Partition Equilibrium of a Solute Between Two Immiscible Solvents 16.8.
Post on 18-Dec-2015
258 Views
Preview:
Transcript
Partition Partition Equilibrium of a Equilibrium of a Solute Between Solute Between Two Immiscible Two Immiscible
SolventsSolvents
16.16.88
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.104)
Partition (Distribution) EquilibriumPartition (Distribution) Equilibrium• The equilibrium established when a
non-volatile solute distributes itself between two immiscible liquids
A(solvent 2) A(solvent 1)
Partition (Distribution) EquilibriumPartition (Distribution) Equilibrium
Water and 1,1,1-trichloroethane are immiscible with each other.
Partition (Distribution) EquilibriumPartition (Distribution) Equilibrium
I2 dissolves in both layers to different extent.
Partition (Distribution) EquilibriumPartition (Distribution) Equilibrium
Strictly speaking, I2 is NOT non-volatile !
Partition (Distribution) EquilibriumPartition (Distribution) EquilibriumWhen dynamic equilibrium is established
rate of movement = rate of movement
Changing the concentrations by the same extent does not affect the quotient
2
2
4
4
3
3
2
2y
x
c y
x
y
x
y
x
y
xK
Suppose the equilibrium concentrations of iodine in H2O and CH3CCl3 are x and y respectively,
y
xKc
When dynamic equilibrium is established
the concentrations of iodine in water and 1,1,1-trichloroethane reach a constant ratio
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.104)
Partition Partition CoefficientCoefficientThe partition law states that:
• At a given temperature, the ratio of the concentrations of a solute in two immiscible solvents (solvent 1 and solvent 2) is constant when equilibrium has been reached
• This constant is known as the partition coefficient (or distribution coefficient)
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.104)
Partition Partition CoefficientCoefficientThe partition law can be represented
by the following equation:
2 solvent in solute of ionConcentrat1 solvent in solute of ionConcentratK
D
2 solvent
1 solvent[Solute][Solute]
(no unit)
Units of concentration : mol dm-3, mol cm-3, g dm-3, g cm-3
Partition Partition CoefficientCoefficient
2 solvent
1 solvent
[Solute][Solute]K
1 solvent
2 solvent
[Solute][Solute]K
The partition coefficient of a solute between solvent 2 and solvent 1 is given by
The partition coefficient of a solute between solvent 1 and solvent 2 is given by
Partition Partition CoefficientCoefficient• Not affected by the amount of solute
added and the volumes of solvents used.
• TAS Experiment No. 12
Partition law holds truePartition law holds true
1. at constant temperature
2. for dilute solutions
For concentrated solutions, interactions between solvent and solute have to be considered and the concentration terms should be expressed by ‘activity’(not required)
Partition law holds truePartition law holds true
3. when the solute exists in the same form in both solvents.
C6H5COOH(benzene) C6H5COOH(aq)
C2 C1
C1 and C2 are determined by titrating the acid in each solvent with standard sodium hydroxide solution.
[C6H5COOH]water(
C1) / mol dm-3
[C6H5COOH]benzene(
C2) / mol dm-3C1/C2
0.06 0.483 0.124
0.12 1.92 0.063
0.14 2.63 0.053
0.20 5.29 0.038
Not a constant
Interpretation : -Interpretation : -
• The solute does not have the same molecular form in both solvents
• Benzoic acid tends to dimerize (associate) in non-polar solvent to give (C6H5COOH)2
Benzoic acid dimer
C
O
O H
C
O
OH
Partition law does not apply
Interpretation : -Interpretation : -
2C6H5COOH(benzene) (C6H5COOH)2(benzene)
[C6H5COOH]total = [C6H5COOH]free + [C6H5COOH]associated
)1(2 C 221 C
C2 C2(1-) C2
= degree of association of benzoic acid
Determined by titration with NaOH
Q.17(a)
The interaction between benzoic acid and benzene molecules are weaker than the hydrogen bonds formed between benzoic acid molecules.
Thus benzoic acids tend to form dimers when dissolved in benzene.
In aqueous solution, benzoic acid molecules form strong H-bond with H2O molecules rather than forming dimer.
Q.17(b)
In aqueous solution, there is no association as explained in (a).
Also, dissociation of acid can be ignored since benzoic acid is a weak acid (Ka = 6.3 10-5 mol dm-3).
Q.17(c)
)1(2 C 221 C
2C6H5COOH(benzene) (C6H5COOH)2(benzene)
22
221
)]1([
C
CK
)1(2 C
C6H5COOH(benzene) C6H5COOH(aq)C1
)1(2
1
C
CKD
Partition coefficient
Applications of partition law
• Solvent extraction
• Chromatography
Two classes of separation techniques based on partition law.
