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Let’s assume I have one liter of distilled $\ce{H2O}$ in an stainless steel container – from the bottom side of the container I do bubble ozone trough the $\ce{H2O}$.

Of cause it is not pure O3, so let’s assume that the gas consists of:

7% $\ce{O3}$

83% $\ce{ O2}$

10%. $\ce{ N2 / NO2 / NO}$

The gas flow is about 5 liters per minute and the corona discharge consumes about 500W.

How long will it approximately take, till full saturation of the $\ce{H2O}$ with $\ce{O3}$ will take place?

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The solubility of ozone in water as per one website is:

enter image description here

First thing is to establish which curve you need. In 1 m$^3$ gas you have 70 L ozone:

$\frac{70\ L\ \ce{O3}}{1\ m^3}\frac{1\ mol}{0.082\cdot 298\ L}\frac{16\cdot 3\ g\ \ce{O3}}{1\ mol}= 138\ g\ \ce{O3}/m^3$

Let us take the curve of 150 g/m$^3$ which has an ozone solubility of 30 mg/L water. Then:

$\frac{30\cdot 10^{-3}\ g\ \ce{O3}}{1\ L\ \ce{H2O}}\frac{1\ mol}{16\cdot 3\ g\ \ce{O3}}\frac{0.082\cdot 298\ L}{1\ mol}\frac{1\ min}{5\cdot 0.07\ L\ \ce{O3}}= 0.044\ min$

Which is about 3 seconds.

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  • $\begingroup$ please explain why time enters your last expression, and there is no mention of the diffusion coefficient of ozone in water. Just because the gas flows in does not mean that it is all absorbed or if so how long this takes. $\endgroup$
    – porphyrin
    Jan 26 '18 at 14:01
  • $\begingroup$ Because the gas flow is 5 L/min with a content of 7% ozone. Of course it is supposing that the mixing is perfect, etc. I have not complicated things because I think that the OP just wants to know if it will take long before the output will stabilize. On basis of the other questions posted by the same OP. $\endgroup$ Jan 26 '18 at 14:07
  • $\begingroup$ I understand what you write. The distance diffused in time $t$ is $L=\sqrt{6Dt}$ and for, say, oxygen in water ,$D\approx 10^{-5}$ cm$^2$/s (which is a typical value) meaning that in 1 sec the gas has diffused only $\approx 0.01$ cm. So I think it will take far longer to dissolve that than you estimate. Of course it will depend on the gas pressure also. $\endgroup$
    – porphyrin
    Jan 26 '18 at 14:50

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