nimgoldman
Hazard to Others
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Registered: 11-6-2018
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Cleaning up sodium bisulfate
I have produced some hydrobromic acid using the following reaction:
1. H2SO4 + NaBr(aq.) -> NaHSO4 + HBr
as a result, I have lots of NaHSO4 left over, which is heavily contaminated with H2SO4 and HBr, with possibly traces of Na2SO4 and bromine.
I am considering recovering NaHSO4 and NaBr since both can be reused later to make more HBr.
My best guess is to dissolve the waste product in water and carefully neutralize with NaOH. This will produce more bisulfate and sodium bromide from
the leftover acids:
2. (unbalanced) NaOH + H2SO4 -> NaHSO4 + H2O
3. (unbalanced) NaOH + HBr -> NaBr + H2O
Then I simply recrystallize the bisulfate from water and filter it out.
The filtrate could be concentrated to precipitate bromide from reaction 3.
However, some Na2SO4 could forming by neutralizing the solution (if too much base is used), leaving me with a mixture of salts (sulfate,
bisulfate, bromide) I am not sure how to separate them efficiently.
The resulting waste will be treated with sodium thiosulfate or metabisulfate (to destroy bromine) and diluted before discarding.
Any ideas for improvements?
[Edited on 18-9-2018 by nimgoldman]
[Edited on 18-9-2018 by nimgoldman]
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macckone
International Hazard
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solubility sequence
NaBr > NaHSO4 > Na2SO4
Sodium sulfate is the least soluble but won't cleanly separate from the sodium bisulfate.
The sodium bromide should remain in solution until last.
BUT
sodium bisulfate is going to become aggressive toward the sodium bromide when you apply heat.
A better method is to make sure you have excess sulfuric acid then heat it to drive off the hydrogen bromide. You can collect the hydrogen bromide in
water then deal with it separately.
That leaves the sodium bisulfate/sulfuric acid mix to deal with.
Dissolve in water then boil down the solution until crystals start to form.
Then allow to cool which will give a good crop of sodium bisulfate as the sulfuric acid was in excess.
Repeat to get a second crop.
If there was too much excess sulfuric acid you will wind up with 80% sulfuric acid with sodium bisulfate crystals that need to be recrystallized.
Recrystallize your crystals until you have to purity you want.
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AJKOER
Radically Dubious
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If a problem relates to any of the products: •Br, •Br + Br- = •Br2- (from photoactive Br2, see https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/200... ), or HOBr,..., I would try adding alkaline H2O2, which creates the radicals
•HO2 and •OH, thereby effectively removing radical species:
•Br2- + H2O2 -> 2Br-+ •HO2 + H+ k=1.0E+05 (Equation reference, see http://binarystore.wiley.com/store/10.1029/2006JD008227/asse... )
•Br2- + OH- -> 2Br- + •OH k=1.1E+10
•Br + OH- -> Br- + •OH
H2O2 + •OH -> H2O + •HO2
Note, H+ + •O2- = •HO2 (pKa = 4.88)
Other reactions employing the above radicals to remove problem •Br, •Br2- radicals:
•Br + •HO2 -> Br- + H+ + O2
•Br + •O2- -> Br- + O2
•Br2- + •HO2 -> 2Br- + H+ + O2 k=4.4E+09
•Br2- + •O2- -> 2Br- + O2 k=1.7E+08
HOBr + •HO2 -> Br- + H2O + O2 k=1.0E+09
HOBr + •O2- -> Br- + OH- + O2 k=3.5E+09
[Edited on 29-9-2018 by AJKOER]
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