Sciencemadness Discussion Board - Strange behaviour of Na2S in a water solution

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]

rockyit98 - 15-7-2019 at 09:00

Quote: Originally posted by teodor2

why buy the $h1T just make it

[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]

teodor2 - 15-7-2019 at 09:53

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]

icelake - 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 669 times

[Edited on 15-7-2019 by icelake]

teodor2 - 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]

icelake - 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.

teodor - 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.

AJKOER - 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]

icelake - 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; C_HS- = 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)
This file has been downloaded 379 times

Attachment: Sodium Sulfide, Nonahydrate.pdf (158kB)
This file has been downloaded 819 times

teodor - 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]

AJKOER - 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]

teodor - 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]

teodor - 18-7-2019 at 02:12

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]

Ubya - 18-7-2019 at 02:54

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

icelake - 18-7-2019 at 05:26

Quote: Originally posted by teodor

Quote:

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

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]

teodor - 18-7-2019 at 07:30

Quote: Originally posted by teodor

Quote:

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

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:

[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]

Tsjerk - 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]

icelake - 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.

teodor - 19-7-2019 at 06:29

Quote:

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?

icelake - 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.

AJKOER - 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]

AJKOER - 21-7-2019 at 07:47

......
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]

Ubya - 21-7-2019 at 12:54

it is even cited on wikipedia, but cadmium sulphide is the catalyst to the decomposition of hydrogen sulphide in solution, it doesn't decompose itself (it was used as a pigment in the 1800, so if it was that unsable it would have never been used as a pigment...) and it does not decompose any sulphide, only hydrogen sulphide is mentioned, on wikipedia is even proposed a mechanism

AJKOER - 21-7-2019 at 14:45

Quote: Originally posted by Ubya

The CdS in paint is not photo-active polycrystalline CdS, and sunlight is not identical to lab light, as apparently CdS is photo-active to select wave lengths of light. Here is a study (link https://aip.scitation.org/doi/abs/10.1063/1.330618 ), to quote:

"The mechanism of photoconductivity in polycrystalline CdS has been studied over the temperature range 100–300 K using Hall‐effect and conductivity measurements in the dark and under white light illumination....The detailed variations of μ and n̄ are interpreted in terms of the Seto model with the added hypothesis that photogenerated holes...."

The creation of electron holes (h+) means in the present of say water vapor:

H2O <--> H+ + OH-

the OH- ion becomes an hydroxyl radical (•OH), and HS- ion becomes the •HS radical which may further react with oxygen creating the powerful reducing radical, •SO2-, as detailed above.

With photo-generated electrons (e-), H+ becomes the hydrogen atom radical. (•H), oxygen in air becomes the superoxide radical anion....

But, you are correct that light itself does not change CdS as h+ cancels e-. There is, nevertheless, possible disruption to intended preparations and associated chemical reactions with the CdS from photo created products.

This is likely implied in my referenced ebook, to quote:

"5.1.2 Cadmium sulfide is not appreciably oxidized even when aspirated with pure oxygen in the dark. However, exposure of a bubbler containing cadmium sulfide to a laboratory or to more intense light sources produces an immediate and variable photo decomposition. Losses of 50 to 90% of added sulfide have been routinely reported by a number of laboratories.”

[Edited on 22-7-2019 by AJKOER]

[Edited on 22-7-2019 by AJKOER]

icelake - 21-7-2019 at 22:25

Quote:

"5.1.2 Calcium sulfide is not appreciably oxidized even ..."

It's Cadmium sulfide not Calcium sulfide.

and more importantly, what does CdS (insoluble in water & used in photoresistors) have anything to do with Na₂S?

AJKOER - 22-7-2019 at 01:37

Quote:

"5.1.2 Calcium sulfide is not appreciably oxidized even ..."

It's Cadmium sulfide not Calcium sulfide.

and more importantly, what does CdS (insoluble in water & used in photoresistors) have anything to do with Na₂S?

Thanks, my advanced logic talk-to-text software I used to reproduce the extracted text from the ebook made a substitution. Apparently, healthy calcium is a much more commonly referenced than toxic cadmium.

I have corrected it in my extract.
-----------------------------------------

Per this extract from 2018 at https://pubs.rsc.org/en/content/articlelanding/2018/cc/c8cc0... :

"shows highly selective photooxidation of sulfides"

implies to me that sulfides, in general, may have associated photo sensitivity properties. Iodide salts also have varying light sensitivity.

