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Author: Subject: Failed HBr distillation...
evil_lurker
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sad.gif posted on 1-7-2006 at 01:33
Failed HBr distillation...


Tryed distilling some HBr tonight and got nada on yields.

Used 4 moles of NaBr dissolved in 400ml of water, 2 moles of H2SO4 in 100ml of water. Proceeded to rig up for simple distillation leaving the sodium sulfate in the flask.

Temperature shot up to 100C and kinda hung there for a few minutes then slowly crept up to 118C... after about 150ml or so was collected, the temp went up to 124C at which time I began collecting the distillate. The whole time the temperature never climbed over 124.5... towards the end though the sodium sulfate was getting pretty thick and there was some slight bumping and so I decided enough was enough. Total fraction caught at 124 degrees was about 150 ml.

When I went to test the finished product with sodium carbonate, nothing. When I went to test the finished product with H2O2, nothing. When I added a small amount of H2O2 to the sodium sulfate remnants in the distillation flask, massive amounts of bromine were quickly liberated.

I thought that surely to god I would getting something at 124 degrees, even if it wasn't the azeotrope.

What the hell went wrong?

[Edited on 1-7-2006 by evil_lurker]
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[*] posted on 1-7-2006 at 03:16


Evil lurker- This project is on my to do list so I do not have practical experience yet. On reading Inorganic Synthesis page 155, the process starts from KBr and sulfuric acid. The bisulfate formed was filtered off BEFORE distillation. What I wonder is did the sodium bisulfate try to precipitate? What was the concentration of your sulfuric acid?



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[*] posted on 1-7-2006 at 04:15


I'd still like to know what distilled off at over 100C.
Can't have been pure water.
Can't have been any Na salt.
Can't have been H2SO4.
Any ideas anyone?
I'd try re-distilling the stuff that came over to see if it gives any better HBr soln.
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[*] posted on 1-7-2006 at 06:48


It could have been water. After a certain concentration, sulfuric acid holds water and doesn't let it go until above a certain point past 100C. Maybe try adding more water to the reaction to dilute the H2SO4 even more. Thats my opinion.



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[*] posted on 1-7-2006 at 07:15


Try taking the boiling point of your distillate. That should shed some light on what it really is.

Wouldn't sulfuric acid only release 1 proton under these conditions?

H<sub>2</sub>SO<sub>4</sub> + NaBr --> NaHSO<sub>4</sub> + HBr

If so, you would obviously need twice as much sulfuric acid.




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[*] posted on 1-7-2006 at 08:55


This old thread from a site long gone from the internet might be helpful:


lugh
(Hive Bee)
05-01-01 02:53
No 188229
Six Laboratory Preparations of Hydrobromic Acid
(Rated as: excellent)

For those who are interested in knowing the time tested methods from "Inorganic Laboratory Preparations":
Hydrobromic Acid
Of the six methods given here, the first three may be used for the production of gaseous hydrogen bromide as well as for aqueous solutions of the gas; IV, V, and VI can be employed only for making the constant-boiling acid.

(I)
H2 + Br2 --> 2HBr
The following apparatus is assembled in order:
1. A gas-washing bottle filled with water.
2. A distilling flask filled with 50 ml of bromine cooled in ice with the inlet tube leading almost to the bottom of the flask.
3. A Pyrex tube, 50 cm long, packed with about 80 cm of activated charcoal which is held in place with glass wool plugs.
4. A U-tube filled with glass beads or porcelain chips coated with moist red phosphorus.
5. Two wash bottles in series filled with 100ml and 50ml of water, respectively; if constant-boiling acid is desired, the second bottle should contain 100ml of water. Both bottles are surrounded by an ice-salt bath. The reaction train is set up in the hood.
A steady flow of hydrogen is started through the apparatus to displace the air. This operation requires about thirty minutes, during which time the bromine is kept in the cooling bath.
The furnace is now slowly warmed to 350-375' C and the flow of hydrogen is increased to a point where the bubbles are just too rapid to count. A bath of water at 40-45' C is now placed under the vessel containing the bromine; under
these conditions, conversion to hydrogen bromide is complete in about 20 minutes. No vapors of bromine should be visible at the outlet of the reaction tube; otherwise the bromine must be cooled somewhat in water.
At the end of the experiment the two wash bottles are removed and the hydrobromic acid absorbed is determined from the gain in weight; if the smaller volume of water was used, the concentration of acid should be 60-65% by weight. If constant-boiling acid is desjred, the solution should be distilled and the fraction boiling at 122-127' C collected as 47-48% hydrobromic acid. A rapid method for determining the concentration of the solution produced is by a rough determination of specific gravity. 65%, 1.78; 60%, 1.68; 55%, 1.60; 50%, 1.52; 45%, 1.44.

