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Nitric acid obtained in laboratory is slightly yellow due to the dissolution of nitrogen dioxide which is produced due to thermal decomposition of a portion of $\ce{HNO3}$.

$$ \ce {4HNO3 -> 2H2O + 4NO2 + O2}$$

Its says here that if $\ce{CO2}$ is bubbled through the acid, it turns colourless- because it drives out $\ce{NO2}$ from warm acid which is further oxidized to nitric acid.

What does it mean by drives out ? How can carbon dioxide drive out nitrogen dioxide that's dissolved in the acid? And also, the oxidation of 'what' is being talked about here?

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  • $\begingroup$ Look up "Henrys law" $\endgroup$
    – Georg
    Sep 20 '12 at 15:47
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I think you can explain this using the equilibrium:

$\ce{NO_2(g) <=> NO_2(dissolved)}$

When you pass $CO_2$ through the warm acid, the partial pressure of $CO_2$ ($P_{CO_2}$) above the gas increases (Since partial pressure generally represents the amount of gas above the solution).

Since

  1. total pressure $P = P_{NO_2} + P_{CO_2}$ (Dalton's Law)

  2. And $P$ remains constant at atmospheric pressure (assuming that the system is not closed).

$P_{NO_2}$ will decrease.

According to Le Chatelier's Principle, the equilibrium proceeds backward to compensate the decrease in $P_{NO_2}$ and thus more $NO_2$ gas is released. The gas flows quickly along with the bubbled gas and this prevents the forward reaction.

This seems like quite a roundabout method but it works. I'm sure there could be a much simpler explanation though.

The oxidation mentioned is most probably the disproportion reaction of $NO_2$ when it comes in contact with water:

$\ce{3NO_2 + H_2O -> 2HNO_3 + NO}$

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