crestind - 26-7-2021 at 10:40
So I see sulfuric, nitric and water can each self ionize, and the stronger of any two will protonate the other, and the reaction also reverses
constantly. I am confused about mixtures of the three.
1. I see that a recommended technique of dehydrating 68pc nitric is by adding an equal mass of anhydrous 98pc+ sulfuric.
https://darets.com/ets/uploadedFiles/wwwjacobscom/20_Learn_A... Best explanation I was able to find. And presumably once distillation is begun, in
theory the vapor should be 95pc nitric. I interpreted this to mean that the sulfuric had essentially reacted or has been saturated so thoroughly with
water that free nitric acid now exists in solution.
2. The above does not seem to be correct because a common ratio given for nitrating mixtures is "concentrated nitric acid" (68pc?) and equal mass
98pc+ sulfuric. This mixture is desirable as the sulfuric is used to form nitronium ions, which disproves my previous assumption.
So does this mean that sulfuric reacts with both components of an equal mass 68pc aqueous nitric? And that it is better expressed in terms of
probability that is, all three are reacting and sulfuric is more likely to have reacted with water rather than nitric at a given time? Or in another
way, does this mean that larger quantities of water reduce the activity of sulfuric?
3. Is there a point at which sulfuric is so hydrated that it no longer reacts with nitric? I see that with copper sulfate hydrate, if 98pc sulfuric is
added, it dehydrates it and forms a white powder. But if water is slowly added at some point the sulfate would rehydrate and tuen the solution back to
blue.
crestind - 26-7-2021 at 17:50
Did manage to find this thread which has a vaguely similar basis.
https://chemistry.stackexchange.com/questions/126442/validit...
https://depts.washington.edu/eooptic/links/acidstrength.html
Looking into it some more and it seems the 50/50 mix has a small excess of H ions, allowing some protonation of the nitric. So this seems to make more
sense now that nitronium would form. But there is still seemingly some nitric remaining, so I am not sure if this would react at all.
50g sulfuric anhydrous = 0.5098mol (Two H per mol)
50g 68pc nitric = 34g nitric and 16g water = 0.5396mol nitric .8881mol water.