Distribution Law B.Sc. 2nd Year

Distribution Law B.Sc. 2nd Year

Distribution Law

Nernst's Distribution Law

When a solute distribute itself between two immiscible solvents in equilibrium, then the ratio of its concentration in two solvents remains constant. This is called Distribution law.
Example: The solute iodine distributes itself in CCl4 and water solvents. Let C1 and C2 be the concentration of I2 in CCl4 and water respectively. Then from Distribution law-
C1 / C2 = KD
where KD is partition coefficient.
The value of KD depends upon the Temperature, Nature of solute, Nature of two solvents and Manner in which the constant is expressed
i.e. C1 / C2 or C2 / C1

Essential Conditions for the Distribution Law

Followings are the essential conditions for that-
1. Temperature must be constant.
2. Solutions are must be dilute.
3. Two solutions should be insoluble or very sparingly soluble and their solubility should not be affected by the presence of solute.
4. Solute must not be undergo dissociation or association or interaction with any of the solvents.

Thermodynamic Derivation of the Distribution Law

The thermodynamic derivation of the distribution law is based upon the principle that if there are two phases in equilibrium (i.e. two immiscible solvents containing the same solute dissolved in them), the chemical potential(μ) of a substance present in them must be same in both the phases.
From thermodynamics, we know that the chemical potential (μ) of a substance is a solution given by-
μ = μo + RT lna
Where μo is the standard chemical potential and 'a' is the activity of the solute in the solution.
Thus for the solute in liquid A, we have-
μA = μoA + RT lnaA
Similarly for the solute in liquid B we have-
μB = μoB + RT lnaB
But as already stated, since the liquids A and B are in equilibrium,
μA = μB
or, μoA + RT lnaA = μoB + RT lnaB
or, RT lnaA − RT lnaB = μoB − μoA
or, ln(aA/aB) = (μoB − μoA)/R -----(equation-1)

At a given temperature, μoAand μoB are constant for given substance in the particular solvents. Hence at constant temperature, equation (1) becomes-
ln(aA/aB) = Constant
or, aA/aB = Constant -----(equation-2)
This is the exact expression of the distribution law. However, if the solutions are dilute, the activates are equal to the concentrations so that the (equation-2) is modified as-
CA/CB = Constant -----(equation-3)
(equation-3) is the original form of the distribution law.

Modified form of distribution law when association of solute occurs in one of the solvent

nA → An
Number of particles decreases after association.
If a solute A present in solvent-I where its concentration is CI and in solvent-II, n molecules of solute A associates to form An and a few molecules of solute A are also present in solvent-II. If the concentration of A and An be CA and CII in solvent-II, respectively, then from distribution law-
CI/CA = KD -----equation(1)

Now equilibrium constant for the reaction nA ⇌ An is-
KC = [An]/[A]n
or, KC = CII/CnA
Taking nth root , we get-
n√CII/CA = n√KC -----equation(2)
Now, dividing equation (1) by equation (2), we get-
(CI/CA) x (CA/n√CII) = (KD/n√KC)
CI/n√CII = KD/n√KC = K -----equation (3)
Equation (3) is a modified form of distribution law when association of solute occurs in one of the solvent.

Modified form of distribution law when dissociation of solute occurs in one of the solvent

Let a solute molecule A which does not dissociate in solvent-I has concentration CI. When it dissociates into x and y in solvent-II having total concentration CII.
If α be the degree of dissociation of solute A in solvent-II, then-
      A       ⇌       x       +       y
CII(1 − α)       CIIα            CIIα
so, the concentration of undissociated molecules of solute A in solvent-II will be CII(1 − α)
Hence, the modified form of distribution law when dissociation of solute occurs in one of the solvent will be-
CI/CII(1 − α) = K

Solvent extraction

The most important application of the distribution law is in the process of extraction, in the laboratory as well as in industry. In the laboratory, it is frequently used for the removal of a dissolved organic substance from aqueous solution with solvents such as benzene, ether, chloroform, carbon tetrachloride, etc. the advantage is taken of the fact that the partition coefficient of most of the organic compounds is very largely in favour of organic solvents.
The process of extraction is more efficient if the solvent is used in a number of small portions than in one whole. This is called Multiple Extraction.

Determination of Equilibrium Constant from Distribution Coefficient

Distribution law helps in the determination of equilibrium constant of a reaction when one of the reactants is soluble in two immiscible solvents.
Example: When KI reacts with I2 to form KI3
KI + I2 ⇌ KI3
This reaction can be carried out in water, while I2 is soluble in both water and benzene.
To find Distribution Coefficient of I2
I2 is shaken with water and benzene in a bottol. The concentration of I2 in the two layers is then determined by titration against standard thiosulphate solution.
So, (Conc. I2 in water/Conc. I2 in benzene) = K

To find Equilibrium Constant using the value of K

A solution of KI of concentration 'a' is shaken with I2 in a bottol, some benzene is also added and shaken. On standing, the mixture seperates into two layers.
The concentration of I2 is determined in the two layers by titration against standard thiosulphate solution.
Let 'b' the concentration of I2 in benzene layer, 'c' be the concentration of I2 in water layer which is really the total of the concentration of free I2 and KI3.
K is the value of Distribution coefficient of I2 between water and I2.
So, (Conc. I2 in water/Conc. I2 in benzene) = K
∴ Concentration of free I2 in water layer = K X b
Hence, Concentration of KI3 in water layer = c-Kb
∴ Concentration of KI in water layer = a-(c-Kb) = a-(c+Kb)

Now, we can write the equilibrium constant of the reaction-
KI + I2 ⇌ KI3
so, Kc = [KI3]/[KI ] [I2]
or, Kc = (c-Kb)/(a-(c+Kb))Kb

Other example of complex formation
CuSO4 + NH3 ⇌ CuSO4.4NH3 or [Cu(NH3)4]+2SO4-2