I have seen the following radical reaction involving bisulfite (which you may have on hand):
.NO3 + HSO3- = NO3- + H+ + .SO3-
Source: See, for example, https://www.google.com/url?sa=t&source=web&rct=j&...
As such, I would similarly expect:
.NO2 + HSO3- = NO2- + H+ + .SO3-
That is, pass some generated NO2 gas over moist bisulfite, or possibly, with a lower yield, into an aqueous oxygen free bisulfite in the presence of
UV. Neutralize with say NaOH after allowing the mix to stand to remove the sulfite radical anion via a self decomposition:
.SO3- + .SO3- = S2O6(2-)
Note, the possibly formed nitrite salt is not pure per the above.
[Edit] To discuss further the aqueous path, the action of NO2 and water produces HNO2 and HNO3. With UV on HNO3, the expected reaction is:
HNO3 + UV ---) .OH + .NO2 (See eq R24 at http://www.ciesin.org/docs/011-457/011-457.html )
where the NO2 could react with HSO3- forming nitrite as before. However, the action of UV on HNO2:
HNO2 + UV ---) .OH + .NO (See, for example, eq 34 at https://www.google.com/url?sa=t&source=web&rct=j&... )
with the possible eventual creation of the NO- anion and HNO. The latter unstable acid decomposes via:
2 HNO ---) H2O + N2O (See Table I at https://books.google.com/books?id=-xbQubw1zbQC&pg=PA105&...
which, while photo active, any formed laughing gas could reduce the yield of nitrite.
[Edited on 30-7-2016 by AJKOER]
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