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I will have water that have been through dichloromethane. Doing so, the dichlorometane (solubility $2~\%$) will contaminate my water by forming an azeotrop which will have a $38~\mathrm{^\circ C}$ boiling point. Would putting that mixture to $40~\mathrm{^\circ C}$ will make that azeotrop leave? Leaving me dichloromethane free water?

If not I could I extract it?

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There are two concepts here which should not be mixed:

The solubility of dichloromethane in water is indeed quite low, around 2% as in the question. But this is not the azeotrope, which is formed when there is 1.5% water - 98.5% dichloromethane mixture. That indeed boils at $38^\rm{o} \rm{C}$. See this. It's worth mentioning that you can't possibly bring the azeotrope to $40^\rm{o} \rm{C}$ - because it will boil at a lower temperature by definition, and the vapour will have exactly the same composition as the liquid.

Because what you have isn't the azeotrope, I'd say boiling your mixture (which will most likely happen near the boiling point of pure water) will remove the dichloromethane content.

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  • $\begingroup$ No. You end up with a lot of distilled water that still contains CH2Cl2. $\endgroup$ – Karl Jul 4 at 9:03
  • $\begingroup$ Boiling with a column on top? Of course you'll always have some minute amount of contaminant remaining but depending on the circumstances I'd think you can get quite far. $\endgroup$ – albapa Jul 5 at 12:55
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    $\begingroup$ No, the methylene chloride will be almost completely in the distillate alright! But if you add a column, the CH2Cl2 has to diffuse through it, that´d also be inefficient. Running the "distillation" with column at 40°C under vacuum would help. Basically sucking the methylene chloride out with a membrane pump. No idea how long that would take. $\endgroup$ – Karl Jul 5 at 16:37
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    $\begingroup$ Given that this was asked almost 5 years ago, by now I would have expected all the CH2Cl2 evaporated anyway :) $\endgroup$ – albapa Jul 6 at 11:02

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