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16.8. Partition Equilibrium of a Solute Between Two Immiscible Solvents. A(solvent 2) A(solvent 1). 16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.104). Partition (Distribution) Equilibrium.

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slide1

16.8

Partition Equilibrium of a Solute Between Two Immiscible Solvents

slide2

A(solvent 2) A(solvent 1)

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
slide3

Partition (Distribution) Equilibrium

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

slide4

Partition (Distribution) Equilibrium

  • I2dissolves in both layers to different extent.
slide5

Partition (Distribution) Equilibrium

Strictly speaking, I2 is NOT non-volatile !

slide6

Partition (Distribution) Equilibrium

When dynamic equilibrium is established

 rate of  movement = rate of  movement

slide8

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

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

When dynamic equilibrium is established

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

slide9

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)
slide10

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:

(no unit)

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

slide11

[Solute]

[Solute]

=

=

K

solvent

solvent

1

2

[Solute]

[Solute]

solvent

solvent

1

2

K

Partition Coefficient

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

slide12

Partition Coefficient

  • Not affected by the amount of solute added and the volumes of solvents used.
  • TAS Experiment No. 12
slide13

Partition law holds true

  • at constant temperature
  • 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)
slide14

C2

C1

C6H5COOH(benzene) C6H5COOH(aq)

Partition law holds true

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

C1and C2 are determined by titrating the acid in each solvent with standard sodium hydroxide solution.

slide16

Benzoic acid dimer

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

Partition law does not apply

slide17

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

Interpretation : -

 = degree of association of benzoic acid

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

C2

C2(1-)

C2

Determined by titration with NaOH

slide18

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.

slide19

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).

slide20

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

C6H5COOH(benzene) C6H5COOH(aq)

Partition coefficient

Q.17(c)

C1

slide22

Applications of partition law

  • Solvent extraction
  • Chromatography

Two classes of separation techniques based on partition law.

slide23

Colourless Hexane

+ hexane

I2 in hexane

I2 in hexane

I2 in KI(aq)

I2 in KI(aq)

I2 in KI(aq)

I2 in KI(aq)

I2 in hexane

I2 in hexane

I2 in KI(aq)

I2 in KI(aq)

Solvent extraction

To remove I2 from an aqueous solution of KI, a suitable solvent is added.

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.

What feature should the solvent have?

At equilibrium,

rate of  movement of I2 = rate of  movement of I2

By partition law,

slide24

Hexane layer

Aqueous layer

After shaking

Before shaking

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

Solvent Extraction

Iodine can be extracted from water by adding hexane, shaking and separating the two layers in a separating funnel

slide25

Determination of I2 left in both layer

Titrated with standard sodium thiosulphate solution

I2 + 2S2O3 2I + S4O62

slide26

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.

slide27

Worked example

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.

slide28

50g X in 40 cm3 ether solution

10g X in 25 cm3 aqueous solution

By partition law,

Worked example : -

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

M is the molecular mass of X

slide29

50g X in 40 cm3 ether solution

10g X in 25 cm3 aqueous solution

Worked example : -

Or simply,

slide30

xg of X in 30 cm3 ether solution

30 cm3 ether

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

5g of X in 30 cm3 aqueous solution

(b)(i)

Determine the mass of X that could be extracted by shaking a 30 cm3 aqueous solution containing 5 g of X witha single 30 cm3 portion of etherat 298 K

slide31

xg of X in 30 cm3 ether solution

30 cm3 ether

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

5g of X in 30 cm3 aqueous solution

(b)(i)

3.79 g of X could be extracted.

slide32

x1g of X in 15 cm3 ether solution

15 cm3 ether

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

5g of X in 30 cm3 aqueous solution

(b)(ii) First extraction

slide33

x2g of X in 15 cm3 ether solution

15 cm3 ether

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

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

(b)(ii) Second extraction

slide34

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

slide35

Important extraction processes : -

  • Products from organic synthesis, if contaminated with water, can be purified by shaking with a suitable organic solvent.
  • Caffeine in coffee beanscan be extracted bySupercritical 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.
slide36

Q.18(a)

Alcohol layer

200 cm3 alcohol

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 with200 cm3 of 2-methylpropan-1-ol;

slide37

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

slide38

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

% of acid extracted = 0.247 + 0.186 = 0.433 = 43.3%

slide39

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.

 7.5 cm3 of solvent X is required

slide40

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

Check Point 16-8A

slide41

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.

slide42

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.

slide43

All chromatographic separations are based upon differences in partition coefficients of the components between a stationary phaseand amobile phase.

slide44

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.

slide45

There are three main types of chromatography

1. Column chromatography

2.Paper chromatography

3.Thin layer chromatography

slide46

Column chromatography

Stationary phase : -

Water adsorbed on the adsorbent (alumina or silica gel)

Mobile phase : -

A suitable solvent (eluant) that seeps through the column

slide47

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.

slide48

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.

slide49

Column chromatography

Suitable for large scale treatment of sample

For treatment of small quantities of samples, paper or thin layer chromatography is preferred.

slide50

X(adsorbed water) X(solvent)

stationary phasemobile phase

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
slide51

Paper chromatography

The solvent moves up the filter paper by capillary action

Components are carried upward by the mobile solvent

Ascending chromatography

slide53

Different dyes have different partition coefficients between the mobile and stationary phases

  • They will move upwards to different extent
slide54

Paper chromatography

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

slide55

filter

paper

spot of coloured

dye

solvent

separated

colours

Using chromatography to separate the colours in a sweet.

slide57

a chromatogram

separated

colours

slide58

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
slide59

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.109)

Paper Chromatography

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

Check Point 16-8B

slide61

Two-dimensional paper chromatography

All spots (except proline) appears visible (purple) when sprayed with ninhydrin (a developing agent)

slide62

X(adsorbed water) X(solvent)

stationary phasemobile phase

Thin layer chromatography

Stationary phase : -

Water adsorbed on a thin layer of solid adsorbent (silica gel or alumina).

Mobile phase : -

A suitable solvent

slide63

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

slide64

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.

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.108)

Check Point 16-8A

Back

slide65

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.110)

Check Point 16-8B

  • 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

slide66

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.110)

Check Point 16-8B

(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.

slide67

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.110)

Check Point 16-8B

(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

slide68

(b)

16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.110)

Check Point 16-8B

Back

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