H+

H+

H+

OH-

OH-OH

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New Way Chemistry for Hong Kong A-Level Book 21

Chapter 23Chapter 23Phase Equilibrium III:

Three-Component Systems

23.1 23.1 Partition of a Solute Between Partition of a Solute Between Two Immiscible SolventsTwo Immiscible Solvents

23.223.2 Solvent ExtractionSolvent Extraction

22.3 22.3 Paper ChromatograghyPaper Chromatograghy

H+

H+

H+

OH-

OH-OH

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New Way Chemistry for Hong Kong A-Level Book 22

Solute = iodine

Solvent 1 = waterSolvent 2

= 1,1,1- trichloroethane

2 immiscible liquids with solute dissolved inside (2 phases)

2 immiscible liquids with solute dissolved inside (2 phases)

3 components3 components

There is a dynamic equilibrium between the iodine dissolved in water and the iodine dissolved in 1,1,1-trichloroethane (phase equilibrium).

There is a dynamic equilibrium between the iodine dissolved in water and the iodine dissolved in 1,1,1-trichloroethane (phase equilibrium).

23.1 Partition of a Solute Between Two Immiscible Solvents (SB p.277)

Partition Law ExampleExample

H+

H+

H+

OH-

OH-OH

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New Way Chemistry for Hong Kong A-Level Book 23

I2 dissolved in H2O

I2 dissolved in CH3CCl3

23.1 Partition of a Solute Between Two Immiscible Solvents (SB p.278)

Some Experimental results

[I2]CH3CCl3

(mol dm-3)

[I2]H2O

(mol dm-3)

KD =

0.002

0.04

0.06

0.08

0.10

2.35 x 10-4

4.70 x 10-4

7.03 x 10-4

9.30 x 10-4

11.40 x 10-4

85.1

85.1

85.3

86.0

87.7

OH2

CClCH2

2

33

][I

][I

Partition coefficientPartition

coefficient

H+

H+

H+

OH-

OH-OH

-

New Way Chemistry for Hong Kong A-Level Book 24

At a given temperature, the concentration ratio of a solute in two immiscible solvents is constant.

Solute dissolved in solvent 1

Solute dissolved in solvent 2

23.1 Partition of a Solute Between Two Immiscible Solvents (SB p.278)

Partition Law

KD =

=

Where KD is the partition coefficient for the system and has no unit.

2solvent in solute ofion Concentrat

1solvent in solute ofion Concentrat

solvent2

solvent1

[Solute]

[Solute]

H+

H+

H+

OH-

OH-OH

-

New Way Chemistry for Hong Kong A-Level Book 25

The Partition Law will NOT hold when there is association or dissociation of the solute in one of the solvents.

RemarkRemark

Example

The distribution of ethanoic acid between water and benzene.

In water, ethanoic acid exists in the form of monomers (it can form intermolecular hydrogen bonds with water molecules).

In water, ethanoic acid exists in the form of monomers (it can form intermolecular hydrogen bonds with water molecules).

In benzene, ethanoic acid exists in the form of dimers .In benzene, ethanoic acid exists in the form of dimers .

23.1 Partition of a Solute Between Two Immiscible Solvents (SB p.278)

H+

H+

H+

OH-

OH-OH

-

New Way Chemistry for Hong Kong A-Level Book 26

Steps involved:

• Aqueous layer containing an organic product is transferred into a separating funnel

• 1,1,1-trichloromethane (or other organic solvent) is added to form 2 immiscible layers

• Apparatus shaken to facilitate phase equilibrium to reach in a short time (most organic product extracted into the organic phase)

• 1,1,1-trichloromethane is distilled off to obtain the organic product

23.2 Solvent Extraction (SB p.279)

Solvent Extraction

H+

H+

H+

OH-

OH-OH

-

New Way Chemistry for Hong Kong A-Level Book 27

23.2 Solvent Extraction (SB p.279)

Example 23-1

An organic compound X has a partition coefficient of 30 in ethoxyethane and water.

= 30

Suppose we have 3.1g of X in 50 cm3 of water and 50 cm3 of ethoxyethane is then added to extract X from water. How much X is extracted by ethoxyethane?

water

neethoxyetha

[X]

[X]

H+

H+

H+

OH-

OH-OH

-

New Way Chemistry for Hong Kong A-Level Book 28

23.2 Solvent Extraction (SB p.279)

SolutionLet a g be the mass of X extracted by 50 cm3 of ethoxyethane, then the mass of X left in water is (3.1 – a) g.[X]ethoxyethane = a/50 g cm-3

[X]water = (3.1 – a)/50 g cm-3

∴ KD =

∴ 30 =

a = 3.0∴3.0g of X is extracted by ethoxyethane.

