1. Partition Equilibrium of a Solute Between Two Immiscible Solvents
16.8
Partition Equilibrium of a Solute Between Two Immiscible Solvents

2. Partition (Distribution) Equilibrium
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)

3. Partition (Distribution) Equilibrium
Water and 1,1,1-trichloroethane are immiscible with each other.

4. Partition (Distribution) Equilibrium
I2 dissolves in both layers to different extent.

5. Partition (Distribution) Equilibrium
Strictly speaking, I2 is NOT non-volatile !

6. Partition (Distribution) Equilibrium
When dynamic equilibrium is established
 rate of  movement = rate of  movement

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

8. When dynamic equilibrium is established
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

9. Partition Coefficient
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)

10. Partition 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:
(no unit)
Units of concentration : mol dm-3, mol cm-3, g dm-3, g cm-3

11. Partition Coefficient
The partition coefficient of a solute between solvent 2 and solvent 1 is given by 1
solvent 2
[Solute] = K
The partition coefficient of a solute between solvent 1 and solvent 2 is given by 2
solvent 1
[Solute] = K

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

13. 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)

14. Partition 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.

15. Not a constant [C6H5COOH]water(C1) / mol dm-3
[C6H5COOH]benzene(C2) / mol dm-3
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

16. 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
 Partition law does not apply

17.  = degree of association of benzoic acid
Interpretation : -
2C6H5COOH(benzene) (C6H5COOH)2(benzene)
 = degree of association of benzoic acid
[C6H5COOH]total = [C6H5COOH]free + [C6H5COOH]associated C2
C2(1-)
C2
Determined by titration with NaOH

18. Thus benzoic acids tend to form dimers when dissolved in benzene.
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.

19. 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).

20. Partition coefficient
Q.17(c)
2C6H5COOH(benzene) (C6H5COOH)2(benzene)
C6H5COOH(benzene) C6H5COOH(aq) C1
Partition coefficient

21.  is a constant at fixed T

22. Applications of partition law
Solvent extraction
Chromatography
Two classes of separation techniques based on partition law.

23. Solvent extraction I2 in KI(aq) I2 in hexane I2 in KI(aq) I2 in hexane
Colourless Hexane
I2 in KI(aq)
I2 in hexane
I2 in KI(aq)
I2 in hexane
+ hexane
I2 in KI(aq)
It dissolves I2 but not KI.
 Organic solvents are preferred.
It is immiscible with water.
 Organic solvents are preferred.
To remove I2 from an aqueous solution of KI, a suitable solvent is added.
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,

24. Solvent Extraction Hexane layer Aqueous layer Before shaking
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.104)
Solvent Extraction
Before shaking
After shaking
Hexane layer
Aqueous layer
Iodine can be extracted from water by adding hexane, shaking and separating the two layers in a separating funnel

25. Determination of I2 left in both layer
Titrated with standard sodium thiosulphate solution
I S2O3  I + S4O62

26. 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.

27. 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

28. Worked example : - 50g X in 40 cm3 ether solution
10g X in 25 cm3 aqueous solution
50g X in 40 cm3 ether solution
By partition law,
(a) Calculate the partition coefficient of X between ether and water at 298 K.
M is the molecular mass of X

29. Or simply, Worked example : - 50g X in 40 cm3 ether solution
10g X in 25 cm3 aqueous solution
50g X in 40 cm3 ether solution
Or simply,

30. (b)(i) xg of X in 30 cm3 ether solution
5g of X in 30 cm3 aqueous solution
30 cm3 ether
(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

31. 3.79 g of X could be extracted.
(b)(i)
xg of X in 30 cm3 ether solution
5g of X in 30 cm3 aqueous solution
30 cm3 ether
(5-x)g of X in 30 cm3 aqueous solution
3.79 g of X could be extracted.

32. (b)(ii) First extraction
x1g of X in 15 cm3 ether solution
5g of X in 30 cm3 aqueous solution
15 cm3 ether
(5-x1)g of X in 30 cm3 aqueous solution

33. (b)(ii) Second extraction
x2g of X in 15 cm3 ether solution
(5-x1)g of X in 30 cm3 aqueous solution
15 cm3 ether
(5-x1-x2)g of X in 30 cm3 aqueous solution

34. total mass of X extracted
= ( ) 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

35. 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.

36. 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 with 200 cm3 of 2-methylpropan-1-ol;

37. 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  = 0.05

38. Q.18(b) 1st extraction 2nd extraction
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 = = = 43.3%

39.  7.5 cm3 of solvent X is required
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

40. 16. 8 Partition Equilibrium of a Solute Between Two Immiscible
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.104)
Check Point 16-8A

41. 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.

42. 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.

43. All chromatographic separations are based upon differences in partition coefficients of the components between a stationary phase and a mobile phase.

44. 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.

45. There are three main types of chromatography
1. Column chromatography
2.     Paper chromatography
3.     Thin layer chromatography

46. Column chromatography
Stationary phase : -
Water adsorbed on the adsorbent (alumina or silica gel)
Mobile phase : -
A suitable solvent (eluant) that seeps through the column

47. 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.

48. 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.

49. Column chromatography
Suitable for large scale treatment of sample
For treatment of small quantities of samples, paper or thin layer chromatography is preferred.

50. 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

51. Paper chromatography
The solvent moves up the filter paper by capillary action
Components are carried upward by the mobile solvent
 Ascending chromatography

53. Different dyes have different partition coefficients between the mobile and stationary phases
They will move upwards to different extent

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

55. Using chromatography to separate the colours in a sweet.
filter
paper
spot of coloured
dye
solvent
separated
colours
Using chromatography to separate the colours in a sweet.

56. Solvent front a d c b

57. a chromatogram
separated
colours

58. 16. 8 Partition Equilibrium of a Solute Between Two Immiscible
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

59. Mixture of phenol and ammonia Mixture of butanol and ethanoic acid
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.109)
Paper Chromatography
Amino 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 16-8B

60. Two-dimensional paper chromatography

61. Two-dimensional paper chromatography
All spots (except proline) appears visible (purple) when sprayed with ninhydrin (a developing agent)

62. 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

63. A variety of different adsorbents can be used.
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

64. 16. 8 Partition Equilibrium of a Solute Between Two Immiscible
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.108)
Check Point 16-8A
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.
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65. 16. 8 Partition Equilibrium of a Solute Between Two Immiscible
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

66. 16. 8 Partition Equilibrium of a Solute Between Two Immiscible
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.

67. 16. 8 Partition Equilibrium of a Solute Between Two Immiscible
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

68. 16. 8 Partition Equilibrium of a Solute Between Two Immiscible
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.110)
Check Point 16-8B
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(b)