(II)
2P + 3Br2 --> 3HBr + H3PO4
In a distilling flask fitted with a dropping funnel are placed 25 g of clean sand and, over this, a mixture of 25g of red phosphorus and l00 g of sand. The mass is moistened with 40-45ml of water, and 50 ml of bromine are introduced into the funnel. A U-tube and absorption bottles as described under (I) are connected to the flask. The reaction flask is cooled in ice while the bromine is added dropwise at a very slow rate; the phosphorus may glow with each addition of halogen. In order to avoid a suck-back of the water in the absorption flasks, it is advisable to place an empty safety trap between the U-tube and the water traps. As the reaction proceeds, the evolution of gas is more readily controlled and the cooling bath may be removed. When all the bromine has been added, the distilling flask is gently warmed to drive off the remaining acid vapors. The working-up of the hydrobromic acid solutions is the same as previously described.
Yield in I and II is over 90% of theory

(III)
C10H12 + 4Br3 ---> 4HBr + C10H8Br4
If anhydrous hydrogen bromide is desired, the procedure is carried out in the absence of water, using dry reagents. The apparatus of method II is used.
Thirty-five grams of tetrahydronaphthalene are placed in the flask with either 150 or 200ml of water, depending on the desired concentration of the acid. At first the flask is cooled in ice while 50ml of bromine are slowly added dropwise; about one gram of iron filings catalyzes the bromination. As the reaction proceeds, the flask may be allowed to warm up to room temperature. After all the bromine has been added the flask should be shaken for some time; the aqueous layer should be colorless. The acid layer is then separated from the organic material and worked up as in I.
Yield about 90%

(IV)
H2SO4 + KBr ---> HBr + KHSO4
A mixture of 120g of potassium bromide and 200ml of water is cooled in ice while 90ml of concentrated sulfuric acid is slowly added. The temperature must not rise over 75' C during this addition; otherwise free bromine may be formed, causing a loss in yield. The reaction mixture is cooled to room temperature and the potassium bisulfate is filtered off by suction through a fritted funnel or a hardened filter paper. The filtrate is then fractionated and the material boiling from 122-127' C is collected as constant-boiling acid.
Yield = 85%
In all cases where a mixture of sulfuric and hydrobromic acids is obtained, a redistillation is necessary to remove about 0.01% of sulfate in the first fractionation; only the acid with a steady boiling point is retained. This operation entails a loss of about 15% in yield.

(V)
Br2 + SO2 + 2H2O --> 2HBr + H2SO4
Fifty milliliters of bromine are covered with 200ml of water and sulfur dioxide is passed into the mixture, under the hood, until a straw-colored liquid results. Fractionation yields about 300g of 47-48 % acid, which is an almost theoretical yield.
As this reaction proceeds, the bromine dissolves in the hydrobromic acid that is formed, yielding a homogeneous liquid into which the sulfur dioxide may be more rapidly introduced.
HBr+Br2 <==> HBr2

(VI)
2S + Br2 --> S2Br3
S2Br3 + 5Br2 +8H2O -->12HBr + 2H2SO4
One hundred and fifty grams of bromine are weighed into a glass-stoppered bottle in the hood and l0g of powdered sulfur are quickly introduced. The bottle is then agitated and the sulfur rapidly dissolves to yield a red oily liquid.
Two hundred grams of ice are placed in a 500ml glass-stoppered bottle and the vessel is immersed in ice. About one-third of the sulfur-bromine mixture is added; over the course of about one hour the red oil disappears. Cooling is maintained throughout the hydrolysis. The second third of the sulfur-bromine compound is now added, followed by the last portion about 30 minutes later. When all the material has dissolved and reacted, a pale yellow liquid remains which is fractionally distilled as usual; b.p. 122-127' C.
Yield about 300g of acid
Hydrobromic acid may be kept colorless for long periods of time by storage in a dark bottle in the refrigerator.
References:
(I) 1. I, vol. 1:152.
(II) 2. FILETI AND CROSA, Gazz. chim. ital., 21:64 (1891); B:71
(III)3. I, vol. 1:151; footnote, p. 152.
(IV) 4. I, vol. 1:155.
(V) 5. Organic Syntheses, Collective Vol. 1, p. 23, Wiley, 1936.
(VI) 6. FARKAS et oL, J. Soc. Chem. Ind., 66:116 (1947).
7. P:464.


