Sciencemadness Discussion Board

Isolation of barium metal from baryte sample

AndrejJoszika - 10-4-2018 at 08:20

Greetings!
I recently got a baryte sample (scientifically known as barium sulfate) and I would like to perform an isolation on it, nothing too complicated. Knowing that BaSO4 is really stable (and also a precipitate) the first method that came up in my mind and the very popular one is reducing the sulfate to sulfide with carbon. After that, I was thinking of adding HCl to BaS, leading to the formation of BaCl2 (and H2S will be produced in gaseous form).
My question is: could barium be isolated from BaCl2 through electrolysis? Or do you have a more efficient method?

Side note: I am fully aware of the formation of carbon monoxide, dioxide, hydrogen sulfide and the chlorine that would be released through electrolysis and I also know that barium itself is able to cause severe damage (making this operation pretty risky), but I can handle the situation with proper laboratory technique.

[Edited on 10-4-2018 by AndrejJoszika]

RawWork - 10-4-2018 at 08:36

Reduce BaCl2 with Al. Heat them together until they melt at around 1000 C. Then distill the AlCl3. Ba will be the leftover.

MrHomeScientist - 10-4-2018 at 10:08

I could have sworn there was a thread here on this topic, but I can't find it right now. If I remember right, barite is a very hard material that's difficult to crush into a powder, and that thread never made it to actual barium metal.

Another route from the sulfate is to convert it to the even less soluble carbonate by extended reflux in a carbonate solution. This can then be dissolved in another acid to make a more friendly barium salt. From there, I'm not so sure. I have a feeling barium is too reactive to be isolated from aqueous solution.

As you know, barium salts are very toxic so take extra care when working with them! Be careful around operations that produce gases, as this creates a mist of fine droplets of solution which could be inhaled or settle on surfaces that you later come into contact with.

AndrejJoszika - 10-4-2018 at 10:27

Quote: Originally posted by MrHomeScientist  
I could have sworn there was a thread here on this topic, but I can't find it right now. If I remember right, barite is a very hard material that's difficult to crush into a powder, and that thread never made it to actual barium metal.

Another route from the sulfate is to convert it to the even less soluble carbonate by extended reflux in a carbonate solution. This can then be dissolved in another acid to make a more friendly barium salt. From there, I'm not so sure. I have a feeling barium is too reactive to be isolated from aqueous solution.

As you know, barium salts are very toxic so take extra care when working with them! Be careful around operations that produce gases, as this creates a mist of fine droplets of solution which could be inhaled or settle on surfaces that you later come into contact with.


Yes, I thought about converting it to carbonate but as you pointed it out, the obtained barium salt would be pretty hard to reduce as barium metal is too reactive. It can be reduced with potassium or caesium, but my current budget does not allow me to purchase any of these metals. I may be able to afford potassium, but there have to be easier and less exothermic methods. This is why I thought about electrolysis.

AndrejJoszika - 10-4-2018 at 10:29

Quote: Originally posted by RawWork  
Reduce BaCl2 with Al. Heat them together until they melt at around 1000 C. Then distill the AlCl3. Ba will be the leftover.


Very interesting advice! I don't mean to prove you wrong, but wouldn't that imply that aluminium is more reactive than barium? Because it isn't. Apologies if what am I saying is wrong.

RawWork - 10-4-2018 at 10:42

We discussed this earlier: http://www.sciencemadness.org/talk/viewthread.php?tid=80545
That's maybe what MrHomeScientist thought.
It's based on equilibrium. You're right that Ba is more reactive (or would easier oxidize) than Al, but it's only relative.
Under specific or extreme conditions, such rule can be reversed.

According to Handbook of Inorganic Chemicals: "4 BaO + 2 Al → BaO•Al2O3 + 3Ba (gas)".
That would be just another example that method works.
AlCl3 will distill much easier than Ba would, as book would reccommend.
You probably don't even need to heat to 1000 C, but somewhat less.

Of course you have to do under inert atmosphere, just like in electrolysis you would. Many if not all chemicals will react with air, moisture etc.

Actually here's more about such equilibriums:
http://www.sciencemadness.org/talk/viewthread.php?tid=80525

[Edited on 10-4-2018 by RawWork]

AndrejJoszika - 10-4-2018 at 10:45

Quote: Originally posted by RawWork  
We discussed this earlier: http://www.sciencemadness.org/talk/viewthread.php?tid=80545
That's maybe what MrHomeScientist thought.
It's based on equilibrium. You're right that Ba is more reactive (or would easier oxidize) than Al, but it's only relative.
Under specific or extreme conditions, such rule can be reversed.

