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I stumbled across this reaction:

$$\ce{BaO2 + H2CO3 -> BaCO3 + H2O2}$$

This reaction seems quite similar to the variant with sulfuric acid

$$\ce{BaO2 + H2SO4 -> BaSO4 + H2O2}$$

which is well known and works.

But as carbonic acid is a lot weaker than sulfuric acid, I started to question this reaction. Why does it even proceed in the first place? I thought that the sulfate/carbonate is really insoluble in water which is why no equilibrium forms (or does it?), however barium peroxide is itself pretty insoluble.

So, a) does the first reaction even happen and b) why?

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  • $\begingroup$ Without going into too much quantitative stuff, barium peroxide is a very strong oxidiser and extremely unstable. It will oxidise just about anything given the chance. $\endgroup$ – Gimelist Jul 8 '17 at 23:09
  • $\begingroup$ Where did you get the first reaction? Mention your source $\endgroup$ – Nilay Ghosh Jul 9 '17 at 5:06
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    $\begingroup$ I found this here: chemieonline.de/forum/showthread.php?t=216287 This is a german chemistry forum, and the guy there also just says he found it "in some old book". The method he describes is: "In a cold solution of $\ce{BaO2}$ a lot of $\ce{CO2}$ is bubbled until enough $\ce{BaCO3}$ is precipitated out. This is separated from the $\ce{H2O2}$ and repeated until the desired concentration is reached." $\endgroup$ – G. Ünther Jul 9 '17 at 6:36
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    $\begingroup$ @G.Ünther In the future, please add the original source to your question: it helps you to stay organized, it helps us to help you, and is overall a good practice:) $\endgroup$ – andselisk♦ Jul 18 '17 at 5:02
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Before the precipitation of $\ce{BaCO3}$ starts, the following equilibrium is arranged:

$$\ce{BaO2 + 2H-O-H <=> H2O2 + Ba^{2+} + 2OH-}$$

$\ce{BaO2}$ is but very slightly soluble in $\ce{H2O}$ because the then occurring hydrolysis produces $\ce{OH^-}$, which raises the pH, and hence stops hydrolysis (and dissolving) of further $\ce{BaO2}$. Simultaneously, by the formed $\ce{OH^-}$, the back reaction is fostered. However, these two blockades may be overcome by adding dilute acid, since this latter neutralizes at once the alkaline product $\ce{Ba(OH)2}(aq)$, and so the equilibrium may be continually shifted to the right side.

Hence, it doesn't matter whether the acid is more or less diluted (nor whether it is strong or weak), provided that acid is continually supplied to the extent that it is consumed during the dissolution/hydrolysis process of $\ce{BaO2}$. An aqueous solution of $\ce{Ba(OH)2}$ is an efficient gas detection or gas scrubbing method for $\ce{CO2}$; hence, by bubbling $\ce{CO2}$-gas through it for a long time, the $\ce{Ba(OH)2}(aq)$ will easily be transformed into $\ce{BaCO3}(s)$. Of course, a gas frit will be used to spread the gas, and the reaction mixture should be constantly stirred.

In the textbook of Sarda/Handa/Arora$^1$ this process with $\ce{CO2}$ is shortly mentioned (without any explanations except that ice-cold water should be used, and that $\ce{BaCO3}$ may be removed by filtration, since the desired product is, of course, $\ce{H2O2}$).

$^1$ Sanda/Harda/Arora, Chemistry, part-2, New Dehli, 2016, p. 885.

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  • $\begingroup$ "which lowers the pH" You made a blunder. $\endgroup$ – Mockingbird Jul 18 '17 at 4:08
  • $\begingroup$ Shouldn't it be $\ce{Ba^{2+}}$? $\endgroup$ – G. Ünther Jul 18 '17 at 16:32

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