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woelen
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Using HCl is not the best option. As I wrote, it is best to overrun the electrolysis in order to keep losses to a minimum, but overruning the
electrolysis also leads to an excess amount of oxidizer. With sulphuric acid or sodium bisulfate this is no problem. The excess oxidizer simply is not
used in that case. With hydrochloric acid, however, the excess bromate is capable of oxidizing chloride to BrCl and this compound mixes extremely well
with Br2. Actually, expect almost all BrCl to end up, dissolved in the Br2. So, you'll end up with Br2 which is contaminated with BrCl, or you'll have
to accept large losses by underrunning the electrolysis (this assures that no Cl2 or BrCl will remain present in the system, because if any were
formed, then the excess bromide will react with this to chloride ions and Br2).
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elementcollector1
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Actually, I ended up using sulfuric acid. This left the solution a deep red color. However, I tried distilling this (my hotplate sucks, and it's
freezing outside here!), and only got a weaker solution of bromine. What am I doing wrong?
EDIT: Added some more sulfuric acid, as well as some H2O2. Solution has turned opaque orange (what?) with a red, almost precipitate-like substance on
the bottom of the flask. Will try re-distilling tomorrow, with sulfuric acid in the receiver as well as the boiling flask.
[Edited on 16-12-2012 by elementcollector1]
Elements Collected:52/87
Latest Acquired: Cl
Next in Line: Nd
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crazyboy
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http://www.youtube.com/watch?v=OB4MmPTOBxg&feature=related
This uses Cl2 to displace Br in KBr to produce bromine. As previously mentioned however this would contaminate the Br2 with BrCl. Is there a good way
of removing the BrCl or is this method simply not viable?
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elementcollector1
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Well, I got some bromine! It's ampouled, and the ampoule is inside a test tube. The stuff's very black, but does have a red tint (and unfortunately,
does wet glass).
I believe plante posted something about keeping the bromine at 10C, and letting the BrCl evaporate off? Read up a few posts.
Elements Collected:52/87
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Endimion17
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It's one of the worst methods available. I suppose one could get rid of most bromine chloride by refluxing the raw product with finely divided
potassium bromide during one day. I'd also employ vigorous stirring. It is possible to purify it just by leaving the bromide inside bromine, but I
think that would require at least a week, maybe a lot more, like a month. Savage stirring and reflux could solve the problem in full 24 hours.
BrCl<sub>(l)</sub> + KBr<sub>(s)</sub> -> Br<sub>2(l)</sub> + KCl<sub>(s)</sub>
(actually BrCl is a gas, but it's inside the mixture; I have no idea in what state, solvated or just there, like most of H<sub>2</sub>S
inside water is)
You can imagine it's a slow reaction - there are almost no free ions except some small quantities due to moisture involved. Those might even serve as
a catalyst, I don't know.
It's much easier simply not to use elemental chlorine or chlorine containing oxidizers for oxidation so all you have to do next is extraction of water
and distillation.
Even if you do that, there will always be some bromine chloride inside because the starting material (analytical purity alkali bromide) always
contains some chlorine in the form of chloride. If there's 0.5% of chloride in your alkali bromide, you'll get around 0.5% BrCl in your bromine, too.
Also don't forget iodine compounds. Chlorine, bromine and iodine always follow each other, just like there's sodium in potassium samples (which is why
its true violet flame can't be visible without cobalt glass).
Leaving BrCl to evaporate can't solve the problem.
[Edited on 16-12-2012 by Endimion17]
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elementcollector1
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Could sulfuric acid suck the BrCl out?
Elements Collected:52/87
Latest Acquired: Cl
Next in Line: Nd
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neptunium
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trying to seperate out each halogen from any element preparation sounds like an isotopic seperation....it aint gonna happen at home unless you invest
in a $90,000 mass spectrometer and run it for a few thousand years..
you ll just have to admit that no matter how pure the sample is, it will always have some Fluoride chloride etc...
the Bromine obtained and dry is usually good enough for element collectors and most reaction at home..
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elementcollector1
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And that's why I went with the sulfuric acid method. No chlorine = no interhalogens.
Elements Collected:52/87
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woelen
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@Endimion17: I think that it is not true that you will always have BrCl in your bromine. Using normal equipment at home and wanting to put some effort
in it you can remove any BrCl and also you can do quickly. Most people (including me), however, do not take the effort to remove the last few tenths
of percent of BrCl.
