Chapter 5
Phase
Equilibrium
• Phase- part of a system which is chemically and physically
uniform throughout
oil
• Symbol  P = number of phase
Phase boundary
• Three phase of state:
▫ Solid
▫ liquid
▫ gaseous
water
Figure: 2 phase system with phase
boundaries -mixture of oil and
water not homogent and have a
phase boundary
• Example;
▫ solution of sugar in water  one phase: liquid
▫ Mixture of gases
one phase: gaseous
▫ Mixture of ice and water
 two phases: solid & liquid)
▫ Oil and water
 two phases: liquid & liquid)
▫ Saturated solution with an excess of insoluble solute
 two phases: solid & liquid
• Plot showing effect on pressure and temperature on
physical state substances@ graphical representation of the
conditions of temperature and pressure at which the solid,
liquid and vapour (gas) exist, either as a single/two or more
phases in equilibrium
• Uses - to predict physical state at specific temperature and
pressure
• Each part in phase diagram represent different things:
area/ region = phase
line = set of temperature and pressure which 2 phase on
either side of line exist in equilibrium
Triple point, O = condition of temperature and pressure
at which all 3 phases of a substances coexist at
equilibrium
Vapour pressure/atm
Liquid
Solid
E
Vapour
Temperature/oC
-
example
• The phase diagram for a substances A is shown. State the
physical state of A at the condition X, Y and Z
x
P
liquid
solid
z
O
y
vapour
T
Solution:
At condition x  compound A exist as solid and liquid in
equilibrium
At condition y  liquid A and gas/ vapour A exist in equilibrium
At condition z  substances A exist as a solid only
The general form of phase diagram and
various changes of state from one phase to
another
Pressure
(atm)
water
C
B
ice
vapour
O
A
Temperature
(ºC)
One Component System
The phase diagram of water
• Water exist as solid (ice), gas (water vapour) and also
liquid (water)
Pressure
(atm)
C
B
200
water
ice
1.0
O
0.006
vapour
D
A
0 0.01
100
374
Temperature
(ºC)
The phase diagram of
water
Pressure
(atm)
C
200
B
water
ice
1.0
O
0.006
vapour
D
A
0 0.01 100 374
OB = vapour pressure curve (show
the vapour pressures of water
changes with T. Also summarised
the conditions of T and P under
which liquid water and water
vapour are in equilibrium)
B
= Critical point which critical
temperature = 374 ºC &
critical pressure = 200 atm.
(The upper limit of OB, beyond
which liquid and vapour phases
cannot be distinguished. There are
no longer two separate phase,
Temperature because densities of the gas and
(ºC)
liquid are equal.
At temperature higher than the
critical temperature, water vapour
cannot be changed into liquid
water even though a high pressure
is used.)
The phase diagram of
water
AO = vapour pressure
curve for ice /
sublimation curve
(show the vapour
pressures of ice varies with
T.
Also summarised the
condition under which
solid ice and water vapour
can be equilibrium)
Pressure
(atm)
C
200
B
water
ice
1.0
O
0.006
vapour
D
A
0 0.01 100 374
Temperature
(ºC)
O = Triple point which
ice, water, water vapour
exist in equilibrium at
0.01ºC & 6.03 x 10-3atm)
The phase diagram of
water
OC
=
melting
temperature curve
(show the effect of
pressure on the melting
point of ice or the
freezing point of water.
Line slope to left : P
high, m.p decreases)
OD
=
Supercooling (vapour
pressure of water below
its freezing point.
Metastable condition
because water is not
liquid below freezing
point.
Pressure
(atm)
C
200
B
water
ice
1.0
O
0.006
vapour
D
A
0 0.01 100 374
Temperature
(ºC)
The phase diagram of Carbon dioxide
• Basically phase diagram for CO2 is the same as phase
diagram for H2O except for two things.
Pressure
(atm)
C
B
217
liquid
solid
5.1
O
vapour
D
1.0
A
-78
-57
374
Temperature
(ºC)
The phase diagram of
carbon dioxide
Pressure
(atm)
C
B
217
liquid
solid
5
.
1.0
1
O
vapour
D
A
-78 -57
374
