RECITATION NOTES FOR EXPERIMENT # 4
LIQUID-LIQUID EXTRACTION
NOTE: In order to follow these notes have your lab textbook available
for quick reference to specific pages (indicated in red).
BASIC PRINCIPLES
The technique of liquid-liquid extraction is used to purify impure substances by
taking advantage of a solubility differential of the substance in different
solvents. It is different from crystallization in that the sample can be solid or
liquid. The impure sample is dissolved in solvent 1 first. Then a second solvent 2
is added in such a way that the sample migrates from the first to the second
solvent without the impurities. This, of course, requires that the two solvents 1
and 2 be immiscible, and ideally that the sample be more soluble in solvent 2 than
in solvent 1.
Solvent 1 + sampl e
+ impuriti es
Ad d
Allow
sol vent 2
equil ibri um
Some of the sampl e
moves to solvent 2
Sampl e becomes di stri buted, or
“ par t i t i oned” bet ween 1 & 2
The process described above is called extraction, and solvent 2 is referred to as
the extracting solvent. Since the sample is soluble in both solvents, it will
“partition,” or become distributed, between the two. The final ratio of the
concentrations of the sample in the two solvents is determined by its distribution
(or partition) coefficient K.
In microscale it is convenient to express the concentration in mg of solute/mL of
solution. After extraction, the final, or equilibrium, concentration of the sample
in the original solvent (1) is given by C1, and the concentration of sample in the
extracting solvent (2) is given by C2. The distribution coefficient is defined as
the ratio of the two concentrations, with C2 as the numerator.
K = C2/C1
Needless to say, the greater the concentration of sample in the extracting
solvent, the greater the distribution coefficient, and the more efficient the
extraction.
EXAMPLE: Calculating the amount of extracted solute in a single extraction,
knowing K.
Refer to fig. 12.2, p. 672 in your lab textbook.
As a point of departure, we have a solution of 50 mg sample in 1 mL aqueous
solution. We extract once by adding 0.5 mL ether. After shaking, an unknown amount
of solute (x mg) will remain in solvent 1 (water), and the rest (50-x mg) will
migrate to solvent 2 (ether). Therefore the equilibrium (or final) concentrations
of solute in the two solvents are:
C1 = x mg/1 mL = x mg/mL
and
C2 = (50-x) mg/0.5 mL = 2(50-x) mg/mL = (100-2x) mg/mL
The distribution coefficient K must be given in order to do the calculations. In
this case it has a value of 10. Therefore:
K = 10 = (100-2x)/x
100-2x = 10x
100 = 10x+2x = 12x
x = 100/12 = 8.3 mg solute remain in the water
50-x = 50 - 8.3 = 41.7 mg sample have moved to the ether
(83.4% or original amount)
Make sure to review the calculations shown on p. 671. These are intended to
demonstrate that it is more effective to extract more times with smaller amounts
of extracting solvent than fewer times with larger amounts of extracting solvent.
Then go back and compare fig. 12.2 and 12.3 (p. 672-673) for a visual illustration
of the same principle.
To test your understanding of these concepts, do the prelab calculation at the top
of p. 34.
The following practical observations apply when performing extractions in the lab:
1. Performing two or three extractions gets most of the sample out of the
original solvent, provided that K is favorable (at least 5 or greater).
2. The more concentrated the solutions, the more efficient the extraction.
That’s another reason for using small amounts of solvent during extractions.
3. If the original solution is too dilute, it’s better to increase its
concentration before the extraction by evaporating some of the solvent.
THE MICROSCALE EXTRACTION TECHNIQUE
The equipment and technique used in microscale extractions is shown in figs. 12.5
and 12.6 on p. 674 and 675. A conical vial or centrifuge tube is used to perform
the operation. A Pasteur pipette is then used to separate the two layers. The
bottom layer can be completely drawn out and transferred to another container.
To perform an extraction, simply combine the original solution and the extracting
solvent. Make sure the vial is closed and shake gently by turning the vial upside
down and vice versa a few times. Do not shake vigorously, or an emulsion might
form, which is difficult to separate. As you shake, pressure might build up inside
the vial. It is important to vent the system by carefully removing the cap once in
a while during the extraction.
