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teodor
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[*] posted on 15-7-2019 at 01:42
Strange behaviour of Na2S in a water solution


I bought 100g of what they say is Na2S * 9H2O. There is no information about purity. It is chewing gum like substance without any odor, white and it dissolves in water very slowly forming a clear solution. The solution also has no odor but smells H2S after some standing (probably, on the next day). Chemically it acts like sulfide : gives a white precipitate with ZnCl2 and a black one with Pb(NO3)2.

When I tried to prepare 2M solution of it I failed. After some concentration the solution instantly turns into suspension similar to a milk of lime. But upon addition few drops of water it also instantly becomes clear. It is around 1.5 M concentration when this transformation happens (rough 37g/100ml) at room temperature.

The data for Na2S solubility (from sciencemadness wiki and wikipedia) is : 18.6 g/100 mL (20 °C) . I believe it is for Na2S part (anhydrous).

So, I have 4 questions:
- does somebody have the same phenomenon with Na2S * 9H2O?
- what is the chemical / physical process which makes so rapid change between 2 forms: solution/suspension?
- is it possible to find some use of it , e.g. for trigger some reaction upon changing of solution concentration

I will delay to make precise analysis of this substance in a pot I bought, I'd like to get some ideas about this strange behaviour.



Na2S.jpg - 97kB


[Edited on 15-7-2019 by teodor]

[Edited on 15-7-2019 by teodor]
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[*] posted on 15-7-2019 at 02:15


Hmmm .. Well Na2S is often yellow, and has the rotten egg odor, so the missing smell is a bad sign for me. i don't have or work with Na2S,9H2O myne is the 60% so its arround 3H2O, has smell, dissolves very nice in water, and its yellow.
Meaby there is sulfide there but definitely not pure. Don't know what causes the suspension-Dissolution. Impurities ?
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[*] posted on 15-7-2019 at 07:13


why buy the $h1T just make it.add charcoal and Na2SO4 heat it to around 750C.
Na2S anhydrous.---->Na2S (aq) is very exothermic be careful!

[Edited on 15-7-2019 by rockyit98]
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[*] posted on 15-7-2019 at 07:31


Quote: Originally posted by rockyit98  
why buy the $h1T just make it



Those are golden words, about making the $h1T.

Just curious, why Brauer suggests to make Na2S by reacting Na and S in liquid ammonia.

Also, why you didn't mention to boil solution of NaOH and S in water with presence of alcohol, which probably is much more convenient method to get the $h1T :)

[Edited on 15-7-2019 by teodor2]
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[*] posted on 15-7-2019 at 09:00


Quote: Originally posted by teodor2  
Quote: Originally posted by rockyit98  
why buy the $h1T just make it



Those are golden words, about making the $h1T.

Just curious, why Brauer suggests to make Na2S by reacting Na and S in liquid ammonia.

Also, why you didn't mention to boil solution of NaOH and S in water with presence of alcohol, which probably is much more convenient method to get the $h1T :)

[Edited on 15-7-2019 by teodor2]

thiosulfate by product contaminate the Na2S

if curious, you can make sodium thiosulfate from sodium bisulfate

sodium bi-sulfate ----(heat 350c)--->Sodium pyrosulfate
Sodium pyrosulfate + charcoal ----(heat 750c)---> Na2S2
Na2S2 + air ----(heat 450c)--->sodium thiosulfate

[Edited on 15-7-2019 by rockyit98]
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[*] posted on 15-7-2019 at 09:53


Quote: Originally posted by rockyit98  

thiosulfate by product contaminate the Na2S


It is just my speculation - Na2S is soluble in alcohol (19.2g in CH3OH at 62 C - http://dx.doi.org/10.1021/je100276c ) but Na2S2O3 is not, so the equilibrium will be shifted.

