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Sodium Ferrocyanide

chemkid - 30-12-2007 at 19:38

I think i found an OTC source for sodium ferrocyanide. Yellow prussiate of soda is the traditional name for sodium ferrocyanide. Kosher salt contains sodium ferrocyanide and salt (NaCl) and costs about a 1.50 a box (which is rather large). By dissolving this in denatured ethanol (or if you've got the money straight ethanol) i would think that the insoluble sodium ferrocyanide would precipitate and NaCl dissolve.

This is only speculation so far. I will try it in the morning.

I know sodium ferrocyanide can be used as an iron indicator. How exactly would such a test be performed?

Chemkid

[Edited on 30-12-2007 by chemkid]

pantone159 - 30-12-2007 at 19:58

I'd be kind of surprised if there was enough ferrocyanide to make it worth the trouble, but hey, give it a shot.

If you mix a solution of ferrocyanide with a solution of ferric Fe(III) ion, you get a very vivid, dark blue precipitate, "Prussian Blue". Although this is a solid, it often forms very fine particles that stay suspended in the liquid, and pass through filter paper.

garage chemist - 30-12-2007 at 20:50

Since when does NaCl dissolve in ethanol? You obviously havent tried that out, because it doesnt work.
Also, the amount of ferrocyanide in salt is really very small.

Add some iron(III)salt solution to a solution of some of your salt in water, and if it doesnt instantly become nearly black with prussian blue then there is no concentration worth extracting in there. The prussian blue reaction is very sensitive.

On the other hand, making prussian blue is a good way to precipitate ferrocyanide ions from solution. If there was a worthwhile aount of ferrocyanide in any product, then precipitation as prussian blue would be the way to extract it.

I think trying to get ferrocyanide from salt is a really dumb idea. Some OTC "sources" of chemicals are damn stupid because of low concentration and/or simply ridiculous, like getting red P from matchbook strikers.

evil_lurker - 30-12-2007 at 23:07

The problem is the ferrocyanide is added like .01% as an anti caking agent... its pretty much worthless as a source as there is very little to extract.

not_important - 31-12-2007 at 01:58

Besides that, not all Kosher salt has sodium ferrocyanide added:

http://www.copia.org/content/node/231

pantone159 - 31-12-2007 at 07:41

'Normal' table salt probably has as much ferrocyanide as any other kind. IIRC, this is where the Morton's salt slogan 'If it rains, it pours' comes from: Even if it is very humid, the salt doesn't clump up, due to the anti-caking effect of ferrocyanide.

I once read somewhere that Kosher salt was more pure than regular salt, maybe this is the lack of iodide though.

not_important - 31-12-2007 at 08:43

Not all table salt uses ferrocyanide, various silicates and magnesium carbonate are common too. Drop some salt into enough water to dissolve it, if there is a milkiness then it's likely a silicate anticaking agent was used.

chemkid - 31-12-2007 at 15:43

I tried it out this morning, and obviously it didn't work, because NaCl is very sparingly soluble in ethanol (stupid Internet). But from the affordability of the salt, it wouldn't be that ridiculous to do and if you have a distillation setup you can recover the solvent you use. I used Morton kosher salt because it was uniodized, and had only salt (and impurities of the salt) and sodium ferrocyanide. Just because somethings is there in low concentrations doesn't make its extraction pointless. Try Marie curie and radium, well i guess were not discovering anything new, but so what.

Chemkid

PS I thought extracting red phosphorus from match boxes sounded like great fun.

[Edited on 31-12-2007 by chemkid]

[Edited on 31-12-2007 by chemkid]

[Edited on 31-12-2007 by chemkid]

evil_lurker - 31-12-2007 at 16:09

Quote:

Originally posted by chemkid
I tried it out this morning, and obviously it didn't work, because NaCl is very sparingly soluble in ethanol (stupid Internet). But from the affordability of the salt, it wouldn't be that ridiculous to do and if you have a distillation setup you can recover the solvent you use. I used Morton kosher salt because it was uniodized, and had only salt (and impurities of the salt) and sodium ferrocyanide. Just because somethings is there in low concentrations doesn't make its extraction pointless. Try Marie curie and radium, well i guess were not discovering anything new, but so what.

