Solvent Extraction is an important tool in the analytical laboratory. It can be used in sample preparation for subsequent analysis and can also show us some of the underlying theory behind chromatographic methods. In addition, the determination of Log P (partitioning) is no more than a solvent extraction experiment. This is an important parameter in drug discovery and development. Partitioning in a fundamental process in biological systems. It is crucial concept in understanding in how cells are able to pass some molecules through their membranes while and yet reject others. Cell membranes can be viewed as an organic solvent into which compounds can partition and a compounds lipophillicity can be a crucial factor in the compounds ability to pass into the cell.
We will want to look at the partitioning of a solute molecule when two immisible liquid phases are brought into contact. An example could be the partitioning of a compound such as aspirin (acetylsalicylic acid) between an aqueous solution and an organic solution such as ether.
The Log P mentioned in the above paragraph is the partitioning of drug molecules between a buffered aqueous solution and noctanol (this compound is chosen as a mimic of cellular membranes). This is such an important basic pharmaceutical parameter that several companies have developed software to calculate this process. Two examples would be ACD from Toronto and Interactive Labs. Links to these company web sites are listed below.
The Interactive Analysis Web Site Allows you to calculate Log P values from smiles or Mol File formats (This will be handy for the first lab)
Lets look at some background.
Let A represent our analyte (solute). The partitioning is an equilibrium
process. By convention the process is looked upon as going from the
aqueous phase (reactant) to the organic phase (product).
We can write an equilibrium expression for this process.
We can express this in terms of moles and volume. We let
m
be the total number of moles in the system. We will then let q
be the fraction of the material that remains in the aqueous phase and then
by difference the fraction in the organic phase must be 1q.
Recall that concentration is moles/volume we can rewrite the above equations
as
Let V_{r} be equal to the volume ratio
We now can make the substitutions and solve for q and we end up with the equation
so for any extraction we can calculate the fraction that will be left
in the aqueous phase after a single extraction.
Since we have a binary system. We also know how much is in the organic phase and that is (1q) which for ease of use we will write as p.
So if the whole system has all the analyte we can readily say that
p + q = 1
And as above the an equation can be written for p
If one knows the value of K then you can find out how much would be
extracted or if one knew how much was extracted then one could find out
K. Let's look at an example
You have 100.0 mL of an aqueous solution that is 100.0 mM in compound C.
This solution is extracted with 50.0 mL of diethyl ether and the aqueous phase
is assayed and it is found that the concentration of compound C that remains
is 20.0 mM. What is the equilibrium constant for this extraction system.
The fraction that remains is 20.0 mM/100.0 mM which is 0.20. (we can take the ratio of the concentrations since the volume is constant)
Vr is 50.0 mL/100.0 mL which is 0.5
Since in many cases we will want to remove a compound from one phase we can
readily determine how to do this from the equations we have covered so far.
If we are removing from the aqueous phase then you could put the required final
fractional amount in for q and then solve for V_{r}.
Often this will give us an unworkable ratio. That is we might have
to extract 1 Liter of organic phase with 0.1 mL of aqueous and this becomes
operational difficult, if not impossible. Another way is to do repeated
extractions. Each extractions removes the same proportion
of material. So one can take the equation for q with our desired
volume ratio and known K and and solve for the number of extractions (n).
Let's look at an example. How many extractions would be required to remove
99.99% of aspirin from an aqueous solution with an equal volume of noctanol?
Since 99.99% must be removed the decimal fraction equivalent of this is 0.9999. This leaves 0.0001 in the aqueous phase. Since we have equal volumes then Vr is 1.00.
We are able to find from the Interactive Analysis Web site that K for Aspirin is 35.5. We plug these values into the q equation and the power is the unknown.
Taking the log of both sides allows us to solve for n which in this case equals 2.56. Since we can not do a partial extraction then we must do next higher number (3) to make sure we get at least the purity that we need.
Here is a problem for you to test your understanding.
Now let's look at another example where we are searching for an optimum volume.
And now lets look at a better way to get the same results as in the last example.
But this brings up another issue. If we look at the structure of Aspirin
we see that is is a carboxylic acid and at higher pH values we might get dissociation
of the proton which will leave the ion. Would you expect the ion to
partition as readily into the organic phase??????
Of course not. Since the fundamental rule of thumb that we will use in chemistry is "likes dissolve likes" we expect that ions might prefer to stay in the very polar aqueous phase. This is in fact the case.
The K value that we derived above is still valid but only applies to the undissociated acid. The equilibrium expression that we have presented above is called the partition coefficient which you will sometime see as K_{p} and since these values can get large we might see them listed as LogP ( P for partition). A more general expression for compounds that undergo other equilibria, such as acid dissociation, dimermzation or ion pairing is called the distribution ratio and is usually denoted with at D or D_{c}
This would be expressed as
Where C_{A }is the formal concentration and represents the concentration
of the compound in all of its possible forms. This can be expressed
as follows for a weak organic acid such as aspirin. In the organic
phase the compound can only be in the undissociated form and in the aqueous
phase the compound can dissociate and we can have both the acid and its
