Sciencemadness Discussion Board - Separation of sodium nitrite from sodium chloride

Separation of sodium nitrite from sodium chloride

McLovin382 - 23-1-2009 at 16:07

Is there a decent way to separate sodium nitrite from sodium chloride? Both of them are relatively soluble in water and thus would be hard (at least it seems that way) to separate this way.

Does anybody have a good way of separating the two?

McLovin382 - 23-1-2009 at 18:15

:-/ it's cheaper to get it as this nitrite/chloride this way than to order it pure, which requires me to both have a credit card and for me to pay for s&h...there's got to be a way to do this, I just can't think of any particular good way.

Acidifying wouldn't work because doing so would form nitrous acid which would of course decompose into nitrogen dioxide. Guess itd be possible to react with an alcohol in acidic conditions to form an alkyl nitrite and a chloroalkane, distill off the chloroalkane, then react the alkyl nitrite with a strong base...but ick - isn't there a way just to separate the two without doing such a procedure?

I noticed that the nitrite's melting point is fairly low, at 270C so maybe it's possible to separate using this but I'd imagine heating the nitrite salt too much would cause decomposition etc..

I tried the search engine, couldn't find any threads on this.

garage chemist - 23-1-2009 at 21:16

Separation via the alkyl nitrite is most likely your best bet.
There will be no chloroalkane formed, since HCl doesn't react with simple alcohols (except tertiary ones) in the cold. HNO2, on the other hand, reacts very readily with alcohols in the cold. Besides that, you can simply use only the amount of acid (HCl) necessary to turn the NaNO2 into HNO2 and leave the bulk of the NaCl unaffected.
Use IPA, the isopropyl nitrite is liquid. Good mixing and cooling is important.

Is your NaCl/NaNO2 mix pickling salt for meat? If so, the percentage of NaNO2 in it is very low (under 1%) and separation via alkyl nitrite will likely give bad yields because of this.
Can't you just synthesize NaNO2 by reduction of NaNO3 followed by purification via alkyl nitrite?

Sedit - 23-1-2009 at 21:45

Iv been kind of on the same line here as mclovin but never really cared to much
I have the drain opener NaOH/sodium nitrate/sodium carbonate mixture that i have been stock piling(not willingly) since i found out that Red devil lye dont exist any more and I have no use for the shit.

Now Mclovin speaks of acidifying but what I was thinking was making a setup to lead the N2O into a Basic solution forming the nitrite salt.

Would this help me and Mclovin at the same time?

Also im a noob for nitric acid here but would feeding this into h2o with absorbtion tower form nitric acid?

help im stupid

Panache - 23-1-2009 at 22:06

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.
Boil the solution down until the sodium nitrite has reached saturation at 100C, which would be around 100mL, but practically it will be closer to 200ml. At this stage you will have an enormous volume of sodium chloride suspended in the sodium chloride/sodium nitrite/water solution. Hot filter very quickly and at temperature.
Your solution now has ~200ml h20, 100g sodium nitrite and ~50g sodium chloride.
Chill to ~2C, the sodium chloride largely remains in solution the sodium nitrite will crystallize, this filtered gives you about 75g of your original sodium nitrite which can be re-crystallized from Methanol/water to give quite pure sodium nitrite.
The remaining solution should be kept and accumulated and re-concentrated.
When i did this some years back i used 10Kg salt at a time in a long column and was all very manageable. I was rather pleased something i figured out worked reasonably well as designed for once. Exploited is the fact the sodium chlorides solubility remains largely consistent over 100C solution temperature whereas the nitrites solubility doubles.
Good luck!

edit--i just found my notes and the curing salt was actually 12.5% nitrite, 87.5% choride, but the concept remains the same.

[Edited on 23-1-2009 by Panache]

McLovin382 - 23-1-2009 at 23:37

Ah thank you all for the intelligent posts

Yeah for some unknown reason on the alkyl nitrite method I was thinking to use sulfuric acid, which obviously HCl would be a much better choice for that

And thanks for noting the use of IPA rather than another alcohol. That sounds much easier to work with.

And yes indeed what I found is used for meat curing - it contains 6.25% sodium nitrite though so it seems a viable (and notably cheap) source of the salt

Guess nitrate reduction is a viable option as well, if I have nitric acid handy which currently I don't.

