Chemistry Alchemist
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Samarium from SmCo5 Magnets
There is a great thread out there on Neodymium from a rare earth magnet but there is another magnet out there that contains Cobalt and Samarium... i
done some tests to find out if i have any of those magnets laying around and my test came out positive.. I Got about 40mls of HCl and placed the
magnet in... it slowly dissolved to form a blue solution indicating Cobalt... but just to make sure i got about a mill after the magnet fully
dissolved and placed it in 2 mls of water... the colour went from blue to red confirming its cobalt chloride.... now how would i get to separating the
Cobalt from the Samarium?
SmCo<sub>5</sub> + 12 HCl = SmCl<sub>2</sub> + 5
CoCl<sub>2</sub> + 6 H<sub>2</sub>
What do you think the next step is?
The picture is after a day of dissolving... the magnet was still dissolving about 4 days later, and the whole magnet wasn't even
SmCO<sub>5</sub> but had a brittile iner part not made of any metal... but the solution had gone really dark... looked a bit purple at
high concerntration
[Edited on 12-11-2011 by Chemistry Alchemist]
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DerAlte
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See my post http://www.sciencemadness.org/talk/viewthread.php?tid=8758&a...
Regards
Der Alte
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blogfast25
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Der Alte: that post doesn't say much about separating the cobalt and samarium though, does it?
Cobalt (II) oxalate is insoluble, so that can't be used here because samarium oxalate is poorly soluble too.
Another cobalt salt that's poorly soluble is cobalt nitrite (cobalt nitrite) but I don't know about Sm(NO2)3...
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IPN
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No offence blogfast25, but did you read the post at all?  Because it clearly
describes the way DerAlte separated the samarium from cobalt by using the solubility difference of the sulphates. Or did I miss something there?
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unionised
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I'm fairly sure that cobalt hydroxide will dissolve in an excess of aqueous ammonia but the Sm hydroxide won't.
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Chemistry Alchemist
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Would sodium hydroxide to anything to the Chloride solution? i know cobalt hydroxide will precipitate but will it still react with the samarium to
form its hydroxide but then im stuck with sodium contamination which then could be separated because most if not all sodium compounds are
soluble...... or dont you think its a viable route... i can get my hands on Sulfuric acid for pH Down from a pet store... would i just use that to
form the sulfates?
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blogfast25
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Quote: Originally posted by IPN  | | No offence blogfast25, but did you read the post at all? Because it clearly
describes the way DerAlte separated the samarium from cobalt by using the solubility difference of the sulphates. Or did I miss something there?
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It does but not at the point Der Alte linked to it. I can see clearly now, thanks!
Yes, Sm2(SO4)3 seems to have low RT solubility.
And Unionised is of course right too: cobalt (II) forms a hexammine complex. Treatment of the mixed hydroxides should leave Sm(OH)3 behind. It should
also dissolve Ni2+ and Cu2+ (if any present) into the bargain.
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Chemistry Alchemist
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Cobalt chloride reacts with aluminum to form cobalt and aluminum chloride and cobalt metal... because the samarium chloride wouldn't react with the
aluminum would it be easier to separate the samarium the aluminum?
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blogfast25
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| Quote: Originally posted by Chemistry Alchemist | | Cobalt chloride reacts with aluminum to form cobalt and aluminum chloride and cobalt metal... because the samarium chloride wouldn't react with the
aluminum would it be easier to separate the samarium the aluminum? |
You mean plating out the cobalt by adding aluminium foil or chips to ‘samarium cobalt chloride’? That’s probably possible but you’re replacing
one separation with another: now you need to separate Al<sup>3+</sup> from Sm<sup>3+</sup>! That’s fairly easy because Al is
highly amphoteric, Sm not at all but it just adds another step.
The methods above based on poor solubility of Sm sulphate or using the soluble cobalt (II) ammine complex are much simpler and adequate.
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Chemistry Alchemist
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What could be the next step then if i were to replace all the cobalt with aluminum? the way suggested above still would mean i would have to go though
another step because id have to convert it to sulfate...
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blogfast25
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| Quote: Originally posted by Chemistry Alchemist | | What could be the next step then if i were to replace all the cobalt with aluminum? the way suggested above still would mean i would have to go though
another step because id have to convert it to sulfate... |
Filter off the cobalt metal and any excess alunium. From the filtrate, precipitate the samarium and aluminium as hydroxides, with NaoH or KOH, then
add a large excess of alkali, which will dissollve the aluminium hydroxide as aluminate:
Al(OH)3.nH2O(s) + NaOH(aq) === > NaAl(OH)4(aq) + n H2O(l)
The samarium remains unaffected as Sn(OH)3.nH2O. Filter it off and wash profusely with hot water till the washwater runs to about pH 10 to 9. Your
filter cake is now 'pure' samarium hydoxide.
It does have the advantage of obtaining the cobalt as metal powder (wash it wil NaOH solution to remove excess aluminium metal: it dissolves easily in
strong alkali).
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Chemistry Alchemist
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The only down side to this is that my magnet was only small so I'll be working with a gram of samarium maybe not even that... I'll need to collect
more and crystilizes that so once I have a good ammount I can try it out...
So when adding the NaOH to turn them into hydroxides, I just add more then I actually need of hydroxide? So it will immediately form a precipitate?
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blogfast25
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The Al and Co2+ react as follows:
3 Co2+(aq) + 2 Al(s) === > 3 Co(s) + 2 Al3+(aq)
So in a sense you’ll go from SmCo5 (assuming yours was a 1:5 magnet, there are other types…) to Sm3+ + 10/3 Al3+, or about Sm3+ + 3 Al3+. That’s
quite a lot of Al. So when you add the hydroxide the Sm and Al will will first precipitate forming quite a thick slurry but on adding more NaOH the Al
starts to dissolve again and the slurry will thin out again. Make sure you use enough alkali and heat the slurry for 15 - 30 mins to ensure full
dissolution of the Al. You need to wash the precipitate firstly with dilute NaOH (to prevent any aluminate in the filter cake from reprecipitating as
Al(OH)3), then with hot pure water…
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