1. C. Y. Yeung (CHW, 2009) p.01 Partition of Solute between 2 Immiscible Solvents “2 phases” in contact with each other … solvent 1 solvent 2 solute X 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.  conc. of X in 1 and 2 will remain constant at constant temperature.

2. p.02 A new K eq for this system: K D (partition coefficient) “distribution of solute in 2 solvents” X (solvent 1) X (solvent 2) CCl 4 and CHCl 3 are the only two organic solvents denser than water. x at start (conc.) 0 x – a a at eqm (conc.) KD =KD =KD =KD = x – a a (no unit) less dense solvent (usually organic solvent) more dense solvent (usually H 2 O) mol dm -3 / g cm -3

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

4. p.04 p. 108 Check Point 16-8A CH 3 CCl 3 (100cm 3 ) A (6g) K D = 15 H 2 O (60cm 3 )  5.77g of A will be extracted. a = 5.77 At eqm: K D = 15 = a/100 (6-a)/60 Let a be the mass of A to be extracted by CH 3 CCl 3,

5. p.05 p. 128 Q. 16 H 2 O (50cm 3 ) lactic acid (8g) K D = 49.3 CHCl 3 (100cm 3 )  7.69g of lactic acid will be extracted. a = 7.69 At eqm: K D = 49.3 = a/50 (8-a)/100 Let a be the mass of A to be extracted by H 2 O, (a)  (7.40 + 0.55) = 7.95 g of lactic acid will be extracted. x = 7.40 At eqm: K D = 49.3 = x/25 (8-x)/100 Let x be the mass of A to be extracted by first 25cm 3 H 2 O, (b) y = 0.55g At eqm: K D = 49.3 = Let y be the mass of A to be extracted by another 25cm 3 H 2 O, y/25 (8-7.4-y)/100

6. p.06 p. 128 Q. 16(c) Solvent extraction is more efficient if the same amount of “extracting solvent” (H 2 O) is added in small portions several times instead of all at once. Solvent extraction is more efficient if the same amount of “extracting solvent” (H 2 O) is added in small portions several times instead of all at once. H 2 O (50cm 3 ) lactic acid (8g) K D = 49.3 CHCl 3 (100cm 3 ) “extracting solvent” Conclusion … ?  The mass of solute extracted by solvent extraction: 50cm 3  1 < 25cm 3  2 < 10cm 3  5 < 5cm 3  10

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

8. p.08 How does Paper Chromatography work? Solvent moves up with the solute. Different solutes have different K D between the mobile phase and stationary phase. Solute with larger K D (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”

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. p.11 R f value: calculated from the Chromatogram d1d1d1d1 d2d2d2d2 In methanol, R f of Caffeine = d 2 /d 1 Rf Rf Rf Rf is always smaller than 1. It is possible to characterize a particular compound separated from a mixture by its Rf Rf Rf Rf value. (ref.: p. 109)

12. p.12 Expt. 11Distribution of ethanoic acid between butan-1-ol and water butan-1-ol (25cm 3 ) water (40cm 3 ) 2M ethanoic acid (10cm 3 ) shaked 10cm 3 sample from organic layer + 25cm 3 H 2 O + phenolphthalein 10cm 3 sample from aqueous layer + phenolphthalein titrated against std. NaOH  V organic titrated against std. NaOH  V aqueous separating funnel

13. p.13 Repeat expt. With different vol. of CH 3 COOH, butan-1-ol and water …. V organic v1v1 v2v2 v3v3 v4v4 V aqueous vivi v ii v iii v iv KD =KD =KD =KD = [CH 3 COOH] organic [CH 3 COOH] aqueous (V organic  [NaOH]) / (10/1000) (V aqueous  [NaOH]) / (10/1000) = V organic V aqueous = Therefore … V organic V aqueous slope!

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