This is interesting topic, rockyit98, but slightly different. I propose to discuss it in threads like "making Na2S" or "purification of Na2S" (I believe there are several of them on SM). Probably I will show you Vogel, page 197 "... at high temperatures sodium sulphide is readily oxidised to sodium sulphate" that probably is something to consider in both of your recipes.

If it is impurities, the behaviour I observed is quite strange, at least for me. Now the question "what are impurities in my sample" is more interesting for me than "how to get pure Na2S".

[Edited on 15-7-2019 by teodor2]

[Edited on 15-7-2019 by teodor2]
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[*] posted on 15-7-2019 at 11:23


Holy crap! (according to Ullmann):


Quote:

Depending on process conditions, the purity of the sodium monosulfide obtained can vary between 50 and 98 wt%.




Quote:

High-purity sodium monosulfide can be obtained by treating the reaction product with methanol to dissolve the sulfide, instead of water.



Attachment: Sulfides, Polysulfides, and Sulfanes.pdf (298kB)
This file has been downloaded 333 times

[Edited on 15-7-2019 by icelake]
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[*] posted on 15-7-2019 at 13:16


Thank you icelake. Nice discovery. I am not surprised with what you wrote, but this is also interesting:

Other aliphatic, aromatic, and polyhydroxy
alcohols can also be used as extractants: the
solubility of Na2S in ethanol is 90 g/L, and in
ethylene glycol >200 g/L. The sodium sulfide
obtained by evaporating the extractant is free of
water, iron, sulfite, sulfate, and thiosulfate.

Anyway, I would not bother myself with home production of sulfides except for the purpose to better study reactions involved. It could be a stinking process even if you can manage H2S safety.



[Edited on 15-7-2019 by teodor2]
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[*] posted on 15-7-2019 at 15:49


After digging into some references [48], I found these patents:

http://www.freepatentsonline.com/4908043.html

http://www.freepatentsonline.com/5173088.html


Quote:

An object of the present invention is to provide high-purity single crystals of anhydrous sodium sulfide which are unlikely to deliquesce and to oxidize.
Another object of the invention is to provide a method of producing high-purity single crystals of anhydrous sodium sulfide with ease.
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[*] posted on 16-7-2019 at 02:24


icelake, I think that your search skills are impressive. Could you search also which process can cause so rapid change between solution and suspension? I think there are at least 2 possibilities: polymerization and hydrate formation.
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[*] posted on 16-7-2019 at 12:09


Some comments with sources, starting with Atomistry.com on sodium sulphide (link: http://sodium.atomistry.com/sodium_monosulphide.html ) to quote:

“The solution of the sulphide in water has an alkaline reaction due to hydrolytic dissociation. Atmospheric oxygen converts the dissolved sulphide into thiosulphate, and electrolytic oxidation yields the sulphate. “

Also, per Atomistry.com on the created sodium thiosulphate (link: http://sodium.atomistry.com/sodium_thiosulphate.html ) to quote:

“Concentrated solutions of sodium thiosulphate are moderately stable, but in dilute solution atmospheric carbon dioxide tends to liberate the unstable thiosulphuric acid, a substance readily changed into sulphurous acid and sulphur. Siebenschuh states that, after being kept for fourteen days, protection from light ensures the stability of the solution.”

So, some of the reported observed end products should apparently include S (as a suspension) and SO3- upon boiling in air (with some carbon dioxide presence) of a sodium sulphide solution:

My take on the underlying chemistry per a review of some sources (noted below), starts with the assumption of some transition metal impurities, like iron, for example, possibly proceeding as follows:

Dissociation: Na2S + 2 H2O ⇌ 2 Na+ + 2 OH- + 2 HS-

And, the suggested reaction system:

Fe(lll) + HS- → Fe(ll) + HS•

O2 + Fe(ll) ⇌ O2•- + Fe(lll)

HS• + O2•- → S + HO2-

HO2- + H+ → H2O2

S + 2 H2O2 → H2SO3 + H2O (see https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3153037/ )

Fe(ll) + H2O2 + H+ → Fe(lll)+ OH• + H2O (fenton reaction with ferrous, fenton-like with other transition metal like Mn, Cu,...)

