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16.8
Partition Equilibrium
of a Solute Between
Two Immiscible
Solvents
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.104)
Partition (Distribution) Equilibrium
• The equilibrium established when a
non-volatile solute distributes itself
between two immiscible liquids
A(solvent 2)
A(solvent 1)
Partition (Distribution) Equilibrium
Water and 1,1,1-trichloroethane are
immiscible with each other.
Partition (Distribution) Equilibrium
I2 dissolves in both layers to different
extent.
Partition (Distribution) Equilibrium
Strictly speaking, I2 is NOT non-volatile !
Partition (Distribution) Equilibrium
When dynamic equilibrium is established
 rate of  movement = rate of  movement
Suppose the equilibrium concentrations of iodine in H2O and
CH3CCl3 are x and y respectively,
x
Kc 
y
Suppose the equilibrium concentrations of iodine in H2O and
CH3CCl3 are x and y respectively,
x
Kc 
y
Changing the concentrations by the same extent does not
affect the quotient
x 2 x 3x 4 x
Kc  



y 2 y 3y 4 y
x
2
y
2
When dynamic equilibrium is established
 the concentrations of iodine in water and 1,1,1trichloroethane reach a constant ratio
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.104)
Partition Coefficient
The partition law states that:
• At a given temperature, the ratio of the
concentrations of a solute in two
immiscible solvents (solvent 1 and solvent
2) is constant when equilibrium has been
reached
• This constant is known as the partition
coefficient (or distribution coefficient)
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.104)
Partition Coefficient
The partition law can be represented
by the following equation:
KD  Concentrat ion of solute in solvent 1
Concentrat ion of solute in solvent 2
[Solute] solvent 1

(no unit)
[Solute] solvent 2
Units of concentration : mol dm-3, mol cm-3, g dm-3, g cm-3
Partition Coefficient
The partition coefficient of a solute between
solvent 2 and solvent 1 is given by
[Solute]solvent 2
K 
[Solute]solvent 1
The partition coefficient of a solute between
solvent 1 and solvent 2 is given by
K
[Solute]solvent 1

[Solute]solvent 2
Partition Coefficient
• Not affected by the amount of solute
added and the volumes of solvents
used.
• TAS Experiment No. 12
Partition law holds true
1. at constant temperature
2. for dilute solutions
For concentrated solutions,
interactions between solvent and
solute have to be considered and the
concentration terms should be
expressed by ‘activity’(not required)
Partition law holds true
3. when the solute exists in the same
form in both solvents.
C6H5COOH(benzene)
C2
C6H5COOH(aq)
C1
C1 and C2 are determined by titrating the acid in
each solvent with standard sodium hydroxide
solution.
[C6H5COOH]water [C6H5COOH]benzen
-3
(C
)
/
mol
dm
(C1) / mol dm-3
e 2
C1/C2
0.06
0.483
0.124
0.12
1.92
0.063
0.14
2.63
0.053
0.20
5.29
0.038
Not a constant
Interpretation : • The solute does not have the same
molecular form in both solvents
• Benzoic acid tends to dimerize (associate)
in non-polar solvent to give (C6H5COOH)2
H
O
O
C
C
O
H
O
Benzoic acid dimer
 Partition law does not apply
Interpretation : 2C6H5COOH(benzene)
C2 (1   )
(C6H5COOH)2(benzene)
1
2 C2
 = degree of association of benzoic acid
[C6H5COOH]total = [C6H5COOH]free + [C6H5COOH]associated
C2
C2(1-)
C2
Determined by titration with NaOH
Q.17(a)
The interaction between benzoic acid and benzene
molecules are weaker than the hydrogen bonds formed
between benzoic acid molecules.
Thus benzoic acids tend to form dimers when dissolved in
benzene.
In aqueous solution, benzoic acid molecules form strong Hbond with H2O molecules rather than forming dimer.
Q.17(b)
In aqueous solution, there is no association as explained
in (a).
Also, dissociation of acid can be ignored since benzoic
acid is a weak acid (Ka = 6.3  10-5 mol dm-3).
Q.17(c)
2C6H5COOH(benzene)
C2 (1   )
(C6H5COOH)2(benzene)
C2
K
[C2 (1   )]2
1
2
C6H5COOH(benzene)
C2 (1   )
C1
Partition
 KD 
coefficient
C2 (1   )
1
2
C2
C6H5COOH(aq)
C1
C2
K
2
[C2 (1   )]
1
2
 is a constant at fixed T
C2
'
 C2 (1   ) 
 K C2
2K
C1
C1
 KD 
 '
C2 (1   ) K C2
C1
'
''

 KD K  K
C2
Applications of partition law
•
Solvent extraction
•
Chromatography
Two classes of separation techniques
based on partition law.
Solvent extraction
+ hexane
I2 in KI(aq)
IColourless
22 in hexane
Hexane
I22 in KI(aq)
To
It
Itdissolves
iscan
remove
immiscible
be recycled
II22from
but
with
not
an
easily
water.
KI.
aqueous
(e.g. by
solution
distillation)
of KI, a
What
feature
should
the
solvent
have?
At equilibrium,
suitable solvent is added.

