Partition Coefficient

Have you heard of DDT?

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DDT stands for Dichlorodiphenyltrichloroethane, an insecticide now banned in many countries because of its toxicity. In Africa, it is still sprayed indoors to deal with mosquitoes. The problem is, it can enter the body through skin, inhalation and food. It stays in the body for a long time, so much so that it has been identified in human breast milk. It can also cross the placenta and reach the foetus. It has serious side effects such as liver impairment and infertility.

The amount of DDT present in breast milk can be identified using RP-HPLC chromatography which relies on the partition coefficient to determine the likelihood of distribution[2] of DDT across biological membranes. Not just DDT, but many other potential contaminants of breast milk can be determined by partition coefficients.

This topic, partition coefficient is important for you as future researchers to understand if a chemical is persistent, if yes, then for how long? The answers to these questions will then help you to make informed decisions on the associated risks and measures to be taken to prevent them.

• We wil start by seeing the definition of partition coefficient.
• Then, we will analyse the equation and formula.
• To finish, we will also see the calculation and interpretation of the partition coefficient.

Partition coefficient: definition

The partition coefficient (Kpc) of a substance is defined as the ratio of concentrations of a soluble solute in two phases of two immiscible solvents at a given temperature when equilibrium is established.[1]

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

The ratio of the concentration of an ionizable solute dissolved in an organic solvent over the concentration of solute in the aqueous phase is called the distribution coefficient. If a substance ionizes in water, the ionized portion of the substance doesn't partition into the organic solvent. In that case, the partition coefficient only takes into account the concentration of the unionized portion of the solute in water because that portion alone can move to the other layer. The ionized portion of water cannot partition or move into the organic layer and vice versa.

Thus, the partition coefficient in the case of ionizable solutes gives the true relative affinities of the solute between the two phases while the distribution coefficient takes into account both the ionized and unionized portions of the solute.

The partition coefficient is of great significance in the pharmaceutical industry. A drug to be taken orally should have affinities for both the organic solvents and water. Organic solvents are lipophilic and can dissolve lipids. So, if a drug is soluble in organic solvents, it can cross the lipid membrane surrounding the cells in our body. Inside the cell, there is water and for the drug to be effective, it should now be soluble in water. If the drug is not soluble at all in organic solvents, it will not be able to cross the lipid membrane of the cell and will be eliminated from the body. Likewise, if the drug is not soluble in water, despite crossing the lipid membrane it will not have any desired and expected effect. Generally, Octanol and water are considered for this purpose of estimating the affinity of drugs in the pharma industry.

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

In a container containing 100 ml of water and 10ml of an organic solvent, 1 g of methylamine (CH3NH2) is added. Upon mixing, 0.6 g of methylamine is transferred to the ether layer. Calculate the partition coefficient between the organic solvent and water.

Solution

Since we have been asked to calculate the partition coefficient between the organic solvent and water, the concentration of solute in ether will be in the numerator since it is mentioned first. Therefore, the formula becomes:

$$K_{pc} = \frac {[CH_3NH_2]_{dissolved~in~organic~solvent}} {[CH_3NH_2]_{dissolved~in~water}}$$

It is given that upon mixing, 0.6 g of methylamine is transferred to the organic solvent.

$$concentration~of~CH_3NH_2~in~organic~solvent = \frac {0.6}{10}~g~cm^{-3}$$

That means the rest of the methylamine is dissolved in water.

$$Therefore~concentration~of~CH_3NH_2~in~organic solvent = \frac {1-0.6}{100} = \frac {0.4}{100}~g~cm^{-3}$$

Now that we know the concentration of the solute in both solvents, we can calculate the partition coefficient:

$$K_{pc} = \frac {\frac {0.6}{10}}{\frac {0.4}{100}}$$

$$K_{pc} = 15$$

Remember that 1 ml = 1 cm3.

You should always take care to use the same units of concentration in the numerator as well as the denominator. As said earlier, notice that the Kpc is a dimensionless quantity

Let us look at the same problem, but with a different angle.

1g of methylamine is shaken with 100ml of water and 5ml of an organic solvent. The partition coefficient, Kpc, between the organic solvent and water is 20. Determine the amount of methylamine dissolved in water and in the organic solvent.