50

a3.150

a

50

a3.150

a

Answer

H+

H+

H+

OH-

OH-OH

-

New Way Chemistry for Hong Kong A-Level Book 29

23.2 Solvent Extraction (SB p.282)

Example 23-4

At 298K, 50cm3 of an aqueous solution containing 6 g of solute Y is in equilibrium with 100 cm3 of an ether solution containing 108g of Y.

(a) 100 cm3 of fresh ether, and

(b) 50 cm3 of fresh ether twice at 298 K.

Solution

[Y]ether = 108g/100cm3 = 1.08 g cm-3

[Y]water = 6g/50cm3 = 0.112 g cm-3

KD = [Y]ether/[Y]water = 1.08 g cm-3/0.12 g cm-3

= 9

Answer

H+

H+

H+

OH-

OH-OH

-

New Way Chemistry for Hong Kong A-Level Book 210

23.2 Solvent Extraction (SB p.282)

(a) Let m g be the mass of Y extracted by 100 cm3 of ether, then the mass of Y left in the aqueous layer is (10-m) g.

KD =

∴ 9 =

m = 9

9 g of Y can be extracted by the first 50 cm3 of ether, then the mass of Y left in the aqueous layer is (10 – m1) g.

100

m10100

m

100

m10100

m

H+

H+

H+

OH-

OH-OH

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New Way Chemistry for Hong Kong A-Level Book 211

23.2 Solvent Extraction (SB p.282)

(b) Let m1 g be the mass of Y extracted by the first 50 cm3 of ether, then the mass of Y left in the aqueous layer is (10 – m1)g.

KD =

∴ 9 =

m1 = 8.182

Mass of Y extracted by the first 50 cm3 of ether = 8.182g

Mass of Y left in water = (10 – 8.182) g = 1.818 g

100

m1050

m

1

1

100

m1050

m

1

1

H+

H+

H+

OH-

OH-OH

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New Way Chemistry for Hong Kong A-Level Book 212

23.2 Solvent Extraction (SB p.282)

Let m2 g be the mass of Y extracted by the second 50 cm3 of ether, then the mass if Y left in the aqueous layer is (1.818 – m2) g.

KD =

∴ 9 =

∴ m2 = 8.182

Mass of Y extracted by the second 50 cm3 of ether = 1.487g

Mass of Y left in water = (1.818 – 1.487) g = 0.331 g

Total mass of Y extracted = m1 + m2

= (8.182 + 1.487) g = 9.669 g

100

m1.81850

m

2

2

100

m1.81850

m

2

2

H+

H+

H+

OH-

OH-OH

-

New Way Chemistry for Hong Kong A-Level Book 213

Paper Chromatography

Filter paper made of cellulose which contains water as a stationary phase.

Filter paper made of cellulose which contains water as a stationary phase.

The solvent moving up serves as a mobile phase.

The solvent moving up serves as a mobile phase.

23.3 Paper Chromatography (SB p.284)

H+

H+

H+

OH-

OH-OH

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New Way Chemistry for Hong Kong A-Level Book 214

As the solvent is moving up, there is a competition between …..

As the solvent is moving up, there is a competition between …..

1. The ability of the dyes to dissolve in the absorbed water (stationary phase).

2. The ability of the dyes to dissolve in the solvent (mobile phase)

As different dyes have different partition between the mobile and the stationary phases, they would be carried forward to different extents.

As different dyes have different partition between the mobile and the stationary phases, they would be carried forward to different extents.

23.3 Paper Chromatography (SB p.285)

H+

H+

H+

OH-

OH-OH

-

New Way Chemistry for Hong Kong A-Level Book 215

23.3 Paper Chromatography (SB p.285)

Retardation factor (Rf)

Rf = solventby travelledDistance

spotby travelledDistance

Rf value of component A = d2/d1

Rf value of component B = d3/d1

H+

H+

H+

OH-

OH-OH

-

New Way Chemistry for Hong Kong A-Level Book 216

RemarkRf value of a substance differs in different solvents and at different temperature.

23.3 Paper Chromatography (SB p.286)

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

Amino acid

Solvent

Mixture of phenol and ammonia

Mixture of butanol and ethanoi

c acid

Cystine

Glycine

Leucine

0.14

0.42

0.87

0.05

0.18

0.62

H+

H+

H+

OH-

OH-OH

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New Way Chemistry for Hong Kong A-Level Book 217

The END