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

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)

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

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

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

Partition (Distribution) Equilibrium When 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,

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

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)

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

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

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

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)

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.

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

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

 = 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

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.

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).

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

 is a constant at fixed T

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

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,

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

Determination of I2 left in both layer Titrated with standard sodium thiosulphate solution I2 + 2S2O3  2I + S4O62

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.

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

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

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,

(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

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.

(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

(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

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

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.

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;

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

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 = 0.247 + 0.186 = 0.433 = 43.3%

 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

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

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

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.

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.

Column chromatography 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

Paper chromatography 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

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

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.

Solvent front a d c b

a chromatogram separated colours

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

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

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

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

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. Back

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

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.

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

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 Back (b)