Solution

Let us write the information given to us in the problem.

$$Weight~of~CH_3NH_2 = 1g$$

$$Volume~of~water = 100 cm^3$$

$$Volume~of~organic~solvent = 5 cm^3$$

$$K_{pc} = 20$$

We have to find out how much of methylamine is dissolved in water, and how much in the organic solvent. We can begin by writing the formula for Ksp. Remember that Ksp is given between the organic solvent and water, so organic solvent will be in the numerator.

$$K_{pc} = \frac {[CH_3NH_2]_{dissolved~in~organic~solvent}}{[CH_3NH_2]_{dissolved~in~water}}$$

Let us assume the amount of methylamine dissolved in the organic solvent to be x. Therefore, the amount dissolved in water becomes 1-x.

$$Concentration~of~methylamine~in~organic~solvent = \frac {x}{5}$$

$$Concentration~of~methylamine~in~water = \frac {1-x}{100}$$

$$\therefore K_{pc} = \frac {\frac {x}{5} }{\frac {1-x}{100} } = 20$$

$$\therefore \frac {20\cdot x}{1-x} = 20$$

$$\therefore x = 1-x$$

$$x = \frac {1}{2} = 0.5$$

Therefore, 0.5g of methylamine is dissolved in the layer of organic solvent, and the other 0.5g is dissolved in water.

Let us look into another example.

A similar model question appeared in one of the previous papers.

Q: Ammonia is soluble in both water and organic solvents. An aqueous solution of ammonia is shaken with the immiscible organic solvent trichloromethane. The mixture is left to reach equilibrium. Samples are taken from each layer and titrated with dilute hydrochloric acid.

• A 25.0cm3 sample from the trichloromethane layer requires 13.0cm3 of 0.100mol·dm–3 HCl to reach the end-point. (PART 1)
• A 10.0cm3 sample from the aqueous layer requires 12.5cm3 of 0.100mol·dm–3 HCl to reach the end-point. (PART 2)

(ii) Calculate the partition coefficient, KPartition, of ammonia between trichloromethane and water.

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,

$$K_{pc}$$ = $$\dfrac {Concentration_{org} } {Concentration_{aq}}$$

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

 $$K_{pc}$$ = $$0.416$$

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

 Kpc < 1 - Hydrophilic/water-loving-more soluble in water-PolarKpc > 1 - Hydrophobic-/water-hating-more soluble in organic solvent-non polar

Partition Coefficient - Key takeaways

• Partition coefficient is the ratio of the concentration of a solute which is dissolved in two immiscible liquids, measured when the solute is in equilibrium across the interface between the two liquids.
• The ratio of concentration of an ionizable solute dissolved in an organic solvent over the concentration of solute in the aqueous phase is called the distribution coefficient.
• Formula of partition coefficient , $$K_pc$$ = $$\dfrac {[Solute]_{org}} {[Solute]_{aq}}$$
• Partition coefficient helps understand how the distribution of a solute takes place between two immiscible liquids.
• Partition coefficients vary with temperature, just like any other equilibrium constant.
• Partition coefficient is dimensionless-doesn't have any units.
• Partition coefficient helps us determine the solubility, polar-non-polar nature and hydrophilicity of a solute.
• If the Kpc is less than 1, it means that the solute is more soluble in water and if the Kpc exceeds 1, it indicates that the solute is less soluble in water and more soluble in the organic solvent

References

2. https://pubmed.ncbi.nlm.nih.gov/21300395/

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What are partition coefficients used for?

Partition coefficients are used in drug design to control how a drug is absorbed into the body. A higher partition coefficient means that a solute will easily permeate from one solvent to the other. A drug with high partition coefficient can easily permeate into the body when it attacks specific sites such as sites with lipid bilayers of cells.

How is partition coefficient useful in extraction?

Pulling a solute out of a solvent using another solvent is called extraction. When there are layers of solvents, the solute has a tendency to dissolve in one solute more than the other. The partition coefficient of solvents can help determine which solute will disolve more in which solvent.

What is the difference between solubility and partition coefficient?

Solubility is the is a measure of the amount of solute that can be dissolved in a given amount of one solvent.

Partition coefficient tells us which solvent the solute will "prefer" to dissolve in when it can permeate between two immiscible solvents.

What is the difference between distribution coefficient and partition coefficient?

Partition coefficient is the concentration ratio of a solute in any two immiscible solvents.

When the ratio is taken between the concentration in an organic solvent and water, it is called distribution coefficient.

What does a high partition coefficient indicate?

A higher partition coefficient means that the solute can easily permeate to that particular solvent from the other solvent.

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