JCM83 - 9-1-2011 at 19:52
Simple reaction:
Cu(s) + 4HNO3(aq) ——> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
So I get this brown gas, nitrogen dioxide, fuming out and trying to poison my fume hood.
Which seems like a lot of hydronium combines with one oxygen of half the nitrate ions, so that
NO3- + 2H+ ——> NO2 + H2O
The nitrogen dioxide then evaporates. My questions are as follows:
Does some of it stay dissolved in the water? If it does so, does it act like CO2 - staying dissolved as carbonic acid- and stay dissolved as nitric
acid?
So is my dissolved copper forming copper nitrate crystals, or copper sulfate crystals with the sulfuric acid that I put in afterward?
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Also, why does CO2 absorption into water lower its pH? Is it eventually converting to Bicarbonate (HCO3-) and thusly absorbing OH to increase the
relative proportion of H+?
[Edited on 10-1-2011 by JCM83]
woelen - 10-1-2011 at 00:29
Your reaction equation is not correct:
NO3(-) + 2H(+) does not give NO2 + H2O. At the left there is a net charge of +1, at the right there is no charge left.
The NO2 is not formed by reaction with H(+) ions, it is a result of the reduction by copper metal.
NO2 does dissolve in water quite well, but it does not act like CO2 giving a reversible reaction in which some equivalent of carbonic acid is formed.
Instead, NO2 reacts with water, giving nitric acid and nitrous acid. The latter is unstable (it falls apart in water, NO2 and NO) and finally, all NO2
reacts giving nitric acid and nitrogen monoxide, NO.
CO2 which dissolves in water forms a small amount of H2CO3, which in turn can give H(+) and HCO3(-). The H(+) makes the liquid somewhat acidic. The
effect, however, is not strong.
Finally, when copper is reacted with nitric acid and sulphuric acid is added afterwards, then you cannot say that it either forms copper nitrate or
copper sulfate. In solution you have copper ions, nitrate ions and sulfate ions (and also H(+) ions) and all of these exist besides each other. Only
when the liquid is evaporated and crystals are formed, then a pure compound might be formed. Given the solubility figures I indeed think that on
evaporation copper sulfate will be formed, but keep in mind that this will be fairly impure.
JCM83 - 10-1-2011 at 09:41
Oh dang you're right about my proposed unbalanced equation.
Reduction by copper metal... so the NO3 gains an electron, is reduced while Cu is oxidized, and then becomes NO2. It lost an oxygen atom which being
more electronegative must've bonded with something else, right?
So 4 released H+, bonding with 2 O(-2) results in 2 H2O, and that leaves NO2 with... a -1 charge?