I didn't know that fact. Actually, my attempt of drying NO2 by passing through conc. H2SO4 was failed because almost no gas (0.2mol as the result of
reaction) was leaving the wash bottle where 30 ml of 96% H2SO4 + ~130mg CuSO4 (mostly undissolved) were placed.
(I've added CuSO4 to detect NO presence - in this case the solution would turn the color to violet).
I didn't make the photo of the whole apparatus but even from this fragment you can see the difference between in- and out- tube. 15 cm after the
outlet the brown color was totally disappeared, so the gas really can't pass the absorber.
Another evidence is the reaction of the liquid in the wash bottle with water. It results instaneously in a big NO2 cloud. But the presence of NO2
isn't noticable in the liquid - no color change, no strong smell.
What could be the chemistry of NO2 absorption by H2SO4?
[Update]
From the article "The sulfuric acid solvent system" by R.J. Gilliespie and E.A. Robinson:
"dinitrogen tetroxide gives a mixture of nitronium and nitrosonium hydrogen sulfates:
N2O4 + 3H2SO4 = NO2+ + NO+ + H3O+ + 3HSO4-
But I have a doubt about NO+ presence because I don't observe any color change of CuSO4 salt.
[Edited on 1-6-2023 by teodor]Lionel Spanner - 1-6-2023 at 05:54
It's most likely reacted to form nitric acid.Raid - 1-6-2023 at 05:58
it dissolves in the H2SO4 to form Nitrosyl sulphuric acid and nitric and nitrous acids.
Raid, I didn't find H2SO4 absorption in the last article (I can access it by sci-hub).
I doubt about the possibility of nitric/nitrous acid presence in the concentrated H2SO4, as I cited nitrogen oxides behaves as bases in regard to
H2SO4 and forms 2 sulfates.
I have a hypotesis that nitrosonium ion converted to nitronium ion by the action of air in the system. So, probably I will try to generate nitrosonium
solution in H2SO4 and will split it into 2 parts: one will keep air-free and the second will allow to react with air.
[Edited on 1-6-2023 by teodor]Raid - 1-6-2023 at 06:25
The other papers said that there would be nitric and nitrous acids in solution. This might not be the case because the paper that said this is about
100 years old.
I would also assume that there would be nitric acid in solution because of the fact that this same sort of process is used to to make nitric acid by
the nitrate salt process or the sulfate salt process.
[Edited on 1-6-2023 by Raid]teodor - 1-6-2023 at 10:22
I think there is a difference between concentrated H2SO4 and H2SO4 + water in regard to NO2. This is about the old articles.
When I compare
KNO3 + 3H2SO4 = K+ +NO2+ + H3O+ + 3HSO4-
and
N2O4 + 3H2SO4 = NO+ + NO2+ + H3O+ + 3HSO4-
it really looks similar except in the second case there is only one H3O+ per 2 nitro[so]nium bases.
Well, I would say in both cases you have nitronium hydrogen sulfate in the solution which is decomposing in the first case during the heating.
[Edited on 1-6-2023 by teodor]
What is attracted my attention is that NO+ ion is absent at the end of absorption because the color of the solution didn't changed comparing to the
case when I pass NO2 + NO mixture.
[Edited on 1-6-2023 by teodor]
teodor - 1-6-2023 at 14:46
Some thoughts ...
Suppose I can distill HNO3 from (NO2)HSO4 + H2SO4 mixture the same way as from KHSO4 + (NO2)HSO4 + H2SO4 mixture (as when making HNO3 by mixing
concentrated H2SO4 and KNO3).
NO2+ + H3O+ + 2HSO4- -> HNO3 + 2H2SO4
That would mean removing H3O+ ions and driving further self-ionization by this equation:
2H2SO4 -> H3O+ + HS2O7-
In the reality at some point HNO3 could stop to form giving off only nitrogen oxides, but it would be interesting if this point could lay beyond 100%
concentration.woelen - 1-6-2023 at 23:34
If NO(+) and NO2(+) are formed, doesn't that perfectly fit your observations? The deep blue complex with copper(II) is a complex with neutral NO, not
with NO(+).
