Question #1d230
1 Answer
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,
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))H2CO3(aq)+NaOH(aq)→Na2CO3(aq)+H2O(l)
The first thing that you should notice here is that you have
"H"_ 2"CO"_ (3(aq)) + color(red)(2)"NaOH"_ ((aq)) -> "Na"_ 2"CO"_ (3(aq)) + "H"_ 2"O"_ ((l))H2CO3(aq)+2NaOH(aq)→Na2CO3(aq)+H2O(l)
At this point, you have a total of
2 xx "H" + color(red)(2) xx "H" = "4 atoms of H"2×H+2×H=4 atoms of H 3 xx "O" + color(red)(2) xx "O" = "5 atoms of O"3×O+2×O=5 atoms of O
on the reactants' side, and only
2 xx "H" = "2 atoms of H"2×H=2 atoms of H 3 xx "O" + 1 xx "O" = "4 atoms of O"3×O+1×O=4 atoms of O
on the products' side. To balance the hydrogen and oxygen atoms, multiply the water molecule by
"H"_ 2"CO"_ (3(aq)) + color(red)(2)"NaOH"_ ((aq)) -> "Na"_ 2"CO"_ (3(aq)) + color(blue)(2)"H"_ 2"O"_ ((l))H2CO3(aq)+2NaOH(aq)→Na2CO3(aq)+2H2O(l)
The reactants' side will now have
2 xx "H" + color(blue)(2) xx "H" = "4 atoms of H"2×H+2×H=4 atoms of H 3 xx "O" + color(blue)(2) xx "O" = "5 atoms of O"3×O+2×O=5 atoms of O
which implies that the chemical equation is now balanced.
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))CO2(aq)+H2O(l)⇌H2CO3(aq)
Since carbonic acid is a weak acid, you can write it as
"H"_ 2"CO"_ (3(aq)) rightleftharpoons "H"_ ((aq))^(+) + "HCO"_ (3(aq))^(-)H2CO3(aq)⇌H+(aq)+HCO−3(aq)
Notice that the reaction can also produce sodium bicarbonate,
"H"_ ((aq))^(+) + "HCO"_ (3(aq))^(-) + "NaOH"_ ((aq)) -> "NaHCO"_ (3(aq)) + "H"_ 2"O"_ ((l))H+(aq)+HCO−3(aq)+NaOH(aq)→NaHCO3(aq)+H2O(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))NaHCO3(aq)+NaOH(aq)→Na2CO3(aq)+H2O(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
