# Why should one perform an extraction with multiple small portions of solvent as opposed to one large portion?

I'm performing a liquid-liquid extraction on a 1 l water sample for GC analysis. The procedure calls for three extractions with 50 ml of dichloromethane. Would one extraction with 150 ml produce a different result?

• Most definitely yes. Qualitatively, you can perform a casual experiment to observe the difference the next time you wash your hands in a bathroom that offers paper towels for drying. Compare how dry your hands get when you (a) pull three paper towels off all at once and dry your hands with them all at the same time; versus (b) pulling three paper towels, one at a time, tossing each paper towel after it feels like it's not getting your hands any drier. (b) will get your hands dry much more effectively. Jun 29 '15 at 2:00

The important thing is the distribution ratio (for the extraction of a metal) and the partition coefficent for the extraction of a given chemical species. This can be used for a first approximation calculation.

The distribution ratio is the total metal concentration in the organic phase (grams of metal per litre) divided by the total metal concentration in the aqueous phase.

If we consider a batchwise extraction where a litre of water bearing x mg of a metal is extracted with an organic solvent. Assume that the distribution ratio is 2. We can do the calculation for the extraction.

The phase ratio (ratio of the volumes of the two phases) is 1, call this theta.

Theta D / (1 + (Theta D)) = the fraction extracted in each stage.

If D is 2, and theta is 1 then we will extract 66 % of the metal.

As an alternative, if we use one litre of water and extract this with 0.5 L of organic phase, then with D = 2 and Theta = 0.5 then we will extract 50 % of the metal.

If we repeat the extraction with another 0.5 L of solvent then we will recover 50 % of the remaining 50 % of the metal. Thus we will have recovered in total 75 % of the metal.

If instead we were to extract the 1 L of water four times with 0.25 L of water then in the first extraction we would leave 66 % of the metal in the lower phase. The easy way of working out the amount left in the aqueous later after n extraction stages with the same D and theta valve is to use

{1 - ( D Theta / {1 + (D Theta)})}^n = fraction left in raffinate. In solvent extraction raffinate is the term for the aqueous layer after an extraction has occured.

This gives me about 20 % left in the aqueous phase after four extractions with 0.25 L of solvent.

It is important to understand that the solubility of the substance is not what you should consider it is the distribution ratio or the partition coefficent which is needed.

If you were to take a dilute solution of DEHPA (150 mM) in aliphatic kerosene and to shake this with 0.1 nitric acid containing a small amount of Fe-55 with no stable iron as iron(III) then you might get a D value of 100 for arguments sake.

Now the organic phase can only extract 50 mM of iron per litre as to extract one iron atom three DEHPA molecules are needed. If we were to repeat the extraction experiment with a solution of 56 grams of iron per litre (roughly 1 moles per litre of iron) spiked with a little Fe-55 then we would measure a far lower distribution ratio.

Equally when you extract a large amount of an organic substance into an organic solvent the solvent properties of the organic layer at the end will be different to those of the pure solvent.

For example if I have 1 litre of water saturated with hexanol with 500 ml of water saturated hexanol floating on the top. Then if I was to add 500 ml of ether and then shake it. I would have a different Kd for hexanol to that which I would have if I was to shake a dilute solution of hexanol in water with ether.

I hold the view that the formula used above for the system with a D value of 2 will only hold perfectly true when the concentration of the metal is zero. This is why radioactive tracers are so useful for academic solvent extraction experiments. This is becuase they allow the measurement of distribution ratios with a zero metal concentration. The concentration in moles per litre of a 1 GBq per litre solution of Fe-55 will be very small.

The half life of Fe-55 is about 2.7 years, thus I have calculated that this very radioactive stock solution will contain 205 nanomoles of iron if the only iron atoms present are Fe-55. The best way to make the iron-55 would be to bombard managense with protons with a cyclotron and then to separate the iron from the manganese by solvent extraction or some other separation method.

I would in a lab expect to be able to run a solvent extraction experiment with only 700 Bq of Fe-55. If we assume that I was to shake 700 microlitres of aqueous phase with 700 microlitres of the organic phase. I would then be able to sample 200 microlitres of the two layers with a pipette.

If we assume that the highest distribution ratio I was to measure was 100 then I would expect that at the end of the shaking that the aqueous layer would contain 10 Bq per ml. If I was to sample 200 microlitres and to put this into a LSC vial and then to count it. Assuming that 100 % of radioactive decays will result in a count in the LSC machine then I would in one minute get 120 counts. If I was to count until I had 10000 counts (1 % error due to count number) I would have to wait about 83 minutes.

For a D value of 10 I would be able get 10000 counts in less than ten minutes for the least radioactive liquid phase. Thus it should be clear that with a radiotracer it is possible to measure the distribution ratio with less than 1 nanomole per litre of metal present.