Liquid liquid extraction sy 2014

LIQUID-LIQUID
EXTRACTION
Dr. Archana Naik, SVB’s College of Pharmacy
1
S.Y.B. Pharm

Introduction
 Liquid-Liquid Extraction or solvent extraction is one
of the separation technique and it is the most
common method adopted in the field of analysis.
 Extraction is the process of transfer of constituent
from one liquid phase to another liquid phase by
contact. This process is also as called Partitioning or
distribution.
 In p’ceutical field solvent extraction method is adopted in
separation and isolation of various ingredients from their
complex formulation.
2
Dr. Archana Naik, Saraswathi Vidya Bhavans College of Pharmacy

Solvent Extraction
 Generally one of the two methods are used for
extraction purpose.
 1. removal of interfering or unwanted substance from
preparation by crystallization, filtration, sublimation or
distillation method.
 2. separation or isolation of active ingredient from
preparation i.e. extraction technique.
There are two types of extraction techniques
 1. Solid phase extracton
 2. Liquid Liquid extraction
3

liquid-liquid extraction
 liquid-liquid extraction has been employed
predominantly and effectively not only for the
pre-concentration and isolation of a ‘single’
chemical entity just before its actual estimation,
but also for the extraction of classes of organic
compounds or groups of metals, just prior to their
usual estimation either by instrumental
techniques.
4

theory
 The concept of “like dissolves like” works well in
LLE.
 A hydrophobic molecules prefers an organic
medium whereas an ionic compound prefers to
remain in aqueous solution.
 When shaken, with two immiscible solvents, the
compound will distribute itself between the two
solvents.
Normally one solvent is water and the other solvent
is a water-immiscible organic solvent.
5

Theory
 Consider a single solute(A) distributed between two
immiscible solvent then the resulting equilibrium can
be written as
 A aq A org or
 Solute in lower phase Solute in upper phase
 At equilibrium the molecules naturally distribute
themselves in the solvent where they are more soluble
 Distribution of solute in upper and lower phase is
depending on the affinity of the drug towards solvents.
6

Nernst distribution law
 Liquid liquid extraction method is based on nernst law
which states that ‘ the ratio of activities of solute material
in the two immiscible solvent at equilibrium is remain
constant.
 Usually concentrations are substituted for activity, then
distribution law is written as
 When solute in dissolved in two solvents which are
immiscible in each other the solute distribute itself in both
solvents and ratio of conc. of two solution remains constant
and it is called as partition coefficient
7

distribution or partition coefficient(KD)
 A quantitative measure of the how an organic compound
will distribute between aqueous and organic phases is
called the distribution or partition coefficient.
 It is denoted by equilibrium constant K
 It is the ratio of the concentrations of the solute in the two
different solvents once the system reaches equilibrium.
 distribution or partition coefficient is also denoted as KD
 KD=
 (Note that K is independent of the actual amounts of the two
solvents mixed but it is dependant on temperature.)
 If KD large , the solute will tend toward quantitative distribution in
[ ]
[ ]a
o
s
s
8

KD
 Benzoic acid is a weak acid in water with a particular
ionisation constant Ka. The distribution constant is given
by
 KD=[HBz]e eq 1
[HBz]a
e is ether solvent and a represents the aqueous solvent.
(note that benzoic acid in aqueous layer exist as Bz-,
depending on magnitude of Ka and pH of the aqueous
layer)
9

Distribution ratio:
 Distribution ratio:
 Is the ratio of concentrations of dissociated and
undssociated form solute in 2 different solvents.
 We can derive the difference between D and KD
from the equilibria involved.
[ ]
[ ] [ ]aa
o
BzHBz
HBz
D
−+
=
eq3
10

 The acidity constant of Ka for ionization of the
acid in the aqueous phase is given by
 Hence
 From eqn 1 [HBz]e=KD[HBz]a

