Sciencemadness Discussion Board

How to separate a bismuth - tin alloy?

khourygeo77 - 10-4-2017 at 11:52

Is it possible to separate the 2 metals in this alloy?

If so, what are the general methods?

Can I melt a metal alone or would the 2 melt together?

Can I perfectly separate one from another somehow with easily available items and chemicals?

Thanks for your time

symboom - 10-4-2017 at 12:15

Nitric acid made from spectracide stump remover potassium nitrate and sulfuric acid drain cleaner nitric acid turns tin into tin nitrate

MrHomeScientist - 10-4-2017 at 12:23

And would also dissolve the bismuth. Then you'd have to find a way of separating the two nitrates in solution, and reducing them back to metals. Possible I'm sure, but undoubtedly a long and arduous process. The first step would be to research the solubility of various bismuth and tin compounds, and see if you can get one to precipitate or crystallize out while the other stays in solution.

Alloys in general can't be separated by melting alone since the whole thing melts at once. There are some exceptions to this (non-eutectic alloys), but your case unfortunately isn't one of them.

PirateDocBrown - 10-4-2017 at 12:34

Once again, these metals can be resolved by exploiting the difference in solubility of the sulfides at different pH. Both sulfides are highly soluble at low pH, but make the solution basic, and the bismuth sulfide will crash out, leaving the tin in solution.

Both will dissolve in HCl, then pass H2S gas through the solution, then add NH4OH.

[Edited on 4/10/17 by PirateDocBrown]

DraconicAcid - 10-4-2017 at 12:43

Vogel separates the two ions by precipitating the two with sulphide ion, then boiling the suspension with ammonium sulphide. The bismuth sulphide stays as a precipitate, and the tin dissolves as ammonium trithiostannate. Not a method I'd care to use.

PirateDocBrown - 10-4-2017 at 17:20

Pretty much the same reaction I described in more practical terms. Ammonium hydroxide added to a sulfide solution will create ammonium sulfide in situ.

[Edited on 4/11/17 by PirateDocBrown]

theAngryLittleBunny - 13-4-2017 at 08:38

Quote: Originally posted by symboom  
Nitric acid made from spectracide stump remover potassium nitrate and sulfuric acid drain cleaner nitric acid turns tin into tin nitrate


Tin nitrate doesn't exist, in nitric acid, tin get's oxidized to the insoluble stannic acid (H2SnO3), which might actually give you a pretty cool advantage. Because then you can just put NaOH in the solution, which would give you insoluble bismuth oxide and soluble sodium stannate (Na2SnO3), which you can then easily seperate by filtration. The filtrate then can be reacidified to precipitate the stannic acid.

chornedsnorkack - 13-4-2017 at 13:43

Quote: Originally posted by theAngryLittleBunny  

Tin nitrate doesn't exist, in nitric acid, tin get's oxidized to the insoluble stannic acid (H2SnO3), which might actually give you a pretty cool advantage.

More precisely, tin nitrate does exist, but is notoriously unstable and will not be forming in nitric acid.
Quote: Originally posted by theAngryLittleBunny  
Because then you can just put NaOH in the solution, which would give you insoluble bismuth oxide and soluble sodium stannate (Na2SnO3), which you can then easily seperate by filtration. The filtrate then can be reacidified to precipitate the stannic acid.


Cannot you simply separate stannic acid (H2SnO3 from dissolved Bi(NO3)3? Meaning that after you precipitate Bi2O3, your filtrate is pure NaNO3?

Also, Bi is not amphoteric... while both Sn(II) and Sn(IV) are.

PirateDocBrown - 13-4-2017 at 14:08

In an alkaline aqueous environment, bismuth nitrate will quickly oxidize to the subnitrate.

theAngryLittleBunny - 13-4-2017 at 14:55

Quote: Originally posted by chornedsnorkack  
Quote: Originally posted by theAngryLittleBunny  

Tin nitrate doesn't exist, in nitric acid, tin get's oxidized to the insoluble stannic acid (H2SnO3), which might actually give you a pretty cool advantage.

More precisely, tin nitrate does exist, but is notoriously unstable and will not be forming in nitric acid.
Quote: Originally posted by theAngryLittleBunny  
Because then you can just put NaOH in the solution, which would give you insoluble bismuth oxide and soluble sodium stannate (Na2SnO3), which you can then easily seperate by filtration. The filtrate then can be reacidified to precipitate the stannic acid.


Cannot you simply separate stannic acid (H2SnO3 from dissolved Bi(NO3)3? Meaning that after you precipitate Bi2O3, your filtrate is pure NaNO3?

Also, Bi is not amphoteric... while both Sn(II) and Sn(IV) are.

Yeah right, I though the Bi(NO3)3 might forum the insoluble BiONO3, but I just looked it up, and Bi(NO3)3 is appearently soluble in dilute nittic acid, do you can do that, with makes it a lot easier. Sorry .-. I should better look stuff up before posting such overly complicated and unnecessary procedures.

Booze - 15-4-2017 at 21:05

You could dissolve it in muriatic acid, and after you wait fifty years for it to dissolve, you can precipetate the bismuth back with aluminum foil , but I don't know how to get the tin out. You could use it as a precious metal detecter.

Σldritch - 21-4-2017 at 14:11

Would not sodium hydroxide be the easiest and possibly cheapest way to dissolve the tin as sodium stannate? To my knowledge bismuth does not dissolve in sodium hydroxide but the dissolution of tin might be too slow to be practical.

clearly_not_atara - 21-4-2017 at 16:58

It appears that a solution of copper sulfate will dissolve tin and precipitate copper. Considering the reactivity series it seems likely that the same occurs to bismuth. The resulting solution contains tin (II) sulfate and bismuth (III) sulfate.

http://lanthanumkchemistry.over-blog.com/article-tin-and-cop...

If the sulfates in solution are then treated with a high concentration of chloride (note: all Cu2+ must be consumed or removed before this stage), bismuth oxychloride precipitates while SnCl2 remains in solution. Such concentration may be obtained by adding a brine of e.g. NaCl or CaCl2.

Note that by generating SnCl2 instead of SnCl4 we were able to avoid worrying about hydrolysis of the latter compound.


[Edited on 22-4-2017 by clearly_not_atara]

chornedsnorkack - 22-4-2017 at 09:50

Quote: Originally posted by clearly_not_atara  
It appears that a solution of copper sulfate will dissolve tin and precipitate copper. Considering the reactivity series it seems likely that the same occurs to bismuth. The resulting solution contains tin (II) sulfate and bismuth (III) sulfate.


From:
https://chem.libretexts.org/Reference/Reference_Tables/Elect...
Bi3+ + 3e− ⇌Bi(s) 0.317
Cu2+ + 2e− ⇌Cu(s) 0.3419
Sn4+ + 2e− ⇌Sn2+ 0.154

clearly_not_atara - 22-4-2017 at 10:50

Yes, that. The experiment linked demonstrates convincingly that Sn will not be oxidized to +4 in the absence of chloride ions. Obviously copper and chloride cannot be present in sol'n at the same time.