Partition Coefficient
If
two immiscible phases are shook in a separatory funnel they will form
two separated layers. The phases can be in the form of gases, liquids
and even solids, but for simplicity only the liquid phase will be used
in this explanation. Suppose liquid one is water (H2O) and the second one is diethyl ether (CH3-CH2-O-CH2-CH3),
the ether will float on top of the water since it has a lower density. A
substance called X, soluble in both liquid was put in before the
shaking took place. Suppose the particles of X are more soluble in the
ether than in the water such that the scenario will resemble the picture
below. A dynamic equilibrium will be established which consists of a
constant rate of particlescrossing the boundary between the two
liquids.
This information could be interpreted in the following equation:
X(in solution in water) 
X(in solution in ether)
X(in solution in ether)
This is a simple equilibrium, hence an equilibrium constant can be found.
Kc = [X in ether] / [X in water]
This particular equilibrium constant is given the name Kcp, which stands for the partition coefficient.
The
partition coefficient is dependent as all equilibrium constants on
temperature but they do have specific limitations as well. For example
the solutions have to be quite dilute for the constant to be accurate,
the solute is required to be in the same molecular state in both
solvents. This means that the solute cannot ionise, react or associate.
Partition Coefficient
The
calculations of the partition coefficient is a simple ratio of two
separate concentrations. This proposes that the units that are used are
irrelevant as long as they are the same in the numerator as in the
denominator. The concentration can be expressed as mol dm-3 - moles per cubic decimetre, but more commonly g cm-3
- grams per cubic centimetre is used. The important distinguishing
factor between the units is that the square brackets are strictly used
for mol dm-3.
This is to say that [X] is only to be used for standard concentration
units, and in other cases “concentration of X” is a viable way.
Lipophilicity and log P
To
summarise; the partition coefficient is the ratio between the
concentration of un-ionized solute in two immiscible solvents. To make
this possible the pH of the water is adjusted to make the largest amount
of compound un-ionized. The logarithm of the Kcp value
is also known as the log P, which is the measure of Lipophilicity.
Lipophilicity is the tendency for molecules to dissolve into fats,
lipids, oils and other hydrophobic solvents such as ether.
log Poctanol/water = log ([solute]octanol / [solute]water)
When
calculating the partition coefficient, the compound dissolved must be
strictly be un-ionized. The same principle can be applied to a compound
that ionizes completely or partially, and in this case it is called the
distribution coefficient. Like the partition coefficient the
distribution coefficient is a ratio between the concentration of solutes
in two immiscible solvents, but in contrast it takes into account
ionized and un-ionized compounds. In practice the aqueous state is
buffered to a specific pH, which will be stable when a compound is
introduced. The distribution coefficient is the logarithm of the ratio
between all the sum of all forms of a solute in two solvents.
log Doctanol/water = log ([solute]octanol / [solute] (ionized) water + [solute] (un-ionized) water)
The
distribution coefficient can be seen as an extended form of the
partition coefficient, taking more factors into account, which implies
that log P = log D when ionization does not occur. As log D is dependent
on pH it must be specified at what value of pH the calculations were
performed.
Practical Applications
The
application of this method its applications can be seen in the chemical
and pharmaceutical sciences. Normally water is chosen as one solvent
while the other one is a hydrophobic, such as octanol. This makes it
possible to determine how hydrophobic or hydrophilic a substance is by
determining the partition coefficient. This in terms of medical studies
allows to determine what distribution a drug will have inside the human
body. Based on this an appropriate compartments of the human body are
chosen to distribute them within. If a drug has a high octanol/water
partition coefficient it is preferably distributed in an area that is
hydrophobic such as the lipid bilayers. On the contrary, drugs that have
a low octanol/water partition coefficient are preferably distributed in
a hydrophilic source such as the blood serum.
Lipophilicity
is crucial in pharmacological practice, not only by how easily a drug
can dissolve into fats as mentioned in the previous paragraph. It helps
determining how strongly it will affect its target and how long it will
stay in the body in its active form.
A
reliable way of determining the partition coefficient is by the “shake
flask method”. The solute is dissolved by shaking the flask containing
the solvents octanol and water. The concentration is then measured in
each phase by UV/VIS spectroscopy
Problem Involving Partition Coefficient
Problem I
A solution of 1.56 g of X in 125 cm3 of water was shaken with 12 cm3 of ether, 0.89 g of X was moved to the ether layer. Now calculate the partition coefficient of X between ether and water.
Solution
The
question asks to calculate the coefficient of X between ether and
water. As the ether is stated first in the question it will come on top
in the expression as follows.
Kcp = concentration of X in ether / concentration of X in water
Concentrations are given in g cm-3 and hence will be calculated with these “non-standard” units.
Concentration of X in ether: 0.89g/12cm3
Concentration of X in water: 0.67g/125cm3
The
concentration of water is not directly given, but can easily be
calculated from the information that 0.89g was transferred to the ether.
Hence (1.56g - 0.89g = 0.67g) is left in the water.
So:
Kcp = concentration of X in etherconcentration of X in water
= (0.89g/12.0cm3) / (30.67g/125cm3)
= 13.8 (3 sig fig)
Note that all units cancel in the expression and a unitless number is the result.
Problem I a)
Now calculate the Lipophilicity (log P) of the previous question.
Solution
The log P value of the previous question is simply the logarithm of the Kcp.
log(13.8) = 1.14 (3 sig fig)
Problem II
In this problem use the same solvents, X and partition coefficient as given or calculated in previous problem.
How much of X would have been extracted to the ether if the original solution (1.56 g of X in 125 cm3 of water) was shaken with half the amount of ether.
Solution II
Like in the previous equation we represent an expression for the concentration of both solvents.
Concentration of X in ether: m/6 g cm-3
The unknown mass will be represented as m.
Concentration of X in water: (1.56 - m)/ 125 g cm-3
Now, the information can be plotted into the expression for Kcp and solved for m.
Kcp = concentration of X in etherconcentration of X in water
13.8 = m/6 g cm-3(1.56 - m)/ 125 g cm-3 Note that all units cancel
13.8 = m/6 (1.56 - m)/ 125
13.8(1.56-m) / 125 = m6
21.528-13.8m / 125 = m6
21.528 = 125m6 + 13.8m
21.528 = 34.633m
m = 21.528 / 34.633
m = 0.622 (3 sig fig)
Hence the mass of X extracted to the ether is 0.622 grams