OB = Liquid & vapour in equilibrium ,
pressure high, temperature high
B
= Critical point which critical
temperature = 374 ºC & critical
pressure = 217 atm.
 temperature high, kinetic
energy, vapour high, pressure
high will change vapour to liquid
 374 ºC is the highest
temperature at which CO2 gas
can be liquefied (critical
temperature)
Temperature 217 atm is highest pressure at
(ºC)
which CO2 gas can be liquefied
(critical pressure)
O
= Triple point ( T= -57ºC, P= 5.1
atm) at which CO2 can exists as
solid , liquid and gas
Comparing the phase diagram for water
and carbon dioxide
Z
Vapour pressure/atm
Y
200
Solid
H2O(s)
0.006
Liquid
H2O(l)
Z
Y
218
Liquid
Solid
E
X
5.0
Vapour
H2O(g)
E
1.0
Vapour
X
B
0.01
374
Temp(oC)
-78
-57
374
Temp/oC
Differences between phase diagram of
CO2 and water
1.
Triple point O is above the atmospheric
pressure (-57ºC, 5.1 atm). So, solid CO2
sublime to gaseous CO2 below 5.1 atm, while
liquid CO2 only exist above than 5.1 atm, with
temp greater than -57ºC
2. OC line slopes to the right for CO2 and for most
substances : pressure high, melting point high
• Phase diagram of water  slope (-ve)
▫ Substance is more dense as liquid than its solid (ice is less
dense than water), liquid is more compact than than the solid
▫ Increasing pressure which will favor the compactness of the
molecules, will thus favor the liquid state
▫ Increasing pressure will thus lower the temperature at which
the solid will melt
▫ At high pressure, melting point decreases
• Phase diagram of most substances, eg CO2  slope (+ve)
▫ Substance is more dense is more dense as a solid than it is as a
liquid
▫ Increasing pressure which will favor the compactness of the
molecules, will thus favor the more dense solid phase
▫ Higher temperature are required to melt the solid phase at
higher pressure
Uses of Dry ice (solid CO2) in industry
Used in food industry as a refrigerant because it is
very cold and does not leave a messy liquid when
sublimes
Dry ice is used for cloud seeding. It is spread into
the cloud to accelerate rain fall
Also used in film or stage performances to
produce special effect involving a cloud or a mist
• The phase rule can enable a chemist to make some prediction
about the possible outcome of changing conditions
• The Phase Rules describes the possible no. degree of freedom or
no of variables (eg: T, P, conc) which can be varied independently
without changing the no. of phases in a (closed) system at
equilibrim.
• It is expressed as:
P+F=C+2
where, P = the number of phases
C = the number of component
F = the number of degrees of freedom/ the minimum
number of variables (e.g pressure, temperature
required to characteristic the phase)
• When C is increase, F will increase
• When P is increase, F will decrease
18
• Constituent- a chemical species (ion or molecule which is
present
• Component (C) - chemically independent constituents of a
system
▫ C = #of independent chemical constituents - # of distinct
chemical reactions
 #of independent chemical constituents = total # of
constituents minus the number of any restrictive
conditions (charge neutrality, material balance etc.)
 Example 1:
CaCO3(s)  CaO(s) + CO2(g)
▫ Phases: P = 2 solid + 1 gas = 3
▫ Component: C = 3 consitiuents - 1 reaction = 2
example
• Calculate number degrees of freedom, F for every part
below:
Pressure
(atm)
C
liquid
B
O
solid
A
Solution:
Area COB : P + F = C + 2
1+F=1+2
F=3–1=2
Line OC : P + F = C + 2
2+F=1+2
F=3–2=1
vapour
Temperature
(ºC)
Point O : P + F = C + 2
3+F=1+2
F=3–3=0
20
One Component Systems
• Phase rule says that you can have at most 3
phases
▫ F = C – P + 2;
C=1
F=3–P
• If P=3, F=0 system is invariant
▫ Specified by temperature and pressure and
occurs at 1 point (called the triple point) 
there is only 1 temperature & 1 pressure where
all 3 phases of water in eqbm
• If one phase is present, F = 2 that is P and T can