I2 in KI(aq)
I2 in hexane
I2 in KI(aq)
I2 in hexane
I2 in KI(aq)
I2 in hexane
I2 in KI(aq)
Colourless Hexane
Solvent extraction
I2 in KI(aq)
+ hexane
I2 in KI(aq)
I2 in hexane
At equilibrium,
rate of movement of I2 = rate of movement of I2
)(2
2
][
][
aqKI
hexane
I
IK
To remove I2 from an aqueous solution of KI, a suitable solvent is added.What feature should the solvent have?It is immiscible with water.
Organic solvents are preferred.
It dissolves I2 but not KI.
Organic solvents are preferred.
It can be recycled easily (e.g. by distillation)
Organic (volatile) solvents are preferred.
By partition law,
Before shakin
g
After shakin
g
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.104)
Solvent Solvent ExtractionExtractionHexane
layer
Aqueous layer
Iodine can be extracted from water by adding hexane, shaking and separating
the two layers in a separating funnel
Determination of I2 left in both layer
I2 + 2S2O3 2I + S4O6
2
Titrated with standard sodium thiosulphate solution
For the hexane layer, starch is not needed because the colour of I2 in hexane is intense enough to give a sharp end point.
Determination of I2 left in the KI solution
For the aqueous layer, starch is used as the indicator.
In solvent extraction, it is more efficient (but more time-consuming) to use the solvent in portions for repeated extractions than to use it all in one extraction.
Worked example
10g X in 25 cm3 aqueous solution
50g X in 40 cm3 ether solution
Worked example : -
M04.0
50
04.0M50
]X[ ether
water
ether
]X[
]X[K By partition law,
M025.0
10
025.0M10
]X[ water
125.3
M025.010
M04.050
K
(a) Calculate the partition coefficient of X between ether and water at 298 K.
M is the molecular mass of X
10g X in 25 cm3 aqueous solution
50g X in 40 cm3 ether solution
Worked example : -
125.32510
4050
K
Or simply,
(b)(i)
Determine the mass of X that could be extracted by shaking a 30 cm3 aqueous solution containing 5 g of X with a single 30 cm3 portion of ether at 298 K
5g of X in 30 cm3 aqueous solution
(5-x)g of X in 30 cm3 aqueous solution
xg of X in 30 cm3 ether solution30 cm3 ether
(b)(i)
5g of X in 30 cm3 aqueous solution
(5-x)g of X in 30 cm3 aqueous solution
xg of X in 30 cm3 ether solution30 cm3 ether
125.3K
79.35
xx
x
305
30
x
x
3.79 g of X could be extracted.
(b)(ii) First extraction
5g of X in 30 cm3 aqueous solution
15 cm3 etherx1g of X in 15 cm3 ether solution
(5-x1)g of X in 30 cm3 aqueous solution
125.3K
05.35
21
1
1
xx
x
30515
1
1
x
x
(b)(ii) Second extraction
(5-x1)g of X in 30 cm3 aqueous solution
15 cm3 etherx2g of X in 15 cm3 ether solution
(5-x1-x2)g of X in 30 cm3 aqueous solution
125.3K
19.105.35
22
2
2
xx
x
305
15
21
2
xx
x
total mass of X extracted
= (3.05 + 1.19) g = 4.24 g > 3.79 g.
Repeated extractions using smaller portions of solvent are more efficient than a single extraction using larger portion of solvent.
However, the former is more time-consuming
• Products from organic synthesis, if contaminated with water, can be purified by shaking with a suitable organic solvent.
• Caffeine in coffee beans can be extracted by Supercritical carbon dioxide fluid (decaffeinated coffee)
• Impurities such as sodium chloride and sodium chlorate present in sodium hydroxide solution can be removed by extracting the solution with liquid ammonia.
Purified sodium hydroxide is the raw material for making soap, artificial fibre, etc.
Important extraction processes : -
Q.18(a)
100 cm3 of 0.5 M ethanoic acid
200 cm3 alcohol
Calculate the % of ethanoic acid extracted at 298 K by shaking 100 cm3 of a 0.50 M aqueous solution of ethanoic acid with 200 cm3 of 2-methylpropan-1-ol;
Alcohol layer
Aqueous layer
Let x be the fraction of ethanoic acid extracted to the alcohol layer
No. of moles of acid in the original solution
= 0.5 0.100 = 0.05
200.0
x05.0]acid[ alcohol 100.0
)x1(05.0]acid[ water
%6.39396.0x
200.0x05.0
100.0)x1(05.0
05.3K
Q.18(a)
Let x1, x2 be the fractions of ethanoic acid extracted to the alcohol layer in the 1st and 2nd extractions respectively.
Q.18(b)
1st extraction 247.0x05.3
100.0x05.0
100.0)x1(05.0
11
1
186.0x05.3
100.0x05.0
100.0)xx1(05.0
22
21
2nd extraction
% of acid extracted = 0.247 + 0.186 = 0.433 = 43.3%
Q.19
Let x cm3 be the volume of solvent X required to extract 90% of iodine from the aqueous solution and y be the no. of moles of iodine in the original aqueous solution.