Work on sensor development (see https://www.mdpi.com/1424-8220/16/3/296/htm ) suggest the use of tin sulfide, in particular, and states more generally:

"The results highlighted the possibility to use metal sulfides as a novel class of sensing materials, owing to their selectivity to specific compounds, stability, and the possibility to operate at room temperature."

This source (https://materion.com/resource-center/product-data-and-relate... ) notes more generally:

"Some sulfides possess semi-metallic character and have potentially valuable electronic properties."
---------------------------------------------------------

Interestingly, a web search cites Wiki as a source claiming that CdS is "sensitive to visible and near infrared light". An MSDS to quote (source: https://www.fishersci.com/store/msds?partNumber=AC310180100&... ):

"Stability - Stable under normal conditions. Hygroscopic. Air sensitive. Light sensitive.
Conditions to Avoid - Exposure to air. Exposure to light. Incompatible products. Exposure to moist air or water."

The current Wikipedia (https://en.wikipedia.org/wiki/Sodium_sulfide ) on Na2S notes oxygen sensitivity, so per Wiki a combination of O2/UVC is not even needed for its decomposition.

Also, as to mechanics for decomposition of the pure product, since aqueous Na2S is alkaline, and given the photo sensitivity, albeit low, of the OH- ion (to UV):

OH- + hv --UV Light--> •OH + e-

as per a source ( https://pubs.rsc.org/en/content/articlelanding/2017/cp/c7cp0... ) to quote: "Irradiation by the light at the Urbach tail excites an electron out of the hydroxide ion, leaving a neutral OH radical behind." Then:

•OH + H2S --> H2O + •HS (Source (link fixed) at p. 7, Eq (9) at https://pdfs.semanticscholar.org/533e/9a0b2e5d938abc555e267f... )

•HS/•S- + O2 --> •SO2- (+ H+) (Source (link fixed) at p. 7, Eq (7) at https://pdfs.semanticscholar.org/533e/9a0b2e5d938abc555e267f... )

which produces but a limited contribution, in my opinion, to so called 'oxygen sensitivity'. In fact, as one source notes, there is no significant O2 reaction until the creation of colloidal sulfur (with the presence of oxygen and say an iron impurity, or e- , generated from the sulfur's colloidal electrostatic charge, or light,..). Then, with associated polysulfides, in the presence of solvated electrons, possible creation of the very light sensitive colorful sulfur radical anions, •Sn-, where n= 2,3 and 4 (see https://pubs.acs.org/doi/abs/10.1021/ic50189a042 ). These colorful sulfur radicals likely, per a modified Eq (7) above, then could lead to •SO2- formation as follows:

•Sn- + O2 --> •SO2- + Sn-1

Ultimately as previously detailed above, all per very recent cited work (2017, 2019,..), it is possible for •SO2- to form S2O4 (2-), which undergoes acid hydrolytic disproportionation to thiosulfate and bisulfite, which are cited decomposition products.

[Edited on 22-7-2019 by AJKOER]

icelake - 22-7-2019 at 12:19

The highlights:

Quote:

Recently, coordination polymers (CPs) have opened up remarkable performance in the growing field of photocatalysis, which is primarily benefited from the apt incorporation of various chromophores and/or functional groups in organic linkers.

Quote:

We describe herein a novel visible-light-photoactive coordination polymer for triggering selective aerobic transformation of sulfides to generate sulfoxides. Orange strip-shaped crystals of 1 were prepared by a solvothermal reaction of Zn(NO₃)₂.6H₂O and 4,40-(anthracene-9,10-diylbis(ethyne-2,1-diyl))dibenzoic acid (ADBEB) in a mixed solvent of H₂O/CH₃CN/DMA (DMA =N,N-dimethylacetamide). 1 can also be synthesized in the mixed solvent of H₂O/DMA with a lower quality and yield, but cannot be synthesized using DMA solely.

Quote:

The capability of 1 serving as a visible-light photocatalyst to carry out aerobic oxidation of sulfides was then investigated, where a typical sulfide of thioanisole (PhSMe) was used as a model substrate and the conversion of thioanisole was monitored via gas chromatography-mass spectrometry (GC-MS).

Quote:

The use of zinc nitrate, i.e. the metal salt for the synthesis of 1, as the catalyst cannot initiate the reaction, which indicates that the photocatalytic activity of 1 results from the organic linker.

TLDR: Photooxidation of R-S-R to R-S=O-R using 1 as a photocatalyst. So much in common with the current discussion (Sodium sulfide), amiright?

Attachment: Selective photooxidation of sulfides mediated by singlet oxygen using visible-light-responsive coordination polymers.pdf (2MB)
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teodor - 22-7-2019 at 12:22