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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.

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C. Y. Yeung (CHW, 2009)

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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 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)


C y yeung chw 2009

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.


C y yeung chw 2009

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


C y yeung chw 2009

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


C y yeung chw 2009

(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.


C y yeung chw 2009

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


C y yeung chw 2009

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 !


C y yeung chw 2009

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?


C y yeung chw 2009

p.09

Chromatogram


C y yeung chw 2009

p.10

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

(methanol as solvent)


C y yeung chw 2009

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)


C y yeung chw 2009

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. 11Distribution of ethanoic acid between butan-1-ol and water


C y yeung chw 2009

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 …


C y yeung chw 2009

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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