be varied independently
▫ This defines an area in a P,T diagram which
only one phase is present
• If two phases are present, F = 1 so only P or T can
be varied independently.
▫ This defines a line in a P, T diagram
▫ Varying the temperature necessitates the
specific changes in the temperature
One component phase
diagrams
Two components system- mixture of two
miscible liquids
• Vapour pressures of solutions
(vapour pressure  pressure of a vapour in
thermodynamic eqbm with its condensed phase in a
closed system)
1. When a pure liquid is placed in a covered container,
the vapour pressure of the pure liquid will increase
gradually.
2. When equilibrium is achieved, the vapour pressure is at
maximum and is called saturated vapour pressure of
the liquid.
3. When the built up vapour pressure of the liquid equals
to the external pressure (ie 1atm or 760 mmHg), the
liquid will boil
Liquid mixtures.
• A miscible solution mixture is a homogeneous
mixture of two different liquids, A and B.
• Examples : mixtures of methyl benzene with
benzene, ethanol and water.
• An immiscible solution mixtures is a solution
when mixed the components remain segregated
in a heterogeneous mixture.
• Example : benzene and water
• carbon tetrachloride and water
Vapour pressure of a solution
• The total vapour pressure of a solution is the sum of the
vapour pressure of each component that make up the
solution
Psolution = PA + PB
where
Psolution = the total vapour pressure
PA = partial pressure of component A
PB =partial pressure of component A
Roult’s law
PA = XAPA *
Ideal solution and non ideal solution
• A solution that obeys Roult’s law is called ideal solution
• An ideal solution is a mixture of two solutions A and B
where
i) the intermolecular forces of attraction between
the A and B molecules are of the same type and of
equal strength. [ forces between A…A is the same
as A…..B or B….B]
ii) the total volume of solutions A and B when mixed
remains unchange [ 50 mL of A when added to 50
mL of B gives 100 mL solution]
Examples: benzene + methylbenzene
2-methylpropanol + propanol
hexane + heptane
VAPOR PRESSURE-COMPOSITION
VAPOR PRESSURE-MOLE FRACTION
Vapour pressure
Po B
PT
Po A
XB
XA
XA
XB
• Component with high vapour pressure means that its
boiling point is low.
• SO REMEMBER!
HIGH VAPOUR PRESSURE = LOW BOILING PT
LOW VAPOUR PRESSURE = HIGH BOILING PT
vapour pressure -composition curves &
boiling point -composition curve
B.p (ºC)
Boiling point.
(at constant P)
Vapour pressure.
(at constant T)
P
XB
XA
XB
XA
The vapour pressure versus mole fraction curve/diagram
is opposite to the boiling point vs mole fraction
• VAPOUR PRESSURE VS COMPOSITION
BOILING POINT VS COMPOSITION
T1
Vapour
Liquid
T2
Liquid
Vapour
Pure A
X3
X2
T2
X3 Pure B
Pure A
X3
X2
X3
Pure B
Deviation from Raoult’s Law
• Raoult's Law only works for ideal mixtures. In
these, the forces between the particles in the
mixture are exactly the same as those in the pure
liquids. The tendency for the particles to escape
is the same in the mixture and in the pure
liquids.
• That's not true in non-ideal mixtures.
31
Deviation from Ideality
• Extreme deviations from ideality
▫ Result in maxima /minima in vapour-pressure
diagrams and temperature- composition
diagrams
AZEOTROPES
a – without
zeo- boil
Trops - change
Means composition does
not change during boiling
Azeotropic Mixture: Mixture of two miscible liquids that has a constant
boiling point and the vapour that exist in equilibrium with the liquid has
the same composition as the liquid
On heating an azeotropic mixture until it boils, the vapor liberated has the
same composition as the original liquid mixture
POSITIVE DEVIATION
• In mixtures showing a positive deviation from
Raoult's Law, the vapour pressure of the mixture
is always higher than that of an ideal mixture,
Explaining positive deviations
