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semiconductive
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Easy sulfite ion in a pinch...
I have sodium sulfite, sodium metabisulfite, and sulfamic acid (no sodium).
I don't have any free source of sulfite ions that don't have a strong alkyl or amino group attached to them.
I got to thinking, maybe I can remove the sodium atoms in a clean way?
eg: Sodium Sulfite salts dissolve only slightly in alcohol and sulfamic acid is soluble in warm alcohol (etoh < 100 C).
But I also read that sulfite ion is very soluble in alcohol by itself.
Could I make free sulfite ions in alcoholic solution by adding finely powdered sodium sulfite or metabisulfite, to a warm beaker of (50 C) denatured
alcohol that has 5% sulfamic acid dissolved in it?
I'm thinking sodium sulfamate is insoluble in alcohol, but both sulfamic acid and sulfite are soluble. So, if I add not quite enough sulfamic acid to
replace all the sodium in the sulfites; I should end up (eventually) with free sulfite ions in alcohol and a mixed precipitate. (perhaps after
chilling to precipitate out as much sodium as possible).
Or would sulfamic acid act as a catalyst, even at 50C temperature, and cause the sulfite ions to react with the alcohol (which sort of defeats the
purpose by producing water and an ester).
Is there an easy way to check / test the result ?
[Edited on 24-11-2025 by semiconductive]
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DraconicAcid
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Quote: Originally posted by semiconductive  | I have sodium sulfite, sodium metabisulfite, and sulfamic acid (no sodium).
I don't have any free source of sulfite ions that don't have a strong alkyl or amino group attached to them. |
Sodium sulphite does not have an alkyl or amino group attached to it. What are you trying to do?
If you react sulphite with an acid, you get sulphur dioxide, which isn't useful in most circumstances.
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semiconductive
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Sigh. alkaline metal (sodium). It's a typo.
In the text I clearly stated what I'm trying to do.
Sulphur dioxide, when dissolved in alcohol, can be ionized according to AI searches.
If I start with a salt (sulfite), then presumably a non-colloidal looking solution in ethanol has molecules of sodium sulfite or (at least) , or of 2
x Na+ and SO₃²⁻ ions at most, in the case of metabisulfite -- I'm not sure what ion I'll get, but it will have sodium and a sulfur oxide
mixture of some kind.
Can I remove the sodium ions while keeping sulfite ions in solution, without it reducing to an ester that isn't ionized ?
AKA: Make a conductive liquid in alcohol, such as I use in my electrodeposition of Nickel thread.
Thanks.
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DraconicAcid
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You cannot have sulphite ions in solution, regardless of the solvent, without a counterion. You could replace the sodium ions with potassium ions,
but you can't get rid of the cation completely.
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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bnull
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Do not trust AI. It tends to select sentences that conform to positions it has defined beforehand. I don't know how this works but that's what I have
seen so far. It also provides sources even when they contradict said position. ChatGPT is very good with translations, and I suppose Grok would do the
same if it wasn't for its stiff right arm. Gemini is like a confused elderly relative. They only provide information, not knowledge, and a good chunk
of it is wrong.
[***]
Sulfur dioxide is soluble in ethanol. As far as I know, it doesn't ionise there. If AI said it does, ask for the sources and carefully check them out
one by one. But beware: I have seen non-existent articles from non-existent publications and authors given as sources by ChatGPT.
Edit: You're not using absolute ethanol, you're using the azeotrope, so perhaps, only perhaps, you may get sulfite ions because of the reaction of
sulfur dioxide with water. But sulfurous acid may be strong enough to catalyze esterification to diethyl sulfite.
[Edited on 24-11-2025 by bnull]
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semiconductive
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Yes.. Yes... balance of charges. Very helpful.
To be extra clear:
Alkyl groups and Alkaline metals are both cations.
I don't want the sodium cation.
Which is why I was thinking about getting the sodium to precipitate with the sulfamic acid ion to become insoluble sodium sulfamate ( perhaps in very
cold alcohol ).
An ioninzed (lysed) alcohol, ethanol, or hydrogen, as a replacement ion for sodium is fine in my book. If the alcohol (ethanol) is reduced to an
alkane without one hydrogen because the oxygen goes away .... I'm totally happy and don't care. I'd say "yaay."
I'm just not happy with ester where *both* of the sulfite charges are neutralized.
