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Author: Subject: Preparation of constant boiling hydrobromic acid
ampakine
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[*] posted on 12-7-2011 at 12:55
Preparation of constant boiling hydrobromic acid


In an article titled "Preparation of constant boiling hydrobromic acid" which I found in the archives on this site, heres the start of the method they give:

Quote:

Add 120 g. of powdered potassium bromide to 200 ml. of water. Place the container in cold water, and slowly add 90 ml. of conc. sulfuric acid (1.7 moles).


I don't understand the stoichiometry involved. If I'm not mistaken KBr reacts with H2SO4 in the following 1:1 reaction:
KBr + H2SO4 -> HBr + KHSO4
therefore there should be 1 mole of H2SO4 for every mole of KBr. Why do they use 1.7 moles of H2SO4 for 1 mole of KBr?? Also I don't understand what they mean by "conc. sulphuric acid". I have 98% H2SO4 and thats what I consider concentrated. 90 mL of that comes out to around 3.4 moles. I'm guessing they're using 50% H2SO4.

[Edited on 12-7-2011 by ampakine]
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AndersHoveland
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[*] posted on 12-7-2011 at 13:01


If concentrated sulfuric acid is used in an attempt to distill out HBr, it is more likely that elemental bromine will be obtained instead. Much better to use phosphoric acid instead, as it will not be oxidizing towards HBr.



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ampakine
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[*] posted on 12-7-2011 at 13:53


If you keep the temperature below 25C then you can prevent the formation of elemental bromine. You can achieve this with an ice bath and dripping the H2SO4 in over a period of a few hours. Is potassium phosphate as insoluble in water as potassium bisulphate?

EDIT: Before distilling, the KHSO4 will be precipitated out to minimise the amount of sulphate ions present when it comes time to distill. A bit of elemental bromine formation is inevitable but it shouldn't drop the yield by more than 5%. This is all pure speculation on my part, I have yet to get this reaction to work, I've just heard of people having success with this method. I'm about to start dripping the H2SO4, when its done I'll post the results here.

[Edited on 12-7-2011 by ampakine]
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UnintentionalChaos
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[*] posted on 12-7-2011 at 13:59


No, they are indeed using 98% sulfuric.

I have no idea how you are fitting 3.4mol of sulfuric into 90ml. Bravo, you should write a paper. :P

Conc. sulfuric is 1.84g/ml and 98g/mol. So at 98%, that's 1mol/100g.

90ml*(1.84g/ml)=166g --> 1.66mol

I suspect that the excess sulfuric drives down the solubility of the KHSO4 and that is its only purpose. I have done this prep using NaBr instead. Chill everything before mixing. Add the sulfuric in two portions. I had no bromine formation at all. Chill in an ice bath between additions. The distilled acid despite my best efforts (and that of other posters) does not seem to be azeotropic, but was nice and clear. The easiest way to concentrate the initial product further is probably to attach a fractionating column and pull distillate (mostly water) off at a very slow rate until the stillpot rises to the b.p. of the constant boiling acid.

[Edited on 7-12-11 by UnintentionalChaos]




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ampakine
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[*] posted on 12-7-2011 at 14:34


lol I'll check my stoichiometry. Dunno where I went wrong but 2nd try I got the same calculations as you. I already went ahead and diluted it down to 50% :( Ah well I'll keep the 50% H2SO4 for something else.

So I should chill the H2SO4 before adding? I was planning on allowing it to drip from my sep. funnel overnight but I'll take your advice and add chilled acid in 2 portions. First attempt at this I added the acid too fast and didn't cool at all and ended up with a beaker of bromine water.

BTW what was the approximate concentration of HBr in the solution you obtained after your attempts to obtain an azeotropic solution?
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ampakine
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[*] posted on 12-7-2011 at 22:42


I forgot to refill the ice bath for a couple of hours and the temperature rose to 50C and generated a bit of Br2 but besides that the reaction went well.
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