Chemkid

PS I thought extracting red phosphorus from match boxes sounded like great fun.

Well you can't exactly go buy radium for $30 (or less) a pound either, can you?

MagicJigPipe - 1-1-2008 at 12:33

Red P from matchbook strikers is only viable if (a) you are a tweaker and can stay up for a week scraping them or dissolving in acetone (works better but wastes acetone and recycling is even more time consuming) (b) if you only need small, impure quantities, like a couple of grams.

I have done it before and it is, most certainly, not fun. I think I got like .2g per box (just an estimate, it was a long time ago) which includes all impurities such as silica.

Also, I use Kosher salt for some reactions/washings because it claims to contain only salt and no iodides. It dissolves clear. I hope it doesn't contain ferrocyanides.

[Edited on 1-1-2008 by MagicJigPipe]

[Edited on 1-1-2008 by MagicJigPipe]

chemkid - 1-1-2008 at 14:07

as states before, it depends on you brand.

I was really looking to purify the salt from the sodium ferrocyanide when i thought i discovered a means for removing it. So far i am yet to find a solvent in which NaCl is soluble and Na ferrocyanide is not or vise versa.

Chemkid

garage chemist - 1-1-2008 at 14:17

You can purify NaCl from almost any contaminants by dissolving in water, filtering until you have a clear solution, and gassing with HCl.
The solubility of NaCl decreases strongly with increasing HCl concentration, and the purified salt precipitates. This removes sodium ferrocyanide completely.
This process is used in industry to prepare high-purity NaCl for various uses.

[Edited on 1-1-2008 by garage chemist]

Tinton - 3-1-2008 at 18:29

Perhaps if you mixed an acid strong enought to create HCN, but not strong enought to create HCL, to the ferrocyanide, the cyanide ion could be liberated from kosher salt.
And then bubbled through NaOH for NaCN...

[Edited on 3-1-2008 by Tinton]

chemkid - 3-1-2008 at 18:51

I e-mailed the company and found out what everyone already pointed out .0013 concentration of pottasium ferrocyanide. Should be pure enough for me. Thats morton kosher salt for all who want to know.

@tinton Not so eager to work with hydrogen cyanide

Xenoid - 3-1-2008 at 18:52

@ chemkid

If you are so desperate for sodium ferrocyanide you are prepared to extract it from salt, why don't you try the "old fashioned way".

""When nitrogenous refuse (blood, horns, leather scraps, etc.) is charred, and the black mass is ignited with potash and iron filings, something is formed which passes into solution when the mass is lixivated with water. The aqueous solution, on evaporation gives yellow crystals of potassium ferrocyanide (K4Fe(CN)6.3H2O).""

""...prepared it by melting dried blood with potash salts and treating the aqueous extract of the mass with ferrous sulphate.""

""...animal flesh... and other animal substances could be used instead of blood.""

To make Na ferrocyanide, use sodium carbonate instead of potash, though Na ferrocyanide is less soluble than K ferrocyanide!

I've always wanted to try this but I've never had the guts (literally)...

Could be a bit smelly, if your'e in an urban area!

This could be a fitting end for the family "moggy" if he/she/it gets hit by a car; converted into yellow sodium ferrocyanide crystals and put in a jar on a shelf!
If you require more, find a dead cow or a friendly butcher!

[Edited on 3-1-2008 by Xenoid]

evil_lurker - 3-1-2008 at 19:21

www.chemsavers.com has potassium ferrocyanide on sale, 2 Kgs for $25, shipped ORM-D.

len1 - 4-1-2008 at 00:44

Quote:

Originally posted by garage chemist
You can purify NaCl from almost any contaminants by dissolving in water, filtering until you have a clear solution, and gassing with HCl.
The solubility of NaCl decreases strongly with increasing HCl concentration, and the purified salt precipitates. This removes sodium ferrocyanide completely.
This process is used in industry to prepare high-purity NaCl for various uses.

