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CHRIS25
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[*] posted on 21-4-2014 at 12:11
Ferric chloride


I am really not sure what to do here because I am a concerned about getting Fe(OH)Cl in the mix and I simply do not want to weaken the concentration too much with constantly having to add weak HCl to prevent the possibility of hydrolysis. I have a flask where half the solution has crystalized between the ferrous and the ferric stage, as expected, and the other half is liquid ferric chloride. I was going to pour out the liquid part and simmer the liquid for a while and then cool it quickly to get the solid. Then add water to dissolve the whole lot so that it all oxidizes into a ferric chloride solution. I never have issues like this with many other solutions that I make it is just that ferric chloride is so difficult to get right and have failed so many times before. I am fed up with messing it up.

[Edited on 21-4-2014 by CHRIS25]




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HgDinis25
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[*] posted on 21-4-2014 at 12:20


Can you give more insight on how you prepaeted the solution? How do you have ferrous contamination?
Anyway, if I understood correctly, you can add Hydrogen Peroxide (3% drug store is good enough) wich will quickly oxidise the ferrous to ferric chloride (in the presence of some HCl). About the hydrolisis, it just simply happens.
AFAIK, you can't solvatate Ferric CHloride. Instead, it gives the following equilibrium:
FeCl3(aq) + 2H2O(l) <--> Fe(OH)Cl(aq) + H3O(aq)
Actualy, that's the reason Ferric chloride solutions are acidic.
Hope it helps in something, but please provide more information.
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[*] posted on 21-4-2014 at 13:06


yep ok, thanks I understand this. And no, no contamination. The beaker has been boiled down successfully, upon cooling, ferrous crystals dropped out of solution which is expected. This will eventually become ferric chloride but half the solution is still liquid, as expected.(no H2O2 will be added it weakens the solution and then you have all the guesswork of boiling down again and I am left wit the same problem as now); I do not want to boil this down any further at the risk of hydrolysis. I want to maintain a fairly concentrated solution, so do I add a weak solution of HCl, say 6M, until all the precipitated matter has dissolved and call it a day, because all I want is a concentrated solution, no solids. Is this the best way to go? Or should one at this point add only water?

[Edited on 21-4-2014 by CHRIS25]




‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)

Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)

The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by precision and law. (me)
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[*] posted on 21-4-2014 at 13:17


Hydrolysis always happen. Hum, If Ferric is all you want you could bubble air directly into the solution.
I think however, that you're missing something: hydrolysys is completrly reversible. Just remove the water and you'll drive the equilibrium to the left, making solid Ferric Chloride. You could then weight and make a solution with the desirable concentration.
Anyway, watch this nurdrage video (it will help you a lot):
http://www.youtube.com/watch?v=43Xsh9J7S-g
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CHRIS25
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[*] posted on 21-4-2014 at 13:36


I appreciate your replies, thankyou. I did not think that hydrolysis was reversible with Ferric hydroxide and the resulting ferric complexes that you can have if you go too far. In fact I have read many times about this on this forum. I made this mistake twice before you see and ended up with a mess.



‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)

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[*] posted on 21-4-2014 at 13:45


The video explains it quite well. And there is no Ferric Hydroxide formation in a Ferric Chloride Solution.
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[*] posted on 22-4-2014 at 08:34


Quote: Originally posted by HgDinis25  
Hydrolysis always happen. Hum, If Ferric is all you want you could bubble air directly into the solution.
I think however, that you're missing something: hydrolysys is completrly reversible. Just remove the water and you'll drive the equilibrium to the left, making solid Ferric Chloride. You could then weight and make a solution with the desirable concentration.
Anyway, watch this nurdrage video (it will help you a lot):
http://www.youtube.com/watch?v=43Xsh9J7S-g


No, hydrolysis doesn't always happen: in acid conditions it is minimised:

[Fe(H<sub>2</sub>O)<sub>6</sub>]<sup>3+</sup> + H<sub>2</sub>O < == > [FeOH(H<sub>2</sub>O)<sub>5</sub>]<sup>2+</sup> + H<sub>3</sub>O<sup>+</sup>

... is an equilibrium that is pushed to the left by addition of acid.

Ferrous salts aren't very prone to hydrolysis but ferric salts can drop out of a solution as Fe(OH)3 if you don't keep the acid reserve of the solution high enough.

In acid conditions, oxidation of Fe(II) to Fe(III) by air oxygen is really slow, impractically so in fact. But freshly precipitated Fe(OH)2 (in alkaline conditions) oxidises quickly in air, more so on heating.

