Question 1d230

Jun 11, 2017

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, ${\text{Na"_2"CO}}_{3}$.

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

${\text{H"_ 2"CO"_ (3(aq)) + "NaOH"_ ((aq)) -> "Na"_ 2"CO"_ (3(aq)) + "H"_ 2"O}}_{\left(l\right)}$

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 $\textcolor{red}{2}$

${\text{H"_ 2"CO"_ (3(aq)) + color(red)(2)"NaOH"_ ((aq)) -> "Na"_ 2"CO"_ (3(aq)) + "H"_ 2"O}}_{\left(l\right)}$

At this point, you have a total of

• $2 \times \text{H" + color(red)(2) xx "H" = "4 atoms of H}$
• $3 \times \text{O" + color(red)(2) xx "O" = "5 atoms of O}$

on the reactants' side, and only

• $2 \times \text{H" = "2 atoms of H}$
• $3 \times \text{O" + 1 xx "O" = "4 atoms of O}$

on the products' side. To balance the hydrogen and oxygen atoms, multiply the water molecule by $\textcolor{b l u e}{2}$.

${\text{H"_ 2"CO"_ (3(aq)) + color(red)(2)"NaOH"_ ((aq)) -> "Na"_ 2"CO"_ (3(aq)) + color(blue)(2)"H"_ 2"O}}_{\left(l\right)}$

The reactants' side will now have

• $2 \times \text{H" + color(blue)(2) xx "H" = "4 atoms of H}$
• $3 \times \text{O" + color(blue)(2) xx "O" = "5 atoms of O}$

which implies that the chemical equation is now balanced.

$\textcolor{w h i t e}{\frac{a}{a}}$
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

${\text{CO"_ (2(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 2"CO}}_{3 \left(a q\right)}$

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

${\text{H"_ 2"CO"_ (3(aq)) rightleftharpoons "H"_ ((aq))^(+) + "HCO}}_{3 \left(a q\right)}^{-}$

Notice that the reaction can also produce sodium bicarbonate, ${\text{NaHCO}}_{3}$, since you can have

${\text{H"_ ((aq))^(+) + "HCO"_ (3(aq))^(-) + "NaOH"_ ((aq)) -> "NaHCO"_ (3(aq)) + "H"_ 2"O}}_{\left(l\right)}$

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

${\text{NaHCO"_ (3(aq)) + "NaOH"_ ((aq)) -> "Na"_ 2"CO"_ (3(aq)) + "H"_ 2"O}}_{\left(l\right)}$

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)))))

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a} \textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{{\text{H"_ 2"CO"_ (3(aq))))) rightleftharpoons color(red)(cancel(color(black)("H"_ ((aq))^(+)))) + color(red)(cancel(color(black)("HCO}}_{3 \left(a q\right)}^{-}}}}$

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))#
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

${\text{CO"_ (2(aq)) + "H"_ 2"O"_ ((l)) + 2"NaOH"_ ((aq)) -> "Na"_ 2"CO"_ (3(aq)) + 2"H"_ 2"O}}_{\left(l\right)}$

which looks very much like what we've got using ${\text{H"_2"CO}}_{3}$ for carbonic acid.