

`"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?