Nernst Distribution Law and factors affecting distribution constant

Nernst Distribution Law, Statement. Distribution constant,
factors affecting distribution constant, validity of distribution law,
modification of distribution law when molecules undergo a)
association b) dissociation. Application of the distribution law

1. Distribution Law
Dr. Mithil Fal Desai
Shree Mallikarjun and Shri Chetan
Manju Desai College Canacona Goa

2. Nernst Distribution Law -statement.
When a solute is distributed between
two immiscible solvents in contact, at
equilibrium, the ratio of its
concentrations in the two solvents is
constant at a given temperature,
provided the solute exists in the same
molecular state in both solvents.
??????
�=
??????
1
??????
2
??????
??????= Distribution constant/Partition coefficient/
Distribution coefficient
C
1 and C
2= Concentration of substance in two
immiscible solvents
Association
No Change
Solvent A
Solvent B
(1891)

3. Factors affecting the distribution
constant
1.Temperature – Higher temperatures may change solubility
2.Nature of solvents – Polarity, hydrogen bonding, dielectric
constant.
3.Nature of solute – Hydrophobic, hydrophilic, tendency to
ionize.
4.Association/Dissociation of solute – If solute aggregates/
ionizes/dissociates in one solvent
5.Presence of common ions/impurities – Alters equilibrium
distribution.

4. Validity of Distribution Law
1.Two solvents are immiscible.
2.Solute exists in the same molecular form in both solvents.
3.Equilibrium is attained.
4.Temperature remains constant.

5. Modification of the distribution law
when molecules undergo
a) association b) dissociation.

6. Association
No Change
HOW IS DISTRIBUTION LAW MODIFIED WHEN
SOLUTE UNDERGOES ASSOCIATION
Solvent A
Solvent B

7. X
nX X
n
HOW IS DISTRIBUTION LAW MODIFIED WHEN
SOLUTE UNDERGOES ASSOCIATION
Let C
1 be concentration of X in solvent A.
C
3 be concentration of X in solvent B
C
2 be concentration of X
n in solvent B.
Solvent A
Solvent B
Applying distribution law to the equilibrium
K
D=C
1 /C
3 ----1
Kc= X
n
/X
n = C
3
n
/ C
2 ----2
Applying Mass Law to the chemical equilibrium
Taking n
th
root on both sides in equation
??????
??????�= C
3 /
??????
C
2 ----3
Dividing equation (1) by (3)
KD/
??????
??????� =(C
1 / C
3 )/ C
3 /
??????
C
2 ----4
KD/
??????
??????� =(C
1 /
??????
C
2) ------5
K=(C
1 /
??????
C
2)
Association
No Change

8. X
X A +B
HOW IS DISTRIBUTION LAW MODIFIED When
Solute undergoes Dissociation
Let C
1 be concentration of X in solvent A.
C
2 be total concentration of X in solvent B.
Solvent A
Solvent B
If the degree of dissociation in solvent B is x,
K
D=C
1 /C
2(1-x) ----1
Applying distribution law to normal molecules
in the two solvents
K=
C
1
C
2(1-x)
Association
No Change
X A + B
x x1-x
Hence, the concentration of the undissociated
molecules in solvent B is C
2(1–x).
A case of this type arises in the distribution of a weak acid (e.g., succinic acid or oxalic
acid) between ether and water.