OH• + H2S → H2O + HS•

Note, HS• is transformed into elemental sulfur and H2O2 (via H+ taken from water plus HO2-) :

HS• + O2•- → S + HO2-

Also, Fe(lll) cations are recycled to Fe(ll) to feed a fenton reaction:

Fe(lll) + HS- → Fe(ll) + HS•

The possible recycling of transition metal ions means that even small amounts of a metal impurity could eventually produce visible side products with oxygen exposure in time.

Peer reviewed source on the above radical chemistry: see, for example, ‘Free Radicals and Chemiluminescence as Products of the Spontaneous Oxidation of Sulfide in Seawater, and Their Biological Implications’, by DAVID W. TAPLEY, at https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4295652/ .

Also, further speculated transient creation of colorful sulfur radical anions •Sn-, where n= 2,3 and 4 (see https://pubs.acs.org/doi/abs/10.1021/ic50189a042 ) from the action of solvated electrons acting on elemental sulfur in limited aqueous conditions:

S + e- ⇌ •S-

(k-1)S + •S- ⇌ •Sk(-) (k=2, 3 or 4, but most readily at k=3) (see also https://pubs.rsc.org/en/content/articlehtml/2016/sc/c6sc0024... )

The basis for this claim is per Wikipedia (https://en.wikipedia.org/wiki/Trisulfur):

"Other methods of production of S−3 include reacting sulfur with slightly dampened magnesium oxide.[12]"

As damp MgO has a high pH (I suspect the pKa of •[S3]- radical anion is high) and as MgO is a photo catalyst forming e- (and also electron holes, h+), this is the basis of my speculation.

However, due to its expected high pKa, in the presence of acid source of H+ (including water), it likely exist in some protonated form:

H+ + •S3(-) ⇌ •HS + 2 S (or •HS3)

And, with excess e-, also possible:

•HS + e- (aq) = -HS (aq)
-------------------------------------------------------------------

Now, the case of no transition metal impurities and just say UVC light, a related reaction system (see ‘Highly efficient method for oxidation of dissolved hydrogen sulfide in water, utilizing a combination of UVC light and dissolved oxygen’ by Yizhak Tzvi and Yaron Paz, in Journal of Photochemistry and Photobiology A: Chemistry, Volume 372, 1 March 2019, Pages 63-70, at https://www.sciencedirect.com/science/article/pii/S101060301... , in particular, Equations (1) to (10)). To quote from the referenced source:

“Results were explained by the following mechanism:
HS− + O2 --Light--> HS• + •O2− (1)
HS• + O2 → HO2• + S (2)
HS• + •O2−→ HO2− +S (3)
HS− + (x−1)S → H+ +Sx2− (4)
where x can be 2-5.
Sx2− +S ⇌ Sx + S2− (5)
Sx2− + 1.5O2 → SO32− + (x−1)S (6)
Sx2− + 1.5O2 → S2O32− + (x−2)S (7)
S2O32− ⇌ S + SO32− (8)
SO32− + 0.5O2 → SO42− (9)
S2O32−+ 2.5O2 → 2SO42− (10)
Once polysulfide is formed, it readily reacts with oxygen, yielding thiosulfate (reaction 7) and sulfite (reaction 6). The elemental sulfur formed in reactions 2 Experimental, 3 Results is consumed by the growing polysulfide chain (reactions 4, 5). This ensures that under the right conditions, HS− is constantly consumed without releasing of elemental sulfur. The thiosulfate and sulfite are further oxidized to give sulfate, the final product (reactions 9, 10). “

[Edited on 17-7-2019 by AJKOER]
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[*] posted on 16-7-2019 at 13:45


This might explain the milkiness of the solution.


