Sulfur reacts with sodium hydroxide in the following way:

$\ce{3S + 6NaOH->2Na_2S + Na_2SO_3 + 3H_2O}$

It is a disproportionation reaction: sulfur's oxidation changes from zero to -2 in sodium sulfide and from zero to +4 in sodium sulfite.

But how to remember the products of this reaction? By rote memorization?

The preparation of sodium sulfide is apparent. Is there a way to understand why sodium sulfite forms? We have the hydroxide ion, and it somehow reacts with sulfur to yield the $\ce{SO3^{2-}}$ ion - but how?

  • 2
    $\begingroup$ Why, that's simple. Sulfide (a reduced form of $\ce{S}$) is apparent, you say. OK, then some $\ce{S}$ is going to be oxidized as well. What is the next stable oxidation state? It is +4. So sulfite it is. Then both react further with neutral $\ce{S}$, and things start to get hairy, with all them polysulfides and thiosulfates. $\endgroup$ May 10, 2016 at 18:20
  • $\begingroup$ see this: chemistry.stackexchange.com/questions/34424/… $\endgroup$ Nov 15, 2017 at 12:56

2 Answers 2


According to webelements, Sulphur reacts with hot aqueous alkali hydroxide, to form trisulphide and thiosulphate species.

$$\ce{S8(s) + 6OH-(aq) → 2S3^2- + S2O3^2- + 3H2O(l)}$$

The primary source of this reaction might be from here:

Chemical reactions of sulfur compounds have been studied in equimolar $\ce{NaOH-H2O}$ melt at 100°C by voltammetry and UV spectrophotometry. Disproportionation of sulfur is fast and quantitative according to:

$$\ce{S8 + 6OH- -> 2S3^2- + S2O3^2- + 3H2O}$$

On the contrary, addition of sulfur to sulfide is not quantitative and gives $\ce{S2^2-}$ besides prevailing sulfur disproportionation.$\ce{SO3^2-}$ reacts with $\ce{S2^2-}$ and $\ce{S3^2-}$ , while $\ce{S2O3^2-}$ decomposes $\ce{S^2−}$. A pseudoequilibrium is observed:

$$\ce{S^2- + S2O3^2- <=> S2^2- + SO3^2-}$$

with concentration quotient Formula Q = $\ce{10^{-0.8}}$.

Also, reasonably stable sodium polysulfide ($\ce{Na2S_x}$) seems to form at high temperature. Check the references given in the researchgate site.

  • $\begingroup$ Sorry, please see my recent EDIT citing work published in Chemical Society of Japan, which clearly puts holes into your reciting this very likely inaccurate (over simplified) chemistry. There may, indeed, be an apparent cited role for oxygen and even radicals! By the way, if you google this question, there is very little support for the above position, and more shocking our 2 answers comes up first! $\endgroup$
    – AJKOER
    Aug 27, 2021 at 16:46

The actual mechanics of the chemistry, I suspect, is advanced citing a recent (2019) reference noting some radical chemistry. My take is as follows:

First, heating in air (an O2 source) a boiling mix of NaOH and water in the presence of sulfur, I would argue may involve the reversal of the following radical reaction:

O2•- + OH• = O2 + OH-

where explicitly the reverse reaction, which was actually proposed as early as 1963 by Schroeter in alkaline conditions (per this Google Book source), is:

O2 + OH- = O2•- + •OH (Source above, Eq 5.98)

which apparently can be accomplished in the presence of high oxygen pressure, heat and alkaline conditions (as employed in oxygen-alkaline bleaching, see comments here. In the current text, the reaction equilibrium may be moved to the right with the consumption of formed radicals with say sulfur/sulfur compounds.

In particular, I suspect, the sulfur in heated water/OH- may acquire a charge upon heating and in accord with the electrostatic properties of a colloidal suspension resulting in the presence of some solvated electrons. Then, the formation of sulfur related radical(s):

S + e-(aq) = S•−

A path to the consumption of the highly reactive and non-selective hydroxyl radical (created above) could be:

S•− + OH• = S + OH-

One could also argue that the action of heat/light (see, for example, this Science Direct compilation) may reverse the above reaction leading to S•− radical.

Next, an interaction with oxygen from a 2019 semantic scholar article:

HS•/S•− + O2 --> SO2•- (+ H+) (Source Page 7, Eq (7))

SO2•- + SO2•- --> S2O4(2-) (Same source, per comment)

Then, per Wikipedia on dithionite which notes that dithionite undergoes an acid hydrolytic disproportionation to thiosulfate and bisulfite (presence of CO2 may assist):

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

[EDIT AUGUST 2021] Here is an interesting partial confirmation of my proposed mechanics. To quote the abstract from this work The Chemical Behavior of Low Valence Sulfur Compounds. I. Oxidation of Elemental Sulfur with Compressed Oxygen in Aqueous Ammonia Solution published in Bulletin of the Chemical Society of Japan, Volume 38, Issue 10, 1965:

  1. The oxidation of sulfur with compressed oxygen proceeds through the following sequence of steps: (a) The disproportionation of sulfur to both pentasulfide and thiosulfate. (b) The oxidation of pentasulfide to thiosulfate. (c) The oxidation of thiosulfate to sulfate and sulfamate through a series of complicated intermediates.
  1. Sulfamate forms only in the decomposition of trithionate.
  1. The overall reaction of the oxidation is prompted by a rise in the temperature, an increase in the ammonia concentration, and the addition of transition metal ions, especially copper ions.

So, a confirmation from a respected journal of my claim of the role of oxygen forming the long known reported products: pentasulfide and thiosulfate.

Also, an interesting reference to an apparent fostering role of transition metal-based radicals, where the citation of Copper is especially interesting as it engages in REDOX reactions over a wider range than other transition metals (like Iron) creating active species.

Now, what could be a possible path, perhaps re-reading my suggested answer will assist.

  • $\begingroup$ Welcome to Chemistry SE! I've forgotten all about this reaction, but I hope that your answer will help others. $\endgroup$ Jul 25, 2019 at 17:10

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