• The fact that the vapour pressure is higher than
ideal in these mixtures means that molecules are
breaking away more easily than they do in the pure
liquids.
• That is because the intermolecular forces between
molecules of A and B are less than they are in the
pure liquids.
• A-A ~ B-B > A-B in which bond A-B is weaker. So,
vapour pressure is greater than predicted by
Raoult’s Law -tendency molecules to escape is
greater
POSITIVE DEVIATION
Boiling point
Vapour
Pressure
Liquid
Azeotrope
PoA =1
TB
=1
Liquid
Vapour
Vapour
XB=1
Composition
TA =1
Liquid
Vapour
PoB
=1
Vapour
Liquid
Azeotrop
e
XA =1
XB=1
Composition
XA =1
Negative deviation
Explaining negative deviation
• These are cases where the molecules break away from
the mixture less easily than they do from the pure
liquids. New stronger forces must exist in the mixture
than in the original liquids
• A-A ~ B-B < A-B in which bond A-B is stronger. So,
vapour pressure is lower than predicted by Raoult’s Law
-tendency molecules to escape into the vapour phase
decreased
NEGATIVE DEVIATION
Vapour Pressure
Boiling point
Po
A =1
Azeotrope
Liquid
Liquid
Vapour
Vapour
Vapour
TB =1
Azeotrope
XB=1
Composition
Liquid
Vapour
Liquid
TA =1
XA =1
XB=1
Composition
XA =1
Distillation of an ideal solution
T1
Vapour
Liquid
Pure A
X3
X2
T2
X1 Pure B
• When the composition of the mixture X1 is heated, it boils at a
temperature T1 to provide a vapour that has a composition X2.
• If this vapour is condensed and then reheated, it will boil at a
temperature T2 and give a vapour whose composition is X3
• Repetition of this process will produce fraction richer in A. This
procedure is called fractional distillation and is useful not only in
laboratory – to purify products of chemical reactions – but also in
industrial sectors such as in the petroleum industry.
• There some solution cannot be totally separated into their
components by fractional distillation (e.g ethanol-water mixture).
Distillation of non ideal mixture
ETHANOL –WATER MIXTURE
• Suppose a mixture of
ethanol and water with
composition C1 is heated.
It will boil at a
temperature given by the
liquid curve and produce a
vapour with composition
C2.
• When that vapour
condenses it will, of
course, still have the
Azeotropic Mixture: Mixture of two miscible
liquids that has a constant boiling point and composition C2. If you
the vapour that exist in equilibrium with the reboil that, it will produce
liquid has the same composition as the liquid
On heating an azeotropic mixture until it boils,
a new vapour with
the vapor liberated has the same
composition C3.
composition as the original liquid mixture
Distillation of solution
Non Ideal solution
-positive deviation
B.p (ºC)
A
B
Boiling point.
(at constant P)
80 ºC
T1
78.5 ºC
T2
X1 X2
100%
benzene
X4
azeotropic
X3
100%
ethanol
• If solution in section A
(composition X1) is
subjected to fractional
distillation, solution will boil
at T1 and the vapour
liberated will have
composition X2.
• This vapour contain a higher
percentage of ethanol than
in original mixture.
• The vapour liberated
undergoes condensation and
vapourisation and finally an
azeotropic mixture is
obtained and distils over.
B.p (ºC)
A
• If solution in section B
(composition X3) is
subjected to fractional
distillation, the solution
boils at T2 and the vapour
that escapes has the
composition X4, that is
richer in benzene.
B
Boiling point.
(at constant P)
80 ºC
T1
78.5 ºC
T2
X1 X2
100%
benzene
X4
azeotropic
X3
100%
ethanol
• The more volatile
component, that is, the
azeotropic mixture with the
lower boiling point, is
distilled first. As
distillation proceeds, the
concentration of ethanol in
the residual liquid
increases. Finally the
temperature of the residual
liquid reaches 78.5ºC and
pure ethanol distills over
Distillation of solution
• If solution in section B (composition
