c. y. yeung (chw, 2009)

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

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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. - PowerPoint PPT Presentation

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C. Y. Yeung (CHW, 2009)

p.01

Partition of Partition of SoluteSolute between between

2 Immiscible 2 Immiscible SolventsSolventsPartition of Partition of SoluteSolute between between

2 Immiscible 2 Immiscible SolventsSolvents

““2 phases” in contact 2 phases” in contact with each other …with each other …

solvent 1solvent 1

solvent 2solvent 2

solute Xsolute X

Solute XSolute X dissolves in both dissolves in both solvents 1 and 2solvents 1 and 2. At . At eqmeqm, t, the rate of diffusion from onhe rate of diffusion from one solvent to another is the se solvent to another is the same as reverse rate.ame as reverse rate.

conc. of X in 1 and 2 willconc. of X in 1 and 2 will remain remain constant at constant temperatureconstant at constant temperature..

p.02

A new KA new Keqeq for this system: for this system:

KKDD (partition coefficient) (partition coefficient)““distributiondistribution of solute in 2 solvents” of solute in 2 solvents”

XX(solvent 1) (solvent 1) XX(solvent 2)(solvent 2)

CClCCl44 and CHCl and CHCl33 are the only two are the only two

organic solvents denser than water.organic solvents denser than water.

xxat start at start (conc.)(conc.) 00

x – ax – a aaat eqm at eqm (conc.)(conc.)

KKDD = =x – ax – a

aa

(no unit)(no unit)

less dense solvent less dense solvent (usually organic solvent)(usually organic solvent)

more dense solvent more dense solvent (usually H(usually H22O)O)

mol dmmol dm-3-3 / g cm / g cm-3-3

p.03At 291K, KAt 291K, KDD of butanoic acid (CH of butanoic acid (CH33CHCH22CHCH22COOH) betCOOH) bet

ween ether and water is 3.5. Calculate the mass of ween ether and water is 3.5. Calculate the mass of butanoic acid extracted by shaking 100 cmbutanoic acid extracted by shaking 100 cm33 of wate of water containing 10g of butanoic acid with 100 cmr containing 10g of butanoic acid with 100 cm33 of et of ether.her.

etherether (100cm (100cm33))

butanoic abutanoic acid (10g)cid (10g)

KKDD (at 291K) = 3.5 (at 291K) = 3.5

HH22OO (100cm (100cm33))

how many grams of butanoic acid how many grams of butanoic acid could be extracted from water?could be extracted from water?

At eqm: At eqm:

KKDD = 3.5 = = 3.5 =aa/100/100

(10-(10-aa)/100)/100

7.78g of butanoic acid will be 7.78g of butanoic acid will be extracted.extracted.

aa = 7.78 = 7.78

Let Let aa be the mass of butanoic aci be the mass of butanoic acid to be extracted by ether,d to be extracted by ether,

1.1. KKDD > 1, i.e. butanoic acid > 1, i.e. butanoic acid

is is more soluble in ethermore soluble in ether t than in water.han in water.

2.2. Butanoic acid Butanoic acid could not could not be completely extractedbe completely extracted f from water by ether.rom water by ether.

At 291K, KAt 291K, KDD of of butanoic acidbutanoic acid (CH (CH33CHCH22CHCH22COOH) betCOOH) bet

ween ween etherether and and waterwater is 3.5. is 3.5.

p.04

p. 108 Check Point 16-8Ap. 108 Check Point 16-8A

CHCH33CClCCl33 (100cm(100cm33))

A (6g)A (6g)

KKDD = 15 = 15

HH22OO (60cm (60cm33))

5.77g of A will be extracted.5.77g of A will be extracted.

aa = 5.77 = 5.77

At eqm: At eqm:

KKDD = 15 = = 15 =aa/100/100

(6-(6-aa)/60)/60

Let Let aa be the mass of A to be be the mass of A to be extracted by CHextracted by CH33CClCCl33,,

p.05p. 128 Q. 16p. 128 Q. 16

HH22OO(50cm(50cm33))

lactic acid lactic acid (8g)(8g)

KKDD = 49.3 = 49.3

CHClCHCl33 (100cm (100cm33))

7.69g of lactic acid will be 7.69g of lactic acid will be extracted.extracted.

aa = 7.69 = 7.69

At eqm: At eqm:

KKDD = 49.3 = = 49.3 =aa/50/50

(8-(8-aa)/100)/100

Let Let aa be the mass of A to be be the mass of A to be extracted by Hextracted by H22O,O,

(a)(a)

(7.40 + 0.55) = 7.95 g of lactic acid will be extracted.(7.40 + 0.55) = 7.95 g of lactic acid will be extracted.

xx = 7.40 = 7.40

At eqm: At eqm:

KKDD = 49.3 = = 49.3 =xx/25/25

(8-(8-xx)/100)/100

Let Let xx be the mass of A to be be the mass of A to be extracted by extracted by firstfirst 25cm 25cm33 H H22O,O,