Organic
Organic
solvents
(volatile)
solvents
preferred.
preferred.
rate of
movement
of I2 =are
rate
of  are
movement
of I2
By partition law,
[ I 2 ]hexane
K
[ I 2 ]KI ( aq )
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.104)
Solvent Extraction
Hexane
layer
Aqueous
layer
Before
shaking
After
shaking
Iodine can be extracted from water by
adding hexane, shaking and separating the
two layers in a separating funnel
Determination of I2 left in both layer
Titrated with standard sodium thiosulphate
solution
I2 + 2S2O3 
2I + S4O62
Determination of I2 left in the KI solution
For the aqueous layer,
starch is used as the indicator.
For the hexane layer,
starch is not needed because the colour of I2 in
hexane is intense enough to give a sharp end
point.
In solvent extraction, it is more efficient (but
more time-consuming) to use the solvent in
portions for repeated extractions than to use it
all in one extraction.
Worked example
Worked example : 50g X in 40 cm3 ether solution
10g X in 25 cm3 aqueous solution
[X ]ether
K  coefficient of X between ether
(a) By partition
Calculatelaw,
the partition
[X ]water
and water at 298 K.
50
50
[X]ether  M 
0.04 0.04M
[X]water
10
10
 M 
0.025 0.025M
50
M is the molecular mass of X
K  0.04M  3.125
10
0.025M
Worked example : 50g X in 40 cm3 ether solution
10g X in 25 cm3 aqueous solution
Or simply,
K
50
40
10
25
 3.125
(b)(i)
30 cm3 ether
5g of X in 30 cm3
aqueous solution
xg of X in 30 cm3
ether solution
(5-x)g of X in 30 cm3
aqueous solution
Determine the mass of X that could be extracted by shaking a
30 cm3 aqueous solution containing 5 g of X with a single 30
cm3 portion of ether at 298 K
(b)(i)
30 cm3 ether
5g of X in 30 cm3
aqueous solution
xg of X in 30 cm3
ether solution
(5-x)g of X in 30 cm3
aqueous solution
K  3.125

x
30
5 x
30
x

 x  3.79
5 x
3.79 g of X could be extracted.
(b)(ii) First extraction
15 cm3 ether
5g of X in 30 cm3
aqueous solution
x1g of X in 15 cm3
ether solution
(5-x1)g of X in 30
cm3 aqueous solution
K  3.125

x1
15
5 x1
30
2 x1

 x1  3.05
5  x1
(b)(ii) Second extraction
15 cm3 ether
(5-x1)g of X in 30 cm3
aqueous solution
x2g of X in 15 cm3
ether solution
(5-x1-x2)g of X in 30
cm3 aqueous solution
K  3.125