145 gms NaCl were dissolved in 500ml H2O giving a 5mol/L solution (about saturation). To half of this 80ml of 32% w/vol HCl was added giving a 10% w/vol HCl solution.

No NaCl precipitated out. Any stronger acid and you should start liberating cyanide.

garage chemist - 4-1-2008 at 00:59

I said *gasssing* with HCl, not adding hydrochloric acid. Adding hydrochloric acid introduces water as well.

However, you can do without the gassing if you slowly add the NaCl solution to the HCl and not the other way round.
Also, you need more HCl in proportion to the NaCl solution- try using equal volumes of HCl and NaCl solutions, or 2/1 HCl/NaCl by volume.

len1 - 4-1-2008 at 01:08

Yes I noticed the gassing - that means you want to achieve soubilities of the order of saturation of HCl - that will precipitate out some NaCl.

But thats just the common-garden common ion effect. Theres nothing special about HCl and NaCl and the effect is rather weak, its certainly no way to eliminate NaCl from solution. Just a way to precipitate out a small amount of chloride salt.

PS If theres ferrocyanide in solution the extremely high H+ concentration will be hazardous.

When dealing with concentrations of 1mol/L or above one should really use activity coefficients - thermodynamics becomes non-linear since the ions no longer form a dilute gas. However the effects can be demostrated in the linear approximation.

The solubility of NaCl at 30C is 6mol/L. Therefore

[Na][Cl]=36(mol/l)^2 is its solubility product - the excess precipitates.

The maximum strength of HCl which can be dissolved in solution is 10 mol/L (though in saline solutions it is less - NaCl has the same effect on it).

x(x+10) = 36 is the amount of NaCl which will dissolve in 37% HCl, we get x ~ 3 mol/L predicting this method will precipitate about half the NaCl in solution.

I did a preliminary investigation and it precipitates a bit more than this due to non-linear effects.

Lets repeat that with the experiment I did above. Equal volumes of 6mol/L NaCl and 10mol/L HCl solution are mixed.

The sodium concetration now is ~ 3 mol/L since the volume ~ doubles

The chlorine concentration now is ~ (10+6)/2 = 8 mol/L, where division by 2 since volume ~ doubled

The solubility product is 8*3 = 24 (mol/L)^2

so nothing precipitates as I observed.

This can be repeated with Na2SO4 and H2SO4 etc..

[Edited on 4-1-2008 by len1]

woelen - 4-1-2008 at 04:19

I think there is more to this, when the HCl is highly concentrated. In 15% HCl quite some table salt can be dissolved, but in 37% HCl it is virtually insoluble. Try it yourself. If you want to heat the solution, then keep it in a closed (pressure withstanding) vessel, otherwise you'll drive off much HCl and then the experiment is not honest anymore.

I think that this has to do with the effect of non-ionized HCl in solution, which makes the solution less suitable for purely ionic compounds. This, combined with the already high concentration of chloride probably results in low solubility of NaCl.

A similar effect exists for water, to which some ethanol or propanol is added. The solubility of NaCl rapidly drops with increasing concentration of the alcohol.

len1 - 4-1-2008 at 04:49

Yeah OK, although I dont know why you mention heating - the solubility of NaCl hardly varies with T and for HCl it decreases.

I think that rather than trying to dissolve salt in 37% HCl equilibrium is more quickly reached by adding a small amount of solution of known concentration to a much larger volume of HCl solution, filtering and weighing. (There is no reason for using HCl gas to test this that I can see).

I have done a preliminary test (without using the analytical ballance) and it seems 1/4 salt remained in solution, where my simple theory predicted 1/2.

However if your hypothesis about the change in the ionic nature of high concentration hydrochloric acid is correct, then this is not the common ion effect, and the results should apply to HCl and any other ionic salt. i.e. adding a small amount of saturated solution of any salt to 37% HCl will precipitate that salt. Len

[Edited on 4-1-2008 by len1]

woelen - 4-1-2008 at 07:47

Your last remark is not correct. Indeed, a precipitate can be formed, but this usually is the metal chloride. E.g. when you add NaBr to concentrated HCl, then it also dissolves with difficulty, but most likely because it becomes covered with a crust of NaCl. If you drip in highly concentrated NaBr solution in concentrated HCl, then you get NaCl as solid precipitate.

