Neodymium Chloride Solution

[Chemistoftheelements]

Good. Before I forget, and this important point I neglected earlier, you may find that the pH range for Pr precititation and that of Nd overlap. By this I mean (and don’t take these figures literally- they’re possibly way out and just to illustrate a point) your Pr(IV) may start precipitating about pH2 and be completely precipitated by 5, whist for Nd(III) it may be 4-7. As you add your NH3 ( which you do in many steps to stop, stir and observe), you will will hopefully reach a point where the precipitate begins to change colour, due to the Nd beginning to drop out of solution (precipitate). Filter as soons as this happens, and not after. You’ve got Pr as free of Nd as you’re going to get using this method ( if it works). Back to your solution and precipitate now; you should find that as you’ve added more NH3 in a handful of steps, each time your precitiate has changed colour slightly. Eventually, with more NH3, the colour will no longer change. This will hopefully tell you that the Pr is now out of solution. Filter now and save the precipitate for a future experiment, as it’s Pr mixed with Nd and useless for this experiment. Add further NH3, gradually and stepwise, and if no colour change occurs, you can neutralise completely (take it to pH 7), and you have your Nd, largely as Nd(OH)₃ if you’ve neutralised all the way.

 

 

 

 

P.S. the formula for your basic Pr sulfate is probably more like Pr₂(OH)₂(SO4)_­3­­.

[John Cuthber]

Most appreciated. And you said Pr and Nd were difficult to separate, John.

As soon as I get back from Canada, I'll get right on this.

 

And I'm saying it again.

So does WIKI.

"Using classical separation methods, praseodymium was always difficult to purify. Much less abundant than the lanthanum and neodymium from which it was being separated (cerium having long since been removed by redox chemistry), praseodymium ended up being dispersed among a large number of fractions, and the resulting yields of purified material were low. "

If you look here you will see a lack of Pr(IV) compounds

http://en.wikipedia.org/wiki/Category:Praseodymium_compounds

The Pr(IV) ion is a very strong oxidant.

If you treat Pr2O (3-and-a-bit) with acid it dissolves to give Pr(III) and O2. The water gets oxidised.

[Chemistoftheelements]

And I'm saying it again.

So does WIKI.

"Using classical separation methods, praseodymium was always difficult to purify. Much less abundant than the lanthanum and neodymium from which it was being separated (cerium having long since been removed by redox chemistry), praseodymium ended up being dispersed among a large number of fractions, and the resulting yields of purified material were low. "

If you look here you will see a lack of Pr(IV) compounds

http://en.wikipedia....ymium_compounds

The Pr(IV) ion is a very strong oxidant.

If you treat Pr2O (3-and-a-bit) with acid it dissolves to give Pr(III) and O2. The water gets oxidised.

 

This is what I feared may be the case, in the same manner Pb(IV) is reduced to Pb(II) in acid solution, so Pr(IV) may be to Pr(III). But 2 questions:

 

If you ignite Pr in air, you get an oxide containing a fair amount of Pr(IV), and I've always considered the highest oxidation state of metals which can be attained by oxidation in air a measure of the stability of that oxidation state in acid solution. PbO2, for example, is way too oxidising to be able to be formed by heating Pb in air, and not surprisingly, PbO2 dissolves in acids. Wouldn't Pr be stable enough to stay in solution in acid at least partially as Pr(IV)?

 

Secondly, Ce(IV) is pretty oxidising as well, but CeO2 doesn't seem to disolve in HCl, (which it would be expected to oxidise, and, becoming itself reduced to Ce(III)2O3, which dissolves in this medium much more readily). If the observations for CeO2 can be extrapolated to Ce(IV) in acid solution, would this not also apply, albeit to a lesser degree, with Pr?

 

Thanks.

[John Cuthber]

"If you ignite Pr in air, you get an oxide containing a fair amount of Pr(IV), and I've always considered the highest oxidation state of metals which can be attained by oxidation in air a measure of the stability of that oxidation state in acid solution."

There are a good number of cases where elements form stable oxides (and fluorides) which are reduced by water or acid. MnO2 is probably the best known.

 

"Wouldn't Pr be stable enough to stay in solution in acid at least partially as Pr(IV)?"

No.

" If the observations for CeO2 can be extrapolated to Ce(IV) in acid solution, would this not also apply, albeit to a lesser degree, with Pr?"

If they could be extrapolated, but they can't.

Apart from anything else, if that technique worked then the Pr wouldn't be there. It would have been removed along with the Ce (which is separatedby oxidation to Ce(IV) during the manufacturing process.

