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SOLVENT EXTRACTIONSOLVENT EXTRACTION
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WHAT IS SOLVENT EXTRACTIONWHAT IS SOLVENT EXTRACTION
Solvent extraction is a selective
separation procedure forisolating,separation, purification and
concentratinga valuable substance
from an aqueous solutions with the aid of
an organic solution.
THE PROCESS OF SOLVENT EXTRACTIONTHE PROCESS OF SOLVENT EXTRACTION
Distribution of a solute
between two immiscible liquid
phases in contact with each
other.
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Organic phase
(phase 1)
Water phase(phase 2)
(
Requirements for extraction phases:
-Immiscible each other
-LLE : differences in dense
Organic solvents less
dense than water: diethylether, toluene, hexane
Organic solvents more
dense than water:
chloroform, CCl4,
dichloromethane.
Phases to be separated: solid,
liquid, gas, supercritical liquid
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[A]o
[A]aq
DISTRIBUTION RATIO OF SOLUTEDISTRIBUTION RATIO OF SOLUTE
PROCESS:
The solved solute A in one phase (ex: aqueous phase,
aq) is extracted into another phase (ex: organic phase,
org) which has lower density than the first phase.
During shaking, solute A is distributed between the two
phases until the equilibrium is reached
When distribution reaches equilibrium, theconcentration distribution of solute A in two phases is
stated by Nernst distribution law:
aqeqAinitialA
oeqA
aq
oAD
VmolmolVmol
A
AK
/)()/(
][
][
,,
,
,
!
!
KD,A is the partition coefficient of the solute A between two phases.
The partition coefficient (KD) is an equilibrium constant and has a fixed
value for the solute
s partitioning between the two phases.
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Example 1:
10 mL of A 0.1 M is extracted to 5 ml of chloroform. The number of
residue A after equilibrium is 0.03 M. Determine the partition coefficient
between two phases:
The initial number of A = 0.1 mmol/mL x 10 mL = 1 mmolThe residue number of A = 0.03 mmol/mL x 10 mL = 0.3 mmol
The number of A distributed in chloroform when equilibrium =
1.0 mmol 0.3 mmol = 0.7 mmol
The A concentration in chloroform = 0.7 mmol / 5 mL = 0.14 M
KD,A = [A]chloroform / [A]aq,eq = 0.14 M / 0.03 M = 4.7
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aqo
aq
aq
o
aq
o
q
qM
Mq
!
!
!
/
/)1(
][
][
q = fraction of A remains in aqueous phase (phase 1) at equilibrium
1-q = fraction of A distributes in organic phase (phase 2) at equilibrium
%100).1(% qE !
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Example 2:
Solute A has a partition coefficient of 4.0 between hexane and water. If
150.0 mL of 0.03 M of aqueous A is extracted with 600.0 mL of hexane,what fraction of A remains in aqueous phase?
Moles of A remains in aqueous phase = 0.05882 x 0.03 M x 150 mL
= 0.2647 mmol
%12.94%
%100).05882.01(%
05882.0
mL150mL)600.0x(4.0
mL150
!
!
!
!
E
E
q
q
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MULTIPLE EXTRACTIONMULTIPLE EXTRACTION
Solute A is extracted to organic phasewith multiple volume
Example 3:
What fraction of A remains in aqueous
phase if A is extracted six times of100 mL of hexane (Use data inexample 2)
So: There is 0.0004115 fraction of Aremains in aqueous phase after sixtime extraction, each with 100 mL
Number of A remains =
0.0004115 x 0.03 M x 150 mL = 0.001852 mmol
n
nq
!
12
1)(
0004115.0
0.150)100(4
0.150)6(
6
!
!
mLmL
mLxq
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Conclusion of ex. 2 vs ex 3
- % Extraction of A after one time extraction of 600 mL of organic
phase = 94.12 %
- % Extraction of A after six times extraction with each 100 mL oforganic phase = 99.96%
It is more efficient carrying out an extraction several times with
small volume of organic phase than performing ones time
extraction with high volume of organic phase
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EXTRACTION OF WEAK ACID/BASEEXTRACTION OF WEAK ACID/BASE
Dissociation of organic acids/ bases in aqueous phase:
HA H+
+A-
Ka
B+H2O BH+
+ OH-
Kb
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Partition of Organic Acid between two immiscible phases
Organic
Aqueous
HA H++A
-a
HA H+
+A-
K
Neutral species are more soluble in organic phase and charged
species are more soluble in aqueous phase
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When acids/ bases can exist in different forms, the distribution coefficient
(D) is used instead of the partition coefficient (KD)
Organic
AqueousHA H
+ + A
-
Ka
HA H+
+ A-
KD
aq
a
D
aq
aaq
o
aq
aqa
aq
o
aqaq
o
H
K
K
H
KHA
HA
H
HAKHA
HA
AHA
HAD
][1
][1][
][
][
][][
][
][][
][
!
!
!
!
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D
aK
H
KD !
][1
a
D
KH
HKD
!
][
][OR
D
pH
K
Mainly HA Mainly A-
[H+] >> Ka
[H+]
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Example 4.