Non Ideal solution
-negative deviation
B.p (ºC)
A
•
B
121
T1•
Boiling point.
(at constant P)
T2
100
78
•
X2
100%
H20
X1 X3
Azeotropic
68.2% HNO3
100%
HNO3
X1) is subjected to fractional
distillation, solution will boil at T1 and
the vapour liberated will have
composition X3.
This vapour contain a higher
percentage of HNO3 than in original
mixture and decreased in the
residual liquid.
The vapour liberated undergoes
condensation and vapourisation and
finally a pure HNO3 is obtained as
first
distillate.
As
distillation
proceeds, the residual liquid in the
distillation flask become richer in
water and its boiling point increases.
This continues until the solution
contain 68.2% HNO3 when the
mixture distils unchanged at 121ºC.
The mixture that contains 68.2%
HNO3 is known as azeotropic
mixture.
Simple distillation
• Would still produce mixture (not effective to separate ideal
solution to pure component)
Fractional distillation
• Separation is better because it can separate A completely from B (for
ideal/approximately ideal solution)
• Processes is in one continuous operation
• Fractionating column is a long tube filled with glass beads or short
pieces of glass tubing so that can provides a large surface area which
condensed liquid and ascending vapour can have maximum contact.
AZEOTROPIC MIXTURE
•
•
•
Azeotropic Mixture: Mixture of two miscible liquids that has a
constant boiling point and the vapour that exist in equilibrium
with the liquid has the same composition as the liquid
On heating an azeotropic mixture until it boils, the vapor liberated
has the same composition as the original liquid mixture
An azeotropic mixture cannot be separated into its component
through fractional distillation. The separation of the two
components in an azeotropic mixture can only be carried out using
the following methods:
a) Using a third component -an azeotropic mixture of ethanol and
water can be separated if benzene is added to the mixture
(benzene/water-pure ethanol)
b) Using a chemical method –usage of CaO to absorb water
c) Using an adsorbent like charcoal or silica gel
d) Using solvent extraction
COOLING CURVE FOR PURE COMPONENT (Pure Pb)
Constructing the phase diagram
PHASE DIAGRAM
Liquid Pb + liquid Sn
Solid Pb +
liquid mixture
Solid Sn +
liquid mixture
Solid Pb + Solid Sn
EUTECTIC SYSTEM
• If a liquid of composition Y (Cy) is cooled, the pure B component
begin to settle out when temperature Ty is reached. The liquid Y
becomes richer in A as the temperature falls and pure B continually
separates out, until the eutectic temperature is reached, when the
remaining liquid crystallizes out without further change in
temperature
• A similar situation occurs with a mixture of composition Z (CZ), but
now, it is pure A that settles out until the eutectic is reached
• If a liquid mixture of the eutectic composition is cooled, it freezes at
the constant eutectic temperatures giving a solid with the same
composition as the liquid (resembling a pure substances)
EUTECTIC SYSTEM
• Solid present as separate phases (e.g ice and salt)
Temperature
X
y
Melting point pure B
z
Ty
liquid
Melting point pure A
Solid A +
liquid
Solid B +
liquid
Eutectic point
Solid A +
eutectic
Pure A
Solid B +
eutectic
CZ
XA
Pure B
Cy
Eutectic
composition
XB
Mole fraction
EUTECTIC SYSTEM
The composition of the liquid mixture changed during the cooling process
due to the increased formation of solid phase
Solid B formed
80 ºC
80 % B
20 % A
25 ºC
40 % B
60 % A
Liquid composition
65 ºC
70 % B
30 % A
50 ºC
60 % B
40 % A
Liquid composition after
B Solid is formed
EUTECTIC SYSTEM
•
The eutectic is not a compound because :
a) The eutectic can be seen to be heterogenous under the
microscope, whereas a compound would be
heterogenous
b) The composition of a eutectic rarely corresponds to that
of a compound
c) The properties of the eutectic, e.g. its heat of solution,
