Question #12933

1 Answer
May 19, 2016

Answer:

#"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.