(b)(b)

yy = 0.55g = 0.55g

At eqm: At eqm:

KKDD = 49.3 = = 49.3 =

Let Let yy be the mass of A to be be the mass of A to be extracted by extracted by anotheranother 25cm 25cm33 H H22O,O,

yy/25/25

(8-7.4-(8-7.4-yy)/100)/100

p.06p. 128 Q. 16(c)p. 128 Q. 16(c)

Solvent extractionSolvent extraction is more efficient if the is more efficient if the same amount of same amount of “extracting solvent” “extracting solvent” (H(H22O) is added O) is added in small in small

portions several timesportions several times instead of all at once.instead of all at once.

HH22OO(50cm(50cm33))

lactic acid lactic acid (8g)(8g)

KKDD = 49.3 = 49.3

CHClCHCl33 (100cm (100cm33))

““extracting solvent”extracting solvent”

Conclusion … ?Conclusion … ?

The mass of solute extracted by solvent extraction:The mass of solute extracted by solvent extraction:

50cm50cm33 1 1 < < 25cm25cm33 2 2 < < 10cm10cm33 5 5 < < 5cm5cm33 1010

p.04

Application of Application of Partition EquilibriumPartition Equilibrium??

Paper Chromatography !Paper Chromatography !distribute distribute betweenbetween

mobile phasemobile phase

stationary phasestationary phase

(stationary phase)(stationary phase)

(mobile phase)(mobile phase)

(solute)(solute)

layer of layer of waterwater adsorbed adsorbed on the filter paperon the filter paper

p.08

How does Paper Chromatography work?How does Paper Chromatography work?

Solvent moves up with the solute.Solvent moves up with the solute.

Different solutes have Different solutes have different Kdifferent KDD between the mobile between the mobile

phase and stationary phase.phase and stationary phase.

Solute with larger KSolute with larger KDD (more soluble in solvent)(more soluble in solvent) will will

move faster on the paper when the solvent is soaking move faster on the paper when the solvent is soaking up.up.

Different solutes could be Different solutes could be separatedseparated on the filter paper. on the filter paper.

““chromotograph”chromotograph”

p.09

ChromatogramChromatogram

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

(methanol as (methanol as solvent)solvent)

p.11

RRff value: calculated from the Chromatogram value: calculated from the Chromatogram

dd11

dd22

In methanol, In methanol,

RRff of Caffeine = d of Caffeine = d22/d/d11

RRff is always smaller than 1. I is always smaller than 1. I

t is possible to characterize t is possible to characterize a particular compound sepaa particular compound separated from a mixture by its Rrated from a mixture by its R

ff value. (ref.: p. 109) value. (ref.: p. 109)

p.12

Expt. 11Expt. 11 Distribution of Distribution of ethanoic acidethanoic acid betw betw

een een butan-1-olbutan-1-ol and and waterwater

butan-1-ol (25butan-1-ol (25cmcm33))

water water (40cm(40cm33))

2M ethanoic acid 2M ethanoic acid (10cm(10cm33))

shakedshaked

10cm10cm33 sample sample from organic layer from organic layer + 25cm+ 25cm33 H H22O + O +

phenolphthaleinphenolphthalein

10cm10cm33 sample from sample from aqueous layer aqueous layer + phenolphthalein+ phenolphthalein

titrated agtitrated against std. ainst std.

NaOH NaOH V V

organicorganic

titrated agatitrated against std. Nainst std. NaOH OH V V

aqueousaqueous

separating funnelseparating funnel

p.13

Repeat expt. With different vol. of CHRepeat expt. With different vol. of CH33

COOH, butan-1-ol and water ….COOH, butan-1-ol and water ….

Vorganic v1 v2 v3 v4

Vaqueous vi vii viii viv

KKDD = =[CH[CH33COOH]COOH]organicorganic

[CH[CH33COOH]COOH]aqueousaqueous

(V(Vorganicorganic [NaOH]) / (10/100 [NaOH]) / (10/100

0)0)(V(Vaqueousaqueous [NaOH]) / (10/100 [NaOH]) / (10/100

0)0)

==VVorganicorganic

VVaqueousaqueous

==

Therefore …Therefore … VVorganicorganic

VVaqueousaqueous

slope!slope!

AssignmentAssignment

p.128 Q.14, 15, 17 p.128 Q.14, 15, 17

p.230 Q.6(b), 12(b), 14 (a), (c) p.230 Q.6(b), 12(b), 14 (a), (c)

[due date: 19/3(Wed)] [due date: 19/3(Wed)]

p.14

Next ….Next ….Acid-Base Eqm:Acid-Base Eqm: Arrhenius Theory & Arrhenius Theory & Bronsted-Lowry Theory, KBronsted-Lowry Theory, Kww & pH & pH

(p. 130-137)(p. 130-137)