# Dissolving Rhodium with potassium disulfate (K2S2O7)



## lazylightning (Jul 11, 2008)

Hi All! 


So I went to the labshop today and instead of potassium bisulfate they accidently sold me something a little different. Actually, on one German chemistry forum I accidently ran accross, has a reference about dissolving Rh with the help of the stuff I accidently got which is K2S2O7 or potassium disulfate. Below is a qoute of the reference: 

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"The dissolution of rhodium in an acidic salt melt functions quite well, if you let react the metal with a mixture of potassiumdisulfate; K2S2O7 and sodium chloride; NaCl. This mixture melts at fairly deep temperatures, so you actually can use pyrex-beakers as reaction containers. Another possibility is, to use porcelain dishes. Reaction is best done in an oven at 500-550o C. Usually in 1-2 hours about 80% of the metal has dissolved in the molten mixture. Prolonged reaction does not lead to significant more dissolution, because the salt melt itself decomposes slowly at these temperatures, probably by releasing gaseous SO3 and/or SO2Cl2. The presence of chloride in abundance leads to the formation of Rh(III)-chloro-complexes, clearly evident by the deep red color of the resolidified melt after cooling down. The combined effects of a molten mixture of NaCl and K2S2O7 on platinum group metals can be compared to the combined effects of HCl and HNO3 in aqua regia on metallic gold. So even metallic iridium, which does not react with molten K2S2O7 alone, is dissolved in a similar way by the molten salt-mixture. 

Here an example for a rhodium batch of 20 grams, which I did myself for a few times: 20 g of rhodium powder, 136 g NaCl and 296 g K2S2O7 are intimately ground together in a mortar. The grey powdery mixture is placed in a porcelain dish or into wide Pyrex or Jenaer-glass beakers. The beaker(s) or dish(es) should only be half filled. They are put in an oven, heated to 500-550oC and held at this temperature for about 1-2 hours. After cooling down the deep red solids are dissolved in a mixture of 1600 ml water and 400 ml of concentrated HCl. Any undissolved rhodium metal, usually about 4 grams, is filtered off and can be reused again. The deep red filtered liquid contains about 16 grams of rhodium and can be treated further to precipitate Rh(OH)3." 

The text is from this forum page: http://www.mathematik-forum.de/forum/showthread.php?t=106496&page=1 
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So judging by the text written above, if I add the table salt to the melt mix then it will dissolve the Ir too. I probably have alot of Ir in my material too. Do I want to do that? Or will Ir precipitate out with the Rh when I add the alchohol? Maybe I should do it without the table salt to insure that the Ir isn't dissolved? They didn't mention whether or not the K2S2O7 will work on the Rh without the table salt, if you look at it carefully. If I add the table salt and it dissolves the Ir too, then it the Ir could precipitate out with the Rh, and then I've done all this for naught? 

Also, I just noticed that they do not refer to using alchohol as a precipitant, perhaps it will not work in this case? Though I dont see why, it's still a Rh sulphate water solution. 

The site is in German though they're posting in English. I'd just join that forum but all the registration process is in German. Ich nicht forschtehen. 

Any ideas? 

The material I have is electroprecipitated sulfides(a red color) which contains Rh, Ir and perhaps other pgm's (but no Ru or Os, they are already removed). There may be base metals also. I can roast them to get oxides, but that will make it harder to dissolve, I assume. Then I will have to reduce the oxides under hydrogen first to make it metal again and easier to dissolve. What about putting these sulfides (ground up) into this melt mix. Again, these sulfides dissolve easier in AR than the metal forms and hot AR seem to dissolve the Rh sulfides pretty fast too. I think they would dissolve pretty easy in any salt melt at all. Then the problem I will face again is the prospect of Ir being in the solution as a water soluble sulfate. Does anyone know if that happens? Again, will it come out of solution as a precipitate along with the Rh when I add alchohol, or will only Rh do that? Rh is my target material of choice because it seems to be highest along with the Ir in concentration. 

Thanks in advance for any thoughts on this.


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## lazersteve (Jul 11, 2008)

Lightning,

Here's an article on separating Rhodium from Iridium from someone at Ledoux.

Separating Rhodium from Iridium

I hope it helps you out.

Steve


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## Lino1406 (Jul 11, 2008)

In addition, I'd try to precipitate
iridium by NH4Cl as the "blue mud".


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## lazylightning (Jul 11, 2008)

Thanks for the ideas! So the method they describe in that pdf document is quite complex. I have no chlorine stream, formic, HF or heat source that will give me 550 C, more over 825C. It looks good for critical analytical work. I have already accepted the fact that no matter how pure I can get it, I have to sell it to a refinery, and they will take their share as if it is only 95% pure Rhodium at best even if it is 99.95% pure. I don't have a name as a refinery that could sell sponge with a guarantee that is credible for anyone involved in that market. 

I guess my main concern is whether or not Ir will precipitate out with the alchohol after the potassium disulfate melt solution is made (I have a new electric eye that gives exactly 350C). I can get some alchohol ;-D I am assuming also, that only a small amount or no Ir will even be attacked by the K2S2O7 melt if I dont put the table salt in it. Of course less Rh will be dissolved, however a considerable amount will be left undissolved regardless, requiring a few melts to get it all dissolved.

I'm also looking for a quartz crucible, however I may have to do with an unglazed porcelain dish or two.

Was the refference for removing Ir with ammonia chloride in regards to removing it from the water soluble sulfate melt solution? Does this mean you think that Ir will also form a water soluble sulfate if it is dissolved using the table salt included in the disulfate melt? -and that it will not precipitate out of the solution together with the Rh when I add the alchohol? Or was this a refference to something in the pdf document about seperating Rh and Ir?


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## Lou (Jul 11, 2008)

Ah, days like this I love having a tube furnace.

By the way, I'll buy your Rh based on an assay. 

Let me know.


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## Lino1406 (Jul 12, 2008)

Precipitates Ir the same way as Pt (yellow mud)


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## lazylightning (Jul 12, 2008)

Lino1406 said:


> Precipitates Ir the same way as Pt (yellow mud)



So the Rh will precipitate out as a fine Rh dioxide grit and the Ir and Pt will be a more soft mud like material?


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## Lou (Jul 12, 2008)

No you won't have RhO2, you'll have Rh2O3; RhO2 is not easy to get from aqueous environment.


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## lazylightning (Jul 13, 2008)

Lou said:


> No you won't have RhO2, you'll have Rh2O3; RhO2 is not easy to get from aqueous environment.



So that's gotten by adding alchohol to the sulfate solution made from the acidic salt melt? Afterwards after filtering out the Rh2O3 I could add the ammonia chloride and drop the other salts out of the solution? Actually though, this material was already boiled in strong AR for four hours that required replentishing the Ar three times, as the 100 ml kept boiling off fast. I suspect that only Rh and Ir could be remaining in it.


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## Lino1406 (Jul 13, 2008)

precipitate Ir 1st, Alcohol later


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## DaniH (Dec 17, 2021)

lazersteve said:


> Lightning,
> 
> Here's an article on separating Rhodium from Iridium from someone at Ledoux.
> 
> ...


Unable to open site. Can you share this document direct link please. 
Thanks


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