are actually the sum of the properties of its constituents.
This is very likely for a true compound
Reference: Advance Physical Chemistry (1976), A.
Holderness, Heinemann Educational Publisher, Oxford
DISTRIBUTION OF A SOLUTE BETWEEN TWO
SOLVENTS
THE DISTRIBUTION LAW
• When a solute is added to two immiscible liquids, it may dissolve in
both liquid. In this case the solute will distribute itself between the
two solvents
• The ratio in which it is distributed is govern by the distribution law.
This states that at constant T, a solutes distribute itself between two
immiscible liquids in a constant ratio of concentration irrespective
of the amount of solute added
solute
Two immiscible
liquids
Shaken-up
The molecules of solutes being
distributed into the two layers
DISTRIBUTION OF A SOLUTE BETWEEN TWO
SOLVENTS
• Consider the distribution of iodine between water and
tetrachloromethane (two immiscible liquids). If we shake up iodine
with these two solvents, some will dissolve in the water and some in
the CCl4. Eventually a dynamic equilibrium is established
I2 (water)
I2(CCl4)
At equilibrium, the rate of I2 passes from CCl4 to water= rate of I2 passes from water to CCl4
• No matter how much iodine we use, the ratio of the final
concentrations is constant. The constant is called the partition
(distribution) coefficient, k
K = [I2 (CCl4) ] or K= [I2 (aq) ]
[I2 (aq) ]
[I2 (CCl4) ]
• The whole equation is called ‘Distribution Law’
DISTRIBUTION OF A SOLUTE BETWEEN TWO
SOLVENTS
•
•
The law only holds under certain conditions:
a) The temperature must be constant
b) The two solutions must be reasonably dilute
c) The solute must not react, associate or dissociate in the
solvents
For example, the distribution coefficient, k for benzoic acid in
benzene and water increases with increasing concentrations in the
two layers (at constant T). This is due to the formation of benzoic
acid dimers in the benzene layer. These dimers are formed as a
result of hydrogen bonding.
O
C-OH
Molecules
associates
O
C-OH
Hydrogen bonding
in organic phase
Benzene (organic)
Water (aques)
APPLICATIONS OF THE
DISTRIBUTION LAW
SOLVENT EXTRACTION
• In this method, two immiscible liquids are used as the selective
solvents for components in a mixture. The mixture is shaken with
the two immiscible liquids. The liquids are then allowed to separate.
Each layer is then subjected to a number of extractions with the
other solvent
• An example of solvent extraction is ether extraction. This is used to
separate the organic compounds from the aqueous solution.
• The ether layers are combined and the solvent (ether) is evaporated
off to leave the organic material
• This technique is particularly useful when the product is volatile or
thermally unstable
APPLICATIONS OF THE
DISTRIBUTION LAW
Shaken-up
Separate
ether
layer
ether
Aqueous
Aqueous containing
organic compound
Aqueous
Aqueous layer
further extract
with fresh ether
No organic
compound
remain in
aqueous layer
fresh ether
Aqueous
MULTIPLE SOLVENT EXTRACTION
•
Repeated extractions using small portions of solvent are the most
efficient than using a single but larger volume of solvent
•
Example:
The product of an organic synthesis 5 g of X obtained in solution in
1000 ml of water. Calculate the mass of X that can be extracted
from the aqueous solution by:
A) 50 ml of ether
B) Two succesive portions of 25 ml of ether
The partition coefficient of X between ether and water is 40 at room
temperature.
a) volume of ether = 50 ml
volume of water = 1000 ml
1 time extraction
mass of X in water = 5 g
Let the mass of X extracted by 50 ml ether  M
X in ether   40
k
X in water  1
(X is an organic compound, therefore it easily dissolved
in ether as compared to water
if M g dissolved in ether, than (5 - M) g remains in aqueous.
Substituti ng into the Distributi on Law :
 M 


