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If normal breathable atmosphere were to be reacted with aqueous sodium hydroxide, what form would the resulting sodium carbonate take, would it be dissolved in the solution or solid?
Note:
I'm defining air here as the atmosphere of earth at between sea level and 200m.
(Not a homework question, just interested.)
This depends on a lot of things: the volume of solution, the time carbon dioxide has to get dissolved and the temperature are probably the most important.
The actual reaction taking place is happening in solution. Thus, carbon dioxide must first dissolve in the solution (1) before dissolved carbon dioxide can react with the hydroxide base (2). You should view these steps as distinct even though the second is much faster than the first.
$$\begin{align}\ce{CO2 (g) &-> CO2 (aq)}\tag{1}\\ \ce{CO2 (aq) + OH- (aq) & -> HCO3- (aq)}\tag{2}\end{align}$$
Since these happen in solution, the counterion — a spectator ion — is irrelevant and the same would happen with $\ce{KOH, Ca(OH)2}$ and others.
At some point you will reach saturation, because either too much carbon dioxide dissolved or (much more likely) too much water evaporated. Thenceforth, sodium hydrogencarbonate or sodium carbonate (I never performed a formal analysis) will precipiate as a white crust. Incidentally in Munich in the inorganic teaching labs, there were 5 litre reservoirs of $6~\mathrm{M}\ \ce{NaOH}$ solution from which smaller bottles could be filled. If there was a spill (and there always was — it was undergrads in that lab after all) it would show up days later as a white crust. When I myself was a TA, part of my job preparing the labs for the next student generation was to clean those areas. Diluted hydrochloric acid worked wonders.