Quote:

Avrahami and Golding [23] observed thiosulfate as the first product of the autoxidation of HS- at temperatures between 25 and 55 °C (pH = 11...13; CHS- = 10-4...10-3 mol/L). After longer reaction times sulfate was formed in addition, but neither polysulfides nor other intermediates were observed, although sulfite was assumed to be a primary product rapidly oxidized to give sulfate. Occasionally colloidal sulfur was formed as concluded from a milkiness of the solutions.


BTW, I double checked the solubility of the Nonahydrate salt in water and there was a huge discrepancy between ACS and FisherSci.

https://www.fishersci.co.uk/shop/products/sodium-sulfide-non...

https://pubs.acs.org/doi/10.1021/acsreagents.4370.20160601 (attached).

Attachment: On the Autoxidation of Aqueous Sodium Polysulfide.pdf (3.5MB)
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Attachment: Sodium Sulfide, Nonahydrate.pdf (158kB)
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[*] posted on 17-7-2019 at 01:23


Thank you Chem Science, rockyit98, icelake and AJKOER for the interest to this topic and the information you shared.


My personal opinion with this sample of Na2S * 9H2O is that it is stabilized with some cellulose-derived substance. I saw some methods of Na2S stabilization where cellulose was mentioned: https://core.ac.uk/download/pdf/82621185.pdf . Both the consistence and phase transition suggest it could be a cellulose, but really I still unable to explain the concentration-dependability.

So, I plan to check it first by dissolving it in ethanol and check whether I will get any residue. And then I will be ready to discuss more.

[Edited on 17-7-2019 by teodor]
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[*] posted on 17-7-2019 at 15:02


Icelake:

Your reported accounts based on your cited reference are in agreement with mine from Atomistry.com (which is based on extracts from old chemical journals), albeit both sources are dated (yours appears to be based on science from the 1980s, based on the paper's references).

Also, another big clue is that the first equation in your referenced paper cites the S(2-) anion, which has been largely proven to be practically non-existent in solution. See, for example, 'Call to erase aqueous sulfide ion from chemistry - Spectroscopic developments question the existence of aqueous sulfide ions', by Alexandra Heaffey, February 2018 at https://www.chemistryworld.com/news/call-to-erase-aqueous-su... . Note, one of my cited sources is as recent as March, 2019.

[Edited on 17-7-2019 by AJKOER]
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[*] posted on 18-7-2019 at 01:38


So, I poured a little methylated spirit (5% MeOH, 4% H2O, 91% EtOH) on a watch glass and stirred a bit of "Na2S * 9H2O" from https://www.deoplosmiddelspecialist.nl into it. The gum was almost immediately converted to a lime-like powder, with sand-over-glass feeling when I rubbed it. I decanted the solution to another watch glass and allowed it to evaporate.

On the morning the alcohor was evaporated completely and I studied the residue under a microscope. There are 2 types of compounds on the second watch glass: the border is the same lime-like (white) substance (I didn't do filtration, so I suspect it is just insoluble residue decanted in small amount as suspension from the first watch glass) and the rest is needle - like crystalls, similar to the paint frost does on windows.
The lime-like substance has a form of balls like balls of ice - they are transparent, just like ice or glass, but the surface is rough. I suspect it is not crystalline but amorphous.


So, I have 2 thoughts:

- the gum->sand like transformation suggest that it is unlikely cellulose, but I will try to do another test (buy 1-naphthol?)
- the impurity in the lime (sand) like form is absolutely insoluble in water - otherwise I would get it in the center of the second watch glass after evaporation of the alcohol (because it contains 4% water and absorbs more upon standing).


My future plans:
- do the same with water-free MeOH and study the residue solubility
- react with HCl (I am aware about H2S), do the cations analysis


[Edited on 18-7-2019 by teodor]
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[*] posted on 18-7-2019 at 02:12


Quote: Originally posted by icelake  
Quote:

High-purity sodium monosulfide can be obtained by treating the reaction product with methanol to dissolve the sulfide, instead of water.






