Sciencemadness Discussion Board - Samarium from SmCo5 Magnets

Samarium from SmCo5 Magnets

Chemistry Alchemist - 11-11-2011 at 22:00

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?

SmCo5 + 12 HCl = SmCl2 + 5 CoCl2 + 6 H2

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 SmCO5 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]

DerAlte - 11-11-2011 at 22:11

See my post http://www.sciencemadness.org/talk/viewthread.php?tid=8758&a...

Regards

Der Alte

blogfast25 - 12-11-2011 at 05:46

Quote: Originally posted by DerAlte

See my post http://www.sciencemadness.org/talk/viewthread.php?tid=8758&a...

Regards

Der Alte

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...

IPN - 12-11-2011 at 06:45

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?

unionised - 12-11-2011 at 06:50

I'm fairly sure that cobalt hydroxide will dissolve in an excess of aqueous ammonia but the Sm hydroxide won't.

Chemistry Alchemist - 12-11-2011 at 07:44

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?

blogfast25 - 12-11-2011 at 13:12

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?

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.

Chemistry Alchemist - 24-11-2011 at 06:26

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?

blogfast25 - 24-11-2011 at 06:40

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 Al3+ from Sm3+! 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.

Chemistry Alchemist - 24-11-2011 at 07:43

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...

blogfast25 - 24-11-2011 at 08:53

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).

Chemistry Alchemist - 24-11-2011 at 09:00

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?

blogfast25 - 24-11-2011 at 09:14

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…