The same effect is present with KCl and KBr. Both dissolve with great difficulty in concentrated HCl.

Not all metal salts suffer from low solubility in conc. HCl, or in solutions with ethanol or acetone. Some salts dissolve well in less-polar solvents also. An example is CuCl2.2H2O. This dissolves in acetone very well, also in ethanol, and in conc. HCl.

Yet another reason, why things may turn out different is complex formation. If copper sulfate is added to concentrated HCl, then it quickly dissolves, giving H2SO4 and a copper complex anion, CuCl4(2-).

Summarizing, there are many different effects, which may have opposite effects on solubility. What the net outcome will be remains a matter of experimentation.

len1 - 4-1-2008 at 12:55

If you are correct regarding the changed nature of the HCl solution then the key test is that a precipitate, on adding a saturated solution of another salt (obviously excluding coordination compounds) will form. You are not correct to say the salt will definitely be a chloride - it depends on the relative solubilities of the chloride vs whatever other ion you have introduced in product with the relevant concentrations. That is not however what we are interested in here - which is to differentiate what you have mentioned from the common ion effect. Len

chemkid - 4-1-2008 at 15:10

That last post was dismissing any attempt at extracting the ferrocyanide. That's pure enough for me referred to the salt

Just in case there was any misunderstanding there. I'm not crazy or desperate enough to extract such an extremely low concentration.

the idea using animal parts does sound interesting.

Chemkid

len1 - 6-1-2008 at 05:22

A saturated solution of NaCl in 30C H2O was formed by dissolving 35gms NaCl in 100mls. 5mls of this solution were added to 55ml of 34% HCl (giving a final HCl concentration of 31%). A precipitate of NaCl formed immediately, the solution was shaken and allowed to stand. It was subsequently rapidly filtered (to allay the possibility of HCl evaporating during filtration) the filter paper was removed, soaked in conc HCl to remove any adhering precipitate and liquids combined. The solution was evaporated and yielded 1.2 gms of NaCl.

This forms 1.2/1.75 ~ 68% of salt in solution precipitated. This compares with 0.5 from the linear analysis from my two posts above, which is reasonably good agreement.

A conc. solution of KNO3 added to conc HCl in the same way produced no precipitate.

This seems to demonstrate

1) What we are looking at here is the common ion effect
2) Theres nothing special about HCl solutions - in particular the polarity of the solution does not alter at high HCl concentrations
3) The effect is good at precipitating high percentages 60-80% of relatively pure chlorides out of solution - other salts do not precipitate

[Edited on 6-1-2008 by len1]

Nicodem - 6-1-2008 at 08:03

What do you mean with your claim No. 1?
Assuming volume additivity, in the final solution you would have got approximately [Na+] = 0,5 mol/L and [Cl-] = 10,3 mol/L, concentrations which do not exceed your Ksp = 36 (mol/L)^2.
I might have calculated it wrong as I was so upset from your terribly unscientific claim No. 2 that thinking straight was not possible. We are not talking about dilute solutions here! You can not use Ksp equations just like that! You seem to have no clue of how ions behave in solutions. There are no free ions in aq. solutions. All ions are solvated. And since they are solvated the [H2O] does not equal c(H2O). The Ksp calculations, which neglect [H2O] by definition, can only be done for salts of low solubility where the part of water molecules occupied with solvation can be neglected. It is this in addition to the common ion effect that causes oversaturation. (In organic chemistry we call this phenomenon "salting out" and by using it, it is possible to severely decrease the solubility of water soluble compounds. For example, water/acetone mixture will become biphasic when you add some brine – acetone is unable to solvatize Cl- ions, so it rather forms a separate layer instead of interfering with the higher energy process of Cl- solvation by H2O molecules.)

