This document discusses the applications of the Nernst distribution law, including determining the degree of association and dissociation of solutes. It provides examples of using the law to calculate the degree of association of benzoic acid in benzene (n=2) and the degree of dissociation of succinic acid in water and ether (33.34%). Solvent extraction applications are also summarized, including simple vs multiple extraction processes.

1. Dr. Y. S. THAKARE
M.Sc. (CHE) Ph D, NET, SET
Assistant Professor in Chemistry,
Shri Shivaji Science College, Amravati
Email: yogitathakare_2007@rediffmail.com
B Sc- II Year
SEM-III
PAPER-III
PHYSICAL CHEMISTRY
UNIT- V -PHASE EQUILIBRIUM
Topic- Application of Nernst Distribution Law
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2. APPLICATIONS OF NERNST DISTRIBUTION
LAW :
• Determination of degree of association
• Determination of degree of dissociation
• Solvent Extraction
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3. Determination of degree of association :
Let in solvent B, ‘n’ number of molecules of ‘X’
associate to form ‘Xn’ molecule. In solvent A, the solute
molecules remain as normal molecules.
1
2
D
n
C
K
C

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4. Let C1 be c onc entration of X in solvent A.
C3 be c onc entr ation of X in solvent B.
C2 be c onc entration of Xn in solvent B.
In this system ther e ar e two equilibr ia.
Fir st equilibr ium is –
X in solvent A X in solvent B.
 K1
1
3
C
C

OR C3
1
1
C
K
 ----- (1)
Sec ond equilibr ium is
nX in solvent B Xn in solvent B.
 Fr om Law of Mass Action
Kc
 
 
 
 
n 2
n n
3
X C
X C
 
OR  n 2
3
C
C
C
K

C2
C1
C3

5. Taking nth
root we get,
n
2
3
n
C
C
C
K
 ----- (2)
From Eq. (1) and (2) we get
1
1
C
K

n
2
n
C
C
K

 OR
𝐶1
𝐶2
𝑛 =
K1
𝐾𝑐
𝑛
But
K1
𝐾𝑐
𝑛 = KD (Constant)
Thus when association occurs in one solvent the
distribution equation is modified as –
𝐶1
𝐶2
𝑛 = KD
As the solute exists largely as associated molecules,
the total concentration of ‘X’ determined experimentally in
solvent B is taken as the concentration of the associated
molecules Xn.

6. Problem : 8 When benzoic acid was shaken with mixtures of
benzene and water at constant temperature, the following results were
obtained.
Concentration of acid in benzene (C1)
Concentration of acid in water (C2)
0.24
0.015
0.55
0.022
0.93
0.029
Determine degree of association of benzoic acid in benzene
Solution : As we known,
1
n
2
C
C
 KD
Consider, C1 = 0.24 C’1 = 0.55
C2 = 0.015 C’2 = 0.022

'
2 2 1
n
'
' '
n n
1
1 2
C C C
C
C C
 
Taking log,
𝑙𝑜𝑔
𝐶2
𝐶2
′ =
1
𝑛
𝑙𝑜𝑔
𝐶1
𝐶1
′
𝑙𝑜𝑔
𝐶1
𝐶1
′ = 𝑛𝑙𝑜𝑔
𝐶2
𝐶2
′
0.24 0.015
log n log
0.55 0.022

 n = 2

7. Determination of degree of dissociation :
Degree of dissociation is the fraction of total
number of moles dissociated at equilibrium.
Let solute undergoes dissociation in solvent
B and it is normal in solvent A.
1
2 (1 )
D
C
K
C 


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8. Degree of dissociation is the fraction of total number of
moles dissociated at equilibrium.
Let solute undergoes dissociation in solvent B and it is
normal in solvent A. The equilibria set up in the two solvents
are shown in figure
Let
C1 = Concentration of ‘X’ in solvent A.
C2 = Concentration of X (dissociation and undissociated) in solvent B.
If degree of dissociation in solvent B is ‘ α’ then.
X Y + Z
Initially 1 0 0
At equilibrium (1 )
   
Hence, the concentration of the undissociated (normal)
molecules in solvent B is C2(1– ).
Applying distribution law to normal molecules in the
two solvents,
1
D
2
C
K
C (1 )

 
This is modified equation in dissociation case.
C1
C2

9. Problem : 9 Succinic acid is distributed in water and ether. In
aqueous solution it undergoes dissociation. The concentration of
acid in aqueous layer was 42.5 x 10-3
kg dm-3
and ether layer
was 7.1 x 10-3
kg dm-3
. If distribution co-efficient of acid is
0.25, calculate the degree of dissociation assuming the normal
molecules in ether.
Solution :
Given That
Concentration in aqueous layer (C2) = 42.5 x 10-3
kg dm-3
Concentration in ether layer (C1) = 7.1 x 10-3
kg dm-3
Distribution coefficient = (KD) = 0.25
We know,
1
D
2
C
K
C (1 )

 
3
3
7.1 10
0.25
42.5 10 (1 )




  
 1 -  = 0.6666
  = 0.3334
%  = 33.34 %

10. EXTRACTION
Simple Extraction Multiple Extraction
There are two types of extraction
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11. Solvent Extraction
Solvent extraction and is
a method to separate
compounds based on their
relative solubilities in two
immiscible liquids, usually
water and an organic
solvent.
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12. • Simple Extraction:
The extraction process when carried out with the
total amount of given organic solvent in a single
step is known as simple extraction.
• Multiple Extraction:
When the extraction is
carried out in more number of
steps by using small amounts
of organic solvent from total
solvent with same aqueous
solution is known as multiple extraction.

14. Applications:
1. In Pharmaceutical Chemistry Separation of many antibiotics,
vitamins and harmones
2. Food Industry
3. In Nuclear chemistry
4. Fiber Industry
5. Extraction of essential oil, flavoring etc.
6. Extraction of Caffeine from coffee and tea.
7. Removal of phenol from the Waste water.
8. Recovery and purification of acetic acid from aqueous
streams.
9. Extraction of antibiotics and organic acids from fermented
broth.
10.Seperation and refining of petrochemicals and petroleum

15. References:
• Instrumental methods of Chemical Analysis
Gurdeep R. Chatwal
Sham K. Anand
• Instrumental Methodology and Chemical Analysis
H. Kaur
• Insrumental methodology of Chemical Analysis
B. K. Sharma
• Textbook of Chemistry for 3rd Semister of Bsc.
• Basic concept of analytical chemistry By S. M. Khopkar