According to Handbook of Inorganic Chemicals: "4 BaO + 2 Al → BaO•Al2O3 + 3Ba (gas)".
That would be just another example that method works.
AlCl3 will distill much easier than Ba would, as book would reccommend.
You probably don't even need to heat to 1000 C, but somewhat less.

Of course you have to do under inert atmosphere, just like in electrolysis you would. Many if not all chemicals will react with air, moisture etc.

[Edited on 10-4-2018 by RawWork]


This is really interesting! Thanks for the advice given.

RawWork - 10-4-2018 at 10:53

Yeah, sodium compound can be reduced by carbon too.

MrHomeScientist - 10-4-2018 at 11:41

Aluminum may not work as you think it would. The change in enthalpy is extremely positive at 1172 kJ/mol. That means this reaction is very unfavorable and might need some impractical conditions. If the aluminum chloride does boil off that would help move it forward, but is it enough to overcome this? I'm not sure.

Reaction:
3BaCl2 + 2Al == 3Ba + 2AlCl3

HoF's:
BaCl2 = -704 kJ/mol
AlCl3 = -860 kJ/mol
Al = Ba = 0 kJ/mol

(delta)H = [products] - [reactants] = [3(0) + 2(-704)] - [3(-860) + 2(0)] = 1172

phlogiston - 10-4-2018 at 14:10

Another possible method is electrolytically, using a mercury cathode.
Barium amalgam will form. The mercury can be distilled off, leaving barium metal.
Obviously, this is risky too, but can be pulled off with proper preparation and equipment.

CobaltChloride - 11-4-2018 at 13:36

Maybe the methods used by amateurs to make sodium might also work here. The ones I'm aware of are electrolysis of the molten hydroxide (but I'm not sure if barium isn't soluble in its molten hydroxide; for example, potassium is soluble in liquid KOH, but sodium isn't soluble in liquid NaOH) and reaction of the hydroxide with magnesium at high temperatures, but this reaction isn't favored in this case according to the standard enthalpy of formation (for Ba(OH)2 it is −944.7 kJ/mol; for MgO it is −601.6 kJ/mol).

ninhydric1 - 11-4-2018 at 15:15

However, if you start with barium oxide, you get the reaction

BaO (s) + Mg (s) --> Ba (s) + MgO (s)

H = Hproducts - Hreactants = H(MgO) - H(BaO) = -601.6 kJ - (-553.5 kJ) = -48.1 kJ

So if you have barium oxide, the reaction itself has a negative enthalpy.

Foeskes - 11-4-2018 at 21:06

Distilling barium is probably nessesary to separate it from MgO
Barium's melting point is way to high for dioxane.

AndrejJoszika - 12-4-2018 at 11:11

Quote: Originally posted by ninhydric1  
However, if you start with barium oxide, you get the reaction

BaO (s) + Mg (s) --> Ba (s) + MgO (s)

H = Hproducts - Hreactants = H(MgO) - H(BaO) = -601.6 kJ - (-553.5 kJ) = -48.1 kJ

So if you have barium oxide, the reaction itself has a negative enthalpy.


This is the method Nurdrage used to isolate sodium, but I didn't do the calculations in this particular case. It seems to be the easiest and it's physically feasible (as you have just proven). I will try it. The conversion from BaSO4 to BaO is pretty easy, once I obtain the carbonate.

phlogiston - 12-4-2018 at 12:30

Another fun path to metallic barium that is via decomposition of barium azide.
You pass hydrazoic acid through a barium hydroxide suspension in water to prepare the barium azide, which you then heat causing it to decompose into metallic barium and nitrogen.

Foeskes - 12-4-2018 at 16:47

Quote: Originally posted by phlogiston  
Another fun path to metallic barium that is via decomposition of barium azide.
You pass hydrazoic acid through a barium hydroxide suspension in water to prepare the barium azide, which you then heat causing it to decompose into metallic barium and nitrogen.

I thought azides especially HN3 are no joke

AndrejJoszika - 12-4-2018 at 20:06

Quote: Originally posted by Foeskes  
Quote: Originally posted by phlogiston  
Another fun path to metallic barium that is via decomposition of barium azide.
You pass hydrazoic acid through a barium hydroxide suspension in water to prepare the barium azide, which you then heat causing it to decompose into metallic barium and nitrogen.

I thought azides especially HN3 are no joke


Well, both of them are explosive as hell.

NeonPulse - 12-4-2018 at 22:27

Ha. I just posted on the energetically board about this very subject.

phlogiston - 13-4-2018 at 02:53

Quote: Originally posted by AndrejJoszika  
Quote: Originally posted by Foeskes  

I thought azides especially HN3 are no joke


Well, both of them are explosive as hell.


And -extremely- poisonous as well.

My comment of this being a 'fun' route was sarcasm. It is prohibitively dangerous.