I want an *ion*.
I'm in a semi-conductive ion mood right now.
Because I don't have sulfur dioxide gas, I don't think sulfur dioxed is ionized, and I don't know how to make it ionized at 10:00PM my time as I
prepare to go to sleep without poisoning myself (again.).
I just want to steal some already ionized sulfite or maybe S₂O₅²⁻ (if that exists as a true radical in solution), AKA: to experiment with
sulfer oxide an-ions. (for they are attracted to the anode).
Then I can have sweet dreams.
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bnull
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| Quote: | I'm just not happy with ester where *both* of the sulfite charges are neutralized.
I want an *ion*. |
There's no ethyl bisulfite. It is all or nothing in this case.
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semiconductive
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duplicate post deleted. Internet problems.
[Edited on 24-11-2025 by semiconductive]
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semiconductive
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| Quote: | | Edit: You're not using absolute ethanol, you're using the azeotrope, so perhaps, only perhaps, you may get sulfite ions because of the reaction of
sulfur dioxide with water. But sulfurous acid may be strong enough to catalyze esterification to diethyl sulfite. |
The internet is really bad here ( Nov 2025 ) -- the sciencemaddness forums have been inaccessible for hours at a time every other day over a week; but
the opening page to the site is just fine. I don't get it.
Note:
I do have access to expensive absolute alcohol, but I'm afraid to open the cap because --- it won't be absolute anymore.
The hardware store stuff is cheap, contaminated with a bit of methanol, and I can put twice baked at 350 [C] magnesium sulfate salt into it which
sinks to the bottom. After first baking, I powder it the MgSO₄, after second baking ... I'm convinced it's dry.
I'm pretty sure the resistivity of the ethanol will rise above what my meter can read (20 mega ohm) once a bunch of salt is sitting at the bottom.
So, I'm guessing there won't be much water left. I generally do try to get rid of water ....
I will repeat experiments with the expensive stuff, after making my obligatory stupid mistakes multiple times with the cheap stuff. 1 gram of
Mg·SO₄ (anhydrous), 3 [CC's] denatured alcohol. I always put kerosene on top to keep moisture out as much as possible...
Yes, Bnull, water is always a problem.
But I think I can at least count on the common ion effect to reduce solubility of sulfates as much as possible ?
maybe not ?
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bnull
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| Quote: | | The internet is really bad here ( Nov 2025 ) -- the sciencemaddness forums have been inaccessible for hours at a time every other day over a week; but
the opening page to the site is just fine. I don't get it. |
It's not the internet. It is AI. The leeches scrape the forum a few times a month, leaving the forum barely functional.
Absolute alcohol may prove to be utter useless in this case. You need a little water to make (bi)sulfite, which is in equilibrium with dissolved
sulfur dioxide. No water means no (bi)sulfite.
An alternative is to use a cation that is inert during electrolysis. Quaternary ammonium, for example. I don't know if it forms a sulfite or
bisulfite.
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semiconductive
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| Quote: | | There's no ethyl bisulfite. It is all or nothing in this case. |
Hmmm.... my thinking:
Sulfite is a triagonal bi-pyramidal ion. It has a resonance structure, which means that the 'oxygens' carrying the two negative charges can shift
around the molecule pretty much instantaneously.
During bonding to a cation, one of the two negative charges of the anion will become fixed (neitralized) in a specific oxygen bond; , which means the
remaining negative ion charge either remains in resonance among two oxygens or else something (unknown) disrupts the symmetry and one oxygen becomes
will more polar negative than the other.
An alcohol can loose either a hydrogen, or the hydroxide group as a whole.
( I've never understood which is more likely, or why. )
Presumably, if the hydrogen leaves then the alcohol is acting as a proton donor (acid). If the hydroxide leaves, the alcohol acts as a base and
becomes an alkyl group; R-⁺ + OH⁻ = C₂H₅⁺ + OH⁻
I know Draconic Acid mentioned some years ago that hydrogen doesn't leave by itself; and I'm thinking, without the presence of water, hydronium
molecules aren't going to form easily.
But, even then, It is possible to imagine two ethanol molecules to collide and one of them loose an 'H' while the other looses an 'OH', thus giving a
temporary situation of R-O-⁻ + H₂O + R⁺.
