Lucid Concept Of Partition Law(Distribution Law)

It is observed that, liquids like chloroform (CHCl₃) and water (H₂O) when shaken together in a test tube, and then the mixture is allowed to stand, the two liquids separate out into two discrete layers. This means they do not mix or dissolve into each other. Such liquids are called immiscible liquids.

Chloroform and water are immiscible liquids. Likewise ether and water are partly miscible, means they are almost insoluble or negligibly soluble into each other.

The phenomenon of distribution is observed in a class of certain liquid substances which are more soluble in one solvent but are less soluble in another solvent. For example, Iodine (I₂) is extremely soluble in organic solvents like: Carbon Disulphide (CS₂), Chloroform (CHCl₃) and Carbon Tetrachloride (CCl₄). But it is poorly soluble in solvent like: Water (H₂O).

When iodine is shaken with a mixture of chloroform and water, it gets dissolved in both the solvents but its concentration in chloroform is more than that in water.

This is called distribution of iodine in chloroform and water.

The phenomenon of distribution was studied first by scientist, "Nernst". After studying this phenomenon in detail he gave the following law which is called, "law of distribution".

“When a solute distributes itself between two immiscible solvents, there exists for each molecular species, at a given temperature, a constant ratio of distribution between the two solvents, and this ratio is independent of any other molecular species which may be present”.

The value of distribution ratio depends on following factors:

(1) Nature of two solvents,

(2) Nature of solute taken, and

(3) Temperature.

Suppose a solute A is added to such two immiscible solvents B & C in both of which it is soluble. On vigorous shaking, the solute gets partitioned between the two solvents in such a way that concentration of solute A in solvent B (say C₂) is directly proportional to its concentration in solvent C (say C₁).

Mathematically:

(concentration of solute A in solvent B) / (concentration of solute A in solvent C) = C₂/C₁

= constant = K_d.

Here, the constant of proportionality "K_d" is known as, "Partition Coefficient" or "Distribution Coefficient" of solute A in solvents B and C respectively.

It should be noted that the ratio: C₂ / C₁ is constant only when the "colligative properties" of both the solutions are similar. It should be noted that the colligative properties of two solutions can remain same only if the dissolved substance does not undergo dissociation or association.

It should be clearly understood that such spontaneous behavior of the solute to get distributed between two immiscible solvents, is a direct consequence of thermodynamic requirement of any system to attain equilibrium, because the state of equilibrium assures the minimum energy level and stability! Therefore, the constant K_d is nothing else, but an equilibrium constant of such system!

The solute can distribute itself between two immiscible liquids only when the following conditions are satisfied.

(a) The solute should not react chemically with either of the solvents.

(b) In both the solvents, molecularity of solute should remain same, i.e. it should neither undergo association nor undergo dissociation, so that the colligative properties of both the solutions remain identical.

(c) The solute must be soluble in both the solvents.

If iodine is shaken with a mixture of two liquids: carbon disulphide (CS₂) & water (H₂O) and then allowed to stand, iodine gets distributed between the two solvents.

Here, a state of equilibrium exists between the concentration of iodine in carbon disulphide and the concentration of iodine in water.

It is also found that when amount of iodine is varied, the ratio of concentration of iodine in two solvents remains constant, provided temperature is kept constant.

Mathematically: (concentration of I₂ in CS₂) / (concentration of I₂ in H₂O) = (C₂/C₁) = K_d.

Constant K_d is known as "distribution coefficient" or "partition coefficient" of iodine in solvents carbon disulphide and water respectively.

This will be clear from following table.

Though the value of distribution coefficient K_d remains constant like any other constant, it may show non-constancy under certain conditions. It is interesting to know about conditions which lead to such anomaly in K_d.

These two conditions are:

(a) If phenomenon of association is taking place among the particles of solute.

(b) If phenomenon of dissociation is taking place among the particles of solute.

If any one of above two phenomena is taking place and if K_d is computed taking total concentration of solute in two phases, then the computed value of K_d will not come to be constant.

