H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book 2 1 Chapter 23 Phase Equilibrium III: Three-Component Systems 23.1 Partition of a Solute Between Two Immiscible Solvents 23.2Solvent Extraction 22.3 Paper Chromatograghy

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book 2 2 Solute = iodine Solvent 1 = water Solvent 2 = 1,1,1- trichloroethane 2 immiscible liquids with solute dissolved inside (2 phases) 3 components There is a dynamic equilibrium between the iodine dissolved in water and the iodine dissolved in 1,1,1-trichloroethane (phase equilibrium) Partition of a Solute Between Two Immiscible Solvents (SB p.277) Partition Law Example

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book 2 3 I 2 dissolved in H 2 O I 2 dissolved in CH 3 CCl Partition of a Solute Between Two Immiscible Solvents (SB p.278) Some Experimental results [I 2 ] CH 3 CCl 3 (mol dm -3 ) [I 2 ] H 2 O (mol dm -3 ) K D = x x x x x Partition coefficient

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book 2 4 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 Partition of a Solute Between Two Immiscible Solvents (SB p.278) Partition Law K D = = Where K D is the partition coefficient for the system and has no unit.

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book 2 5 The Partition Law will NOT hold when there is association or dissociation of the solute in one of the solvents. Remark 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 benzene, ethanoic acid exists in the form of dimers Partition of a Solute Between Two Immiscible Solvents (SB p.278)

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book 2 6 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+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book 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 cm 3 of water and 50 cm 3 of ethoxyethane is then added to extract X from water. How much X is extracted by ethoxyethane?

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book Solvent Extraction (SB p.279) Solution Let a g be the mass of X extracted by 50 cm 3 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 ∴ K D = ∴ 30 = a = 3.0 ∴ 3.0g of X is extracted by ethoxyethane. Answer

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book Solvent Extraction (SB p.282) Example 23-4 At 298K, 50cm 3 of an aqueous solution containing 6 g of solute Y is in equilibrium with 100 cm 3 of an ether solution containing 108g of Y. (a)100 cm 3 of fresh ether, and (b)50 cm 3 of fresh ether twice at 298 K. Solution [Y] ether = 108g/100cm 3 = 1.08 g cm -3 [Y] water = 6g/50cm 3 = g cm -3 K D = [Y] ether /[Y] water = 1.08 g cm -3 /0.12 g cm -3 = 9 Answer

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book Solvent Extraction (SB p.282) (a) Let m g be the mass of Y extracted by 100 cm 3 of ether, then the mass of Y left in the aqueous layer is (10-m) g. K D = ∴ 9 = m = 9 9 g of Y can be extracted by the first 50 cm 3 of ether, then the mass of Y left in the aqueous layer is (10 – m 1 ) g.

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book Solvent Extraction (SB p.282) (b) Let m 1 g be the mass of Y extracted by the first 50 cm 3 of ether, then the mass of Y left in the aqueous layer is (10 – m 1 )g. K D = ∴ 9 = m 1 = Mass of Y extracted by the first 50 cm3 of ether = 8.182g Mass of Y left in water = (10 – 8.182) g = g

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book Solvent Extraction (SB p.282) Let m 2 g be the mass of Y extracted by the second 50 cm 3 of ether, then the mass if Y left in the aqueous layer is (1.818 – m 2 ) g. K D = ∴ 9 = ∴ m 2 = Mass of Y extracted by the second 50 cm 3 of ether = 1.487g Mass of Y left in water = (1.818 – 1.487) g = g Total mass of Y extracted = m 1 + m 2 = ( ) g = g

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book 2 13 Paper Chromatography Filter paper made of cellulose which contains water as a stationary phase. The solvent moving up serves as a mobile phase Paper Chromatography (SB p.284)

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book 2 14 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 Paper Chromatography (SB p.285)

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book Paper Chromatography (SB p.285) Retardation factor (R f ) R f = R f value of component A = d 2 /d 1 R f value of component B = d 3 /d 1

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book 2 16 Remark R f value of a substance differs in different solvents and at different temperature Paper Chromatography (SB p.286) R f values of some amino acids in two solvents at a given temperature Amino acid Solvent Mixture of phenol and ammonia Mixture of butanol and ethanoic acid Cystine Glycine Leucine

H+H+ H+H+ H+H+ OH - New Way Chemistry for Hong Kong A-Level Book 2 17 The END