In that case, the very presence of the water molecule is what prevents the system of two ions from being an ester immediately. ( Shortly thereafter,
it might become one if the water is removed or kinetically leaves due to heat. )
If the two alcohol collision arose with a sulfite ion nearby , I don't see why the R⁺ would not be attracted to it while the R-O-⁻ was repelled
by it.
I don't quite get why would a single positiviely charged R⁺ ion would not be attracted to a sulfite ion (-2) ?
I'm not asking that an ethyl bisulfite molecule be isolated from the solution. I'm fine if it's a so called 'phantom' molecule. I merely asking why
this 'phantom' molecule can't exist as a loose association of ions that is never isolatable.
eg:
A wandering sodium ion, when it does hit a sulfamic acid anion ( negatively charged ) , will form a very stable structure; eg: otherwise, it would be
easily soluble in alcohol -- and -- well, it isn't. So sodium sulfamate has to be pretty stable compared to alcohol.
I'm thinking, the big issue here is whether the stability of the sodium sulfamate molcule is enough to remove the sodium from solution or not. Isn't
this just a matter of solubility ? The less soluble it is, the stronger the bond must be ?
Please elucidate, what exactly prevents a liquid ethyl bisulfate from existing in solution (dissolved only) if sodium is removed by sulfamic acid
precipitation.
Your knowledge is beyond mine, or I've forgotten something.
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DraconicAcid
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Hmmm.... my thinking:
Sulfite is a triagonal bi-pyramidal ion. It has a resonance structure, which means that the 'oxygens' carrying the two negative charges can shift
around the molecule pretty much instantaneously.
The charge is delocalized around the three oxygens. It's not shifting.
During bonding to a cation, one of the two negative charges of the anion will become fixed (neitralized) in a specific oxygen bond;
No. The anion does not form a covalent bond to the cation.
An alcohol can loose either a hydrogen, or the hydroxide group as a whole.
( I've never understood which is more likely, or why. )
Presumably, if the hydrogen leaves then the alcohol is acting as a proton donor (acid). If the hydroxide leaves, the alcohol acts as a base and
becomes an alkyl group; R-⁺ + OH⁻ = C₂H₅⁺ + OH⁻
Again, no. The alcohol can act as an acid and lose H+, or it can act as a base and gain H+. ROH + ROH <==> RO- +
ROH2(+). The equilibrium constant for that reaction is very small (several orders of magnitude lower than the autoionization of water). While one
could imagine the two ions then reacting to give an ether and a molecule of water, it's not actually going to happen.
I don't quite get why would a single positiviely charged R⁺ ion would not be attracted to a sulfite ion (-2) ?
They would be, but you're not going to form any extremely electrophilic and unstable carbocations in alcohol
solution.
I'm thinking, the big issue here is whether the stability of the sodium sulfamate molcule is enough to remove the sodium from solution or not. Isn't
this just a matter of solubility ? The less soluble it is, the stronger the bond must be ?
What cation are you going to replace the sodium with? The hydrogen ion from the acid? In that case, you're going to get
H2SO3, which will decompose to give sulphur dioxide.[color]
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semiconductive
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| Quote: | | The charge is delocalized around the three oxygens. It's not shifting. |
Um. I'm not a big fan of the "Wanted: Schrodinger's cat both dead and alive." interpretation. I find it often makes people claim to be impossible
some some things which are normally possible.
eg: When a sodium ion Na⁺ approaches a sulfite ion (2-), if the delocalized charge had to *Stay* delocalized, then I suspect the only place the
sodium ion could approach the atom would be the top or bottom of the trigonal pyramid ( symmetrically spaced from all the charges ).
Either the charge can shift or it can't, or maybe it's both shifted and not shifted??
For the bonds that eventually happens could include sodium near one of the oxygens -- and not necessarily sodium at the top of the pyramid and
equi-distant from all the oxygens.
I'm trying to be inclusive of possibilities rather than exclusive, when I don't know enough to be sure.
| Quote: |
No. The anion does not form a covalent bond to the cation.
|
I'm thinking: Covalent vs. Ionic is a matter of degree.
So, you've re-enforced the notion that the anion charge remains de-localized even when a sodium cation is in very close proximity to one of the
sulfite ion's oxygens while still being relatively far away from the Sulfur atom.
So, you're giving me new data. (for me).