5.7x
100y1.0
xy9.0
]I[
]I[120K
water2
solventX2
7.5 cm3 of solvent X is required
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.104)
Check Point 16-8ACheck Point 16-8A
Chromatography
A family of analytical techniques for separating the components of a mixture.
Derived from the Greek root chroma, meaning “colour”, because the original chromatographic separations involved coloured substances.
Chromatography
In chromatography, repeated extractions are carried out successively in one operation (compared with fractional distillation in which repeated distillations are performed) which results, as shown in the worked example and Q.18, in an effective separation of components.
All chromatographic separations are based upon differences in partition coefficients of the components between a stationary phase and a mobile phase.
The stationary phase is a solvent (often H2O) adsorbed (bonded to the surface) on a solid.
This may be paper or a solid such as alumina or silica gel, which has been packed into a column or spread on a glass plate.
The mobile phase is a second solvent which seeps through the stationary phase.
There are three main types of chromatography
1. Column chromatography
2. Paper chromatography
3. Thin layer chromatography
Column chromatography
Stationary phase : -
Water adsorbed on the adsorbent (alumina or silica gel)
Mobile phase : -
A suitable solvent (eluant) that seeps through the column
Column chromatography
Partition of components takes place repeatedly between the two phases as the components are carried down the column by the eluant.
The components are separated into different bands according to their partition coefficients.
Column chromatography
The component with the highest coefficient between mobile phase and stationary phase is carried down the column by the mobile phase most quickly and comes out first.
Column chromatography
Suitable for large scale treatment of sample
For treatment of small quantities of samples, paper or thin layer chromatography is preferred.
Paper chromatography
• Stationary phase : -
Water adsorbed on paper.
• Mobile phase : -
A suitable solvent
The best solvent for a particular separation should be worked out by trials-and-errors
X(adsorbed water) X(solvent)
stationary phase mobile phase
Paper chromatography
The solvent moves up the filter paper by capillary action
Components are carried upward by the mobile solvent
Ascending chromatography
• Different dyes have different partition coefficients between the mobile and stationary phases
• They will move upwards to different extent
Paper chromatography
The components separated can be identified by their specific retardation factors, Rf , which
are calculated by
solventby d travelledistancespotby d travelledistance
fR
filter paper
spot of coloured dye
solvent
separated colours
Using chromatography to separate the colours in a sweet.
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.109)
Paper Paper ChromatographyChromatography• The Rf value of any particular substance
is about the same when using a particular solvent at a given temperature
• The Rf value of a substance differs in different solvents and at different temperatures
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.109)
Paper Paper ChromatographyChromatographyAmino acid Solvent
Mixture of phenol and ammonia
Mixture of butanol and
ethanoic acid
Cystine 0.14 0.05
Glycine 0.42 0.18
Leucine 0.87 0.62
Rf values of some amino acids in two different solvents at a given
temperature Check Point 16-8BCheck Point 16-8B
Two-dimensional paper chromatography
All spots (except proline) appears visible (purple) when sprayed with ninhydrin (a developing agent)
Thin layer chromatography
Stationary phase : -
Water adsorbed on a thin layer of solid adsorbent (silica gel or alumina).
Mobile phase : -
A suitable solvent
X(adsorbed water) X(solvent)
stationary phase mobile phase
Q.20
Suggest any advantage of thin layer chromatography over paper chromatography.
A variety of different adsorbents can be used.
The thin layer is more compact than paper, more equilibrations can be achieved in a few centimetres (no. of extraction ).
A microscope slide can be used as the glass plate
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.108)
Let m be the mass of A extracted using 100 cm3 of 1,1,1-
trichloroethane, then the mass of A left in 60 cm3 of aqueous
layer is (6 – m).
m = 5.77 g
5.77 g of A is extracted using 100 cm3 of 1,1,1-
trichloroethane.
60m6
100m
KD
60m6
100m
15
Back
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.110)
(a)A student wrote the following explanation for the different Rf values found in the separation of two amino acids, leucine (Rf value = 0.5) and glycine (Rf value = 0.3), by paper chromatography using a solvent containing 20% of water.
“Leucine is a much lighter molecule than glycine.”
Do you agree with this explanation? Explain your answer.
Answer
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.110)
(a) The difference in Rf value of leucine and glycine is due to
the fact that they have different partition between the
stationary phase and the mobile phase. Therefore, they
move upwards to different extent. The Rf value is not
related to the mass of the solute.
16.8 Partition Equilibrium of a Solute Between Two Immiscible Solvents (SB p.110)
(b) Draw a diagram to show the expected chromatogram of a mixture of A, B, C and D using a solvent X, given that the Rf values of A, B, C and D are 0.15, 0.40, 0.70 and 0.75 respectively. Answer
top related