There are several other threads on this site that discuss the preparation of HBr, perhaps using the search engine would be helpful :cool:

[Edited on 1-7-2006 by leu]




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[*] posted on 1-7-2006 at 09:47


evil_lurker,

I have recently made 48%HBr using the method from Brauer in our library. (Thanks to S.C. Wack! ;) ) I also substituted NaBr for KBr, mole for mole. I used Rooto sulfuric, making sure to add extra (factor for me = 1.14) since my Rooto is below the reagent grade "con. sulfuric acid" strength of 18M.

I filtered out the NaHSO4 crystals using a Buchner funnel prior to distillation, per procedure.

My intial temperature also first came up to 100C. I switched out the receiver at 125C.

The product seemed perfectly fine to me - very dense and
acidic.

According to Gmehling this is a maximum boiling azeotrope. So if you are rich in water you would expect the intial product to boil at 100C, right?

[Edited on 1-7-2006 by Magpie]




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[*] posted on 1-7-2006 at 12:10


I was using .5 moles of ACS 98% sulfuric to reduce 1 mole of salt. It is my understanding that both H+ should be given up on the H2SO4.

I think the key difference is the fact that I left the sulfate in the distillation flask which for all intents and purposes "bunged it all up".
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[*] posted on 1-7-2006 at 13:13


Evil lurker, were the temperatures the boiling points of the liquid in the flask or the vapour coming off? Where was the thermometer bulb?
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[*] posted on 1-7-2006 at 15:05


Thermometer bulb was placed exactly where it was supposed to be (correct depth) on the distillation adaptor. The finished product had no smell like that of hydrochloric acid...

I managed to find another couple moles of NaBr and did the reaction again using the same molar amount of reagents as before and let stir for 10 minutes. This time though I dropped the temp rapidly in the flask from around 50C to about 25C using an ice bath and magnetic stirring. Took my thermometer and placed it in front of a fan to dry which produced Na2SO4 seed crystals. As soon as it was placed in the cool flask tiny Na2SO4 crystals began rapidly forming. I let it go some more, and then put the flask in the freezer which the whole bottom layer is full of fine crystals, and now some larger ones are forming in the upper solution that look more like glaubers salt. The solution itself is fairly viscous.

So I figure whatever is in there now has got to be fairly pure HBr. The weird thing is, the solution has no smell.

Really has me stumped.

I'll try a reacting a small amount with H2O2 and see what happens.
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[*] posted on 1-7-2006 at 15:19


Definately HBr. Took about 5 ml and added some H202 to it. Apparently the 40 volume hair developer is buffered all to hell. It took a few minutes then all of a sudden began to give off lots of Br2 vapor.

Also the crystals in the flask look like they might be re-arranging themselves to larger "ice" crystals....

Definatly weird.
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[*] posted on 1-7-2006 at 15:25


Why donĀ“t use H3PO4 and avoid a lot of the hassle?



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[*] posted on 1-7-2006 at 15:43


Got a point there.. as of yet I do not have an OTC source of H3PO4... but that might be somethign ot look into the future.. if I'm correct you can make HI out of it as well by subsituting the idodide salt, correct?
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[*] posted on 1-7-2006 at 16:05


Sure. Concentrated H3PO4 + NaI or KI gives HI.



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[*] posted on 1-7-2006 at 18:32


I also used NaBr and didn't filter out the NaHSO<sub>4</sub> crystals and I didn't have a problem, lots of HBr<sub>(aq)</sub>.

https://sciencemadness.org/talk/viewthread.php?tid=2169




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[*] posted on 2-7-2006 at 01:09


I think this is the problem.
"Apparently the 40 volume hair developer is buffered all to hell."
I think the first experiment produced HBr but rather dilute. The junk in the H2O2 stopped the reaction that should have detected it.
With the thermometer bulb in the still head we already know what the boling point of the distillate was 100C--->118C--->124C and since it was (mostly) over 100C it's fair to assume it had HBr in it.

H3PO4 attacks glass. I don't know how quickly, but I'd want to find out before I tried using it for this sort of reaction.
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[*] posted on 2-7-2006 at 06:33


Remember that adding large amounts of a solute to a liquid raises its boiling point. For example, I remember from experience that a saturated calcium chloride solution boils at well over 100*C, somewhere in the 120-130*C range.

This could have been the same effect. To shed some light on this, please find the boiling points of the distillate fractions and post them.




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[*] posted on 2-7-2006 at 09:57


I remember it OK, that's why I asked where the thermomter was.
Any sodium sulphate etc present didn't get to the thermometer bulb because it isn't volatile. In every case of a distillation I have seen there are drops of liquid present on the thermometer and continuously dripping back into the boiling pot. That means the thermometer is at the temperature where the liquid and vapour are at equilibrium (other wise the stuff would evaporate or condense until it was cooled or heated to the eqm temperature). That means we already know the boling point of the liquid that distilled over.
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