Is this a 'totally' no change in delocalized charge denisty -- or is it a shifty 1% change which is not covalent -- but still, not zero ?? ( How
would I know? )
| Quote: | | Again, no. The alcohol can act as an acid and lose H+, or it can act as a base and gain H+. ROH + ROH <==> RO- + ROH2(+).
|
Ok, let's correct my misconception; for this may help me in the future make better guesses: Before, I said I don't know if there is an analog to the
hydronium ion, but you seem to be saying there is an alochol analog to the hydronium ion. It's ROH₂⁺. Correct ?
| Quote: | | The equilibrium constant for that reaction is very small (several orders of magnitude lower than the autoionization of water). While one could
imagine the two ions then reacting to give an ether and a molecule of water, it's not actually going to happen. |
Again, I wasn't even trying to say an isolatable ether gets formed.
But, I think you're giving me quantitative argument and not a qualitative one.
I don't have quantiative data -- especially when not talking grossly aqueous soltuions ; and I don't have your experience.
eg: As far as I know -- The relative re-ionizable magnitude of sodium sulfamate precipitate in alcohol may also be orders of magnitude smaller than
autoionization of water. But, if the ability to ionize the precipitate exists at all and is smaller than that of the ether reaction -- the reaction
could proceed ( but possibly very slowly).
This is the kind of thought that was going through my head before falling asleep last night.
AKA: Without your data (with no general citation) -- I have no rule of thumb to estimate the relative liklihood or make better predictions in the
future. Which I would like to be able to do.
| Quote: |
What cation are you going to replace the sodium with? The hydrogen ion from the acid? In that case, you're going to get H2SO3, which will decompose to
give sulphur dioxide.
|
I'm not sure of your point.
I'm experimenting, I will try many things.
I know H₂SO₃ does not exist in isolation; but I was not trying to isolate it.
H₂SO₃ has a neutral leaving group of H₂O -- and it follows that the molecule could split into SO₂ + H₂O.
I Agree.
But -- Is the mere presence of hydronium ions in solution enough to cause a nearby sulfite ion to decompose and leave solution ?
For, then SO₂ gas ought to be produced in proportion to the probability of the presence of hydronum squared in all sulfite solutions. I don't smell
much sulfur dioxide and I am not sure how big of a number I should assign to that as an estimate.
I'd like to learn how to predict based on what I can measure or detect at home.
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semiconductive
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Bnull:
One of the top three chemicals I thought of try as an 'cation' replacement was Choline, isolated from choline chloride. I think that's a quaternary
ammonium compound. I have a bag of it. So, yes... that's possible.
It's the same bag I bought to try choline chloride + Urea that Draconic recommended -- and which turned black instead of plating.
I've got all of urea, thiourea, choline chloride, oxalic acid, and a computer controlled thermometer with soldering iron and something called a
'schiff base' sitting on my desk. I even have a glass coated electrode that can be charged to +1126volts DC in order to attract anions in a thin
layer near it's surface.
Sorry, I've been trying to post this partial answer for over an hour. The AI stuff is basically starting to lock me out ... I can't even read the
site other than the login page. I'm quite frustrated right now! 
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chornedsnorkack
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| Quote: Originally posted by bnull |
Absolute alcohol may prove to be utter useless in this case. You need a little water to make (bi)sulfite, which is in equilibrium with dissolved
sulfur dioxide. No water means no (bi)sulfite.
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"Bi"sulfite or hydrogen sulphite?
Hydrogen sulphite has extra decay paths - more options to eliminate compared to sulphite
C2H5OH+SO2 <-> C2H5OSO2H
but I suspect the equilibrium would be on the left, towards elimination of SO2
| Quote: Originally posted by bnull |
An alternative is to use a cation that is inert during electrolysis. Quaternary ammonium, for example. I don't know if it forms a sulfite or
bisulfite. |
With excess of SO2, I suspect "bi"sulphite. But the equilibrium of these would depend on solubilities.
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bnull
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| Quote: | | "Bi"sulfite or hydrogen sulphite? |
Both are the same. And "(bi)sulfite" is shorthand for "bisulfite or sulfite or a mixture of both".
I don't know if $$C_2H_5OH+SO_2 \leftrightarrow C_2H_5OSO_2H$$ happens (zero indication so far and the only possible source I found is paywalled). I
was thinking of $$SO_2+H_2O \leftrightarrow H_2SO_3.$$ Sulfurous acid is unstable and decomposes to sulfur dioxide and water. Alcohol would work as a
dehydrating agent in this case.