 Substitution of equation 5 and 6 in eqn3 gives
[ ] [ ]
aH
aHBzKa
aBz
][ +
=−
[ ]
[ ]aHBz
aBzaH
Ka
−+
=
][ eq4
eq5
eq6
[ ]
[ ] [ ] [ ]aHaHBzKaHBz
aHBzK
D
a
D
++
=
/
[ ]aa HK
K
D
D
++
=
/1
eq7
eq8
Distribution ratio:
11

 Eq 3
 put value of [HBz]o asKD[HBz]a (calculatedineq5)
 put value of [Bz-]oas (calculatedineq6)
 eq 7
12
[ ]
[ ] [ ]aa
o
BzHBz
HBz
D
−+
=
[ ] [ ]
aH
aHBzKa
aBz
][ +
=−
[ ]
[ ] [ ] [ ]aHaHBzKaHBz
aHBzK
D
a
D
++
=
/
[ ]aa HK
K
D
D
++
=
/1

Distribution ratio
 This equation predicts that when [H+]a >Ka, D is nearly
equal to KD , and if KD is large the benzoic acid will be
extracted into the ether layer; D is maximum under
these condition.
 If on the other hand, [H+] < Ka, then D reduces to KD
[H+]a/Ka, which will be small and the benzoic acid
will remain in the aqueous layer.
 i.e. in alkaline solution the benzoic acid is ionized and
cannot be extracted, while in acid solution it is largely
undissociated. In solvent extraction the separation
efficiency is usually independent of the
13

Separation factor
 The separation factor is one distribution ratio divided
by another; it is a measure of the ability of the system
to separate two solutes. For instance, if the distribution
ratio for nickel (DNi) is 10 and the distribution ratio for
silver (DAg) is 100, then the silver/nickel separation
factor (SFAg/Ni) is equal to DAg/DNi = SFAg/Ni = 10.
14

Extraction of drug
 The distribution ratio D is a constant independent of
the volume ratio. However the fraction of the solute
extracted will depend on the volume ratio of the
two solvents.
 If larger volume of organic solvent is used, more
solute must dissolve in this layer to keep the
concentration ratio constant and to satisfy the
distribution ratio.
 The fraction of the solute extracted is equal to the
milimoles of the solute in the organic layer divided by the
total no of milimoles of solute.
15

Extraction
 If the partition coefficient for a solute between
two solvents is known, it is possible to calculate
the fraction of the solute that is present in each
phase at equlibrium.
 Let P be the fraction of that solute in the upper
phase and q is the fraction in the lower phase at
equilibrium.
 The quantity p is defined by
P= Amount of solute in upper phase
Total amount of solute
16

Extraction in upper(p) and lower
phase(q)
Amount is express in terms of concentration C and volume V
 Amount of solute in upper phase=CuVu
 Total amount of solute=CuVu +CLVL
 Ratio of phase volume U=Vu/VL
 Eqn K=Cu/CL Cu=KCL 2
 Eqn U=Vu/VL Vu=UVL 3
 Therefore eq 1 becomes P= CuVu /CuVu +CLVL
 and P= KCL.UVL / KCL.UVL + CLVL
 P= CLVL KU/ CLVL(KU +1)
17

 The equation relates the fraction of solute extracted
into upper phase to the partition coefficient and ratio
of phase volumes since P is the fraction extracted into
phase volume ,
 100P is % extracted into upper phase.
 P+q =1, q=1-p, q=1- [KU/KU+1]
q= 1/KU+1
q= (KU +1) –KU /KU+1
q = 1/KU+1
The larger the partition coefficient, large amount of
solute will be the extracted in upper phase.
Fraction of drug in lower phase
18

%P
 A solute is known to have a partition coefficient of 4
between water and ether. If 15 ml of an aqueous
solution of the compound is extracted with one 20 ml
portion of ether what % of original solute will be found
I the ether
 K=4 U=20/15= 1.33
 P=KU/KU+1= 4 x1.33/4x 1.33+ 1 =5.32/5.32+1= 0.842
 In percentage it is 84.2%
19