conjugate base.
.
Now we might expect that [A^{}] has a partition between both phases and it does. But since the ion has such a small affinity (i.e. none will go there) for the organic phase we may ignore it. So the only other equilibrium we must be concerned with is the acid dissociation in the aqueous phase. For a weak acid we can write in general terms as
We can now derive an expression for D_{c} in terms of K_{p} and K_{a}.
We have the formula for a weak monoprotic acid such as aspirin or acetic acid.
So from this expression we can see that the [H^{+}] will effect
how the compound distributes itself between the two phases. In systems
that can undergo these other equilibrium processes we will use D in place
of K for our equations of q and p.
Lets look at the expression above. When the [H^{+}] is large compared to K_{a}_{ }we can ignore K_{a}. The denominator is left with just the [H^{+}] term and this will cancel the [H^{+}] term in the numerator leaving just K_{d}. Therefore, at high [H^{+}] (low pH) the D term equals K. When Ka is large compared to [H^{+}] then we are left with the expression.
This shows us that at low [H^{+}] the D decreases as the same
rate as [H^{+}].
We can plot this relationship
This plot shows that at higher pH values the compound has a high affinity
for the aqueous phase.
We also have an expression for for a weak monoprotic base. Many pharmaceutical
compounds are weak nitrogen bases and this must be considered when designing
new drug compounds. What would a plot of log D vs pH look
like for these compounds?
In addition to the pharmaceutical applications for this relationship there have been many many analytical procedures developed using solvent extraction. Harris covers several examples in this area.
Well we can now move a solute (analyte) from one phase to another. This can be very useful when extracting a compound that has significant chemical differences from other compounds in solution. As a matter of fact this has been used as an interview question for prospective coops when I worked in industry.
The question would go like this. You have carried out a series
of reactions and it is now time to work up the product which currently
sits in an organic solution (methylene chloride). Your expected product
is a primary amine. Which of the following solutions would you extract
this methylene chloride solution with to isolate your amine.
Your choices are:
A) Toluene.
B) 0.1 N NaOH (aq)
C) 0.1 N HCl (aq)
D) I never wanted to work here anyhow.
But that is an organic application and now lets look at some more analytical
applications
What happens if you have two analytes A and B. The equations we have already shown are still valid but me must now use them for both compounds. We will start them off in aqueous solution since that is our convention. We will select a solvent pair that will extract one and not the other, or if these are acids or bases we could select a pH that would optimize our separation. Do we need to worry about more than just the difference of the two D values?
Let's look at an example:
System I
D_{a} = 32
D_{b} = 0.032
V_{r}= 1
Let's recall our equations
q _{(fraction in aqueous)} = 1 / (DV_{r} + 1)
p _{(fraction in organic) } = DV_{r} / (DV_{r} + 1)
Vr (volume ratio) = V_{o} / V_{a}
So
p_{a} = 32*1 / (32*1 + 1) = 0.97
p_{b} = 0.032*1/ (0.032*1 + 1) = 0.03
If we assume that we have equal moles of A and B to start then what is the purity of A in the Organic Phase?
Purity = moles A / (moles A + moles B)
Purity = 0.97 / (0.97 + 0.03) = 0.97 or 97 %
Let's look at another system that has a ratio of 1000 for the two D's and see if we get the same result.
Case II
D_{a} = 1000
Db_{} = 1
p_{a} = 1000*1 / (1000*1 + 1*1) = 1000/1001 = 0.999 Aha we got more a into the organic, as we would expect with a higher D value.
Now
p_{b} = 1*1 / ( 1*1 +1) = 1/2 = 0.5
ohoh
What do we get for purity of compound a now?
purity = 0.999 / (0.999 + 0.50) = 0.666
Yuck!
How might one get around this?
Once we have selected the solvent and pH, then there is little
that we can do to change D. What else do we have
in our control?????
Let's look
p = DV_{r} / (DV_{r} + 1)
Not much here except V_{r} and in fact that is the key to this problem. Is there an optimum V_{r} value for the values of D that we have? Yes!
Our equation for this is V_{ r(opt)} = (D_{a}*D_{b})^{0.5}
So let us look at our two cases and see which will give us the optimum values.
Case I
D_{a} = 32 and Db_{} =
0.032
V_{ r(opt)} = (32 * 0.032)^{0.5} = ( 1 )^{0.5} = 1
Case II
DA_{} = 1000 and D_{b} = 1
Vr_{ (opt)} = (1000*1)^{0.5} = 1000^{0.5} = 0.032
Which mean that when we do our extraction we will extract _______ mL of organic for each _______ mL of aqueous.
What is our purity for this system?
p_{a} = 1000*0.032 / (1000*0.032 + 1) = 32/33 = 0.97
and
p_{b } = 1*0.032 / (1*0.032 + 1) = 0.032/1.032
= 0.03 which will give us the 97% purity we had for Case with with
the V_{r} of 1.
OK, What if we need better purity than 97%???? A ratio of 1000 for the D values is actually very good and very few systems will be so fortunate. We have shown how to optimize Vr .
We can do one more step and really help ourselves.
Let's look back to Case I.
If we extract our original aqueous solution with fresh organic phase then we
would extract more of what was there. This is the procedure we covered
above. We removed more for the aqueous phase but each added extraction
would end up spreading the extracted substance into more and more volume and
the ratios of a two compound system would stay fixed and the relative amounts
of A and B would remain the same. Each extraction would get p worth
more.
We would have an improved yield. Yield is the amount of product you get from your operations.
In our case the yield would be the #moles extracted over the total moles to start in our system.
However since the same proportion of each compound is extracted we would
still we would have the same purity.
What if we take our organic phase and extract it with fresh aqueous phase.
We are extracting back into the original phase and this would be called a extraction.
Let's look at the numbers.
D_{a} = 32
D_{b} = 0.032
V_{r} = 1
p_{a} = 0.97
p_{b} = 0.03
q_{a} = 0.03
q_{b} = 0.97
We will pour an equal volume of fresh aqueous (since our V_{r }is
1) into the isolated organic phase and let's look at a little chart to
see where things are.









