Partition Equilibrium of a Solute Between Two Immiscible Solvents 16.8.

Partition Partition Equilibrium of a Equilibrium of a Solute Between Solute Between Two Immiscible Two Immiscible

SolventsSolvents

16.16.88

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.104)

Partition (Distribution) EquilibriumPartition (Distribution) Equilibrium• The equilibrium established when a

non-volatile solute distributes itself between two immiscible liquids

A(solvent 2) A(solvent 1)

Partition (Distribution) EquilibriumPartition (Distribution) Equilibrium

Water and 1,1,1-trichloroethane are immiscible with each other.


I2 dissolves in both layers to different extent.


Strictly speaking, I2 is NOT non-volatile !

Partition (Distribution) EquilibriumPartition (Distribution) EquilibriumWhen 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,

y

xKc

Changing the concentrations by the same extent does not affect the quotient

2

2

4

4

3

3

2

2y

x

c y

x

y

x

y

x

y

xK

Suppose the equilibrium concentrations of iodine in H2O and CH3CCl3 are x and y respectively,

y

xKc

When dynamic equilibrium is established

the concentrations of iodine in water and 1,1,1-trichloroethane reach a constant ratio


Partition Partition CoefficientCoefficientThe 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)


Partition Partition CoefficientCoefficientThe partition law can be represented

by the following equation:

2 solvent in solute of ionConcentrat1 solvent in solute of ionConcentratK

D

2 solvent

1 solvent[Solute][Solute]

(no unit)

Units of concentration : mol dm-3, mol cm-3, g dm-3, g cm-3

Partition Partition CoefficientCoefficient

2 solvent

1 solvent

[Solute][Solute]K

1 solvent

2 solvent

[Solute][Solute]K

The partition coefficient of a solute between solvent 2 and solvent 1 is given by

The partition coefficient of a solute between solvent 1 and solvent 2 is given by

Partition Partition CoefficientCoefficient• Not affected by the amount of solute

added and the volumes of solvents used.

• TAS Experiment No. 12

Partition law holds truePartition 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 truePartition law holds true

3. when the solute exists in the same form in both solvents.

C6H5COOH(benzene) C6H5COOH(aq)

C2 C1

C1 and C2 are determined by titrating the acid in each solvent with standard sodium hydroxide solution.

[C6H5COOH]water(

C1) / mol dm-3

[C6H5COOH]benzene(

C2) / mol dm-3C1/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 : -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

Benzoic acid dimer

C

O

O H

C

O

OH

Partition law does not apply

Interpretation : -Interpretation : -

2C6H5COOH(benzene) (C6H5COOH)2(benzene)

[C6H5COOH]total = [C6H5COOH]free + [C6H5COOH]associated

)1(2 C 221 C

C2 C2(1-) C2

= degree of association of benzoic acid

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 H-bond 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)

)1(2 C 221 C

2C6H5COOH(benzene) (C6H5COOH)2(benzene)

22

221

)]1([

C

CK

)1(2 C

C6H5COOH(benzene) C6H5COOH(aq)C1

)1(2

1

C

CKD

Partition coefficient

22

221

)]1([

C

CK

2'2

2 2)1( CK

K

CC

2'

1

2

1

)1( CK

C

C

CKD

'''

2

1 KKKC

CD

is a constant at fixed T

Applications of partition law

• Solvent extraction

• Chromatography

Two classes of separation techniques based on partition law.

I2 in KI(aq)

I2 in hexane

I2 in KI(aq)

I2 in hexane

I2 in KI(aq)

I2 in hexane

I2 in KI(aq)

Colourless Hexane

Solvent extraction

I2 in KI(aq)

+ hexane

I2 in KI(aq)

I2 in hexane

At equilibrium,

rate of movement of I2 = rate of movement of I2

)(2

2

][

][

aqKI

hexane

I

IK

To remove I2 from an aqueous solution of KI, a suitable solvent is added.What feature should the solvent have?It is immiscible with water.

Organic solvents are preferred.

It dissolves I2 but not KI.

Organic solvents are preferred.

It can be recycled easily (e.g. by distillation)

Organic (volatile) solvents are preferred.