No, it is not more should be considered true. According to http://dx.doi.org/10.1021/je100276c

3/4 parts of dissolved substance will be CH3ONa, so these old recipes should be revised.



[Edited on 18-7-2019 by teodor]

[Edited on 18-7-2019 by teodor]
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[*] posted on 18-7-2019 at 02:54


Quote: Originally posted by AJKOER  


Now, the case of no transition metal impurities and just say UVC light, a related reaction system (see ‘Highly efficient method for oxidation of dissolved hydrogen sulfide in water, utilizing a combination of UVC light and dissolved oxygen’ by Yizhak Tzvi and Yaron Paz, in Journal of Photochemistry and Photobiology A: Chemistry, Volume 372, 1 March 2019, Pages 63-70, at https://www.sciencedirect.com/science/article/pii/S101060301... , in particular, Equations (1) to (10)). To quote from the referenced source:

“Results were explained by the following mechanism:
HS− + O2 --Light--> HS• + •O2− (1)
HS• + O2 → HO2• + S (2)
HS• + •O2−→ HO2− +S (3)
HS− + (x−1)S → H+ +Sx2− (4)
where x can be 2-5.
Sx2− +S ⇌ Sx + S2− (5)
Sx2− + 1.5O2 → SO32− + (x−1)S (6)
Sx2− + 1.5O2 → S2O32− + (x−2)S (7)
S2O32− ⇌ S + SO32− (8)
SO32− + 0.5O2 → SO42− (9)
S2O32−+ 2.5O2 → 2SO42− (10)
Once polysulfide is formed, it readily reacts with oxygen, yielding thiosulfate (reaction 7) and sulfite (reaction 6). The elemental sulfur formed in reactions 2 Experimental, 3 Results is consumed by the growing polysulfide chain (reactions 4, 5). This ensures that under the right conditions, HS− is constantly consumed without releasing of elemental sulfur. The thiosulfate and sulfite are further oxidized to give sulfate, the final product (reactions 9, 10). “

[Edited on 17-7-2019 by AJKOER]


since when UVC reaches earth soil? UVC is totally absorbed by the thermosphere and mesosphere, maybe some reaches the stratosphere but none reaches the ground, in a sealed plastic container





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[*] posted on 18-7-2019 at 05:26


Quote: Originally posted by teodor  
Quote: Originally posted by icelake  
Quote:

High-purity sodium monosulfide can be obtained by treating the reaction product with methanol to dissolve the sulfide, instead of water.






No, it is not more should be considered true. According to http://dx.doi.org/10.1021/je100276c

3/4 parts of dissolved substance will be CH3ONa, so these old recipes should be revised.



[Edited on 18-7-2019 by teodor]

[Edited on 18-7-2019 by teodor]


Sodium Sulfide, Carbonate and Cyanide may react with alcohols - no surprises there - but that 3/4 (Methoxide) has nothing to do with the solid phase (Na2S and NaHS).


Quote:

Only two components, namely, sodium sulfide and sodium hydrosulfide, were found in the solid phase of the heterogeneous Na2S + ROH reaction systems. The absence of sodium alkoxide in the solid phase was confirmed in a separate trial.


[Edited on 18-7-2019 by icelake]
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[*] posted on 18-7-2019 at 07:30


Quote: Originally posted by icelake  
Quote: Originally posted by teodor  
Quote: Originally posted by icelake  
Quote:

High-purity sodium monosulfide can be obtained by treating the reaction product with methanol to dissolve the sulfide, instead of water.






No, it is not more should be considered true. According to http://dx.doi.org/10.1021/je100276c

3/4 parts of dissolved substance will be CH3ONa, so these old recipes should be revised.



[Edited on 18-7-2019 by teodor]

[Edited on 18-7-2019 by teodor]


Sodium Sulfide, Carbonate and Cyanide may react with alcohols - no surprises there - but that 3/4 (Methoxide) has nothing to do with the solid phase (Na2S and NaHS).


