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?

  • 1
    $\begingroup$ 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. $\endgroup$ – Ivan Neretin May 10 '16 at 18:20
  • $\begingroup$ see this: chemistry.stackexchange.com/questions/34424/… $\endgroup$ – Nilay Ghosh Nov 15 '17 at 12:56

The reaction products depends on the conditions(temperature, pressure etc.) and stoichiometric ratio of reactants taken. Depending on that, you will get various products in each case if you slightly change the conditions or the ratio of sodium hydroxide to sulfur taken.

  1. $$\ce{4S + 8NaOH ->[600 C] Na2SO4 + 3Na2S + 4H2O}$$

Sulfur react with sodium hydroxide to produce sodium sulfate, sodium sulfide and water. This reaction takes place at a temperature of over 600°C.(source)

  1. $$\ce{4S + 6NaOH → Na2S2O3 + 2Na2S + 3H2O}$$

Sulfur react with sodium hydroxide to produce sodium thiosulfate, sodium sulfide and water. Sodium hydroxide - concentrated solution. The reaction takes place in a boiling solution.(source)

  1. Your reaction

When you compare the three equation, you can see that though stoichiometric ratio of the reactants in the 1st and 3rd equation is same, the moles of sulfur and sodium hydroxide is different and thus gives different products in each case.

Also, from reaction 2 we can say that sodium thiosulfate may have disproportionate into sodium sulfite and sulfur.

$$\ce{Na2S2O3 ->[220-300 C] Na2SO3 + S}$$

and thus one mole sulfur cancel from each side and gave reaction 3 (your reaction). This site explains the disproportionation reaction.

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 = $\ce{10^{-0.8}}$.


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 (see https://books.google.com/books?id=57noBQAAQBAJ&pg=PT717&lpg=PT717&dq=ozonide+anion&source=bl&ots=l2q2T3F5M1&sig=vN96PvvfnVU_NMsWppZ9w35_HYw&hl=en&sa=X&ved=0ahUKEwjKysOjxLnKAhVF8x4KHXeZBJY4ChDoAQgwMAg#v=onepage&q=ozonide%20anion&f=false ), 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 at: https://www.lindeus.com/en/processes/cleaning_polishing_grinding/bleaching/oxygen_reinforced_alkaline_extraction/index.html ). 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, https://www.sciencedirect.com/science/article/pii/S1010603018309729 ) may reverse the above reaction leading to S•− radical.

Next, an interaction with oxygen per a 2019 source (https://pdfs.semanticscholar.org/533e/9a0b2e5d938abc555e267f2f9b1a6a29f720.pdf ):

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

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

Then, per Wikipedia on dithionite (see https://en.wikipedia.org/wiki/Dithionite ), 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-

  • $\begingroup$ Welcome to Chemistry SE! I've forgotten all about this reaction, but I hope that your answer will help others. $\endgroup$ – CowperKettle Jul 25 '19 at 17:10

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