khourygeo77
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How to separate a bismuth - tin alloy?
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
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symboom
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Nitric acid made from spectracide stump remover potassium nitrate and sulfuric acid drain cleaner nitric acid turns tin into tin nitrate
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MrHomeScientist
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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.
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PirateDocBrown
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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]
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DraconicAcid
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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.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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PirateDocBrown
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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]
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theAngryLittleBunny
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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
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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.
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chornedsnorkack
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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.
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PirateDocBrown
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In an alkaline aqueous environment, bismuth nitrate will quickly oxidize to the subnitrate.
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theAngryLittleBunny
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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.
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Booze
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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.
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Σldritch
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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.
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clearly_not_atara
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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]
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chornedsnorkack
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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
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clearly_not_atara
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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.
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