Now we shake our flask and let the system come to equilibrium.
How much goes to the Aqueous phase
q which is 0.03 for A and 0.97 for B
How much goes to the Organic phase
p which is 0.97 for
A and 0.03 for B









Now what is the purity for A in the organic phase???
Purity = Amount A / (Amount A + Amount B) = 0.97*0.97 / (0.97*0.97 + 0.03*0.03) = 0.94/(0.94 + 0.0009) = 99.9%
Aha ! Much better. The yield now is less but the purity is greatly improved.
What is the Yield?
We could continue to do this to improve our purity.
Lets look at what we get with a second fresh aqueous solution extraction.
After second Back Extraction  Amount A  Amount B 
Organic Phase  0.94*0.97  0.0009*0.03 
Aqueous Phase  0.94*0.03  0.0009*0.97 
Purity A = 0.913 / (0.913 + 0.000027) = 99.997%
But our yield has dropped to 91.3%, there is a price
to pay for the added purity.
Now can this process be expanded? There is a method that is called Craig
Counter Current Extraction that can do repeated extractions. You might
not see this in the course of your career but it does serve as a good example
of extraction and will also serve as a springboard for our discussion in
chromatography.
It is also a viable alternative for preparative work.
Here is what such a system looks like.
From, D. G. Peters, J. M. Hayes and G. M. Hiefje, Chemical Separations and Measurements ( Philadelphia, W.B Saunders, 1974)
Let's Rock
Equal amounts of organic (red) and aqueous (blue) solvents with the analyte(s) are added to the A arm of the tube via port O. Fresh Aqueous Solvent is added to each of the tubes down the apparatus.
Rock the system back and forth and to establish equilibrium.
Allow the system to stand for the layers to separate.
Rotate the apparatus counter clockwise about 90^{o} to 100^{o}.
The top organic layer will dribble into the upper arm and the aqueous layer will settle as shown
When the organic layer has been transferred to the upper arm the apparatus will be rotated back to the horizontal position and we will transfer the organic phase out of arm C and out at point E and into the next tube at point D.
Fresh organic phase is introduced to the first tube at point O and and the process is repeated.
This can be done many times and some set ups can have hundreds of tubes. Let's look at how a compound will distribute itself through such a system.
We will show a five tube system. The values shown in each box represent the amount in that phase in that tube.
Starting Conditions