Try magnesium nitrate.
$$Mg(NO_3)_2(H_2O)_2 + 4H_2O \rightarrow Mg(NO_3)_2(H_2O)_6$$
Drying the magnesium nitrate at around 300c will restore the dihydrate form and keeping the gas stream below 89c might help to absorb water
Dividing the molar mass of the hexahydrate by the dihydrate gives a weight ratio of 1.39g water absorbed per gram of dihydrate usedteodor - 2-6-2023 at 02:40
Woelen, when I asked my question I was unaware about basic properties of NO2. I was expecting they are attributed only to NO. During investigation
I've also made conclusion it should form nitronium and nitrosonium ions based on their basic character but your idea about H2O which is bound by H2SO4
put some new light on it. And what is unusual way of water formation in acid-base reaction, not by H+ + OH- but 2H+ + O-!
I didn't know the blue (for my sight rather violet) compound is forming only with neutral NO. Reading there and there I saw the formula somewhere
which I can recall as:
4 NO + 2 H2SO4 = N2O + 2 (NO)HSO4 + H2O
But on further investigation I've made a conclusion it is not so simple as that so the character of NO solution in H2SO4 is not quite clear for me. In
Gmelin (1936) there is the formula:
Thanks for the idea of the usage of ~anhydrous nitrate.
Have you the data for Mg(NO3)2 * 2H2O absorption power? We are dealing with the compound (NO2) which is suspected (by my speculation in one of the
previous posts) to be capable to dehydrate 100% H2SO4, and here we have 2 mols of water for it in bar's menu.
[Edited on 2-6-2023 by teodor]Tsjerk - 2-6-2023 at 03:12
Dividing the molar mass of the hexahydrate by the dihydrate gives a weight ratio of 1.39g water absorbed per gram of dihydrate used
Isn't it absorbing 18 x 4 = 72 gram per 184 gram magnesium nitrate dihydrate? So 0.39 gram water per gram of dihydrate?teodor - 2-6-2023 at 04:04
I found something written by Paul Sabatier
about the "blue Cu complex".
Attachment: blue_acid_discovery.pdf (543kB) This file has been downloaded 193 times
Attachment: blue_complex.pdf (778kB) This file has been downloaded 221 times
[Edited on 2-6-2023 by teodor]
[Edited on 2-6-2023 by teodor]Rainwater - 2-6-2023 at 04:10
Half asleep when i worked the math.
Its worth double checking.
Molar mass copied from Wikipedia
184.35 g/mol (dihydrate) Mg(NO3)2(H2O)2
256.41 g/mol (hexahydr.) Mg(NO3)2(H2O)6
1g of dihydrate = 0.0054347826 mols
Add 4 times the molar about if water = 0.0217391304 mols or 0.3913043472g of water.
Yep. I $<"ed up. Sorry.
[Edited on 2-6-2023 by Rainwater]teodor - 2-6-2023 at 12:19
Oh, I also see my error.
It is not Cu(NO)[SO3]2 , because NO as a cation has the oxidation number +1.
This complex has the same anion as Fremy's salt
So, it is actually Cu[ON(SO3)2] where Cu has the oxidation number +2. NO is neither +1 nor 0 here. Is it compound of N in +4 oxidation or did I miss
something ?
[Edited on 2-6-2023 by teodor]teodor - 3-6-2023 at 00:35
I found some information, that compound (NO)HSO4, nitrosonium hydrosulfate, is not possible to separate from concentrated H2SO4 by distillation. By my
understanding, either compound itself or N2O3 forms azeotrope with concentrated H2SO4, because I don't understand how it may be another way. Did
somebody see the boiling curve of the mixture?woelen - 5-6-2023 at 23:24
It is not Cu(NO)[SO3]2 , because NO as a cation has the oxidation number +1. [...]
