Sciencemadness Discussion Board

Solubility in two immiscible solvents?

Fyndium - 27-7-2020 at 12:13

If we have a substance that is dissolved in a polar solvent A, and it is washed with immiscible non-polar solvent B and the substance is soluble in both of them, although more so in the A, will it dissolve enough that it could be mostly or even fully extracted into solvent B if washed with it multiple times?

The instance is that solvent A has fairly high boiling point, hence being more difficult to remove by distillation, while solvent B has a low boiling point.

Another method would be to crash the substance from A with simple water addition, but in this instance the solvent would like to be preserved for reuse.

ThoughtsIControl - 27-7-2020 at 12:52

I would be happy to help answer your question but you're missing a few details. For starters, I don't know whether or not your solvent is going to form an azeotrope with your solute. It would help if you were to describe the situation you've encountered with the names of the solvents. However, I understand where you're coming from with your question. Performing a wash with your nonpolar solvent B which has less affinity for the substance as compared to your solvent A may perhaps extract small amounts of the substance. However, since the solvent A has a stronger affinity for the substance as compared to B, then this wash won't be very effective.

Feel free to correct me if I'm wrong

mackolol - 27-7-2020 at 13:02

I described in some post, when I was deethylating ethylvanillin and was trying to extract the resulting compound with ethyl acetate, I was only able to extract half of it, rest remained still in water.

Fyndium - 27-7-2020 at 13:37

This question is more of a general matter in a chain of thinking how solvents could be used to support each other and a process. I've faced this kind of possibility in many synthesis, but it usually is not viable if the reaction is not for continuous process or in a lab scale. I wasn't thinking of azeotropes, but they could hinder the process at least in some cases, unless the solutes are of high bp (200-300C). Acetic acid and hexane should not form any form of solutions though, being an example of immiscible solvents.

If it is not possible to extract all of the solute, but let's say half of it, would it be possible to wash the solvent to the full extent, and then reapply the partially regenerated solution in a process? The remaining substance would remain as a spectator, being inert to all of the reaction compounds, including possible catalysts that are involved. I understand this kind of application is used in industry in a continuous process cycle where the starting charge would be 1+n, when the reaction is finished, it would contain 1+n of finished substance and after the wash, 0.5+n, and then a new precursor amount of 0.5 units would be added to create a buffer that enables the sustaining extraction of the upper half.

I think I just should test it and see how it works.

dawt - 27-7-2020 at 20:27

Quote: Originally posted by Fyndium  
If we have a substance that is dissolved in a polar solvent A, and it is washed with immiscible non-polar solvent B and the substance is soluble in both of them, although more so in the A, will it dissolve enough that it could be mostly or even fully extracted into solvent B if washed with it multiple times?


Yep! If your compound isn't all too heat sensitive this sounds like a perfect use case for a perforator/continuous liquid-liquid extractor. Depending on the type they are not cheap, unfortunately, and rarely show up on the used market.

outer_limits - 27-7-2020 at 21:50

Quote: Originally posted by Fyndium  
If we have a substance that is dissolved in a polar solvent A, and it is washed with immiscible non-polar solvent B and the substance is soluble in both of them, although more so in the A, will it dissolve enough that it could be mostly or even fully extracted into solvent B if washed with it multiple times?


It is possible. Nerst's distribution law states that solute distributes itself between two immiscible solvents only in a particular ratio.
k = c1/c2

So, in standard extraction when the we use the solvent in which the substance has better solubility we can extract effectively using much smaller amount of solvent. In the opposite situation, to keep the concentration at lowest level in less-solvating solvent you should use much more volume and it depends on the equilibrium value.



[Edited on 28-7-2020 by outer_limits]

[Edited on 28-7-2020 by outer_limits]

Fyndium - 28-7-2020 at 12:33

Sounds good. How about washing the solvent with multiple smaller batches in order to always introduce a new solvent into which the solvate migrates? I've seen this "wash with 5x200mL of chloroform" (for example) in many syntheses and apparently this is the purpose? It wouldn't result in 100% recovery, but 50% is good for a start, and anything above would be even better.

Then, how about second step of solute precipitation. If we take for example hexane, into which chloroform should be miscible but a solute x is only 1g/L at ntp, shouldn't it mostly crash out? This would enable a very fast and efficient recovery and recycle of the solvent. Stripping the chloroform is easier than hexane, and it's volume wouldn't necessarily be equal or larger than hexane.

