# Question #c4ae1

May 19, 2017

I don't think that you would be able to actually observe the solid.

#### Explanation:

You won't get a solid simply by exposing the sodium hydroxide solution to air, but the strength of the solution will decrease when exposed to air.

The idea here is that the carbon dioxide present in the air will dissolve in water to produce carbonic acid, ${\text{H"_2"CO}}_{3}$, which will neutralize some of the hydroxide anions present in solution.

You have

${\text{CO"_ (2(g)) rightleftharpoons "CO}}_{2 \left(a q\right)}$

followed by

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

As you know, sodium hydroxide is a strong base, which implies that it ionizes completely in aqueous solution to produce sodium cations and hydroxide anions

${\text{NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

At this point, you have

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

A more accurate description of this reaction would be

${\text{OH"_ ((aq))^(-) + overbrace("H"_ 2"O"_ ((l)) + "CO"_ (2(aq)))^(color(purple)("H"_ 2"CO"_ (3(aq)))) -> "HCO"_ (3(aq))^(-) + "H"_ 2"O}}_{\left(l\right)}$

The sodium cations are spectator ions, meaning that you have

$\textcolor{b l u e}{{\text{Na"_ ((aq))^(+)) + "OH"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) + "CO"_ (2(aq)) -> "HCO"_ (3(aq))^(-) + "H"_ 2"O"_ ((l)) + color(blue)("Na}}_{\left(a q\right)}^{+}}$

At this point, if you were to heat the solution in order to evaporate the water, you will get sodium bicarbonate, ${\text{NaHCO}}_{3}$, as a white solid, along with solid sodium hydroxide.

So your mystery solid would be sodium bicarbonate, but you wouldn't be able to observe it in solution because it is soluble in aqueous solution, i.e. it will exist as sodium cations and bicarbonate anions, ${\text{HCO}}_{3}^{-}$, in aqueous solution.