Panache that is a wonderful post - thanks for your insight, I'll definitely give that a try! It will serve fine for my purposes which are purely experimental and don't require super purity of the nitrite anyways, although both your method and garage chemist's method seem to be both good ways of going about this!

(Ftw do you happen to remember what curing salt you found that contains 12.5% nitrite?)

JohnWW - 24-1-2009 at 00:25

Nitrite used as a preservative in meats is suspected of causing cancer, particularly colon cancer, due to its formation of nitrosamines by reaction with the -NH2 groups in proteins.

not_important - 24-1-2009 at 00:58

Some suggestions for the hot filtering, for those without funnels with build in heating/cooling support.

Use a funnel with a wide stem, even a powder funnel. This reduces the chances of clogging.

Put some water into the receiving flask, put the funnel and filter paper into place, and heat the receiving flask enough that steam escapes through the funnel, but not very quickly. Allow the steam to escape until the funnel has warmed enough that steam doesn't seem to be condensing on it.

Even better is making a small hot water bath out of a can or small pot, adding a hole in the bottom for the funnel stem. Ordinary cork can be used as a stopper there. A small immersible heat, such as for making a cup of boiling water, can be used as a heat, with a lamp dimmer to control the heat level/boiling rate.

Panache - 2-2-2009 at 05:13

Quote:

Originally posted by not_important
Some suggestions for the hot filtering, for those without funnels with build in heating/cooling support.

Use a funnel with a wide stem, even a powder funnel. This reduces the chances of clogging.

Put some water into the receiving flask, put the funnel and filter paper into place, and heat the receiving flask enough that steam escapes through the funnel, but not very quickly. Allow the steam to escape until the funnel has warmed enough that steam doesn't seem to be condensing on it.

Even better is making a small hot water bath out of a can or small pot, adding a hole in the bottom for the funnel stem. Ordinary cork can be used as a stopper there. A small immersible heat, such as for making a cup of boiling water, can be used as a heat, with a lamp dimmer to control the heat level/boiling rate.

NI hit the nail on the head as far identifying the difficult part of the process. To be honest i was a little lazy in the write up and should have informed that the concentration was performed in two steps, making the hot filtration less critical. Explaining this succinctly was going to be difficult and anyone performing it would likely have quickly realized the limitations. However i should not have assumed this. By performing it in two stages i refer to step where the volume of water in reduced by boiling away and precipitating sodium chloride. Instead of taking the volume of water right down to it theoretical solubility limit with respect to the nitrite (or close), you stop well short of it and filter some sodium chloride off, then continue to concentrate etc etc.

I also remember now that i lifted the crux of the idea from a patent concerned with concentrating nitrite from brine waste streams in food manufacturing. So it wasn't entirely my idea as i may have indicated (and honestly thought it was, nice thing human memory, always trying to make you seem better than you were)

Separation of Sodium Nitrite from water.

littlepop - 27-5-2009 at 03:14

I have a different problem. I bought a partial drum of Sodium Nitrate of ebay (cheap). I picked it up myself so shipping was no problem. It ended up being a sealed (full) 30 gal. drum and almost 200 pounds. I got three drums of other stuff while I was there and was so busy I didn't check until I got home.
When I looked it was Sodium Nitrite and in solution, not crystals as I expected.
I got some ideas how to get it out of solution from. If I cool the solution down some of it should crystalize and I can filter it, right?
Also, does anyone have a recipe for hotdogs? Or, what else can I do with the Sodium Nitrite?

hissingnoise - 27-5-2009 at 04:58

Quote: Originally posted by littlepop

Also, does anyone have a recipe for hotdogs? Or, what else can I do with the Sodium Nitrite?

First off, nitrites in foods are potential (or actual) carcinogens, as JohnWW has already pointed out!
Filter your solution and evaporate to dryness (30 gal. requires a bit of work?) and recrystallise from a minimum of hot water.
The initial precipitate will have the highest purity and these crystals can be removed as evaporation proceeds.
The final crop will have the lowest purity and be most in need of recrystallisation.
Losses should be small anyway, and purity reasonable. . .
DDNP, a useful primary, requires nitrous acid for its synthesis and some members here might be interested.
Using it to generate NO is a daft idea, IMO, so I'd forget that!
Oh, and welcome to SciMad. . .

