Partition Coefficient

Partition coefficient: definition

Let us understand this definition with an example:

Let us consider equal volumes of ether, an organic solvent and water. Ether and water are two immiscible liquids and when taken in a separation funnel, they separate into two distinct layers. The top layer will be the organic ether layer, less dense than the bottom aqueous layer.

Now, let us take a solute, ‘S’ which has some degree of solubility in both the solvents. The funnel is given a thorough shake and left to settle. The two layers separate with the solute distributed in both of them. At a certain point, equilibrium is established which means that the rate at which the ammonia molecules move into the organic layer from the aqueous layer is equal to the rate at which they move from the aqueous layer to the organic layer.

This can be represented by the equation

\(S_(org)\rightleftharpoons S_(aq)\)

Partition Coefficient: Equation and Formula

Based on the definition, we will now write the equation/formula of partition coefficient.

$$Kpc=\frac{[solute]_{Organic solvent}}{[solute]_{Water}}$$

The square brackets represent the concentration of the substance, solute in this case. An easy way to remember this is that the organic layer floats on the top; hence you can write it down on the numerator.

Have a look at the diagrammatic representation of what we just discussed:

Fig. 1: Separatory Funnel, Archana Tadimeti, StudySmarter

Things to be noted about the partition coefficient -

• Partition coefficients vary with temperature, just like any other equilibrium constant.
• The solute should not react with either solvents, neither ionize in it.
• The solute must be present in the same physical state in both the solvents.
• Partition coefficients are more accurate for dilute solutions.

Partition coefficient calculation

SOLUTION

Before attempting to answer such questions, you should carefully look at the units of all the physical quantities. They should all agree with each other.

Here the volumes are given in cm³ while the concentration is given in mol dm^-3. You should first convert either the volume into dm³ or the concentration to mol cm^-3. Either way works fine.

For convenience, we indicate the trichloromethane layer as (org) and the aqueous layer as (aq)

Here, we will approach this question by converting the units of volume from cm³ to dm³.

PART- I: TRICHLOROMETHANE/ORGANIC LAYER

\(V_{NH_3(Org)}\) = \(25.0 cm^3\) = \(25 \times 10^{-3} dm^3\qquad 1 cm^3 = 10^{-3} dm^3\)

\(V_{HCl}\) = \(13.0 cm^3\) = \(13.0 \times 10^{-3} dm^3\)

For ease of calculation, let us convert the concentration of \(HCl\) from real number to scientific notation.

\(0.100~ mol~ dm^{-3} = 100 \times 10^{-3} dm^{-3} \)

Let us now find the concentration of from the results of titration.

\(NH_3+HCl\rightarrow NH_4Cl\)

one mole of ammonia reacts with one mole of HCl. Therefore, the number of moles is equal. If we find the number of moles of HCl from the concentration and volume of HCl, we will know the number of moles of Ammonia . From this information, we can deduce the concentration of Ammonia in the organic layer.

Concentration of HCl = number of moles of HCl / volume of HCl used for reaching the endpoint

\(100 \times 10^{-3} mol~dm^{-3} \) = \(\dfrac {n_{HCl} } {13 \times 10^{-3}dm^3} \)

\(n_{HCl}\) = \(13 \times 10^{-3} \cancel {dm^3} \times 100 \times 10^{-3} mol~\cancel {dm^{-3}} \)

On simplifying...

\(n_{HCl}\) = \(1.3 \times 10^{-3} mol\)

∴ \(n_{HCl}\) = \(n_{NH_3}\) = \(1.3 \times 10^ {-3} \)

Now, using this number of moles and volume given in the question, let us find the concentration of ammonia in the organic layer (chloroform/trichloromethane).

\( \mathrm{Concentration_{org}}\) = \(\dfrac {n_{org}} {V_{org}}\)

= \(\dfrac {1.3 \times 10^ {-3}mol} { 25 \times 10^{-3} dm^{3}}\)

= \(\dfrac {1.3 \times \cancel {10^{-3}}mol} { 25 \times \cancel {10^{-3}}dm^{3}}\)

= \(0.052~mol~dm^{-3}\)

∴ \( \mathrm{Concentration_{org}}\) = \(0.052~mol~dm^{-3}\) -------------------------------1

Part II : AQUEOUS LAYER

Converting all the units

Data:

Volume of the sample in aqueous layer \(V_{aq}\) = \(10 cm^3 \) = \(10 \times 10^{-3} dm^3\)

Volume of HCl required to neutralise ammonia in aqueous layer = \(12.5 cm^3\) = \(12.5 \times 10^{-3}dm^3\)

Concentration of HCl used = \(0.100~mol~dm^{-3}\) = \(100 \times 10^{-3} mol~dm^{-3}\)

From the data, let us first calculate the number of moles of HCl reacted with ammonia in aqueous layer. The number of moles obtained will be equal to the number of moles of ammonia.(hint: equation)

Concentration of HCl = number of moles of HCl / volume of HCl used for reaching the endpoint

\( \mathrm {Concentration_{aq}}\) = \( \dfrac {n_{HCl}} {V_{HCl}}\)

\(100 \times 10^{-3} mol~dm^{-3}\) = \( \dfrac {n_{HCl}} {12.5 \times 10^{-3}dm^3}\)

\(n_{HCl}\) =\(12.5 \times 10^{-3} \cancel {dm^3} \times 100 \times 10^{-3} mol~ \cancel {dm^{-3}} \)

\(n_{HCl}\) = \(1.25 \times 10^ {-3} mol\)

Now that we have the number of moles, we can deduce the concentration of ammonia in water.

\( \mathrm Concentration_{NH_3~in~aq}\) = \( \dfrac {n_{NH_3}} { V_{NH_3}}\)

= \( \dfrac { 1.25 \times \cancel {10^{-3}}mol} {10 \times \cancel {10^{-3}} dm^3}\)

= \(0.125 mol~dm^{-3}\)

∴ \( \mathrm Concentration_{aq} \) = \(0.125 mol~dm^{-3}\) ------------------ 2

FINDING PARTITION COEFFICIENT OF AMMONIA:

From results 1 and 2,

\(K_{pc}\) = \( \dfrac {Concentration_{org} } {Concentration_{aq}}\)

= \( \dfrac {0.052 \cancel {mol~dm^{-3}}} {0.125 \cancel {mol~dm^{-3}}}\)

Thus, the partition coefficient of ammonia is 0.416. Notice that the units cancelled out and the obtained ratio has no units.

Partition coefficient: Interpretation

Why do we need to know the partition coefficient of a solute?

Well, the Partition coefficient of the solute tells us the affinity of the solute towards the water and organic solvents. If a solute is more soluble in water than the organic solvent, then the concentration of the solute will be more in water. This means that the solute is hydrophilic-water loving.

Let us say, the concentration of a solute in water is 8 mol/dm3 and the concentration of a solute in an organic solvent is 2 mol/dm3. The partition coefficient will be

$$Kpc=\frac{2}{8}=0.25$$

Notice that the partition coefficient has no units, it is dimensionless since it is a ratio.

The partition coefficient value less than 1 indicates that the solute is hydrophilic, hence has more affinity for water. On the other hand, if the Kpc value exceeds 1, it indicates that the solute is hydrophobic (and lipophilic), hence more soluble in the organic solvent.

Therefore, it is interesting to note that if