Besides, the recrystallization of crude NaCl out of conc. HCl standard way of purifying it (the solubility of NaCl at room temperature is low, while it is much higher at boiling point). It is also used to grow large NaCl monocrystals.

woelen - 6-1-2008 at 10:46

Quote:

1) What we are looking at here is the common ion effect

You take 5 ml of a concentrated solution of NaCl and mix this with 55 ml of conc. HCl.

Let's do a rough computation, according to the standard theory about common ion effect:
Solubility of NaCl in water is 300 to 350 g/l, which is 5 to 6 mol/l (I do rough calculations, just to get the gist of what I want to say and do not want to go into detail about numbers). Hence, the solubility product of NaCl would be appr. 30 mol2/l2.

Now, if we take 55 ml of conc. HCl, 34%, which is around 11 mol/l.
Now we add 5 ml of your concentrated solution of NaCl, let's say 5.5 mol/l NaCl. The new solution, after thorough mixing, will have appr. 10 mol/l of chloride ion, and appr. 0.5 mol/l of sodium ion (your 5 ml of solution is diluted by mixing it with 55 ml of acid). So, the product of [Na(+)]*[Cl(-)] is appr. 0.5*10 = 5 mol2/l2, which is well below the number of 30 mol2/l2.

So, there really is more to say about this than simple common ion effect. If that were the only effect, then there would be no precipitate of NaCl at all. You have done the measurement of the precipitated NaCl and according to your measurement, only 0.5 grams of NaCl remain in solution at the 60 ml of 31% HCl. Your 0.5 grams of NaCl in 60 ml of liquid is only roughly 0.15 mol/l. With a chloride ion concentration of appr. 10 mol/l, your theory would predict a solubility of around 3 mol/l of NaCl, which is 20 times as much as what one find in a real situation.

[Edited on 6-1-08 by woelen]

len1 - 6-1-2008 at 13:30

1) Woelen, you are right. I made a mistake, the solubility product if one assumes activity coefficients of 1 is well below the 36mol/L^2.

2) Thats still not too bad because activity coefficients in very concentrated solutions can differ substantially from 1. So if the activity coeff is about 16 for 10mol/L Cl- solutions the solubility product holds, using molalities. The coefficient is very large though - suggesting non-linear effects [Cl]^2 etc cant be neglected, its a useful first approx though.

3) My point about this not being a way to eliminate NaCl from solution also still holds - 1/3 of the salt remained in solution using liquid HCl - and ~ 0.14mol/L minimum NaCl remnant using HCl gas.

4) It has been shown that this is the common ion effect. I could get no precipitate with KNO3 is an example I gave. Woelen if you want anyone to believe you, you must be able to precipitate other salts in HCl, that is the consequence of what you are saying. It would then run counter to what garage chemist is claiming, that this is method is used to precipitate pure NaCl from solution. My experiments (irrespective of the numerical error) show that he is right in this - though not as a way of clearing the soln of NaCl.

I emphasise that using activities to 'patch up' the mistake I made in the calculation is not some slight of hand way to maintain what I was saying. That is because my assertion that this is the common ion effect is based entirely on 2) - not being able to get a precipitate with non-chloride salts. The calculation is useful for judging the strength of the effect - a factor of six for the activity means that its actually quite strong, that is the import of the numerical correction.

@Nicodem. With respect to the solubility product you are right. As for the rest, read up on activities. When a person shows hes prepared to cheat using obfuscation as cover I dont argue with them - its a waste of time. If people want to read your posts in this thread each can draw their own conclusions http://www.sciencemadness.org/talk/viewthread.php?tid=9717&a...

[Edited on 7-1-2008 by len1]

woelen - 6-1-2008 at 14:33

You could try to repeat exactly the same experiment as you did with NaCl, but now with 55 ml of 34% HCl and 5 ml of a 6M solution of NaBr instead of 6M NaCl. I'm quite sure that the precipitate will be near 100% of NaCl and only contains a small amount of bromide. I am inclined to think that the same is true for 5ml of 6M NaNO3.