 

Don't you think you should have checked this before suggesting it?

[Chemistoftheelements]

The starting point for this project is a magnet, composed of, and this a a presumption, Pr as well as Nd, along with Fe, etc. The Pr and Nd by this presumption are intentionally there, and presumably the rest of the rare earth's aren't, so this is a completely seperate issue, is it not, to dealing with separating all of the rare earths?

 

If you mean that the commercial extraction processes for the rare earths typically don't take out Pr with Ce, could this not be something to do with the influence of the so many other lanthanoids as are present in rare-earth ores, which aren't capable of being oxidised to a 3+ or higher oxidation state, with which the Pr may be carried over? Your point is valid only if the Ce is taken out first, then the rest of the rare earth elements are treated sepparetely.

 

Lastly, I've said all along that this is project is full of uncertainties, many of which I've highlighted, if you read the whole discussion from when I entered it, so I don't understand your last quip, if that's how it was intended.

[John Cuthber]

" this is a completely seperate issue, is it not, to dealing with separating all of the rare earths? "

No, not really. Ce and Eu which have relatively stable oxidation states other than (II) are easily removed. All the others are difficult to split. Even if only two of them are present, the fact is that those two are still very similar and so it's hard to get either of them pure.

 

"If you mean that the commercial extraction processes for the rare earths typically don't take out Pr with Ce, could this not be something to do with the influence of the so many other lanthanoids as are present in rare-earth ores, which aren't capable of being oxidised to a 3+ or higher oxidation state"

How, exactly, does an ion know what other ions are present?

Imagine a Pr+++ ion in solution wondering whether or not to be oxidised to Pr++++, is there some way where it can say to itself "Oh, I see there's some Ce(IV) over there- that will lower my redox potential"?

(and incidentally, you seem to have your oxidation states muddled, they can all be oxidised to "+3 or higher" because they can all be oxidised to +3.)

 

 

" Your point is valid only if the Ce is taken out first, then the rest of the rare earth elements are treated sepparetely."

It is, so it is.

Again, shouldn't you have checked?

 

 

"Lastly, I've said all along that this is project is full of uncertainties"

Exactly my point.

There is no uncertainty. It won't work.

 

It remains difficult to separate these two elements. I still don't think it's practical as a home experiment.

[Chemistoftheelements]

It was late, I meant that the rest of the lanthanides can’t be oxidised to +4 or greater.

 

I am here to discuss where the practice and theory of chemistry meet, which can only be done by experimenting, otherwise it isn’t much fun, and you don’t learn as much, even if what you are attempting doesn’t work. If I am wrong on anything, I will gladly be corrected, but on the other forums of which I am a member, this has always been conducted in a friendly and constructive manner, even by the occasional leading experts who graciously give their time to these forums- this is a first for me. Politeness costs nothing and often, I find, produces more rewarding discussions. I really hope that this discussion is not turning into a confrontation, as I am not interested in this, not my game. Thank you in advance for your cooperation.

 

 

[John Cuthber]

In the end any confrontation will be with reality, and it's a lot less polite than me.

 

I'd still like to know how you think the presence of the Nd will affect the oxidation potential of Pr.

[Chemistoftheelements]

Indeed reality may well prove impolite.

 

On the second point, if you were referring to this:

 

"1) At the oxide stage, the Pr would be in solid solution in the Nd2O3, and probably being only a minor constituent , would be prevented from oxidising to higher than Pr(III)"

 

then this was at the stage when elementcollector1 was contemplating oxidising the mixed Nd/ Pr oxides with H₂O­_2. I was thinking that if the Pr was the less abundant constituent by some way, then, being in solid solution, it would be "shielded"- any oxidant would be in contact mostly with Nd oxide, and only the Pr oxide which was right at the surface could be oxidised. The more Pr in the mix, I figure, would allow more oxidation, and this method may become more feasible, especially with fine grounding.

 

Sorry about the irregular line spacing, I'm trying to figure out why it's doing this.

 

 

 

 

P.S. when i say "right at the surface", I mean of the crystals/particles of mixed Nd/Pr oxide.

[elementcollector1]

P.S. when i say "right at the surface", I mean of the crystals/particles of mixed Nd/Pr oxide.

Please explain. When did the oxide crystals form again?

John, no offense but you seem to be a bit pessimistic about this. It might not work, but it'd be great fun to try.

[Chemistoftheelements]

Hi elementcollector1

 

"Please explain. When did the oxide crystals form again?"