50
ml

  40
1
 5M 


 1000ml 
M  3.33 g
The mass of M that can be extracted by using all
the ether at once is 3.33 g 50ml ether
After
1000ml
water
5
extraction
M
5-M
b) volume of ether = 25 ml
volume of water = 1000 ml
mass of X in water = 5 g
2 time extraction
Let M  mass of X extracted by first 25 ml ether
1
M  mass of X extracted by second 25 ml ether
2
k  X in ether   40
X in water  1
25 ether
 M 

1 
 25 ml 
1000



  40
water
1
 5M 


1
 1000ml 




M  2.50 g
1
if 2.50 g of X are extracted by ether, 2.50 g remain in the aqueous.
Therefore :
 M


25 ether
2 
 25 ml 



  40
1000
1
 2.5  M 


2
water
 1000ml 




M  1.25 g
2
Total mass of X extracted by ether in two portions is
(2.50  1.25 )  3.75 g.
So, repeated extraction s using small portions of solvent (25 ml x 2)
are more efficient than using a single but larger vol ume of solvent
(50 ml)
After
extraction
5
M1
5 – M1
2.5
After
extraction
2.5
M2
2.5 – M2
• V cm3 of a solution which contains W0 g of a solute is extracted repeatedly
with equal portions of S cm3 fresh organic solvent which is immicible in
water
S cm3
V cm3
W0
After
extraction
W0-W1
W1
S cm3
V cm3
Let W1  mass of solute remains in the aqueous layer after the first extraction
W1
gcm 3
v
W - W1
concentrat ion of the solute in the organic phase  0
gcm 3
S
By Distributi on Law :
concentrat ion of the solute in the aqueous phase 
 W1 


Concentrat ion of the X in aqueous phase
v
 k
 
Concentrat ion of the X in organic phase  W0 - W1 


 S

 W1 


 v 
 k
 W0  W1 


S


W1S
 k
v(W 0  W1 )
W1S  Kv(W0  W1 )  kvW 0  kvW1
W1S  kvW1  kvW 0
W1 (kv  S)  kvW 0
kv


W1  
 W0
 kv  S 
Let W 2  mass of the solute that remains in the aqueous layer
after second extraction
 W2 


v 
kv



k 
 W2  
 W1
kv

S
 W1  W2 




S


substituti on W1 above in W 2 :
2
kv
kv
kv





W2  

 W0  
 W0
 kv  S   kv  S 
 kv  S 
if the mass of the solute that remains in the aqueous layer after n times
extraction is Wn , it follow that :
Wn
kv




 kv  S 
n
W0
For multiple extraction
if k 
concentrat ion of organic phase 
concentrat ion of aqueous phase 
n
 v 
Wn 
 W0
 v  ks 
where;Wn  mass of solute remain in aqueous layer after n time extraction
W0 initial mass of solute in aqueos layer before extraction
n  number of extraction
v  volume of aqueous phase
s  volume of organic phase
k
concentrat ion of aqueous phase 
concentrat ion of organic phase 
n
 kv 
Wn 
 W0
 kv  s 
a) volume of ether, s = 50 ml
volume of water, v = 1000 ml
mass of X in water, W0 = 5 g
1 time extraction, n = 1
if k 
X in ether   40
X in water  1
n
 v 
Wn  
 W0
 v  ks 
1


1000
 5
 
 1000  (40)(50) 
 1.6 g
so, mass of solute in organic (ether) layer  5 - 1.6g
b) volume of ether, s = 25 ml
volume of water, v = 1000 ml
mass of X in water, W0 = 5 g
2 time extraction, n = 2
if k 
 3.33 g
X in ether   40
X in water  1
n
 v 
Wn  
 W0
v

ks


2


1000
 5
 
 1000  (40)(25) 
 1.25 g
so, mass of solute in organic(et her) layer  5 - 1.25 g
 3.75 g