After One Equilibrium















Transfer Step 1
















Second Equilibrium
How much is in tube #1 before we start to shake.
________
How much is in tube #2 before we start to shake.
________
How much goes to Aqueous upon equilibrium (p or q)
________
How much goes to Organic upon equilibrium (p or q)
________
Now here is what in each tube/phase after equilibrium is reached.
















Now we Transfer 2

















And now we are ready for another equilibrium step.
How much is in tube #2 now???? Remember to add both phases.
___________
Answer
Now here is what we have in each tube after the next equilibrium.. The total in each tube times either p or q as appropriate.

















Transfer Step 3
We transfer again.


















Now how much is in tube #2
_________ Answer
Shake Again Equilibrium 4


















Transfer 4


















See a trend????
How about a binomial expansion?
(q + p)^{n } = 1
Where n is the number of transfers. The power of q would start at n and decrease
by one and power of p would start at zero and increase by one. The coefficients
would be from Pascal's triangle.
1
1 1
1 2 1
1 3 3
1
1 4 6
4 1
1 5 10
10 5 1
1 6 15
20 15 6
1
1 7 21
35 35 21
7 1
Or from the formula.
F_{r,n} = n!/((nr)!r!) p^{r} q^{(nr)}
Where n is the number of transfers carried out and r is the tube number (start counting at zero).
Now for each compound in the system we will have a p and q based on the D values. The volume ratio is usually fixed for these systems and is usually 1. Each analyte will distribute independently and so we can find the yield and purity in each tube in a system.
Let's look at and example.
D_{a} = 3
p = 0.75
q = 0.25
Db = 0.333
p = 0.25
q = 0.75
What would be the purity and yield of Compound A if collected from the last
in our above example.
Amount of A
p^{4} or 0.75^{4} =
0.3164
Amount of B
p^{4} or 0.25^{4} =
0.0039
Purity of A
0.3164 / (0.3164 + 0.0039) = 0.9878 or 98.78%
Yield of A
We collect a fraction of 0.3164 or 31.64%
Horrible Yield!
What if we collect the last two tubes??
Amount of A
p^{4} and 4p^{3}q or 0.75^{4}
+ 4*(0.75)^{3}(0.25) = 0.3164 + 0.4219
= 0.7383
Amount of B
p^{4} and 4p^{3}q or
0.25^{4} + 4*(0.25)^{3}(0.75) =
0.0039 + 0.0469 = 0.0508
Purity of A
(0.3164 + 0.4219) / (0.3164 + 0.4219 + 0.0039 + 0.0469 ) =
0.9356 or 93.56%
Yield of A
We collect a fraction of 0.3164 + 0.4219 = 0.7383
or 73.83 %
Purity still ok and yield is much better.
To collect pure B we would just look to the first tubes.
How else might we enhance purity from this experiment???????????
Without doing additional transfers. We could collect the phase that favors
the solute we are trying to isolate.
Why this headache? Could it be just a rite of passage for
students in separations???
We have just developed plate theory that is used throughout chromatography.
Lets look at a Craig system that has undergone 200 transfers. We will select D values for compounds A and B to show our example. D_{a} of 2.0 and D_{b} of 4.0
This will give us a p_{a} of 0.666 and a p_{b} of 0.800.
If we plot the result we can see.
Which looks like what we will see much of in this course.
Now to find the tube with the maximum amount of a component then you would use:
r_{max} = np = nDV_{r}/(DV_{r} +1)
To find the separation between two peaks we would use.
Dr_{max} = (r_{max})_{a}
 (r_{max})_{b} = n(p_{a}p_{b})
The Gaussian distribution approximation for our binomial expansion would be (when n>24)
Fr,n = (2p)^{0.5}(npq)^{0.5} exp^{[(npr)2/2npq]}
The width of the distribution through the system would be:
w = 4s = 4(npq)^{0.5}
Resolution would be
R = Dr_{max}/w = Dr_{max}/4s
or
R = nDp/(4(npq)^{0.5})
= n^{0.5} Dp / 4(pq)^{0.5}
^{Finis}