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I'm told that when sodium chlorite $\ce{(NaClO2)}$ is mixed with aqueous acid, chlorine dioxide ($\ce{ClO2}$) is produced and remains mostly dissolved in water.

I would like to know how the strength of the acid (concentration of acid and/or the type of acid) affects this reaction. I would also like to know what happens (afterwards) to the $\ce{ClO2}$ over time. For example, if undisturbed, does the $\ce{ClO2}$ enter the air, convert into another compound, or something else?

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When acid is added to sodium chlorite in solution (note: $\ce{ClO2-}$ is the chlorite ion), chlorous acid ($\ce{HClO2}$) is transiently formed which then goes on to decompose. However, you need hydrochloric acid, not just any acid, because chloride ion ($\ce{Cl-}$) itself acts as a catalyst for the decomposition. What happens is that chlorous acid disproportionates; that is, some of it gets oxidized while the rest gets reduced.

Specifically, out of five $\ce{Cl}$ atoms in chlorous acid (oxidation state +3), one will end up as chloride in the oxidation state $-1$, and four will end up as $\ce{ClO2}$ in the oxidation state +4:

$$\ce{5HClO2 -> 4ClO2 + Cl- + H+ + 2H2O}$$

Pretty much any concentration will do. Calculating the proportions is easy if you know a bit of stoichiometry: you need 5 moles of chlorite for 4 moles of hydrochloric acid. The reaction proceeds to completion, and you get 4 moles of $\ce{ClO2}$.

The $\ce{ClO2}$ remains in solution and is a good bleaching agent; see Wikipedia. It can undergo further disproportionation but if the solutions are kept in the dark, this process is slow.

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  • $\begingroup$ sorry, I messed that one up real bad. $\endgroup$ – orthocresol Sep 8 '17 at 11:09

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