Quote:

Only two components, namely, sodium sulfide and sodium hydrosulfide, were found in the solid phase of the heterogeneous Na2S + ROH reaction systems. The absence of sodium alkoxide in the solid phase was confirmed in a separate trial.



[Edited on 18-7-2019 by icelake]


I don't get you. "Sodium methylate is a white amorphous powder" - https://pubchem.ncbi.nlm.nih.gov/compound/Sodium-methoxide . I believe you will get it together with Na2S after evaporation of MeOH.


[Edited on 18-7-2019 by teodor]
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[*] posted on 18-7-2019 at 08:25


Wouldn't that suggest a mechanism where sulfide pulls a proton from methanol and evaporate as H2S? I doubt sulfide is a strong enough base to do that... But I have been corrected before.

And ofcourse everything is in some sort of a equilibrium, if enough methanol is added at some point almost pure NaOCH3 would be left wouldn't it? If above is correct.

Edit:
I guess the above will stop at the point of NaHS... So you do get a mixture but it will be impossible to separate.

[Edited on 18-7-2019 by Tsjerk]
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[*] posted on 18-7-2019 at 09:22


Excerpt from U.S. Patent 2278550:


Quote:

2.5 parts of solid anhydrous NaHS precipitates from such a solution and the remaining 1.9 parts of NaHS and the 15.6 parts of Na2S are precipitated from the Solution by concentration in a manner similar to that disclosed in Example 1A.


You'll need to evaporate some of the methanol to precipitate NaHS and Na2S, evaporating all of it doesn't make any sense.
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[*] posted on 19-7-2019 at 06:29


Quote: Originally posted by icelake  

Quote:

2.5 parts of solid anhydrous NaHS precipitates from such a solution and the remaining 1.9 parts of NaHS and the 15.6 parts of Na2S are precipitated from the Solution by concentration in a manner similar to that disclosed in Example 1A.



Well, I don't understand this patent. How we can precipitate carbonate (in the example 1A) from saturated solution of methoxide without getting methoxide also.

Let's assume it works somehow. Do you know how to check that precipitation (carbonates, sulfides, doesn't matter) doesn't contain methoxides? Do you know some specific qualitative test for methoxides?
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[*] posted on 19-7-2019 at 10:14


Well, the authors (Kurzin, et al.) have relied heavily on this crap (fifth reference, aka, "U.S. Patent 2278550") and this is one of reasons you should take everything with a grain of salt.

Quote:

Do you know how to check that precipitation (carbonates, sulfides, doesn't matter) doesn't contain methoxides?

According to literature CH3ONa is insoluble in benzene but the authors have decided to use benzene to extract the methoxide from Na2S/NaHS:
Quote:
After filtering the suspension, the precipitate was extracted by dry benzene, the solvent was removed in a vacuum, and the residue was treated with water and analyzed by titrimetry. No strong base was discovered.


Essentially they extract the methoxide from precipitate, evaporate the benzene in vacuum, hydrolyze the residue in water and analyze the hydrolysis products by titration.

Quote:
Do you know some specific qualitative test for methoxides?

They catch fire and char easily on a hot plate.
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[*] posted on 20-7-2019 at 05:04


Quote: Originally posted by Ubya  


since when UVC reaches earth soil? UVC is totally absorbed by the thermosphere and mesosphere, maybe some reaches the stratosphere but none reaches the ground, in a sealed plastic container


True, but lab light appears to produce remarkable decomposing ability on some sulfides, some of which that appear to be seemingly impervious to even pure oxygen treatment alone (hence the use of a combined O2/light protocol).

