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 n-octanol (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 ACD Labs Web site
 
 

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 1-q.  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 (1-q) 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 n-octanol?
 

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.

 Example 1

Now let's look at another example where we are searching for an optimum volume.

 Example 2

And now lets look at a better way to get the same results as in the last example.

 Example 3

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 co-ops 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.

 Answer

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

oh-oh

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.

 Answer

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 BACK 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.
 
 

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.
 
 

Or from the formula.

F_r,n  =  n!/((n-r)!r!) p^r q^(n-r)

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                        p⁴   or  0.75⁴   =   0.3164
Amount of B                        p⁴   or  0.25⁴   =   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                        p⁴   and  4p³q  or  0.75⁴ +  4*(0.75)³(0.25)  =   0.3164 + 0.4219 = 0.7383
Amount of B                        p⁴   and  4p³q   or  0.25⁴ +  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^{-[(np-r)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