ekstraksi asam lemah dan logam

8/7/2019 Ekstraksi asam lemah dan logam

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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.

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