 

This is at the stage after you ignite your oxalate, which you were wondering how you should convert, in terms of it's praseodymium content, to PrO2, and you suggested H2O2._.<br style="mso-special-character:line-break"> <br style="mso-special-character:line-break">

 

You wrote:

 

“ Hah, speaking of sodium bicarbonate, I just lost a large beaker to the stuff. Poor thing has a massive chunk missing.

Anyway, I know oxalic acid is used as wood bleach, so I'll look into that.

I'd like to go all the way to Nd and maybe Pr metals if I could, but we'll see.

Process of getting to Nd from neomagnets, in short:

1) Dissolve in acid.

2) Add oxalic acid or salt to precipitate insoluble oxalates.

3) Ignite in air to produce oxides.

 

(for separation of Pr)

4) Somehow convert the Pr₆O₁₁ that formed into PrO₂.

5) (for separation of Pr) Place in 5% acetic acid? I read below on Sciencemadness that PrO₂ is insoluble in 5% acetic acid, and apparently Nd₂O₃ is.”<br style="mso-special-character:line-break"> <br style="mso-special-character:line-break">

 

Then later on:

 

“My main problem with this is how to convert the Pr₆O₁₁ to PrO₂. Would hydrogen peroxide work? Excess heating? From the same forum as earlier, members have mentioned that on ignition, praseodymium oxalate turns brown, but I'm not sure if this is PrO₂ (in which case, yay!) or something else. Nd (III) cannot be oxidized any further by igniting in air, so that's safe.”

 

 

This was before I came up with my prodedure suggestion. Good luck!

 

 

"<br style="mso-special-character:line-break"> <br style="mso-special-character:line-break"> "

 

This bit, of course, is complete nonsense. I think this sort of thing is happening because I'm copying from microsoft word before I post.

[elementcollector1]

Right, right, now I remember. So, what you're saying is I'd get a group of Nd-rich precipitate and Pr-rich precipitate, purifying the Pr by oxidizing it to PrO₂ (if that works) and separating by acetic acid (if THAT works), and purifying the Nd if necessary. It's not really needed because I can fairly easily find a Pr-free magnet, so I guess the Pr would have priority here.

EDIT: It looks like I might have a bit of trouble with spacing as well.

 

EDIT #2: Acetic acid does NOT work. Also, to convert the Pr(III) oxide to Pr (IV) oxide, temperatures of at least 600 degrees Celsius are required (fairly easy with a blowtorch). Should turn black or brown, presence of another oxidizer such as KClO3 helps.

Edited January 5, 2012 by elementcollector1

[Chemistoftheelements]

Hi elementcolloctor1, yes, I know the feeling, it can be a little hard to keep track of these things. I think we're getting a little confused her, so I think it might be helpfull if I re- post a conversation we had about 3 days ago. Note that as far as I can tell, whether the 8- point method I suggest works is a matter, I think, of whether you have any Pr in there in the first place, and just how long Pr(IV) will hang around in acid solution; you MIGHT, might be lucky and get a small amount of Pr(IV) precipitate, but as John Cuthber has rightfully pointed out, it is pretty unstable. Here was our ealier conversation:

 

elementcollector1, on 31 December 2011 - 12:45 AM, said:

 

Hah, speaking of sodium bicarbonate, I just lost a large beaker to the stuff. Poor thing has a massive chunk missing.

Anyway, I know oxalic acid is used as wood bleach, so I'll look into that.

I'd like to go all the way to Nd and maybe Pr metals if I could, but we'll see.

Process of getting to Nd from neomagnets, in short:

1) Dissolve in acid.

2) Add oxalic acid or salt to precipitate insoluble oxalates.

3) Ignite in air to produce oxides.

 

(for separation of Pr)

4) Somehow convert the Pr₆O₁₁ that formed into PrO₂.

5) (for separation of Pr) Place in 5% acetic acid? I read below on Sciencemadness that PrO₂ is insoluble in 5% acetic acid, and apparently Nd₂O₃ is.

 

 

 

 

Elementcollector1,

 

1) Happy new year!

 

2) To answer this and your latest question:

There’s some good logic to this, but I’m not sure that this would work, for several reasons:

 

1) At the oxide stage, the Pr would be in solid solution in the Nd2O3, and probably being only a minor constituent , would be prevented from oxidising to higher than Pr(III). In other words, the Nd(III) would stabilise the Pr(III), in the same way that Th(IV) stabilises U( IV) in high ( and low) U thorianites.