Two organic acids with pKa = 4 (acid 1) and pKa = 8 (acid 2) are
extracted into organic phase at pH 6. Draw the scheme of extractionand determine which acid is more extracted into organic phase
D
pH4 6 8
K2
K1 At pH =6, acid 1 exist mainly
as A-
(pH > pKa) and acid 2exist mainly as HA (pH < pKa),
so acid 2 is more extracted into
organic phase and acid 1 stays
in aqueous phase
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TASK
0,5 M asam HA 25 mL (Ka = 10-4) akan
diekstrak kedalam 10 mL fasa organik Jika
nilai Kd adalah 5,0:
Dapatkan HA diekstraksi ke fasa organikpada pH 10? (Jelaskan disertai
perhitungan)?
Berapakah % HA yang terekstraksi kedalam
fasa organik jika ekstraksi dilakukan padapH 3?
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43
3
1010
)10(5
][
][
!
!
D
KH
HKD
a
D
ao
a
VDV
Vq
!
%100).1(% qE !
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METAL EXTRACTION
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METAL COMPLEX COMPOUNDS
Metal/metal ions
Ligand: electrons donor
One or more donor atoms: atom with pairs ofelectron, lone pair electrons, covalent bonding
electrons.
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Monodentat ligands: one bonding atom
Polydentat ligands (chelating ligands): more than one bondingatoms
? A? A? A
!
ACu
CuA21
F ? A? A ? A
! ACu
CuA12F
? A? A? A
? A? A? A_ a? A
!
!
AACuCuA
ACuA
CuA
2
1
22
22
FF
F ? A? A ? A
2
212
2
! A
Cu
CuAFF
? A
? A? A !
22
2
ACu
CuAnF
!21FFFn
1. Cu2+ + A- ' CuA+
2. CuA+ + A- ' CuA2
Total: Cu2+ + 2A- ' CuA2
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SELECTIVE EXTRACTION OF METAL IONS USING A
CHELATING AGENT
Most chelating agents have a limited solubility in water/ easily hydrolyzed/air oxidation chelating agent is added to the organic solvent
org
Aq
TYPE 1:
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If the ligands concentration is much greater than the metal ions concentration the
distribution ratio is :
Where:
D = distribution coefficient metal complex
F = formation constant of metal complex
KD,C = partition coefficient of metal complex
CL = initial ligand concentration in the organic phase before extraction
KD,L = partition coefficient of ligand
Ka = acid dissociation constant
n = number of charge
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A divalent metal ion, M2+, is to be extracted from an aqueous solution into an organic
solvent using a chelating agent, HL, dissolved in the organic solvent. The partition
coefficients for the chelating agent, KD,L, and the metalligand complex, KD,c, are 1.0 104
and 7.0 104, respectively. The acid dissociation constant for the chelating agent, Ka, is5.0 105, and the formation constant for the metalligand complex, , is 2.5 1016.
Calculate the extraction efficiency when 100.0 mL of a 1.0 106 M aqueous solution of
M2+, buffered to a pH of 1.00, is extracted with 10.00 mL of an organic solvent that is 0.1
mM in the chelating agent. Repeat the calculation at a pH of 3.00.
the fraction of metal ion remaining in the aqueous phase is:
Thus, at a pH of 1.00, only 0.40% of the metal is extracted. Changing the pH to
3.00, however, gives an extraction efficiency of 97.8%.
Example 5.
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Advantages of using a
chelating agent is the high
degree of selectivity of metal
ions extraction.
The extraction efficiency for a
divalent cation increases from
approximately 0%100% over
a range of only 2 pH units.
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? A? A
aq
oMM
MD
! 2
? A? A ? A
aqaq
oM
MM
MD
2
2
2
!
2
2][
1
F
!
L
KD DDivided by
[ML2]aq
? A? A
aq
oD
L
LK
2
2!
TYPE 2:
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TASK
Calculate the extraction efficiency when 50.0 mL of an aqueous solution that is
0.15 mM in M2+ and 0.12 M in L is extracted with 25.0 mL of the organic
phase. Assume that KD is 10.3 and 2 is 560.
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Sample containing an analyte in a matrix
To separate an analyte from its matrix, its distribution ratio must be significantly
greater than that for all other components in the matrix. When the analytes
distribution ratio is similar to that of another species, then a separation becomes
impossible.
LIMITATION
The problem with a simple extraction is that the separation only occurs in one
direction. In a liquidliquid extraction, for example, we extract a solute from
its initial phase into the extracting phase.
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A single liquidliquid extraction transfers 83% of
the analyte and 33% of the interferent to the
extracting phase. If the concentrations of A and I in
the sample were identical, then their concentration
ratio in the extracting phase after one extraction is
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A second extraction actually leads to a poorer separation.
After combining the two portions of the extracting phase, the concentration ratio
decreases to
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We can improve the separation by first extracting the solutes into the extracting
phase, and then extracting them back into a fresh portion of the initial phase
(Figure 12.2). Because solute A has the larger distribution ratio, it is extracted to a
greater extent during the first extraction and to a lesser extent during the second
extraction. In this case the final concentration ratio of
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The process of extracting the solutes back and forthbetween fresh portions of the two phases, which is
called a counter-current extraction, was developed by
Craig in the 1940s.
The same phenomenon
forms the basis of modern chromatography.