Most likely the solubility of KCl in conc. HCl is somewhat better than the solubility of NaCl in conc. HCl and that is why no precipitate is formed.

I never said that I believe that I can precipitate other salts in conc. HCl. If you have concluded that from my previous writings, then I just want to say sorry, I might have been somewhat unclear

.
But I do believe that ANY Na-salt of sufficiently high solubility does lead to precipitation of NaCl (and not of that salt) under the conditions of relatively small amount of salt, added to large excess of HCl (your 5 vs. 55 ml experiment is a good way of testing that).

I hope to find some time for this, tomorrow or Tuesday. Now it is too late for me, we have a baby crying in the night at least once and then there are tasks with higher priority

(such as getting sufficient sleeping time).

[Edited on 6-1-08 by woelen]

len1 - 6-1-2008 at 14:52

Yeah well, my experiment has shown that Na+ concentrations in excess of about 0.14mol/L in conc. HCl lead to NaCl precipitation, but thats just using standard thermodynamics. So I dont need to repeat the exp. with NaBr, Im pretty sure of the result. BUT If there is some substantial change in the nature of conc. HCl solutions, making the solvent non-polar or whatever, this should give a precipitate with salts such as KNO3. You can not make exceptions willy nilly, else Ocams razor comes into play.

Good luck with the kid - ours have fortunately stopped crying now.

[Edited on 6-1-2008 by len1]

garage chemist - 6-1-2008 at 21:52

I once made some perchloric acid by mixing conc. NaClO4 solution and a large amount of HCl followed by filtration of the NaCl and workup via distillation. It worked well, only a small amount of sodium salt remained as distillation residue.
That your KNO3 solution did not precipitate KCl surprises me a bit.
Perhaps KCl behaves different than NaCl here?

len1 - 6-1-2008 at 23:27

Yes, KNO3 gives no precipitate under similar circumstances.

This is a very interesting phenomenon. For HCl up to about 20% what I got is roughlly in accord with the common ion effect (NaCl gives precipitate NaNO3 doesnt). However above that there is definitely more than just the common ion effect, I have to take back my earlier statement. The activity coefficient grows much faster than expected, so that at 37% the solution holds maximum sodium ion concentrations of the order of fractions of mol/L. With such a small figure you have to say that sodium ion concentrations will be small no matter what the anion it comes with.

I disagree with woelen as to the nature of the effect however. I think the evidence is unequivocal. If there was a drastic change in the nature of the solvent the solubility of salts such as KNO3 would be affected.

Interestingly KCl does precipitate from 32% HCl, but to a smaller extent than the sodium salt. It would appear the nitrate ion, despite its low concentration has an effect in stabilizing the K+ in solutions of KNO3. NaNO3 does produce a precipitate - but less than NaCl. We have a clear picture here. At very high [HCl] almost all water molecules are tied-up to the ions, and so the ion with greatest difference between solvation energy and lattice energy stays in solution. Potassium solvates easier than sodium, while the nitrate anion acts the other way, increasing solubility for both sodium and potassium. Based on this we expect KNO3 to be the most soluble salt of the four combinations.

I have repeated the exp as to the solubility of Na+ ions in 34% HCl, it is > 0.14mol/L as I earlier determined. I have determined this by adding solutions of lower concentration to conc. HCL solutions and observing that it precipitates nothing.

PS There is a PS to this. You can get Na+ concentrations of 0.2 mol/L or less by gassing concentrated solutions of NaCl and perhaps NaBr, NaI. Salt of Na with other anions, such as SO4- or NO3- PO43- will not be precipitated to nearly this degree from conc solutions - the common ion effect if you like.

[Edited on 8-1-2008 by len1]

woelen - 7-1-2008 at 23:45

Another interesting experiment could be to take a precipitate of KCl from conc. HCl and add a few drops of conc. HNO3 and mix. It would be interesting, if indeed the precipitate of KCl redissolves in this case. Of course, the total volume also increases, so one must be sure that the total amount of dissolved KCl is (much) larger than the amount one would expect because of the increase of the volume.