By partition law,

Before shakin

g

After shakin

g


Solvent Solvent ExtractionExtractionHexane

layer

Aqueous layer

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

I2 + 2S2O3 2I + S4O6

2

Titrated with standard sodium thiosulphate solution

For the hexane layer, starch is not needed because the colour of I2 in hexane is intense enough to give a sharp end point.

Determination of I2 left in the KI solution

For the aqueous layer, starch is used as the indicator.

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

10g X in 25 cm3 aqueous solution

50g X in 40 cm3 ether solution

Worked example : -

M04.0

50

04.0M50

]X[ ether

water

ether

]X[

]X[K By partition law,

M025.0

10

025.0M10

]X[ water

125.3

M025.010

M04.050

K

(a) Calculate the partition coefficient of X between ether and water at 298 K.

M is the molecular mass of X

10g X in 25 cm3 aqueous solution

50g X in 40 cm3 ether solution

Worked example : -

125.32510

4050

K

Or simply,

(b)(i)

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

5g of X in 30 cm3 aqueous solution

(5-x)g of X in 30 cm3 aqueous solution

xg of X in 30 cm3 ether solution30 cm3 ether

(b)(i)


(5-x)g of X in 30 cm3 aqueous solution

xg of X in 30 cm3 ether solution30 cm3 ether

125.3K

79.35

xx

x

305

30

x

x

3.79 g of X could be extracted.

(b)(ii) First extraction


15 cm3 etherx1g of X in 15 cm3 ether solution

(5-x1)g of X in 30 cm3 aqueous solution

125.3K

05.35

21

1

1

xx

x

30515

1

1

x

x

(b)(ii) Second extraction

(5-x1)g of X in 30 cm3 aqueous solution

15 cm3 etherx2g of X in 15 cm3 ether solution

(5-x1-x2)g of X in 30 cm3 aqueous solution

125.3K

19.105.35

22

2

2

xx

x

305

15

21

2

xx

x

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

• 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.

Important extraction processes : -

Q.18(a)

100 cm3 of 0.5 M ethanoic acid

200 cm3 alcohol

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;

Alcohol layer

Aqueous layer

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

200.0

x05.0]acid[ alcohol 100.0

)x1(05.0]acid[ water

%6.39396.0x

200.0x05.0

100.0)x1(05.0

05.3K

Q.18(a)

Let x1, x2 be the fractions of ethanoic acid extracted to the alcohol layer in the 1st and 2nd extractions respectively.

Q.18(b)

1st extraction 247.0x05.3

100.0x05.0

100.0)x1(05.0

11

1

186.0x05.3

100.0x05.0

100.0)xx1(05.0

22

21

2nd extraction

% 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.

5.7x

100y1.0

xy9.0

]I[

]I[120K

water2

solventX2

7.5 cm3 of solvent X is required


Check Point 16-8ACheck 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


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.


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.


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


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


The components separated can be identified by their specific retardation factors, Rf , which

are calculated by

solventby d travelledistancespotby d travelledistance

fR

filter paper

spot of coloured dye

solvent

separated colours

Using chromatography to separate the colours in a sweet.

b

dc

a

a

bblueR f )(

a

credR f )(

a

dgreenR f )(

Solvent front

separated colours

a chromatogram


Paper Paper ChromatographyChromatography• 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


Paper Paper ChromatographyChromatographyAmino acid Solvent

Mixture of phenol and ammonia

Mixture of butanol and

ethanoic acid

Cystine 0.14 0.05

Glycine 0.42 0.18

Leucine 0.87 0.62

Rf values of some amino acids in two different solvents at a given

temperature Check Point 16-8BCheck 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


Let m be the mass of A extracted using 100 cm3 of 1,1,1-

trichloroethane, then the mass of A left in 60 cm3 of aqueous

layer is (6 – m).

m = 5.77 g

5.77 g of A is extracted using 100 cm3 of 1,1,1-

trichloroethane.

60m6

100m

KD

60m6

100m

15

Back


(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


(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.


(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


(b)

Back

Partition Equilibrium of a Solute Between Two Immiscible Solvents 16.8.

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