I have some sodium chlorite and Chinese mystery powder which are meant to be mixed with water, then mixed together to form chlorine dioxide. I wanted to try a more food safe (and less mysterious) route for making small amounts of chlorine dioxide, so I diluted the sodium chlorite, then mixed in some bleach and a tiny bit of citric acid. I don't have a firm understanding of the reactions, but know $\ce{Cl2}$ and hypochlorous acid will react with sodium chlorite, but plain bleach may not, so the acid is to acidify the bleach and shift its equilibrium away from being hypochlorite. It worked, as evidenced by a pale yellow color. (It was very dilute, which is not a bad thing, for this chemical).

What I don't understand is that when I add more citric acid or more bleach, the color turns clear again. Either the chlorine dioxide is reacting or coming out of solution. What are the reactions with excess citric acid or bleach that might eliminate $\ce{ClO2}$?

  • $\begingroup$ How will we know if it's 'your' chlorine dioxide haha $\endgroup$
    – dr.drizzy
    Commented Mar 6, 2018 at 16:40

1 Answer 1


The "mystery powder" was proposed to be sodium bisulfate, in a recent answer to another question. Sulfuric acid is used in one process to turn sodium chlorite into $\ce{ClO2}$; bisulfite would be slower, but should do the same.

Citric acid will acidify the sodium chlorite and sodium hypochlorite, but will itself be oxidized, perhaps all the way to $\ce{CO2}$. $\ce{HCl}$ is used to convert $\ce{NaClO2}$ and $\ce{NaOCl}$ to $\ce{ClO2}$ and citric acid could well do the same (be oxidized at the same time).

$$\ce{2 HCl + NaClO + 2 NaClO2 -> 2 ClO2 + 3 NaCl + H2O}$$ $$\ce{4HCl + 5NaClO2 -> 4ClO2 + 5NaCl + 2H2O}$$

Chlorite ion and hypochlorite ion are known to react rapidly to form chloride and chlorate (Lister, Can. J. Chem., 30, 879-889, 1952).

The unknown amounts of oxidizing agents and citric acid (a reducing agent) and the complex reactions make the pale yellow color a poor indicator of what might be going on. The presence of $\ce{ClO2}$ does not seem to be positively identified and might not even be needed to explain the observations.

  • $\begingroup$ And regarding the redox reactions, an excess of citric acid would use up the ClO₂ as it is oxidized? That makes sense. About the excess bleach, the hypochlorite could be being oxidized to chlorate (and using up the ClO₂)? Is the chlorate ion more stable so the reaction isn't very bidirectional? Because I was unable to correct it by adding more of either of the other two substances. $\endgroup$
    – piojo
    Commented Mar 7, 2018 at 0:55
  • 1
    $\begingroup$ I found a reference to an alkaline equilibrium: 2 OH- + 2 ClO2 = ClO2- + ClO3- + H2O, and another in acid: 4 HClO3 = 4 ClO2 + O2 + 2 H2O (but it went on: If concentrated sulfuric acid is added to solid chlorate, the above reaction takes place, and usually the ClO2 formed explodes with great violence.) I didn't mean to say that no ClO2 was present, but that there could have been lots of compounds, especially from oxidation of the citric acid. $\endgroup$ Commented Mar 7, 2018 at 5:27
  • $\begingroup$ James please edit and correct your cited reaction as it is NOT correct:2HCl+5NaClO⟶2ClO2+5NaCl+H2O. The action of HCl on OCl- forms HOCl + Cl-. With excess HCl, HOCl + HCl = Cl2 + H2O (sources: See en.wikipedia.org/wiki/Hypochlorous_acid ), The action of HCl, or Cl2, or HCl + OCl-, acting on ClO2- are known paths to ClO2 (see en.wikipedia.org/wiki/Chlorine_dioxide). $\endgroup$
    – AJKOER
    Commented Jun 12, 2020 at 19:38
  • $\begingroup$ James, if you do not correct, I will downgrade as it is an egregious error to those intimately acquainted with this chemistry. $\endgroup$
    – AJKOER
    Commented Jun 12, 2020 at 19:43
  • $\begingroup$ @AJKOER: Thank you for the correction! I must have messed up something from Lister. Interestingly, my edit was accepted by one but rejected by two referees, so you and I made the deciding votes. (My edit was posted to the referees as from "anonymous". Perhaps I should have expanded my reason for editing to include your comment.) $\endgroup$ Commented Jun 14, 2020 at 20:37

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