After shaking, allow the mixture to rest undisturbed until the solvent layers
separate.
TELLING THE LAYERS APPART
It is frequently the case that the original solvent is water and the extracting
solvent is an organic one such as ether, benzene, or methylene chloride. All of
these solvents are colorless, so the issue comes up of how to tell the solvents
apart once the layers separate. The first thing to keep in mind is that the
heavier layer will go the bottom and the lighter layer will stay at the top. It is
therefore the relative densities of the two solutions that determine their
relative positions. Now, most of the times you can guide yourself by the relative
densities of the pure solvents, which you can find in tables. But keep in mind
that in some cases the density of the actual solution can be substantially
different from that of the pure solvent if large amounts of heavy solutes are
dissolved in it. A table of densities for common extracting solvents is given on
p. 673. Thus, if the original solvent is water and the extracting solvent is
methylene chloride, the organic layer (methylene chloride) is expected to be the
bottom layer because it is “heavier” than water. The following guidelines may
apply:
1. Halogenated solvents such as methylene chloride (CH2Cl2), chloroform (CHCl3),
and carbon tetrachloride (CCl4) are heavier than water and go to the bottom.
2. Low molecular weight hydrocarbons such as ligroin, petroleum ether, and
benzene are lighter than water and go to the top.
3. Diethyl ether (frequently called simply “ether”) typically goes to the top.
4. If possible test the layers by adding a drop of either solvent and see which
layer dissolves it. This procedure can be tricky due to the effect of the
surface tension.
DEALING WITH EMULSIONS AND “THIRD LAYERS”
An emulsion is different from a true solution in that it consists of two or more
phases, but clear boundaries between the phases are not obvious because they
coexist intimately. As an example consider a mixture of water and oil. This
mixture is typically heterogeneous and the boundary between the oil and the water
is clearly visible. However, if some soap is added and the mixture stirred, an
emulsion forms and the boundary becomes blurred. But this emulsion is not a true
solution.
Emulsions form because:
1. The relative densities of the layers become too similar, taking longer for
the layers to separate.
2. The relative polarities of the two layers become too similar due to the
nature of the solutes dissolved in them.
3. Sometimes emulsions form because the layers were shaken too long and/or too
vigorously.
A “third layer” is an emulsion that forms between the two layers with no clear
separation. To avoid these problems shake gently and allow enough time for the
layers to separate (patience!).
HOW TO BREAK UP EMULSIONS
1. Change the concentration of one of the solutions by adding more of that
solvent (preferably the original one). Shake gently again and allow
separation.
2. Change the density of the aqueous layer by adding a neutral salt such as
NaCl or KCl. Shake gently again and allow separation. This will have the
added benefit of decreasing the solubility of the sample in water.
3. Place the emulsion in a centrifuge tube and spin for about 3 min.
4. Filter the emulsion through filter paper using gravity filtration. However
this technique works better for macroscale extractions and it can be tricky
to perform).
See note on emulsions on p. 684.
ADDITIONAL GUIDELINES
1. With a favorable distribution coefficient (5 or greater, as is the case with
most of the solvents used in this course), it is not necessary to perform
more than 2 or 3 extractions to obtain a good percent recovery of the
sample. More extractions will actually start to bring out the impurities
we’re trying to get rid of.
2. “Washing” the organic layer means to shake it with distilled water and then
discard the water.
3. “Drying” the organic layer means to remove trace amounts of water from it by
treating it with a drying agent such as anhydrous sodium sulfate (a table is
shown on p. 681). This can be done by adding the drying agent with a
spatula, or by filtering the solution through a column containing the drying
agent. This method is not effective if the solvent contains large amounts of
water (see table 12.3 on p. 683). Drying agents pick up water from the
atmosphere. Do not leave the containers open for long periods of time.
4. Never discard any layers until you’re sure you’ve recovered your sample and
the experiment is finished.
TO INCLUDE IN YOUR LAB REPORT:
As part of your lab report, include items 1 and 2 under the heading “Experiment
4A” on p. 41 of the textbook. Also include also items 1 and 2 under the heading
“Experiment 4B” on the same page.
These items should be part of the results and discussion sections.