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C. Y. Yeung (CHW, 2009)

Partition of Solute between 2 Immiscible Solvents. Partition of Solute between 2 Immiscible Solvents. p.01. “2 phases” in contact with each other …. solvent 1. solute X. solvent 2.  conc. of X in 1 and 2 will remain constant at constant temperature.

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C. Y. Yeung (CHW, 2009)

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  1. Partition of Solute between 2 Immiscible Solvents Partition of Solute between 2 Immiscible Solvents p.01 “2 phases” in contact with each other … solvent 1 solute X solvent 2  conc. of X in 1 and 2 willremain constant at constant temperature. Solute X dissolves in both solvents 1 and 2. At eqm, the rate of diffusion from one solvent to another is the same as reverse rate. C. Y. Yeung (CHW, 2009)

  2. A new Keq for this system: KD (partition coefficient) p.02 “distribution of solute in 2 solvents” x 0 at start (conc.) X(solvent 1) X(solvent 2) a x – a at eqm (conc.) less dense solvent (usually organic solvent) a KD = mol dm-3 / g cm-3 x – a more dense solvent (usually H2O) (no unit) CCl4 and CHCl3 are the only two organic solvents denser than water.

  3. At 291K, KD of butanoic acid (CH3CH2CH2COOH) between ether and water is 3.5. Calculate the mass of butanoic acid extracted by shaking 100 cm3 of water containing 10g of butanoic acid with 100 cm3 of ether. p.03 how many grams of butanoic acid could be extracted from water? ether (100cm3) Let a be the mass of butanoic acid to be extracted by ether, At eqm: KD = 3.5 = a/100 H2O (100cm3) (10-a)/100 1. KD > 1, i.e. butanoic acid is more soluble in ether than in water. a = 7.78  7.78g of butanoic acid will be extracted. 2. Butanoic acid could not be completely extracted from water by ether. At 291K, KD of butanoic acid (CH3CH2CH2COOH) between ether and water is 3.5. butanoic acid (10g) KD (at 291K) = 3.5

  4. p.04 CH3CCl3(100cm3) Let a be the mass of A to be extracted by CH3CCl3, At eqm: KD = 15 = A (6g) a/100 KD = 15 H2O (60cm3) (6-a)/60 a = 5.77  5.77g of A will be extracted. p. 108 Check Point 16-8A

  5. (a) Let a be the mass of A to be extracted by H2O, p.05 At eqm: KD = 49.3 = H2O(50cm3) a/50 (8-a)/100 lactic acid (8g) a = 7.69 KD = 49.3  7.69g of lactic acid will be extracted. CHCl3 (100cm3) Let y be the mass of A to be extracted by another 25cm3 H2O, (b) Let x be the mass of A to be extracted by first 25cm3 H2O, At eqm: KD = 49.3 = At eqm: KD = 49.3 = y/25 x/25 (8-x)/100 (8-7.4-y)/100 x = 7.40 y = 0.55g p. 128 Q. 16  (7.40 + 0.55) = 7.95 g of lactic acid will be extracted.

  6. p.06 “extracting solvent” Solvent extraction is more efficient if the same amount of “extracting solvent” (H2O) is added in small portions several times instead of all at once. H2O(50cm3) lactic acid (8g) KD = 49.3 CHCl3 (100cm3) p. 128 Q. 16(c) Conclusion … ?  The mass of solute extracted by solvent extraction: 50cm3  1 < 25cm3  2 < 10cm3  5 < 5cm3  10

  7. Application of Partition Equilibrium? p.04 mobile phase distribute between stationary phase layer of water adsorbed on the filter paper (stationary phase) (mobile phase) (solute) Paper Chromatography !

  8. p.08 Solvent moves up with the solute. Different solutes have different KD between the mobile phase and stationary phase. Solute with larger KD(more soluble in solvent) will move faster on the paper when the solvent is soaking up. Different solutes could be separated on the filter paper. “chromotograph” How does Paper Chromatography work?

  9. p.09 Chromatogram

  10. p.10 Chromatography is used by the ‘Horse Racing Forensic Laboratory’ to test for the presence of illegal drugs in racehorses. (methanol as solvent)

  11. Rf value: calculated from the Chromatogram p.11 d1 d2 In methanol, Rf of Caffeine = d2/d1 Rf is always smaller than 1. It is possible to characterize a particular compound separated from a mixture by its Rf value. (ref.: p. 109)

  12. p.12 2M ethanoic acid (10cm3) titrated against std. NaOH  Vorganic 10cm3 sample from organic layer + 25cm3 H2O + phenolphthalein butan-1-ol (25cm3) shaked titrated against std. NaOH  Vaqueous 10cm3 sample from aqueous layer + phenolphthalein water (40cm3) separating funnel Expt. 11 Distribution of ethanoic acid between butan-1-ol and water

  13. p.13 (Vorganic [NaOH]) / (10/1000) Vorganic Vorganic = (Vaqueous [NaOH]) / (10/1000) Vaqueous slope! [CH3COOH]organic = [CH3COOH]aqueous Vaqueous Repeat expt. With different vol. of CH3COOH, butan-1-ol and water …. KD = Therefore …

  14. p.14 Next …. Acid-Base Eqm: Arrhenius Theory & Bronsted-Lowry Theory, Kw & pH (p. 130-137) Assignment p.128 Q.14, 15, 17 p.230 Q.6(b), 12(b), 14 (a), (c) [due date: 19/3(Wed)]

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