What I don't know is if quaternary ammonium forms bisulfite or sulfite. Maybe both, maybe one of them, maybe none depending on the specific radicals.
I'm in the dark here.
[Edited on 25-11-2025 by bnull]
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chornedsnorkack
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Alkyl cations are very hard to get and very active.
| Quote: Originally posted by semiconductive |
Which is why I was thinking about getting the sodium to precipitate with the sulfamic acid ion to become insoluble sodium sulfamate ( perhaps in very
cold alcohol ).
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You´re considering very cold alcohol?
The liquid range of neat ethanol at 1 bar is from +78 to -114.
The liquid range of neat sulphur dioxide also at 1 bar is from -10 to -75.
| Quote: Originally posted by semiconductive |
Because I don't have sulfur dioxide gas, I don't think sulfur dioxed is ionized, and I don't know how to make it ionized at 10:00PM my time as I
prepare to go to sleep without poisoning myself (again.).
I just want to steal some already ionized sulfite or maybe S₂O₅²⁻ (if that exists as a true radical in solution), AKA: to experiment with
sulfer oxide an-ions. (for they are attracted to the anode).
Then I can have sweet dreams.
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You need a suitably polar solvent.
At about 20 Celsius, the dielectric permittivity of ethanol is 25.
At -10, the dielectric permittivity of sulphur dioxide is 16.
Not awfully good for ions but not quite intolerable either.
Certainly sodium ethoxide C2H5ONa has high solubility in ethanol (20%). Do sodium ethoxide solutions in dry ethanol conduct electricity and
electrolyze?
Would sodium ethoxide react with dry acidic oxides? Like
C2H5O-+SO2=C2H5SO3-
C2H5O-+CO2=C2H5CO3-?
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bnull
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Liquid sulfur dioxide dissolves quite a few salts, especially quaternary ammonium salts. See p. 40 of Waddington, Non-aqueous Solvent Systems (can be
borrowed at https://archive.org/details/nonaqueoussolven0000unse).
| Quote: | Would sodium ethoxide react with dry acidic oxides? Like
C2H5O-+SO2=C2H5SO3-
C2H5O-+CO2=C2H5CO3-? |
The second reaction is well known, it is a way to make alkyl carbonates. The first reaction is the problem. As far as I could find, reactions
involving sulfur dioxide and alkoxides result in dialkyl sulfites, which do not produce ions. If it happens the way you wrote, then sulfur dioxide is
essentially lost as the alkyl sulfonates are stable. Esylic acid (ethanesulfonic acid) is strong and stable.
| Quote: | | Certainly sodium ethoxide C2H5ONa has high solubility in ethanol (20%). Do sodium ethoxide solutions in dry ethanol conduct electricity and
electrolyze? |
S. Tijmstra wrote a paper about the conductivity of sodium methoxide and ethoxide in alcoholic solutions for the Zeitschrift für physikalische
Chemie, Volume 49, beginning from page 345 (https://archive.org/details/sim_zeitschrift_physikalische_ch...). My German is not that good and I'd probably miss a few words and mistranslate
the whole thing.
Edit: Typo.
[Edited on 25-11-2025 by bnull]
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DraconicAcid
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You might find sodium lauryl sulphonate (a common detergent ingredient) to be sufficiently soluble in alcohols.
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semiconductive
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| Quote: | The liquid range of neat ethanol at 1 bar is from +78 to -114.
The liquid range of neat sulphur dioxide also at 1 bar is from -10 to -75. |
I can get dry ice at the local store. That goes to -78.5 [°C].
I've tried it on ethanol and acetone before and got them in the -40's, I seem to recall them becoming slightly hetrogenous. But, that might be
because I didn't dry them before cooling.
Hmmm..
Water becomes ice which is a solid. But, it's highest density is at 4 [°C].
The spacing of atoms has to move very slowly in both situations. But, that means the math will be very close (qualitatively) to a solid making the
QM/Boltzmann math identical to semiconductors physics.
I'm going to try computing the auto-ionization of water from ice data, and see if I get an accurate approximation. If I do, then I'll assume alcohol
is less complicated (because of larger masses), and repeat the same procedure to approximate the properties of alcohol.