See, for example, commentary on Cadmium sulfide under lab light at https://books.google.com/books?id=GClzU2-Rj18C&pg=PA487&... with reports in some labs of 50% to 90% decomposition! The effect of the light is described as 'immediate' and 'variable'.
----------------------------------------------

Also, the cited author's reaction with oxygen should more likely be viewed as a simplification, say for example, with x=2 the reaction:

S2 (2−) + 1.5 O2 → SO3 (2−) + S

as shown more generally in Reaction (6). The sulfur is actually in the form, I suspect, of a colloidal suspension, so S(s). As such, likely in the presence of light and charge attributes of colloids, a possible presence of aqueous electrons, e-, which with dissolved oxygen, my take on some of the reactions:

O2 (aq) + e- (aq) = •O2- (aq) (see Eq (1) at https://pubs.acs.org/doi/full/10.1021/acs.chemrev.5b00407 )

S (s) + •O2- (aq) = •SO2- (aq)

•SO2- + O2 <--> SO2 + •O2- (see Eq (48) at https://pubs.acs.org/doi/full/10.1021/acs.chemrev.5b00407 )

•SO2- + •SO2- = S2O4 (2-) (same source, per comment)

And, per Wikipedia (https://en.wikipedia.org/wiki/Dithionite ):

"Dithionite undergoes acid hydrolytic disproportionation to thiosulfate and bisulfite:[citation needed]

2 S2O4 (2-) + H2O --> S2O3 (2−) + 2 HSO3− "

.............

[Edited on 20-7-2019 by AJKOER]
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[*] posted on 21-7-2019 at 07:47


Quote: Originally posted by AJKOER  

......
Also, Fe(lll) cations are recycled to Fe(ll) to feed a fenton reaction:

Fe(lll) + HS- → Fe(ll) + HS•

The possible recycling of transition metal ions means that even small amounts of a metal impurity could eventually produce visible side products with oxygen exposure in time.

Peer reviewed source on the above radical chemistry: see, for example, ‘Free Radicals and Chemiluminescence as Products of the Spontaneous Oxidation of Sulfide in Seawater, and Their Biological Implications’, by DAVID W. TAPLEY, at https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4295652/ .

Also, further speculated transient creation of colorful sulfur radical anions •Sn-, where n= 2,3 and 4 (see https://pubs.acs.org/doi/abs/10.1021/ic50189a042 ) from the action of solvated electrons acting on elemental sulfur in limited aqueous conditions:

S + e- ⇌ •S-

(k-1)S + •S- ⇌ •Sk(-) (k=2, 3 or 4, but most readily at k=3) (see also https://pubs.rsc.org/en/content/articlehtml/2016/sc/c6sc0024... )

The basis for this claim is per Wikipedia (https://en.wikipedia.org/wiki/Trisulfur):

"Other methods of production of S−3 include reacting sulfur with slightly dampened magnesium oxide.[12]"

As damp MgO has a high pH (I suspect the pKa of •[S3]- radical anion is high) and as MgO is a photo catalyst forming e- (and also electron holes, h+), this is the basis of my speculation.

However, due to its expected high pKa, in the presence of acid source of H+ (including water), it likely exist in some protonated form:

H+ + •S3(-) ⇌ •HS + 2 S (or •HS3)



Came across a new 2019 reference on some of the radical chemistry with sulfur (which I likely will reference in the future). In particular, the sulfiyl radical is described as HS•/S•-(see page 6 at https://pdfs.semanticscholar.org/533e/9a0b2e5d938abc555e267f... ) due to possible protonation:

HS• = H+ + S•- pKa = 3.4 (+/- 0.7) (Source: Page 436, at https://www.bnl.gov/isd/documents/92710.pdf )

Also, importantly the associated reaction of interest with oxygen is given by:

HS•/S•− + O2 → SO2•- (+ H+) (Source: See Page 7, Eq (7) at https://pdfs.semanticscholar.org/533e/9a0b2e5d938abc555e267f... )

So, please disregard my prior, not sourced, reaction above leading to SO2•- (namely, S (s) + •O2- (aq) = •SO2- (aq) ).

Then, as previously detailed, it is possible for •SO2- to form S2O4 (2-), which undergoes acid hydrolytic disproportionation to thiosulfate and bisulfite.

[Edited on 21-7-2019 by AJKOER]
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