In such a case, the value of K_d can be worked out to be constant if the concentration of a common species in both of solutions is considered for calculations.

To understand this, consider the distribution of benzoic acid (C₆H₅COOH) between water (H₂O) and chloroform (CHCl₃).

In this case, both dissociation as well as association takes place simultaneously as explained below.

(1) In water, the acid is partially dissociated in the form of benzoate anion having chemical formula: C₆H₅COO^- and hydronium cation having chemical formula: H₃O⁺).

(2) In chloroform, the acid is partially associated to form bi-molecule having chemical formula: [(C₆H₅COOH)₂].

(The corresponding equations for both the phenomena are given below.)

In such a case, there is no resemblance of K_d if total concentration of benzoic acid is taken for calculation.

However, if the concentration of common species in both of the phases is considered for calculation, the computed value of K_d computed shows constancy!

Which is the common species in both of the phases?

It is unassociated or un-dissociated single molecule of benzoic acid having chemical formula: C₆H₅COOH.

This is clear from the experimental details given in following two tables.

(a) Dissociation of benzoic acid In water: C₆H₅COOH + H₂O → C₆H₅COO^-+ H₃O⁺.

(b) Association of benzoic acid in chloroform: 2 C₆H₅COOH → (C₆H₅COOH)₂.

In practice, the law of distribution is useful in:

(a) Removal of a particular substance from given solution

(b) To carry out some qualitative analysis

(c) To determine the degree of hydrolysis of certain salts

(d) To determine the constitution of complex halide ions

(e) To prove the existence of “Tetraamminecopper (II) ion in aqueous ammoniacal solution of copper sulphate. (The chemical formula of tetraamminecopper (II) ion is: [Cu(NH₃)₄]²⁺.)

The detailed information on above said applications is given below.

The distribution coefficient of iodine in carbon disulphide (CS₂) / water (H₂O) system is very high, having value of about 410. Hence, when aqueous solution of Iodine is shaken with carbon disulphide, concentration of iodine in the layer of carbon disulphide is about 400 times that in water.

The CS₂ layer rich in iodine is isolated with a separating funnel and the process is repeated till concentration of I₂ becomes negligible in aqueous layer.

(Theoretically, complete removal of I₂ by this process is not possible, however practically such removal may be considered as almost complete).

[Note: The extraction of Iodine from its aqueous solution using other solvents like chloroform or carbon tetrachloride is less efficient compared to that with carbon disulphide. This is because at laboratory temperature, the value of distribution coefficient for I₂, in CHCl₃ / H₂O system is 109 and that in CCl₄ / H₂O system is 85. Both of these values are much less than that of CS₂.]

Example: 10 milligrams of Iodine is dissolved in 12 milliliters of water & then shaken with 2 milliliters of carbon tetrachloride till equilibrium is reached. Calculate remaining weight of Iodine in aqueous layer. (Given: K_d: CCl₄ / H₂O = 85).

Solution: Assume milligrams of remaining Iodine in aqueous layer to be x mg, hence milligram of Iodine in the layer of CCl₄ can be taken as 10-x mg.

Hence, the concentration of I₂ in water = (x/12) while that in CCl₄ = [(10-x)/2].
(Note: Concentration is expressed in terms of milligrams per liter).

As, K_d = 85;

[(10-x)/2] / [(x/12)] = 85;

this gives, x = 0.66 mg.

Thus, remaining weight of I₂ in aqueous layer will be 0.66 mg.

[If 12 milliliters of residual aqueous layer is separated and shaken with further 2 milliliters of CCl₄ to attain equilibrium, the mg of I₂ remaining left in aqueous layer can be calculated to be 0.043 mg! This means 9.957 mg of I₂ (99.57 %) can be extracted using only 4 milliliters of CCl₄!].

