Question

Question #1d230

Answer 1

Here's what I got.

Explanation:

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.