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I have a protocol for separating an inorganic component from an organic mixture, which asks for the mixture to be suspended in methanol and water and then doing an extraction with hexane. I understand that this allows separation of polar components from non-polar components. I am wondering what use methanol in water has in liquid-liquid extractions compared to just using water? My only ideas are: lowering the polarity of water, increasing the density, or somehow allowing some compounds to collect at the interface.

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  • $\begingroup$ Lowering the surface tension, maybe? $\endgroup$ – Ivan Neretin Jan 15 '16 at 5:43
  • $\begingroup$ @IvanNeretin what does a lower surface tension do? $\endgroup$ – k-- Jan 15 '16 at 8:04
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    $\begingroup$ Lower surface tension would make it easier to get good phase intermixing when agitating the mixture. Droplets of water/MeOH in hexanes would be smaller, last (slightly) longer, and better facilitate the equilibrium partitioning of solutes between the two phases. $\endgroup$ – Curt F. Jan 15 '16 at 14:29
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    $\begingroup$ Another idea: the methanol is easier to remove than water, so after the extraction is done and the inorganic component is in the aqueous phase, the time & energy required to remove all the solvent is lower than if 100% water was used instead. $\endgroup$ – Curt F. Jan 15 '16 at 14:30

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