----
In solid state semiconductors the intrinsic carrier concentration function is well known:
Ni = C₁ · T^(3/2) · e^(- E/( 2·k·T ))
The energy gap (E) is a smooth function that decreases with temperature and usually has a linear and an Arrehnius factor in it. ( A rational
polynomial can easily model it given three data points. )
E ≈ a·T / ( c + b/T )
Looking up a bunch of data points from different authors, I curve fit a
linear log model of water auto-ionization near the freezing point of water:
(I have no way of knowing how accurate this is).
Bounding Kw ⪝ 3.008·( 3.661 - 1000/T )-14.94
Converting the intrinisic carrier concentration to a log10 formula (like pH):
Ni = C₁ · T^(3/2) · e^(- E/( 2·k·T ))
Assume E is in electron volts, just like a semiconductor:
Ni = ( Const + T^(3/2·ln(T) - E/( 2·k·T ) ) / ln(10)
log₁₀( Ni ) ≈ ArbitraryConst + 0.6514*ln( T ) - E·2553/T
The energy gap between H⁺ and OH⁻ should just be the difference in ionization potentials between neutral water and released ions.
Since hydrogen gas electrode is the standard reference at 0V for oxidation/reduction tables; I think I can get away with the E = energy required to
convert H₂ gas with aqueous hydroxide ions into water.
so Eg ≈ 0.83 volts at 25 [°C].
If I'm wrong we'll soon know:
2553·0.83 ≈ 2119
A crude theoretical calculation just assumes Energy gap (E) is constant from freezing to standard temperature 25 [°C]:
At freezing:
Kw ≈ -14.94
Kw = log₁₀( Ni ) ≈ ArbitraryConst + 0.6514*ln( 273.15 ) - 2119/273.15
-14.95 ≈ ArbitraryConst + 3.6543 - 7.7576
ArbitraryConst ≈ -10.85
Therefore, I have created a crude model for temperaturem in celcius:
crude Kw ≈ -10.84 + 0.6514·ln( 273.15 + Tc ) - 2119/( 273.15 + Tc )
This is slightly lower than the upper bound calculation. I am confident that the math is qualitatively correct. The slope is lower but not even off
by half, so I'm confident I can compute a refined model that will fit very well.
A refined model will modify the values of the first term (constant) and the last term (Arrhenius physics) to get the correct slope of ionization ; and
my experience with solids is that extrapolation is usually pretty accurate at higher temperatures. If it's also accurate in liquds we can estimate.
I'm going to just use freezing and standard lab temperature since the standard ionization potentials are known precisely. I'll post a refined
equation tomorrow. (see post below.)
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DraconicAcid
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For methanol, pK(autoionization) = 22.67
https://pubs.acs.org/doi/10.1021/acs.jpca.5c03979
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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clearly_not_atara
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| Quote: Originally posted by bnull | Liquid sulfur dioxide dissolves quite a few salts, especially quaternary ammonium salts. See p. 40 of Waddington, Non-aqueous Solvent Systems (can be
borrowed at https://archive.org/details/nonaqueoussolven0000unse).
| Quote: | Would sodium ethoxide react with dry acidic oxides? Like
C2H5O-+SO2=C2H5SO3-
C2H5O-+CO2=C2H5CO3-? |
The second reaction is well known, it is a way to make alkyl carbonates. The first reaction is the problem. As far as I could find, reactions
involving sulfur dioxide and alkoxides result in dialkyl sulfites, which do not produce ions. If it happens the way you wrote, then sulfur dioxide is
essentially lost as the alkyl sulfonates are stable. Esylic acid (ethanesulfonic acid) is strong and stable. |
I think you are confusing the alkyl sulfonates with the alkyl sulfites. What he hopes to produce is "monoethyl sulfite (-1)". The corresponding
hydrogen ethyl sulfite is probably strongly disfavored (sulfurous acid is basically not observed in solution, while carbonic acid is present to a
small extent). The SMILES CCOS(=O)O- is a plausible result of SO2 + EtO-, but the following rxn may destroy it:
2 CCOS(=O)O- >> CCOS(=O)OCC + SO3(2-)
But there are just no sulfite ions without counterions, and any variant of this would risk exposure to large quantities of SO2 gas. I am concerned
about recommending any procedure to someone who does not understand why.