[Edited on 27-5-2009 by hissingnoise]

littlepop - 27-5-2009 at 06:53

Alright hissingnoise, everyone knows hotdogs are hazardous to your health regardless of sodium nitrite content.

Won't sodium nitrite change to sodium nitrate on exposure to air? That's where I wanted to be in the beginning but whatever I have I want to be pure.

Where do I get picramic acid?

I think NO is a laughable idea.

Thanks for the welcome.

hissingnoise - 27-5-2009 at 10:03

Quote: Originally posted by littlepop

Won't sodium nitrite change to sodium nitrate on exposure to air?

Yes it will littlepop, but at too slow a rate to be of use; complicated by the fact that it's also very hygroscopic---practically deliquescent in humid air. . .
IIRC, there's a thread on picramic acid---UTFSE!
Picramic acid, BTW, is thought to be carcinogenic and mutagenic!
The synthesis of DDNP and tetracene is in COPAE in the forum library.
Nitrites are harder to get than nitrates and you probably have several lifetimes supply. . .

littlepop - 27-5-2009 at 23:56

Yes, I do have several lifetime's supply. I just bought it because it was reasonable. Is there a way to make some of this available to others. I'm new here and don't know the protocol.
Also, what kind of containers can I transfer portions to. It's in a 30 gal. steel drum (weighing 192 pounds).
And, what about shipping, can this be mailed. I live in South Jersey.
Should I have started a new thread on this, or is it too late to ask that now?

littlepop - 27-5-2009 at 23:58

Thanks for the heads up on the health risks of the picramic acid. I'll be sure not to put any of that in my hotdogs.

hissingnoise - 2-6-2009 at 11:33

Hotdogs?
Oh I get it littlepop---great euphemism!

littlepop - 6-6-2009 at 15:47

Okay, after preliminary testing, the solution I have is about 5% sodium nitrite. I have about a half a cup of crystals. But, I have one more issue now. Will caramel coloring affect the chemical properties of the Sodium Nitrite? Apparently this 20 gallon drum of 5% sodium nitrite was intended for food processing. It has a beautiful caramel color to it. A logical follow on question, is there an easy way to extract the coloring?

Lanthanide - 8-7-2009 at 18:22

This is a great thread, the only place I've found on the internet discussing separating meat cures like prague powder #1 to get the constituent sodium nitrite out.

Most cures at 6.25% nitrite, however I have found a few mentions of double-strength cures that contain 12.5% around. I found this site that mentions 12.5% kuritkwik, which has a link to here where you can buy it online from a store in Australia (in the cures - Ham & Beef section). They probably only ship to Oz though, and I'm on the wrong side of the Tasman for that :/

Panache's method looks great, however I decided to investigate the maths behind it and I believe he may have a few mistakes, firstly in saying to use 1222mls to dissolve the nitrite (only 122 mls should be required), and secondly saying that the nitrite will crysallise out when chilling the 200ml solution to 2ºC - I believe only the sodium chloride would precipitate out of this solution.

All working is done using a theoretical 10% NaNO2 mix as per the original post. I will include an appendix with values for 6.25% mixes at the bottom of the post. The following solubility values are used, from Wikipedia:
0ºC: 71.2g/100g NaNO2, 35.7g/100g NaCl
20ºC: 80.8g/100g NaNO2, 35.9g/100g NaCl
100ºC: 160g/100g NaNO2, 39.2g/100g NaCl

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. So in this case you might see say 10g of NaNO2 dissolved and 4g of NaCl dissolved, but you would *not* see 7g of NaNO2 and 7g of NaCl dissolved. If this assertion is incorrect, then the rest of my working in this post is probably also wrong as a result.

What this means, however, is that 1222ml, topped up to 1400ml, is far more water than is required to guarantee that all 100g of NaNO2 has dissolved out of the 1kg of the original mix. As there is 100g of NaNO2 present, this amount can be dissolved to super-saturation in just 124ml (100/0.808) of 20ºC water. Obviously it is mechanically/physically difficult to pass 1kg of powder through just 124ml of water and get all of the NaNO2 to dissolve, so using a volume of 1L is simply easier. Being able to do this with 1L or less or water (700-800ml might also be feasible) means that the later boiling step down to a solution of 200mls is much faster/easier as you don't have to boil quite as much water.

