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Chapter 23

Chapter 23. Phase Equilibrium III: Three-Component Systems. 23.1 Partition of a Solute Between Two Immiscible Solvents 23.2 Solvent Extraction 22.3 Paper Chromatograghy. 3 components. 2 immiscible liquids with solute dissolved inside (2 phases).

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Chapter 23

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  1. Chapter 23 Phase Equilibrium III: Three-Component Systems 23.1 Partition of a Solute Between Two Immiscible Solvents 23.2 Solvent Extraction 22.3 Paper Chromatograghy

  2. 3 components 2 immiscible liquids with solute dissolved inside (2 phases) 23.1 Partition of a Solute Between Two Immiscible Solvents (SB p.277) Partition Law Example Solute = iodine Solvent 1 = water Solvent 2 = 1,1,1- trichloroethane There is a dynamic equilibrium between the iodine dissolved in water and the iodine dissolved in 1,1,1-trichloroethane (phase equilibrium).

  3. 23.1 Partition of a Solute Between Two Immiscible Solvents (SB p.278) Partition coefficient I2 dissolved in H2O I2 dissolved in CH3CCl3 Some Experimental results

  4. KD = = Where KD is the partition coefficient for the system and has no unit. 23.1 Partition of a Solute Between Two Immiscible Solvents (SB p.278) Partition Law 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

  5. 23.1 Partition of a Solute Between Two Immiscible Solvents (SB p.278) Example The distribution of ethanoic acid between water and benzene. Remark The Partition Law will NOT hold when there is association or dissociation of the solute in one of the solvents. 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 .

  6. 23.2 Solvent Extraction (SB p.279) Solvent Extraction 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

  7. 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?

  8. 23.2 Solvent Extraction (SB p.279) Solution Let 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. Answer

  9. 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. • 100 cm3 of fresh ether, and • 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

  10. 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.

  11. 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

  12. 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

  13. 23.3 Paper Chromatography (SB p.284) Paper Chromatography Filter paper made of cellulose which contains water as a stationary phase. The solvent moving up serves as a mobile phase.

  14. 23.3 Paper Chromatography (SB p.285) 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.

  15. 23.3 Paper Chromatography (SB p.285) Rf = Retardation factor (Rf) Rf value of component A = d2/d1 Rf value of component B = d3/d1

  16. 23.3 Paper Chromatography (SB p.286) Rf values of some amino acids in two solvents at a given temperature RemarkRf value of a substance differs in different solvents and at different temperature.

  17. The END

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