Indeed, normal commercial grade NaBr or KBr certainly will contain a few tenths of percent of chloride as well. Using my method of electrolysing such
that 1/6 of the bromide is converted to bromate and using slight overrunning of the electrolysis step in order to get decent yield results in
formation of BrCl in appr. the same percentage as you had chloride in your bromide species.
Removal of this BrCl is not really hard though, but it leads to losses. One way of doing this is dissolving some NaBr in water (make the solution
highly concentrated) and add 1 ml of this solution to each 10 ml or so of your Br2. Shake the mix well and after a few minutes all BrCl will be
converted to Br2.
After that step, however, you have another impurity (water, NaBr and NaCl). You need to distill the bromine first to get rid of the NaBr and NaCl and
you then need to dry the bromine with conc. H2SO4 to get rid of the water (the distillation is not enough to leave all water behind) and then you need
to pipette away the bromine from under the H2SO4. All these steps lead to additional losses (mostly mechanical) and especially if you work on a small
scale of milliliters then the mechanical losses may be considerable.
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Endimion17
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Few tenths of percent is still "some". 
Of course, it's impossible to remove all BrCl from a macroscopic sample, but 0.5% is not a small amount. Consider 100 g of the sample. It
would contain 0.5 g of BrCl. That's 0.23 ml of the BrCl gas at STP stuffed into roughly 32 ml of liquid. Now it looks as a nuisance, at least to me.
And if you look at it from an organic chemist's view, it's a pain in the ass.
Shaking with alkali bromide (salt or conc. solution - that works, too, but introduces water) could push that below detectable levels, at least
detectable using usual analytical methods.
Mechanical losses are the worst, I agree on that. The stuff just evaporates. If I had lots of crude bromine, I'd try to do this purification, but it's
just a stupid thing to do when you've got few mililiters. The losses are incredibly high. For display purposes, it's important to get rid of the water
so it doesn't wet the glass and looks nice.
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bromobagel
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I tried and documented a similar prep, and thought I'd post it here.
The yield I obtained was disappointing, but the reason for this is obvious and the problem is easily fixable (see Discussion).
(1) 2KBr + MnO2 + 3H2SO4 ---(heat)---> Br2 + 2KHSO4 + MnSO4 + 2H2O
Source : http://143.239.128.67/academic/chem/dolchem/html/elem035.html
Experimental
In a 500 mL 24/40 RBF was loaded 40.0 g (0,336 mol) KBr and this was dissolved in a minimal amount of distilled water (The dissolution is endothermic,
at the end the solution temperature was 5 C). 100 mL dilute H2SO4 [30% or 5,625M] (0,5625 mol, a slight excess) is added to the cold KBr solution with
swirling of the flask. Finally 16.3 g (0,182 mol, a slight excess) MnO2 and a few boiling chips were added and the flask was fitted for simple
distillation. The receiving flask (150 mL RBF) was submerged in an ice bath, and ice was also added regularly to the condenser water tank. The vac.
outlet of the distillation apparatus was fitted with a conc. NaOH bubbler to trap bromine vapours. The reaction flask was then gently heated by means
of an oil bath. When the bath temperature reached 44-45C bubbles evolved and the air in the flask took an orange tint. When the bath temp. reached 60C
the whole apparatus was orange but no bromine had condensed. When the bath temperature reached 75C bromine started coming over at a rate of one drop
about every 4 seconds. After 1h I noticed that the pressure fluctuated wildly, and that a large amount (~60 mL) of NaOH solution had been sucked back
into the receiver flask. The distillation was continued anyway over the course of 2,5 additional hrs, watching the pressure closely, until no more
bubbles formed in solution and the air inside the apparatus was almost colourless. At this point the bath temperature was 107C. The oil bath was
removed and the NaOH bubbler disconnected. The bromine was pipetted into a 50mL Erlenmeyer and 20mL of concentrated H2SO4 were added. This was swirled
around a few times and the layers allowed to separate. The bromine was pipetted into a tared graduated cylinder and the volume and weight were
recorded. It was the transferred into a 25mL vial, the taps were wrapped in Teflon tape and the lid was tightly closed. The vial itself was put in a
Mason canning jar along with a few mL of sodium thiosulphate solution. Final yield : 12,3g and 3,90mL
(46% of theoretical based on KBr, d=3,15 g/mL)
Discussion
It seems obvious that the low yield was a result of the NaOH being sucked back into the collection flask and neutralizing part of the product. Much
better yields would be obtained simply by placing a guard flask between the trap and the distillation apparatus. Overall, I think this method is
better than the H2O2 one (if it can be repeated with better yields) because MnO2 is much more easily obtainable than concentrated peroxide (it is
easily found by opening alkaline batteries and washing the crude MnO2 with water to get rid of the electrolyte). Also, the fact that the reaction is
not exothermic and does not start at room temperature avoids the possibility of a thermal runaway, disastrous when working with Br2.
