Question #1d230

You're actually dealing with a neutralization reaction that takes place between carbonic acid, a weak acid, and sodium hydroxide, a strong base.

A complete neutralization will produce water and aqueous sodium carbonate, #"Na"_2"CO"_3#.

So, start by writing the unbalanced chemical equation that describes this reaction

#"H"_ 2"CO"_ (3(aq)) + "NaOH"_ ((aq)) -> "Na"_ 2"CO"_ (3(aq)) + "H"_ 2"O"_ ((l))#

The first thing that you should notice here is that you have #2# atoms of sodium on the products' side and only #1# on the reactants' side, so start by multiplying the sodium hydroxide by #color(red)(2)#

#"H"_ 2"CO"_ (3(aq)) + color(red)(2)"NaOH"_ ((aq)) -> "Na"_ 2"CO"_ (3(aq)) + "H"_ 2"O"_ ((l))#

At this point, you have a total of

• #2 xx "H" + color(red)(2) xx "H" = "4 atoms of H"#
• #3 xx "O" + color(red)(2) xx "O" = "5 atoms of O"#

on the reactants' side, and only

• #2 xx "H" = "2 atoms of H"#
• #3 xx "O" + 1 xx "O" = "4 atoms of O"#

on the products' side. To balance the hydrogen and oxygen atoms, multiply the water molecule by #color(blue)(2)#.

#"H"_ 2"CO"_ (3(aq)) + color(red)(2)"NaOH"_ ((aq)) -> "Na"_ 2"CO"_ (3(aq)) + color(blue)(2)"H"_ 2"O"_ ((l))#

The reactants' side will now have

• #2 xx "H" + color(blue)(2) xx "H" = "4 atoms of H"#
• #3 xx "O" + color(blue)(2) xx "O" = "5 atoms of O"#

which implies that the chemical equation is now balanced.

#color(white)(a/a)#
SIDE NOTE This is actually a bit of an oversimplification of what's going on with this reaction.

For starters, carbonic acid is highly unstable in aqueous solution, meaning that it is best represented as

#"CO"_ (2(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 2"CO"_ (3(aq))#

Since carbonic acid is a weak acid, you can write it as

#"H"_ 2"CO"_ (3(aq)) rightleftharpoons "H"_ ((aq))^(+) + "HCO"_ (3(aq))^(-)#

Notice that the reaction can also produce sodium bicarbonate, #"NaHCO"_3#, since you can have

#"H"_ ((aq))^(+) + "HCO"_ (3(aq))^(-) + "NaOH"_ ((aq)) -> "NaHCO"_ (3(aq)) + "H"_ 2"O"_ ((l))#

At this point, if you continue to add sodium hydroxide until the neutralization is complete, you will end up with

#"NaHCO"_ (3(aq)) + "NaOH"_ ((aq)) -> "Na"_ 2"CO"_ (3(aq)) + "H"_ 2"O"_ ((l))#

If you put all this together into one equation, you will get

#color(white)(aaaaaaaaaaa)"CO"_ (2(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons color(purple)(cancel(color(black)("H"_ 2"CO"_ (3(aq)))))#

#color(white)(aaaaaaaaaaaaaaaaa)color(purple)(cancel(color(black)("H"_ 2"CO"_ (3(aq))))) rightleftharpoons color(red)(cancel(color(black)("H"_ ((aq))^(+)))) + color(red)(cancel(color(black)("HCO"_ (3(aq))^(-))))#

#color(red)(cancel(color(black)("H"_ ((aq))^(+)))) + color(red)(cancel(color(black)("HCO"_ (3(aq))^(-)))) + "NaOH"_ ((aq)) -> color(blue)(cancel(color(black)("NaHCO"_ (3(aq))))) + "H"_ 2"O"_ ((l))#

#color(white)(aaaaa)color(blue)(cancel(color(black)("NaHCO"_ (3(aq))))) + "NaOH"_ ((aq)) -> "Na"_ 2"CO"_ (3(aq)) + "H"_ 2"O"_ ((l))#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#

#"CO"_ (2(aq)) + "H"_ 2"O"_ ((l)) + 2"NaOH"_ ((aq)) -> "Na"_ 2"CO"_ (3(aq)) + 2"H"_ 2"O"_ ((l))#

which looks very much like what we've got using #"H"_2"CO"_3# for carbonic acid.