x2
15
5 x1  x2
30
2 x2

 x2  1.19
5  3.05  x2
total mass of X extracted
= (3.05 + 1.19) g = 4.24 g > 3.79 g.
Repeated extractions using smaller portions of
solvent are more efficient than a single
extraction using larger portion of solvent.
However, the former is more time-consuming
Important extraction processes : •
Products from organic synthesis, if contaminated with
water, can be purified by shaking with a suitable organic
solvent.
•
Caffeine in coffee beans can be extracted by Supercritical
carbon dioxide fluid (decaffeinated coffee)
•
Impurities such as sodium chloride and sodium chlorate
present in sodium hydroxide solution can be removed by
extracting the solution with liquid ammonia.
Purified sodium hydroxide is the raw material for making
soap, artificial fibre, etc.
Q.18(a)
200 cm3 alcohol
Alcohol layer
Aqueous layer
100 cm3 of 0.5 M
ethanoic acid
Calculate the % of ethanoic acid extracted at 298 K by shaking
100 cm3 of a 0.50 M aqueous solution of ethanoic acid with 200
cm3 of 2-methylpropan-1-ol;
Q.18(a)
Let x be the fraction of ethanoic acid extracted to the alcohol layer
No. of moles of acid in the original solution
= 0.5  0.100 = 0.05
[acid ]alcohol 
0.05x
0.200
[acid ]water 
0.05(1  x )
0.100
0.05(1  x )
K  3.05  0.100  x  0.396  39.6%
0.05x
0.200
Q.18(b)
Let x1, x2 be the fractions of ethanoic acid extracted to the alcohol
layer in the 1st and 2nd extractions respectively.
1st extraction
2nd extraction
0.05(1  x1 )
0.100  3.05  x  0.247
1
0.05x1
0.100
0.05(1  x1  x 2 )
0.100
 3.05  x 2  0.186
0.05x 2
0.100
% of acid extracted = 0.247 + 0.186 = 0.433 = 43.3%
Q.19
Let x cm3 be the volume of solvent X required to extract 90% of
iodine from the aqueous solution and y be the no. of moles of
iodine in the original aqueous solution.
[I 2 ]solventX
K  120 
[I 2 ]water
0.9 y
 x  x  7.5
0.1y
100
 7.5 cm3 of solvent X is required
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.104)
Check Point 16-8A
Chromatography
A family of analytical techniques for separating
the components of a mixture.
Derived from the Greek root chroma, meaning
“colour”, because the original chromatographic
separations involved coloured substances.
Chromatography
In chromatography, repeated extractions are
carried out successively in one operation
(compared with fractional distillation in which
repeated distillations are performed) which
results, as shown in the worked example and
Q.18, in an effective separation of components.
All chromatographic separations are based
upon differences in partition coefficients of
the components between a stationary phase
and a mobile phase.
The stationary phase is a solvent (often H2O)
adsorbed (bonded to the surface) on a solid.
This may be paper or a solid such as
alumina or silica gel, which has been
packed into a column or spread on a glass
plate.
The mobile phase is a second solvent which
seeps through the stationary phase.
There are three main types of chromatography
1.
Column chromatography
2.
Paper chromatography
3.
Thin layer chromatography
Column chromatography
Stationary phase : Water adsorbed on the
adsorbent (alumina or silica gel)
Mobile phase : A suitable solvent (eluant) that
seeps through the column
Column chromatography
Partition of components takes
place repeatedly between the two
phases as the components are
carried down the column by the
eluant.
The components are separated
into different bands according to
their partition coefficients.
Column chromatography
The component with the highest
coefficient between mobile phase
and stationary phase is carried
down the column by the mobile
phase most quickly and comes
out first.
Column chromatography
Suitable for large scale
treatment of sample
For treatment of small
quantities of samples, paper
or thin layer chromatography
is preferred.
Paper chromatography
• Stationary phase : -
Water adsorbed on paper.
• Mobile phase : A suitable solvent
The best solvent for a particular
separation should be worked out
by trials-and-errors
X(adsorbed water)
X(solvent)
stationary phase
mobile phase
Paper chromatography
The solvent moves up the filter
paper by capillary action
Components are carried
upward by the mobile solvent
 Ascending chromatography
• Different dyes have
different partition
coefficients between
the mobile and
stationary phases
• They will move
upwards to different
extent
Paper chromatography
The components separated can
be identified by their specific
retardation factors, Rf , which
are calculated by
Rf  distance travelled by spot
distance travelled by solvent
filter
paper
separated
colours
spot of
coloured
dye
solvent
Using chromatography to separate the colours in a sweet.
Solvent front
a
d
c
b
b
R f (blue ) 
a
c
R f (red ) 
a
d
R f ( green ) 
a
a chromatogram
separated
colours
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.109)
Paper Chromatography
• The Rf value of any particular substance
is about the same when using a
particular solvent at a given temperature
• The Rf value of a substance differs in
different solvents and at different
temperatures
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.109)
Paper Chromatography
Amino acid
Cystine
Glycine
Leucine
Solvent
Mixture of
Mixture of
phenol and
butanol and
ammonia
ethanoic acid
0.14
0.05
0.42
0.87
0.18
0.62
Rf values of some amino acids in two
different solvents at a given temperature
Check Point 16-8B
Two-dimensional paper chromatography
Two-dimensional paper chromatography
All spots (except proline)
appears visible (purple)
when sprayed with
ninhydrin (a developing
agent)
Thin layer chromatography
Stationary phase : -
Water adsorbed on a thin layer
of solid adsorbent (silica gel or
alumina).
Mobile phase : A suitable solvent
X(adsorbed water)
X(solvent)
stationary phase
mobile phase
Q.20
Suggest any advantage of thin layer chromatography
over paper chromatography.
A variety of different adsorbents can be used.
The thin layer is more compact than paper,
more equilibrations can be achieved in a few
centimetres (no. of extraction ).
A microscope slide can be used as the glass plate
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.108)
Let m be the mass of A extracted using 100 cm3 of 1,1,1trichloroethane, then the mass of A left in 60 cm3 of aqueous
layer is (6 – m).
m
K D  100
6m
60
Back
m
15  100
6m
60

m = 5.77 g
5.77 g of A is extracted using 100 cm3 of 1,1,1trichloroethane.
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.110)
(a) A student wrote the following explanation for the
different Rf values found in the separation of two amino
acids, leucine (Rf value = 0.5) and glycine (Rf value = 0.3),
by paper chromatography using a solvent containing
20% of water.
“Leucine is a much lighter molecule than glycine.”
Do you agree with this explanation? Explain your
answer.
Answer
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.110)
(a) The difference in Rf value of leucine and glycine is due to
the fact that they have different partition between the
stationary phase and the mobile phase. Therefore, they
move upwards to different extent. The Rf value is not
related to the mass of the solute.
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.110)
(b) Draw a diagram to show the expected chromatogram of a
mixture of A, B, C and D using a solvent X, given that
the Rf values of A, B, C and D are 0.15, 0.40, 0.70 and 0.75
respectively.
Answer
16.8 Partition Equilibrium of a Solute Between Two Immiscible
Solvents (SB p.110)
Back
(b)