So, it is actually Cu[ON(SO3)2] where Cu has the oxidation number +2. NO is neither +1 nor 0 here. Is it compound of N in +4 oxidation or did I miss
something ?
I hardly can believe it has a formula like the one you cite. You have a strongly oxidizing environment, with all the NO2 and conc. H2SO4 around. You
can assume that sulfur is in oxidation state +6 in this environment. There are no peroxo-species, so oxygen is in oxidation state -2. Copper is in
oxidation state +2. So, the entity [ON(SO3)2] must have formal charge -2. This only is possible if N has oxidation state 0. I do not believe that.teodor - 6-6-2023 at 00:16
The blue acid is not stable in the stronly oxidizing environment, but if there is constant NO or SO2 supply, it stays in the solution. Also there are
some other cases. I observed the forming of the blue acid on walls of a test tube when I put a drop of water in H2SO4 saturated with NO2.
Hm, but I don't quite agree about "strong oxidizing" because there are both NO+ and NO2+ ions here.
There is description of electrolizing NOHSO4 solution in H2SO4 resulting in NO2HSO4 + H2SO4, this one I would name "oxidizing".
[Edited on 6-6-2023 by teodor]woelen - 6-6-2023 at 02:08
Both NO(+) and NO2(+) are strongly oxidizing. They contain nitrogen in oxidation states +3 and +5.
You also must distinguish between a purple/blue complex, formed with Cu(2+) and the purple acid, formed from sulfite (or SO2), nitrite and conc.
H2SO4. The latter is a weird compound with sulfur in oxidation state +4, which quickly decomposes, leading to formation of sulfate (sulfur in
oxidation state +6). The copper complex is more stable. These different compounds have very similar colors.Bedlasky - 7-6-2023 at 04:35
Purple acid is [N2O2]HSO4. There isn't any sulfur in IV oxidation state.teodor - 7-6-2023 at 04:57
Woelen, Bedlasky, do you have any references to description of compounds you mentioned as "blue" or "purple" acid?
My reference is the article "blue_acid_discovery.pdf" which I have attached. It's a quite old source, so let's put something more fresh here about the
topic.Bedlasky - 7-6-2023 at 06:41
Thanks Bedlasky. Oh, I see, the literature prior 1953 is completely wrong about this compound. From the pdf attached there:
"The determination of the substance’s constitution was accompanied by considerable difficulties because
the compound is only present in solution and therefore a simple sum formula can’t be analytically
determined. According to P. Sabatier (C. r., 122, [1896], 1417/9, 1419, 1479/82, 1481, Bl. Soc Chim.
[3] 17 [1897] 782/91, 783) the purple acid is supposed to be identical with nitroso bis(sulfuric) acid
(HSO3)2NO (see p. 1628). He concludes from the remarkable match in color, that the first "blue acid"
first observed by him had to be the same substance that was first described by E. Frémy (Ann. Chim.
Phys. [3] 15 [1845] 408/88; Lieb. Ann. 56 [1845] 315/54, 337) as the yellow potassium salt of sulfazilic
acid (acide sulfazilique) that gives a deep purple color. M. Trautz (Z. phys. Ch., 47, [1904], 513/610,
599) adopts the description of P. Sabatier.
The correctness of P. Sabatiers formula, which was only based on analogy reasoning, is doubted by
T. Haga (J. chem. Soc., 85, [1904], 78/107, 93). According to F. Raschig (Ber., 18, [1905], 1281/1323,
1305) the formula (HSO3)2NO is already made improbable by the fact that the dissolution of copper sheet
in a solution of NOHSO4 in H2SO4, which first forms and then decomposes the purple compound, no
SO2 is produced ..."
But its amazing, Sabatier's fromula was doubted in 1904 but in Gmelin "stickstoff" (nitrogen) published 1936 it is still used.
So, this formula cited by me from Gmelin "Nitrogen" is incorrect:
