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franklyn
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separation of NaNO2 from NaCl
I just buy NaNO2 and have not tried separation from meat salt.
I like the method Panache describes since it obviates the need for DMSO
which is the answer for separation, !00 cc dissolves 20 grams NaNO2 @ 25 ºC
Solubility of NaCl in DMSO is just 1/50 th as much ~ 0.4 grams @ 25 ºC
Pour it hot through the mix in filter paper 3 times to assure saturation then cool in
your refrigerator. Filter the crystals that settle ( relatively pure NaNO2 ) and save ,
warm the solvent and pour again through the mixture to remove more nitrite.
Repeat this until the content has been exhausted. Yes it's tedious but results
in a reliable product. The purity obtained depends on what the original impurities
of the meat salt is. Nitrate is inevitably present in miniscule amounts and is as
soluble as the Nitrite in DMSO.
Next dissolve the collected total in water and re-crystalize. This should rid it of
any NaCl residue. Finally dissolve the total collected in Methylol and re-crystalize.
!00 grams Methylol dissolves 4.4 grams NaNO2 at 19.5 ºC
At 25 ºC NaNO2 will dissolve even more, but NaNO3 is dissolved only an
1/11 th as much ~ 0.4 gram
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Jor
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I would add the mixture to isopropyl alcohol acidified with HCl, and leave to stand. A layer of isopropyl nitrite seperates (do not inhale this, it's
highly volatile), seperate, wash with water.
Next hydrolyse this in NaOH-solution and evaporate to dryness, to obtain sodium nitrite.
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ExXOlon
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I try to dissolve Nitrite in DMSO (like franklin) but when i cool the solution Nothing Crystallize ...except DMSO  Think too much solvent. What can i do now ? Distill DMSO under vaccum ? is it the only Solution ? TY
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franklyn
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Sorry for the trouble you're having , there is always more to this than meets the eye.
DMSO becomes solid at 18.5 ºC , it boils at 189 ºC , instead of evaporative distillation ,
slight dilution with a miscible nonpolar organic chlorocarbon reduces NaNO2 solublility
and lowers the freezing point to precipitate the solute. The lower boiling point will also
allow easier fractional distillation to recover the DMSO. Try perchloroethylene
CRC 05089 " Brakleen " parts cleaner at automotive stores, here on the left ~ $ 4.00
http://www.jcwhitney.com/jcwhitney/product/images/large/G_17...
google " brakleen brake parts cleaner "
http://www.crcindustries.com/faxdocs/msds/5089.pdf
DMSO Reaction solvent guide
http://www.gaylordchemical.com/bulletins/Bulletin105B/Bullet...
DMSO Physical properties
http://www.gaylordchemical.com/bulletins/Bulletin101B/Bullet...
How to Dry DMSO
http://www.gaylordchemical.com/bulletins/Bulletin109B/Bullet...
DMSO Solubility data
http://www.gaylordchemical.com/bulletins/Bulletin102B/Bullet...
Solvent miscibility
http://www.erowid.org/archive/rhodium/chemistry/equipment/pi...
Attachment: Solvent MiscibilityTable.pdf (273kB)
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Attachment: Solvent Miscibility table.pdf (327kB)
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franklyn
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Although DMSO is miscible with water, given that Perchloroethylene is not miscible with water
but is miscible with DMSO, the NaNO2 which is very soluble in water might be extracted with
water forming a separate phase. Worth a try with a small test sample.
http://chemicalland21.com/industrialchem/solalc/PERCHLOROETH...
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cnidocyte
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Quote: Originally posted by Lanthanide  |
Now NaNO2 is much more readily soluble than NaCl, and I believe this means that, for example, given 10g of NaNO2 and 10g of NaCl, when adding a
quantity of water this means you will see all of the NaNO2 dissolve before *any* of the NaCL dissolves at all.
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Is that really the case though? I've been wondering this for years. If this is the case then separation of salts would be ridiculously easy, you just
add enough to fully dissolve the more soluble salt. It makes sense that the more soluble salt might force the less soluble one out but at the same
time I don'y know how it works.
| Quote: Originally posted by not_important | Starting assumption is wrong - if the solubility of A is given as 100 g/100 cc H2O, and that of B is given as 200 g/100 C, then 100 CC of water will
dissolve 100 g of A plus 200 g of B unless a common ion effect comes into play. All of the water goes to dissolving both substances,
so more NaCl will remain in solution than you've calculated.
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If this is the case then adding just to dissolve the nitrite will make NaCl the minority. For example NaCl = 359 g/L, NaNO2 = 820g/L. If I add 1L of
water to an excess of the mixture and filter I end up with 820g of NaNO2 and 359g of NaCl. Is that how it works?
[Edited on 17-2-2011 by cnidocyte]
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cnidocyte
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| Quote: Originally posted by Panache | i have done this many times and it is relatively simple, the mixture was a 10%sodium nitrite, 90% sodium chloride. I will give an example for a 1kg
mixture.
Place mixture into a column and run enough water through the column to dissolve all the sodium nitrite at room temp. This is 1000/815=1222ml, then add
10%, so around 1400ml.
You now have a solution of 1400ml water, 100g NaNO2 and around 350g (x1.4)=490g sodium chloride, its solubility in cold water. |
I don't understand your calculations. At room temp the solubility of NaNO2 is 820g/1000mL. If your using 1000g of a 10% mix then you want to know how
much water it takes to dissolve 100g NaNO2 therefore
(1000mL/820g) * 100g = 121.95 mL
is the volume of water required to dissolve 100g NaNO2.
I want to try this now with 6.25% mix but I'm not sure about the chemistry behind it. If I add 100mL of water (a huge excess) to 100g of 6.25% and
filter out the solid will I be left with a solution containing 6.25g (all of it) NaNO2 and 35.9g (the solubility of it in 100mL) of NaCl? If so I can
do this to remove most of the NaCl then repeat this process using 10mL of water and finally recrystallise from a suitable solvent.
[Edited on 17-2-2011 by cnidocyte]
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