ch23---phase-equilibrium-III.ppt

5/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 21Chapter 23Phase Equilibrium III:Three-Component Systems23.1 Partition of a Solute Between Two Immiscible Solvents23.2 Solvent Extraction22.3 Paper Chromatograghy5/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 22Solute=iodineSolvent 1=waterSolvent 2=1,1,1-trichloroethane2 immiscible liquids with solute dissolved inside(2 phases)3 componentsThere is a dynamic equilibrium between the iodine dissolved in water and the iodine dissolved in 1,1,1-trichloroethane(phase equilibrium).23.1 Partition of a Solute Between Two Immiscible Solvents (SB p.277)Partition LawExample5/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 23I2 dissolved in H2OI2 dissolved in CH3CCl323.1 Partition of a Solute Between Two Immiscible Solvents (SB p.278)Some Experimental resultsI2CH3CCl3(mol dm-3)I2H2O(mol dm-3)KD=0.0020.040.060.080.102.35 x 10-44.70 x 10-47.03 x 10-49.30 x 10-411.40 x 10-485.185.185.386.087.7Partition coefficient5/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 24At a given temperature,the concentration ratio of a solute in two immiscible solvents is constant.Solute dissolved in solvent 1Solute dissolved in solvent 223.1 Partition of a Solute Between Two Immiscible Solvents (SB p.278)Partition Law KD=Where KD is the partition coefficient for the system and has no unit.5/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 25The Partition Law will NOT hold when there is association or dissociation of the solute in one of the solvents.RemarkExampleThe 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.23.1 Partition of a Solute Between Two Immiscible Solvents (SB p.278)5/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 26Steps involved:Aqueous layer containing an organic product is transferred into a separating funnel1,1,1-trichloromethane(or other organic solvent)is added to form 2 immiscible layersApparatus 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 product23.2 Solvent Extraction (SB p.279)Solvent Extraction5/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 2723.2 Solvent Extraction (SB p.279)Example 23-1An organic compound X has a partition coefficient of 30 in ethoxyethane and water.=30Suppose 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?5/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 2823.2 Solvent Extraction (SB p.279)SolutionLet 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.Xethoxyethane=a/50 g cm-3Xwater=(3.1 a)/50 g cm-3 KD=30=a=3.03.0g of X is extracted by ethoxyethane.Answer5/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 2923.2 Solvent Extraction (SB p.282)Example 23-4At 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.(a)100 cm3 of fresh ether,and(b)50 cm3 of fresh ether twice at 298 K.SolutionYether=108g/100cm3=1.08 g cm-3Ywater=6g/50cm3=0.112 g cm-3KD=Yether/Ywater=1.08 g cm-3/0.12 g cm-3 =9Answer5/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 21023.2 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=99 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.5/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 21123.2 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=8.182Mass of Y extracted by the first 50 cm3 of ether=8.182gMass of Y left in water=(10 8.182)g=1.818 g5/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 21223.2 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 =8.182Mass of Y extracted by the second 50 cm3 of ether=1.487gMass of Y left in water=(1.818 1.487)g=0.331 gTotal mass of Y extracted =m1+m2 =(8.182+1.487)g=9.669 g5/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 213Paper ChromatographyFilter paper made of cellulose which contains water as a stationary phase.The solvent moving up serves as a mobile phase.23.3 Paper Chromatography (SB p.284)5/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 214As 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.23.3 Paper Chromatography (SB p.285)5/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 21523.3 Paper Chromatography (SB p.285)Retardation factor(Rf)Rf=Rf value of component A=d2/d1Rf value of component B=d3/d15/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 216RemarkRf value of a substance differs in different solvents and at different temperature.23.3 Paper Chromatography (SB p.286)Rf values of some amino acids in two solvents at a given temperatureAmino acidSolventMixture of phenol and ammoniaMixture of butanol and ethanoic acidCystine GlycineLeucine0.140.420.870.050.180.625/21/20245/21/2024H+H+H+OH-OH-OH-New Way Chemistry for Hong Kong A-Level Book 217