Proceeding with using 1L of water to as the solute for the powder, with the 100g NaNO2 dissolving in 124ml, this leaves 876ml of water leftover to dissolve as much NaCl as possible, which works out to 314.5g (876*0.359). From our 1kg of mix, we have dissolved 100g of NaNO2 and 314.5g of NaCl, meaning that the remaining 585.5g of NaCl will sit on the bottom of the beaker as an undissolvable precipitate. Filtering this off gives us 1000ml of solution that weighs a total of 1414.5g (the dissolved NaNO2, NaCl and 1000ml of water). This gives us concentrations of:
NaNO2: 7.07%
NaCl: 22.24%
H2O: 70.69%

Now with the precipitated NaCl removed, if we heat this solution up to 100ºC and start evaporating/boiling off the water, we essentially reduce the amount of water present, resulting in a concentration of the remaining compounds. At 100ºC the solubility of NaNO2 has increased dramatically (a smidge under double), while NaCl has only changed a little. This gives us a scenario where the 100g of NaNO2 can be dissolved in just 62.5ml of water (100/160), leaving 437.50mls to dissolve 171.5g of NaCl (437.5*0.392). As we now have only 171.5g of NaCl dissolved, it means 143g from our original 1L solution has now precipated out onto the bottom of the beaker. The concentrations are now:
NaNO2: 12.96%
NaCl: 22.23%
H2O: 64.81%

Continuing to boil down to 200ml at 100ºC, and filtering off all precipated NaCl, we now have a situation where 62.5ml of water dissolve 100g of NaNO2 and the remaining 137.5ml dissolves 53.9g of NaCl. This is similar to Panache's estimate of 100g NaNO2 / 50g NaCl in the 200ml solution. The concentrations are:
NaNO2: 28.26%
NaCl: 15.23%
H2O: 56.51%

If we chill this solution down to 0ºC (don't have solubility values for 2ºC but they'd be largly the same), we can still dissolve 100g of the NaNO2 into the solution of 200mls, it just requires 140.5ml (100/0.712) to do so, leaving 59.5ml to dissolve 21.26g of NaCl. As we now have only 21.26g of NaCl dissolved, it means cooling from 100ºC to 0ºC has resulted in 32.64g of NaCl precipitating out, but all of the NaNO2 is still dissolved. This is in contradiction to what Panache stated when he said the NaNO2 would crystallise out at this point. The concentrations are now:
NaNO2: 31.13%
NaCl: 6.62%
H20: 62.25%

I believe that in order to precipitate NaNO2 out of the solution from 100ºC cooling it to 0ºC you would need to have a super-saturated solution of NaNO2, which means boiling/evaporating down to just 62.5ml of water at 100ºC. This would result in 88.8g of NaNO2 precipitating out (160-71.2) per 100ml of water, but because we had only 62.5ml of water we would recover 55.5g of our original 100g of NaNO2. Of course if you had a pure super-saturated solution, simply evaporating/boiling all of the water away would give you theoretically all of your original 100g back.

Alternatively you could boil down to 140.5ml at 100ºC and then chill that down to 0ºC, which should result in all of the NaCL precipitating out and being left with a super-saturated solution of just NaNO2 and water, at 41.5% concentration (100/240.5).

Can anyone spot anything wrong with my calculations and assumptions here? Am I in fact wrong when I assume that you will achieve 100% dissolution of NaNO2 before any of the NaCl were to dissolve?

Appendix: Values for 6.25% NaNO2 mixture:
Original 1L solution: 62.5g NaNO2 4.48%, 331g NaCl 23.77%
200ml at 100ºC: 62.5g NaNO2 19.20%, 63g NaCl 19.38%
200ml at 0ºC: 62.5g NaNO2 20.66%, 40g NaCl, 13.24%

Reducing temp from 100ºC to 0ºC of 200ml will precipitate 23g of NaCl and 0g of NaNO2.

not_important - 8-7-2009 at 20:14

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.

Lanthanide - 8-7-2009 at 23:58

A common ion effect should come into play, as both of these compounds have sodium in them.

NaCl and NaNO2, both have 1 sodium atom (ion?) each.

I'd also like to point out, that assuming my above maths/procedure is correct, there is no need to perform hot-filtering at any stage (unless you boil down to between 140.5ml and 62.5ml), because the NaNO2 will always be preferentially dissolved to the NaCl, even as the temperature of the solution drops.