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davidfetter
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STOP RIGHT THERE
If you're consulting AI for literally anything, you do not have the judgment needed to mess with chemistry. Doing so is a sign that you need to do
some pretty large reassessments of what you're doing with your life, what sources of information you trust, and what you use to establish that trust.
Chemistry can be extremely unforgiving, and AI will happily tell you to do things in that field that will kill you and could kill people near you.
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bnull
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No, I'm not confusing them. What @chornedsnorkack wrote was a reaction between an ethoxy group and sulfur dioxide with ethanesulfonate as product
(ethoxy loses its oxygen to sulfur and a bond is formed between sulfur and the carbon). As far as I know, it doesn't work that way. What
@semiconductive wants is monoethyl sulfite (ethyl bisulfite), or some source of sulfite ions in nonaqueous medium that do not contain or generates
alkaline cations. As far as I know, ethyl bisulfite does not exist. If it did exist, it wouldn't dissociate to ethyl and sulfite ions.
H. F. van Woerden wrote a review about organic sulfites (https://doi.org/10.1021/cr60226a001). I haven't read it yet. Maybe there's something there about monoalkyl sulfites.
As this whole thing has to do with nickel plating in non-aqueous solutions, the question that should be asked is, is nickel sulfite soluble in any
non-aqueous solvent? The SDS below has a few references that may lead somewhere.
Attachment: SDS-26-pages_258.pdf (58kB)
This file has been downloaded 53 times
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semiconductive
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@DraconicAcid, Thank you for the Methanol link. That looks very promising as a second test and to build intuition about alcohol vs. water. I
wouldn't have been able to find it myself.
General question:
I am looking at a NIST page on standard potentials in water:
https://www.nist.gov/system/files/documents/2019/04/02/jpcrd...
There are a few problems with NIST data, such as no mention of isotope blends, etc., so I can't get really be sure what conditions the experiment was
done under. But, this is where I normally get engineering data....
In the NIST paper, I see the hydrogen gas vs. hydroxide ions reaction listed as:
E°=-0.828 , ΔE=-0.0008360 For H₂(g).OH⁻ ⟷ H₂O(liq)
I know from electronics that the energy 'band-gap' in semiconductors is affected by whether the chip is packaged in epoxy and under compression -- or
the die is bare and exposed to air ; the same should happen in electrolytes. So, I suppose it's possible to reverse the trends of the energy gap by
putting it in pressure container. But, normally I would expect the magnitude of E to get smaller with increasing temperature.
But, the sign convention of the energy change is negative, and the document equation (2) on page (2) shows the coefficient as added to the initial E
in proportion to temperature.
That doesn't make sense to me. The analogy seems broken.
Does hydroxide to water conversion create more voltage as the solution gets hotter?
If I naively apply the NIST equation (2) -- I get energy gaps that indicate a voltage magnitude increase:
['-0.8071', '-0.8155', '-0.8196', '-0.8238', '-0.8280', '-0.8322', '-0.8489', '-0.8907']
Therefore:
I grabbed some auto-ionization of water values vs. temperature from online searches, and converted them to exponents and averaged them to compare to
do a sanity check.
Tc =[ 0, 10, 15, 20, 25, 30, 50, 100 ]
log₁₀(Kw)=[ -14.94, -14.54, -14.35, -14.17, -14.00, -13.83, -13.26, -12.30 ]
If I use the NIST values with increasing voltage magnitude, I get pretty bad agreement:
Eg = ['-0.8071', '-0.8155', '-0.8196', '-0.8238', '-0.8280', '-0.8322', '-0.8489', '-0.8907'
log Kw= ['-14.51', '-14.30', '-14.19', '-14.10', '-14.00', '-13.91', '-13.56', '-12.86']
On the other hand, if I use the wrong sign for equation (2), the energy gap goes:
['-0.8489', '-0.8405', '-0.8364', '-0.8322', '-0.8280', '-0.8238', '-0.8071', '-0.7653']
['-14.90', '-14.52', '-14.34', '-14.17', '-14.00', '-13.84', '-13.23', '-12.00']
It's clear for small deviations (5 degrees) that using the 'wrong' sign agrees very closely with experimental data that can be found online.
Before I curve fit the correction -- does anyone know why the sign of the potential change is negative instead of positive in the NIST paper? Is it a
convention, or a typo?
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