 

2) Oxidation states of Pr higher than 3 would be unstable in acid solution- think CeO2, so adding acetic acid would probably cause what little Pr in a >III oxidation state to revert pretty quickly to PrIII and go into solution.

 

3) Lanthanoid (and related oxides) don’t easily dissolve in acid once they’ve been ignited. Working on a theory that whether a metal oxide dissolves or not depends on the ionic radius, the coordination number of the metal, and possibly it’s affinity for oxigen, I produced some Yb2O3, Sc2O3 and Sm2O3 and found that all 3 were very reluctant to dissolve; after ignition at 1000 degrees C (and then cooling), only the Sm2O3 dissolved, but very slowly. If, however you heat them only minimally ( just enough to ignite the oxalate in your case), then you will probably succeed.

 

4) By using H2O 2, you may end up with a product containing O-O bonds, which would be bad, because such a compound may prove thermally dangerously unstable. You would have to work on very small quantities to find this out, or alternatively oxidise with something else.

 

What I would suggest is the following, which circumvents problem 1, but not the others, so it’s not guaranteed to work:

 

1) Dissolve in acid

 

2) Add H2C2O4 to precipitate Nd and Pr and separate them from Fe, etc. Add NH3 to raise pH to increase precipitation if necessary, or dilute with water.

 

3) Gently heat precipitate to create Nd/Pr oxides.

 

4) Dissolve oxides in H2SO4, NOT HCl, as this is reducing.

 

5) Oxidise this resulting solution to produce Pr2 O(SO4)3(?), a higher basic sulfate of praseodymium which will be insoluble.

 

6) Filter. You have now recovered Pr.

 

7) Ignite your P2 O(SO4)3

 

7)To remaining solution, add NH3 to precipitate Nd(OH)3.

 

8) Ignite the Nd(OH)3.

 

From here, you’ll have to figure out a way to reduce the Pr and Nd oxides. Good luck, and let us know how you get on. In the mean time, I’ll try to figure out how to create subscripts!

[Chemistoftheelements]

OK, I goofed whilst numbering the points- there are infact 9 of these!

 

Of course, if Pr(IV) reduces instantly, i.e. is extremely unstable relative to Pr(III), then you will get only one precipitate, containing both metals. If this is the case, which it may well be, then you are no worse off than you were to begin with.

[John Cuthber]

Unless you can tell me what oxidant you plan to use for step 5 ("Oxidise this resulting solution to produce Pr2 O(SO4)3(?)")

you are talking nonsense.

Also "you are no worse off than you were to begin with. " is only true if you don't value your time (strictly speaking, someone else's time) and didn't pay for the materials you wasted.

[elementcollector1]

Unless you can tell me what oxidant you plan to use for step 5 ("Oxidise this resulting solution to produce Pr2 O(SO4)3(?)")

you are talking nonsense.

Also "you are no worse off than you were to begin with. " is only true if you don't value your time (strictly speaking, someone else's time) and didn't pay for the materials you wasted.

 

I'd guess H₂O₂.

"Someone else's time?" "materials you wasted?" Brother, I'm experimenting with my stuff, not someone else's. I paid for everything, and do this in my free time.

Even if I do get a precipitate containing both metals, then it would still be in the (III) oxidation state. If that's true, then I can probably just re-dissolve in HCl or H₂SO₄.

Again, no matter how much speculating is done on whether or not it will work, it all comes down to the actual experiments, which I'm perfectly willing to sacrifice my materials and free time to do. Now, it's just a matter of coming up with the easiest way to do it.

[elementcollector1]

Just tested the oxalic acid + magnet chloride today, and found myself with a greenish-gray precipitate. This is not neodymium oxalate, but could it be the oxide? Something else entirely?

[elementcollector1]

Decided to restart this experiment with a more knowledgeable approach to the problem. From previous tests, I have concluded that:

-There is no praseodymium in my magnets.

-I can get to neodymium oxalate, but I can't calcine to oxide without serious equipment.

-I think I'll follow "TheChemLife"'s procedure, outlined here: http://www.youtube.com/watch?v=0hLEGMufP78

-OTC source of MEK (methyl-ethyl ketone)? We have "MEK Substitute" at the local McLendon's, which is presumably ethyl acetate. It appears lacquer thinner, if I can find any, would be a good option to distill for MEK.

 

Plans for the purified NdCl3:

1) React with soluble fluoride (probably sodium fluoride) to precipitate insoluble NdF3.

2) React with Mg or Ca under argon blanket to (hopefully!) create Nd metal.

3) Ampoule under argon, and store!