NR's video is correct and easy to practically follow. Nitric acid can be used instead of hydrogen peroxide but most prefer H2O2.

Quote: Originally posted by HgDinis25  
And there is no Ferric Hydroxide formation in a Ferric Chloride Solution.


A pure ferric chloride solution doesn't even exist: all but the most dilute FeCl3 solutions would immediately produce Fe(OH)3, due to hydrolysis. So you need to add extra acid to keep pH low (3 - 4, depending on FeCl3 concentration). You can work out the acid reserve in NR's FeCl3 solution from the stoichiometry, if you like...

Quote: Originally posted by CHRIS25  
I appreciate your replies, thankyou. I did not think that hydrolysis was reversible with Ferric hydroxide and the resulting ferric complexes that you can have if you go too far. In fact I have read many times about this on this forum. I made this mistake twice before you see and ended up with a mess.


Unless there is an actual complexing agent present (like oxalates) no ferric complexes can form, only ferric hydroxide. Usually, if you don't leave the ferric hydroxide to age for too long, that is reversible: add extra acid and heat gently and the freshly precipitated Fe(OH)3 will re-enter solution.

Simply put: Fe(OH)3 + 3 HCl === > FeCl3 + 3 H2O

[Edited on 22-4-2014 by blogfast25]




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[*] posted on 22-4-2014 at 08:58


Got a 4M ferric chloride solution, hardly needed any H2O2 (about 20mLs of 6%)which I added only to the ferrous crystals then mixed them in with the already mostly oxidized ferric solution. It is a very deep black yellow, could be brown, but dip a spatula in and you have a very bright yellow. So I think this is perfect. But am interested in finding out how to calculate the free acid in this solution.



‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)

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[*] posted on 22-4-2014 at 11:04


I've used ferric chloride many times for PCBs.

This thread has inspired me to actually make some ! Thanks !
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[*] posted on 22-4-2014 at 11:51


Quote: Originally posted by CHRIS25  
But am interested in finding out how to calculate the free acid in this solution.


Tell me the exact amounts of everything used, including peroxide and any extra acid you may have added and I'll show you.

You say you added the peroxide 'to the ferrous crystals' but that's very confusing language. Did you mean 'to the solution containing ferrous chloride'?

[Edited on 22-4-2014 by blogfast25]




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[*] posted on 22-4-2014 at 12:09


blogfast25, despite the acid amount used, there will always be hydrolysis. Of course, even a low amount of H3O ionds added will significantly decrease the process.

CHRIS25, did you filter the crystals of Ferrous Chloride and then added the peroxide to make a seperated solution?
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[*] posted on 22-4-2014 at 12:27


Quote: Originally posted by HgDinis25  
Just remove the water and you'll drive the equilibrium to the left, making solid Ferric Chloride. You could then weight and make a solution with the desirable concentration.


Have you ever tried that? I have: it's very hard to do because the hexahydrate of FeCl3 is extremely water soluble and doesn't crystallise out easily. I'm fairly sure industrially vacuum is used to remove the last bit of solvent. A waxy yellow mass is obtained.




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[*] posted on 22-4-2014 at 12:37


Quote: Originally posted by blogfast25  
Quote: Originally posted by HgDinis25  
Just remove the water and you'll drive the equilibrium to the left, making solid Ferric Chloride. You could then weight and make a solution with the desirable concentration.


Have you ever tried that? I have: it's very hard to do because the hexahydrate of FeCl3 is extremely water soluble and doesn't crystallise out easily. I'm fairly sure industrially vacuum is used to remove the last bit of solvent. A waxy yellow mass is obtained.


My mistake, it was scientifical incorrect. I was talking basing myself that the reader would watch nurdrage's video and assume I was talking about the hexahydrated crystals. Sorry for the confusion. I've done it and the worst problem I got into was oxychloride production.

BTW, AFAIK Thionyl Chloride is used to prepare the anhydrous from the hydrate (like in many other cases of salts where heat doesn't lead to anhydrous forming).
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[*] posted on 22-4-2014 at 13:22


HgDinis25: I filtered the ferrous chloride crystals, added peroxide then mixed it back in with the ferrous so as to maintain my 0.5mole solution. Also I saw that video last year and but it did not answer any of my questions and is only a quick demo really. Don't get me wrong, I like nurdrage and he has helped me understand quite some things, but the ferrous and ferric chloride crystalization, and getting specific molarities is not something that is answered by the video. And for example out of interest I could find nowhere guides or opinions on roughly how much excess acid should be kept in reserve in this sort of synthesis. I just guess this one.


