Gooferking Science
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ClBr from chlorine replacement?
I posted a video on youtube about making bromine by this method.
NaBr + HCl --> HBr + NaCl
HBr + Ca(ClO) --> CaCl2 + H2O + Br2
Is it possible that some ClBr will be produced during the time when the chlorine takes the bromine's spot in NaBr? I don't think it is but someone
asked me if it was possible in the comments and wanted to be for sure.
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kristofvagyok
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Nope.
HBr is a stronger acid than HCl so you just simply dissolved the NaBr in some hydrochloric acid, only a traces of HBr is generated in that solution.
Why? Ka of HBr is 2.2 x 10^10, while the HCl only has a Ka 1.2 x 10^8, it means that HBr is 200 times stronger acid than HCl.
So how the bromine is generated?
Chlorine is produced from the hydrochloric acid and the hypochorite and this chlorine, what is stronger oxidizer than bromine will "oxidize out" the
bromine from it's salt.
Cl2 + 2NaBr = 2 NaCl and Br2.
This Br2 is NOT pure, so please do not write on youtube that you have produced high purity bromine.
ClBr is also generated, so if you want to make ClBr free, relative pure bromine, than use a high excess of NaBr compared to the generated chlorine.
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Gooferking Science
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Well I distilled it so it is a little higher purity.
But It is as pure as I need it.
I also dried it with sulfuric acid to rid it of any extra water. It is in ampoules.
[Edited on 31-7-2013 by Gooferking Science]
[Edited on 31-7-2013 by Gooferking Science]
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Oscilllator
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kristofvagyok couldn't you also remove the ClBr (as well as any Cl) by refluxing the bromine over a bromide salt?
ClBr + NaBr -> NaCl + Br2
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AndersHoveland
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be careful with that reaction, in dilute solutions bromate can form
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yuno
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Can you explain?
What do you propose for removing ClBr from elemental bromine?
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violet sin
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I think oscilllator meant the dried bromine w/ Cl impurities refluxed over dry bromide salt. Cl portion replaces bound Br and lock NaCl. Not in
solution
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Oscilllator
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Correct violet, that is what I meant. I think woelen suggested it first, in another thread.
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AJKOER
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As a reference, see "Equilibrium and kinetics of bromine chloride hydrolysis" by Liu Q, Margerum DW, Department of Chemistry, Purdue University, West
Lafayette, Indiana 47907-1393, USA. To quote:
"Abstract
Aqueous-phase halogen reactions play an important role in tropospheric ozone depletion that is observed during Arctic sunrise where bromine chloride
is a key intermediate. The temperature dependencies of BrCl(aq) equilibration with BrCl2-, HOBr(aq), Br2(aq), Cl2(aq), HOCl(aq), Br-, and other
species (Br3-, Br2Cl-, Cl3-, OBr-, and OCI-) are determined as a function of Cl- concentration from pH 0 to pH 7. Values for K1
(=[BrCl2-]/([BrCl(aq)][Cl-])) at mu = 1.0 M are 3.8 M(-1) at 25.0 degrees C, 4.7 M(-1) at 10.0 degrees C, and 5.5 M(-1) at 0.0 degrees C, with deltaH1
degrees = -9.9 kJ mol(-1) and deltaS1 degrees = -22 J K(-1) mol(-1). BrCl(aq) hydrolysis equilibria have little or no temperature dependence with Kh1
(=[HOBr(aq)][Cl-][H+]/[BrCl(aq)]) = 1.3 x 10(-4) M2 from 25.0 to 5.0 degrees C, mu = 1.0 M. When conditions are adjusted to give a rapid partial
hydrolysis of BrCl in equilibrium with HOBr and Cl- at p[H+] 4.31, a relatively slow reaction (kobsd = 2.4 s(-1)) to form HOCl and Br- is observed.
This takes place via BrCl reaction with Cl- to form Cl2, which hydrolyzes in the rate-determining step to give HOCl. On the other hand, the rate of
complete BrCl hydrolysis to form HOBr and Cl- at p[H+] 6.4 is extremely rapid with a first-order rate constant of 3.0 x 10(6) s(-1) at 25.0 degrees C.
The reverse reaction between HOBr, Cl-, and H+ has a rate constant of 2.3 x 10(10) M(-2) s(-1), so that in seawater, where [Cl-]/[Br-] = 700, the
formation of BrCl is much faster than the formation of Br2 from HOBr, Br-, and H+. Rapid formation of BrCl(aq) and its subsequent reaction with Br- is
a viable pathway to give Br2(aq). Photolysis of Br2(g) is believed to initiate the reactions associated with ozone depletion."
Link: http://www.ncbi.nlm.nih.gov/pubmed/11347924
Now, with the formation of HOBr and the right conditions (concentration, temperature,..) a disproportionation creating some bromate could follow:
3 HOBr --> 2 HBr + HBrO3
[Edited on 20-8-2013 by AJKOER]
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woelen
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In the context of removing chlorine or BrCl from bromine by reacting with NaBr, the formation of bromate hardly is relevant. I think that you can
state that the loss of bromine to bromate will be less than 0.1%, in other words, you can neglect that reaction completely.
The formation of bromate is an important reaction at high pH, but in neatral or acidic solutions it hardly occurs and it would require contact of the
bromine with water for weeks before a noticeable loss occurs.
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violet sin
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So does the dry reflux of crude BrCl + Br over dry NaBr work? I didn't see a response to that above. Normally I would look it up but I'm stuck with
an iTouch for a couple weeks and the tiny letters kill my eyes. I was wondering if there was enough kinetic energy to push that at reflux? Sorry for
the laziness there. I'm not doing the process or any thing just curious. Thanks
-violet sin-
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woelen
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I am not sure whether this will work. NaBr is a solid and I'm quite sure that it will not dissolve in Br2 (or just a tiny little bit). Normally, you
make a very concentrated solution of NaBr in water and you shake your impure Br2 with this solution. I would take 1 ml of solution of highly
concentrated solution of NaBr or KBr for each 10 ml of bromine and shake this solution with the bromine vigorously to get it well mixed. Next step is
to allow the layers to separate and separate the bromine from the aqueous layer. Losses will be low when only a small amount of solution of NaBr is
used (keep in mind that Br2 dissolves quite well in a solution containing bromide ions). The bromine now contains some water. This can be removed by
shaking it with conc. H2SO4 and separate the layers again. Even better purity is reached by distilling the bromine from the H2SO4/Br2 layers.
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