Sciencemadness Discussion Board

Hot Concentrated Sulfuric Acid?

primezero - 22-5-2018 at 20:01

I want to dissolve Tin with sulfuric acid. Online articles declare that Tin will only react with "hot concentrated sulfuric acid."

Unfortunately I couldn't find anything on if a passivation layer will form, what concentration is ideal and what temperature is ideal to dissolve 1 gram of tin.

I'm hoping this doesn't require watching all my H2SO4 boil away in my fumehood... :)

TheMrbunGee - 22-5-2018 at 20:18

Quote: Originally posted by primezero  
I want to dissolve Tin with sulfuric acid. Online articles declare that Tin will only react with "hot concentrated sulfuric acid."

Unfortunately I couldn't find anything on if a passivation layer will form, what concentration is ideal and what temperature is ideal to dissolve 1 gram of tin.

I'm hoping this doesn't require watching all my H2SO4 boil away in my fumehood... :)


Dissolve it it in copper sulfate.

AJKOER - 23-5-2018 at 20:45

You need an oxidizing acid (like HOCl or HNO3) added to your H2SO4 (concentrated not required) to dissolve the tin. My understanding is that an acid like HOCl can create an oxide coating on the Sn which is apparently more readily dissolved by the H2SO4.

References: http://tin.atomistry.com/stannous_sulphate.html and https://books.google.com/books?id=tjIAAAAAMAAJ&pg=PA134&... .

My preferred path to HOCl, avoiding organic acids subject to attack by the hypochlorous acid:

2 NaOCl + CaCl2 = 2 NaCl + Ca(OCl)2
Ca(OCl)2 + 2 NaHCO3 --> Na2CO3 + CaCO3 (s) + 2 HOCl

Net: 2 NaOCl + CaCl2 + 2 NaHCO3 --> 2 NaCl + Na2CO3 + CaCO3 (s) + 2 HOCl

Or with more CaCl2:

2 NaOCl + 2 CaCl2 + 2 NaHCO3 --> 4 NaCl + 2 CaCO3 (s) + 2 HOCl

I freeze the mix to remove the suspension of CaCO3 and decant the HOCl/NaCl solution.

Now, the NaCl presence is a problem if adding H2SO4 as chlorine will form, so, in this special case, just add NaOCl to the H2SO4 and freeze out the Na2SO4.xH2O as a path to HOCl, assuming the chlorine bleach has no NaCl (should test as generally it does being prepared by the action of Cl2 on aqueous NaOH). Simple solution, boil off half the HOCl/NaCl aqueous mix and pass vapors into the H2SO4 (see http://www.sciencemadness.org/talk/viewthread.php?tid=71058#... ). Other solutions are harder like converting NaOCl/NaCl to Mg(OCl)2/MgCl2 by adding MgSO4 and freezing out Na2SO4.10H2O. Then, to quote a source (http://magnesium.atomistry.com/magnesium_chloride.html):

"2MgCl2.CaCl2.12H2O crystallises, above 22° C., from a solution of its constituent salts in appropriate proportions."

which may still leave a chloride presence in the HOCl with too much CaCl2. Note: Hypochlorous acid is not very stable in the presence of chloride, so keep cool and use promptly after preparation.

[Edited on 25-5-2018 by AJKOER]

VSEPR_VOID - 23-5-2018 at 21:39

I would add half its volume of 3% H2O2 and raise the temperature to 100C. Do not let it go higher as you want to prevent the acid from boiling dangerously. This solution would be a weak piranha solution and should dissolve this tin. If this fails play with the acid-peroxide ratio.

AJKOER - 2-6-2018 at 07:02

Here is a possible poor man's attempt at dissolving tin starting with SO2, water and passing air/O2 through the formed H2SO3 in the presence of a hydroxyl radical source (like by action of sunlight near to an ocean, large lake,.. on aqueous N2O, or aqueous nitrite or a photo active oxide like TiO2,..).

For example, per one source: "Rate Constants for Reactions of Inorganic Radicals in Aqueous Solution" by P. Neta and Robert E. Huie, and also this reference: https://pubs.acs.org/doi/abs/10.1021/j100143a031?journalCode... . To quote from Neta:

"3.11. Sulfite Radical
The .SO3− radical is most commonly produced by the reaction of .OH with sulfite or bisulfite.
.OH + SO3(2−) → OH− + .SO3−"

And further:
"3.13. Peroxomonosulfate Radical
The SO5.− radical is produced by the reaction of .SO3− with O2.
.SO3− + O2 → SO5.−"

Also:
"The SO5.− radical is a stronger oxidant than .SO3−, with an estimated redox potential of 1.1 V at pH 7.39. It is still quite selective in its reactions, oxidizing hydroxybenzenes at high pH rapidly but only slowly or not at all at low pH."

The radical apparently reacts by oxygen-atom transfer with sulfite or bisulfite, which is not desirable for the current application, so avoid an excess sulfite as HSO4- is not the target reagent to dissolve the tin:

SO5.− + SO3(2−) → SO4.− + SO4(2−)

Instead, at this point, I suggest try adding CO2 forming a bicarbonate ion to react with the SO5.- radical:

SO5.− + HCO3- --?→ HSO5− + CO3.- (paralleling Eq. R6a in reference above)

Or, add methanol or ethanol (based on reported reaction with SO4•- radical):

SO5•- + CH3CH2OH --?→ HSO5- + CH3•CHOH

where it is hoped that sufficient HSO5- is created to dissolve Sn like the suggested mix of dilute H2O2 and strong H2SO4 (which may not be available).

For this AOP approach, there is perhaps some support, occurring at pH 3, provided in the case of copper (nano), in place of Sn which is higher in the reactivity series than Cu, but as a less reactive fine powder, reacting with acidic HSO5-, per this reference: Water Sci Technol. 2016 Oct;74(8):1946-1952, 'Radicals induced from peroxomonosulfate by nanoscale zero-valent copper in the acidic solution'., by Zhou P, et al, at https://www.ncbi.nlm.nih.gov/pubmed/27789895.

Also, interestingly, there has been some discussion previously on SM as to whether 'acidified oxone and piranha solution (H2SO4 + H2O2 mix) are indeed equivalent', see https://www.sciencemadness.org/whisper/viewthread.php?tid=71... .

[Edited on 2-6-2018 by AJKOER]