Cou - 28-7-2020 at 18:09

Quote: Originally posted by Fyndium  
Sounds good. How about washing the solvent with multiple smaller batches in order to always introduce a new solvent into which the solvate migrates? I've seen this "wash with 5x200mL of chloroform" (for example) in many syntheses and apparently this is the purpose? It wouldn't result in 100% recovery, but 50% is good for a start, and anything above would be even better.


i thought it would be fun to calculate this with partition coefficients.

Imagine a partition coefficient of 0.1 between solvent 1 and solvent 2. starting out with 1000 mL of 1 M in solvent 2.

if you extract once with 1000 mL of solvent 1, the final result is 1000 mL of 0.0909 M in solvent 1, and 1000 mL of 0.909 M in solvent 2. Only 9.1% of solute was extracted.

If you extract five times with 200 mL of solvent 1... the final result is 1000 mL of of 0.906 M in solvent 2, so only 9.4% extracted, not a big difference. did I calculate right?


HeYBrO - 28-7-2020 at 21:41

just read this....



[Edited on 29-7-2020 by HeYBrO]

Fyndium - 29-7-2020 at 01:06

Ah, finally some actual numbers. Thanks!

Cou - 29-7-2020 at 08:46


Fyndium - 29-7-2020 at 09:08

Now it would make more sense that the washing solvent has much higher solubility for the solute than the standing solvent. We'll say that the reaction solvent dissolves up to 3mol, but the washing solvent can, at least when warm, dissolve probably up to 10mol.

I'd also like to ask another question regarding other instance.

A reaction contains an aromatic compound immiscible with water dissolved in acetone-water solution, and it is washed with brine to extract most of the water, and this washing seems to be quite effective since when distilling off the residual acetone, only little to no water separates from the aromatic when the acetone content falls so low it cannot hold the three compounds together anymore.

The question is, how does acetone not completely dissolve into the brine and the aromatic separate right away? The washing does seem to dissolve some of the acetone, but it forms two neat layers, one with brine-water and one with acetone-aromatic. The aromatic separates when a much larger portion of water is used. Apparently the aromatic compound itself acts as a solvent in this case, keeping the water out and holding the acetone. At some point the equilibrium seems to turn over to water-acetone and the aromatic drops out.

What I'd like to happen is drying the acetone-aromatic solution to as anhydrous as possible. How can the water content be determined, or should it just be dried over CaSO4 or other suitable desiccant until it no more clumps?

Cou - 29-7-2020 at 18:36

If a water insoluble compound (e.g. isopentyl acetate) is dissolved in a solvent that both dissolves the ester and is misible with water (e.g. acetic acid), adding water causes a bit of ester to drop out into a sep layer, and the ester becomes less soluble in the AcOH/water aqueous layer as more water is added to reduce fraction of acetic acid. This is why a large volume of water is used to workup a fischer esterification (isolate ester from an ester/carboxylic acid mixturr)

[Edited on 7-30-2020 by Cou]

Fyndium - 16-2-2021 at 06:06

A synthesis calls for extraction of reactant with solvent in a similar situation, where two water miscible solvents are used in a reaction. Reaction is diluted with large amount of water, and this causes the reactant to completely crash out of solution, as it is immiscible with water.

Washing the solution with extraction solvent seems unnecessary to me, so is there a catch why the synthesis suggests this? It would be more straightforward just to dilute it with water to crash it all out and separate the reactant. After separation, adding more water to the reaction leftover solution does not precipitate any more of the product, so I expect full recovery. Extraction solvent will also extract some of the reaction solvents, which needs to be washed off again, creating a lot of extra work, instead of just processing the pure reactant itself without having to distill it off, etc? Possible trace losses in yield would well justify skipping it.

PS. As a sidenote, reading the upper messages prior, I thought a continuous liquid solvent extraction could be carried out with dean-stark style trap where new solvent is continuously introduced, and depending on structure it would either drip over or flow through the extractable phase. Conceptually talking on an industrial setting here, pretty much. Just a thought. This sort of extraction system could allow reuse of the main batch for new reaction immediately after extraction phase and effective use of small volume of recyclable solvent batch. Green chemistry, I suppose.

[Edited on 16-2-2021 by Fyndium]