Blogfast: Tell me the exact amounts of everything used, including peroxide and any extra acid you may have added and I'll show you

28g lab grade iron filings + 165mLs 11.6M (SG 1.18) HCl + 100mLs water
boiled down gently to about 70mLs
ferrous chloride dropped out of solution whilst the beaker was still half full with liquid, stopped heating, filtered. Collected the brown yellow liquid and separated the solids;
Added 30mLs 6% H2O2 to ferrous crystals solid; added nothing to the liquid; the ferrous crystals solid turned a lovely bright reddish.
Mixed this oxidized mass back in with the rest of the liquid so as to have a saturated as possible solution with no precipitation.
The deep brown/black yellowish solution is now at 125mLs
This is 4M




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[*] posted on 23-4-2014 at 04:34


C25:

Stoichiometrically what happens is:

Fe + 2 HCl === > FeCl2 + H2

FeCl2 + ½ H2O2 + HCl === > FeCl3 + H2O

Or added up:

Fe + 3 HCl + ½ H2O2 === > FeCl3 + H2 + H2O

So effectively 3 mol HCl are used per mol Fe. So that from the total amount of HCl used and the amount of Fe, you can easily calculate how much HCl is left.

Unfortunately, you also lost an unknown quantity of HCl, due to:

Quote: Originally posted by CHRIS25  

boiled down gently to about 70mLs


That makes any calculation of residual HCl very inaccurate because we don’t know how much HCl you blew off when boiling down.




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[*] posted on 23-4-2014 at 05:23


Ah yes, I realize this, I lost some HCl. But I suppose I can at the least make a rough estimate from what you have said. The original required only 1 mole HCl, since I used a total of 1.9 moles, it's just a matter of discovering how much HCl can be lost at a constant temperature of 80 - 90 degrees c in one hour. If there is such a formula for this?



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[*] posted on 23-4-2014 at 08:39


Quote: Originally posted by CHRIS25  
If there is such a formula for this?


Kind of but it isn't really worth doing, IMHO.

One can reasonably assume what you boiled off was so-called azeotropic HCl (about 20 w%). If you knew the initial volume (before boiling in) and the final volume (after boiling in), then the difference would be roughly the amount of ml of azeotropic HCl that got 'lost'. Factoring that into your mass balance for HCl would then tell you how much HCl should be left in your final product. Basically: final moles HCl = initial moles of HCl - 3 x moles of Fe - moles HCl lost by boiling in.

But that's quite a lot of assumptions and inaccuracies to be of much use. It could give a rough idea...

Going by your initial recipe:

Quote: Originally posted by CHRIS25  

28g lab grade iron filings + 165mLs 11.6M (SG 1.18) HCl + 100mLs water
boiled down gently to about 70mLs
ferrous chloride dropped out of solution whilst the beaker was still half full with liquid, stopped heating, filtered. Collected the brown yellow liquid and separated the solids;
Added 30mLs 6% H2O2 to ferrous crystals solid; added nothing to the liquid; the ferrous crystals solid turned a lovely bright reddish.
Mixed this oxidized mass back in with the rest of the liquid so as to have a saturated as possible solution with no precipitation.
The deep brown/black yellowish solution is now at 125mLs
This is 4M


... you had roughly 0.9 moles of HCl 'spare', after the dissolution of the iron. Another 0.5 mol was spent on oxidising Fe (II) to Fe (III), leaving about 0.4 mol. And then some was lost during the boiling in. So your acid reserve is less than 0.4 mil / 0.125 L = 3.2 M. Probably about 2 M in reality. That's about what you'd expect to keep a 4 M FeCl3 in solution without Fe(OH)3 forming...


[Edited on 23-4-2014 by blogfast25]




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[*] posted on 23-4-2014 at 09:36


Appreciative of your tips here blogfast. Thanking you.
Is the miL here a mistake, should it be mole? ".. less than 0.4 mil / 0.125 L.." If not then please explain why miL and not mole.

[Edited on 23-4-2014 by CHRIS25]




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[*] posted on 23-4-2014 at 13:12


Yes that is supposed to be moles, 0.4mols/0.125L = 3.2M = 3.2mols/L.
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[*] posted on 23-4-2014 at 13:22


Quote: Originally posted by CHRIS25  
Appreciative of your tips here blogfast. Thanking you.
Is the miL here a mistake, should it be mole? ".. less than 0.4 mil / 0.125 L.." If not then please explain why miL and not mole.

[Edited on 23-4-2014 by CHRIS25]


Yes, mol, not 'mil'. Typo.




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