[Note: It is interesting to note that if the original aqueous solution was shaken with 4 milliliters of CCl₄ (instead of taking 2 milliliters of CCl₄ two times as discussed above), then usual calculation shows that the aqueous layer will contain 0.34 mg of I₂ (instead of 0.043 mg of I₂).

This proves that, it is more efficient and also economical to carry out number of successive extractions taking small quantity of solvent rather than a single extraction taking large quantity of solvent.]

When acidic solution of chromate is treated with hydrogen peroxide, a deep blue colored solution of perchromic acid is formed, which is a test to confirm presence of either chromate or hydrogen peroxide.

However, perchromic acid is highly unstable hence soon decomposes to give oxygen and green solution of chromic salt. Hence deep blue color obtained disappears soon. This makes the detection difficult.

Here law of distribution can become helpful as explained below.

Perchromic acid is more soluble and also more stable in amyl alcohol than in water. Hence, to avoid above difficulty, the testing solution is immediately shaken with amyl alcohol, so that blue color of perchromic acid lasts longer.

This makes easily to detect the presence of either chromate or hydrogen peroxide.

On hydrolysis, salt like aniline hydrochloride, yields aniline and hydrochloric acid.

Here, if concentration of aniline produced can be found out, degree of hydrolysis can be calculated. To achieve this, the mixed solution is shaken with benzene, K_d of which is known.

Aniline distributes itself between water and benzene in the ratio of K_d. The concentration of aniline in benzene is determined by suitable experimental technique. Now, by subtracting concentration of aniline in benzene from initial concentration of aniline hydrochloride, the concentration of aniline in aqueous medium can be calculated.

This enables us to calculate degree of hydrolysis of aniline hydrochloride. (By equation, 1 mole aniline hydrochloride is equivalent to 1 mole of aniline).

Iodine is much more soluble in an aqueous solution of potassium iodide (KI) than in water. This is due to the formation of potassium triiodide (KI₃).

The following equilibrium exists in such solution:

KI + I₂ ↔ KI₃ or I^- + I₂ ↔ I₃^-.

If the solution is shaken with CCl₄, in which I₂ alone is soluble appreciably, most of I₂ passes in CCl₄.

By determining the concentration of I₂ in CCl₄, the concentration of I₂ in aqueous solution can be calculated from the known value of K_d. Summing of these two will give total concentration of I₂ present at equilibrium. Subtracting this from the initial concentration of I₂ will give concentration of combined iodine having formula: KI₃. Subtracting the concentration of KI₃ produced from initial concentration of KI will give concentration of KI at equilibrium.

Let, initial concentration of KI = a;
and initial concentration of I₂ = b;
concentration of I₂ determined in CCl₄ = c;
concentration of I₂ calculated in H₂O = d = (c) / (K_d);
total concentration of I₂ at equilibrium = [c+{c/K_d}];
concentration of KI₃ produced = (initial concentration of I₂) - (total concentration of I₂ at equilibrium)
= b - [c+{c/K_d}].
Concentration of KI left at equilibrium = (initial concentration of KI) - (concentration of KI₃ formed);
= a - b- [c+{c/K_d}].
The computation of equilibrium constant as per equation will be,

K_eq = [KI] [I₂] / [KI₃]; indicates that complex halide is in the form of KI_3.

Existence of tetraamminecopper(II) Ion in aqueous ammoniacal solution of copper sulphate can be proved by studying the distribution of free ammonia between chloroform and water as shown below:

[Cu(NH₃)₄]²⁺ = Cu²⁺ + 4NH₃.

(1) Qualitative Inorganic Analysis, Arthur I Vogel, (formerly Head Of Chemistry Department, Woolwich Polytechnic, London.)

(2) Principles Of Physical Chemistry, By: Samuel H. Maron & Carl F. Prutton.
Published by: Oxford & IBH Publishing Co., New Delhi (Original Publisher: The Macmillan Company, New York).

(3) Advanced Chemistry, Physical & industrial; By: Philip Matthews.
(Cambridge University Press); Published in India by: Foundation Books Pvt. Ltd., New Delhi.