Details about experimental : the original prep mentioned dilute sulphuric acid, for the exact concentration I just took a guess and figured 30% was
dilute enough. It seemed to work okay. The temperature at the distilling head was not recorded because my thermometer was broken, and I couldn’t use
my bath thermometer because it doesn’t fit in my 24/40 thermometer adaptor.
General notes about working with bromine : Wear full PPE and nitrile gloves throughout. Workup of the product and cleaning of the glassware were
extremely unpleasant. I don’t have a sink in my fume hood, so I had to wipe the glassware with thiosulphate-soaked towels before washing them, and
some Br2 still escaped, bathing the general area with the irritant and persistent smell of pool water. I don’t think I will be doing this again
until I renovate my fume hood with plumbing and better ventilation.
I hope this can help someone...
-- Bromobagel
[Edited on 24-12-2012 by bromobagel]
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AJKOER
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It may be interesting to note without the use of a strong acid, the following quoted reactions from http://infoscience.epfl.ch/record/33222/files/EPFL_TH2746.pd... page 122 :
"Cl2O(g) + H2O(a) → 2HOCl(a) (4.6.6)
HOCl(a) + K-Br(s) → KOCl(s) + HBr(a) (4.6.8)
HOCl(a) + HBr(a) → BrCl(a) + H2O(a) (4.6.9)
BrCl(a) + K-Br(s) → K-Cl(s) + Br2(g) (4.5.4 and 4.5.5)
Net: Cl2O(g) + 2K-Br(s) → K-Cl(s) + KOCl(s) + Br2(g)"
From which I would infer:
2 HOCl + 2 KBr --> KCl + KOCl + H2O + Br2
Note, it is important to avoid a large excess of HOCl as:
Br2 + 5 HOCl + H2O --> 2 HBrO3 + 5 HCl
However, the above process has been described by those performing Bromine extraction as most likely sub optimal 
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AJKOER
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Has anyone tried the original path? To quote:
"Bromine was discovered by A J Balardin 1826AD, by the action of Chlorine on the residues (i.e. Bromide salts) after the crystallisation of the salt
from the salt-marshes of Montpellier.
NaBr + Cl2 ==> NaCl + Br2 "
Source: http://www.ucc.ie/academic/chem/dolchem/html/elem035.html
However, per a more recent reference given above (link: http://infoscience.epfl.ch/record/33222/files/EPFL_TH2746.pd... ), per page 92, an updated surface chemistry rendition, to quote:
"The fact that the uptake coefficient does not significantly change whereas the yield of Br2 increases after an exposure to H2O(g) may lead to the
conclusion that 1) the presence of H2O(a) enhances the conversion of surface adsorbed chlorine to surface-adsorbed bromine or polyhalide species, 2)
bromine species are not removed from the surface under ambient conditions, 3) the adsorbed bromine species may be displaced by chlorine adsorbing on
the surface. 1) and 2) explain the yield of Br2 of almost 100% of the lost Cl2 and 3) may explain that the uptake coefficient remains almost
unchanged after an exposure of the KBr sample to ambient conditions, which means that a “bromine site” is kinetically almost equivalent to a
“free site”.
This may be explained by the following reaction scheme:
Cl2 + S = Cl2-S
Cl2-S + KBr → BrCl-S + KCl
BrCl-S + KBr → Br2-S + KCl
Br2-S = Br2 + S
Net: Cl2 + 2KBr → Br2 + 2KCl "
[Edited on 25-12-2015 by AJKOER]
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Texium
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Yes, that method does work fine, but it also leads to the formation of interhalogens, so it isn't great for preparing pure bromine.
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