[Edited on 9-7-2009 by Lanthanide]

Lanthanide - 9-7-2009 at 21:35

Thinking some more about the common ion effect, if I understand it correctly, shouldn't this mean that NaCl will not dissolve in a solution that has an appropriate level of Cl ions, while the remaining NaNO2 will dissolve, because it does not have a Cl ion?

The obvious candidate here is HCl, which is easily obtainable.

I worked out the Ksp of NaCl to be 37.74:
359g/1L at 20ºC
6.143 mol/L
Ksp = [Na+][Cl-]
Ksp = 6.143^2 = 37.74

So in order to prevent any NaCl from dissolving in the solution, the solution needs to contain a Cl- ion concentration of 6.143 mol/L or higher. HCl has a molar mass of 36.46g and completely dissociates in water, so to get 6.143 mol/L of Cl ions from HCl I would need 224g/L of HCl, or a solution at 18.3% strength. I can buy 16.6% solution from a hardware store so this is very close to meeting my needs off-the-shelf.

Given a hydrochloric acid solution at that strength, I should then be able to pour the NaCl/NaNO2 mixture through it, in which only the NaNO2 will dissolve and all of the NaCl will settle as precipitate on the bottom of the beaker.

Is there anything faulty with my reasoning? Surely it can't be this easy?

How would you then go about extracting the pure NaNO2 from this solution? Evaporating/boiling off all the solution has to result in something happening to the Cl- ions present? Would the Na+ ions from the NaNO2 combine with the Cl- from the HCl?

I always hated doing this solubility stuff in chemistry at high school and uni, however now that I have a particular use for it myself it is quite interesting.

UnintentionalChaos - 9-7-2009 at 21:42

NaNO2 plus acid equals clouds of NO2 and NO. You need a neutral source of Cl-

Lanthanide - 9-7-2009 at 22:35

Oh duh, wasn't thinking about that. I knew it couldn't be that simple, thanks.

Are there any readily available bulk sources of Cl- ions other than NaCl? KCl perhaps?

UnintentionalChaos - 10-7-2009 at 06:07

You will never make all of the NaCl precipitate and KCl would be even more problematic to remove than NaCl. This completely defeats the purpose of this exercise in futility.

The fastest way to effect the seperation is to convert the nitrite to an ester and distill it off, followed by hydrolysis in NaOH or KOH solution with a very slight excess of nitrite ester.

The NaCl will remain at least somewhat soluble despite your greatest efforts. There is a limit to how many moles of reagents you can pack into a liter of solution.

[Edited on 7-10-09 by UnintentionalChaos]

Lanthanide - 10-7-2009 at 15:00

Yeah, I realized later after I posted it that substituting the KCl for the NaCl doesn't really get me anywhere.

So the best way to do it looks to be the simple dilution in water. So what I want to know is whether my points here are correct - that you don't need 1222ml to dissolve the NaNO2, really just 122ml will do (but higher volume is physically/mechanically easier), and that cooling the 200ml solution down to 0C will actually only precipitate NaCl and not NaNO2 as Panache indicated.

I don't actually really need the NaNO2 in solid/pure form itself, just a solution at 35% concentration will do for my purposes (even with remaining NaCl in solution), but given that my source is only 6.25% and I need about 210g, increasing the yeild by even a few grams per kg of mixture is of interest to me.

[Edited on 10-7-2009 by Lanthanide]

separation of NaNO2 from NaCl

franklyn - 5-12-2009 at 19:21

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

.

Jor - 12-12-2009 at 14:50

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.

ExXOlon - 15-12-2009 at 17:37

I try to dissolve Nitrite in DMSO (like franklin) but when i cool the solution Nothing Crystallize ...except DMSO :mad:

Think too much solvent. What can i do now ? Distill DMSO under vaccum ? is it the only Solution ? TY

franklyn - 18-12-2009 at 09:12

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)
This file has been downloaded 974 times

Attachment: Solvent Miscibility table.pdf (327kB)
This file has been downloaded 1074 times

franklyn - 20-12-2009 at 06:00

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...

.

cnidocyte - 17-2-2011 at 05:12

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.

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.

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]

cnidocyte - 17-2-2011 at 15:12

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]