Question

Question #12933

Answer 1

`"HNO"_ (3(aq)) + "LiOH"_ ((aq)) -> "LiNO"_ (3(aq)) + "H"_ 2"O"_((l))`

Explanation:

You're dealing with a neutralization reaction in which nitric acid, `"HNO"_3`, reacts with lithium hydroxide, `"LiOH"`, to produce aqueous lithium nitrate, `"LiNO"_3`, and water.

The balanced chemical equation that describes this reaction looks like this

`"HNO"_ (3(aq)) + "LiOH"_ ((aq)) -> "LiNO"_ (3(aq)) + "H"_ 2"O"_((l))`

Nitric acid is a strong acid, which means that it ionizes completely in aqueous solution to form hydronium cations, `"H"_3"O"^(+)`, and nitrate anions, `"NO"_3^(-)`

`"HNO"_ (3(aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "NO"_(3(aq))^(-)`

Lithium hydroxide is a strong base, so it too will ionize completely in aqueous solution to form lithium cations, `"Li"^(+)`, and hydroxide anions, `"OH"^(-)`

`"LiOH"_ ((aq)) ->."Li"_ ((aq))^(+) + "OH"_((aq))^(-)`

The complete ionic equation for this reaction will look like this

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

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

`"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"_((l))`

This is equivalent to

`"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2"O"_((l))`

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.