Multiple extraction
 Unless K is extremely large, a significant portion of
solute will not get in both phases after extraction.
 Therefore at that time we have to reextract the
solute for maximum recovery.
 If very less amount of solute is extracted in 1st
extraction so to remove all the remaining solute
we have to reextract with additional portion of
fresh extractant.
20

 During every extraction calculate the fraction of solute
extracted or calculate no of extraction required to
achieve desired extent of extraction. State of extraction
after one equilibrium is mention below.
 1=p+q total fraction extracted (p) =1-q
 In multiple extraction p=1-qn
Sr no Fraction of total
extracted in nth
extraction
Total fraction
extracted
Fraction remaining
1 P p q [1-p]
2 Pq1
p+pq q2
[1-(p+pq)]
3 Pq2
p+pq+pq2
q3
[1-(p+pq +pq2
)]
n pq(n-1)
∑n
n=1 pq(n-1)
qn
21

Percent efficiency
 Consider K=4 in ether water system
 U=1 so no of extraction required for extraction of
maximum drug can be calculated by using formula given
above.
 (out of 1, [ku/ku+1 = 4/5]0.80 of the solute is extracted in the first
step. In second extraction fraction extracted is 0.80 of remaining
0.20 [pq= 0.8x0.2=0.16] so total extraction is 0.8+.16= 0.96)
No of extraction n Total extracted (%)
1 80.00
2 96.00
3 99.20
4 99.84
5 99.97
22

Comparison of single and multiple
extraction
Compare the efficiency of extraction of 10 ml aqueous
solution of compound with a) one 40 ml portion of ether b)
four 10 ml portions of ether if K is 4 in ether water system.
 Single extraction
a) K=4, U= ether layer vol/aq layer vol= 40/10=4
P=KU/KU+1 =4x4/4x4+1=16/17=0.9412 =94.12%
b) Multiple Extraction
K=4,U=ether layer vol/aq layer vol= 10/10=1
(q =1/KU+1=1/4x1+1=0.2)
So total fraction extracted in 4th
extraction= 1-q4
=
1-(0.2)4 =1-0.0016=0.9984 in percentage it is 99.84%
23

Comparison of single and multiple
extraction
Results indicate that more efficient extraction is
achieved with several extraction than single
extraction utilizing same total volume of
extractant.
In multiple extraction Only 20 ml (2 portions
of10-10 ml ) can extract more than 96% than
that of 40 ml used in single extraction.
24

Types of Liquid Liquid extraction:
A. Single extraction (Batch)
 Batch extraction is carried out using set of separating funnel.
 A solution from which a substance is to be extracted and
immiscible solvent are introduced into separating funnel. Two
phases are mixed thoroughly in order to extract the
substance from one phase to other phase.
 For more efficiency fresh extractant is added to raffinate and
extraction is continued several times.
 Extraction by this method is carried out when partition
coefficient of solute is high. Method is simple and quick.
Hence widely used for extraction on small scale. For
extraction on a large scale, a continuous extraction procedure
is used.
25

Single extraction (Batch)
26

B.Continuous Extraction
 In some cases, it is difficult to efficiently remove a
solute unless a large number of extractions are
conducted.
 An alternate approach is a continuous extraction.
 With an appropriate setup, an efficient extraction can
be conducted with a minimum of extracting solvent.
 Advantages
 Only uses a small amount of solvent
 Can remove a high percent of a solute
 Can work unattended for long periods
27

Continuous Extraction
 The continuous extraction method is carried out
when the partition coefficient of solutes is low.
 An immiscible extracting liquid is kept flowing
continuously through the solution from which a
solute is to be extracted.
 Although there is not enough time for the equilibrium
to reach, a solute is extracted continuously in this
extraction method.
 This method requires a special kind of extractor,
depending on whether the solvent used for extraction
is lighter or heavier than sample solution.
28

i) Extracting solvent lighter than the sample solution
 Apparatus used for this type of extraction is shown in fig
 It is similar to that of the Soxhlet apparatus but instead of one
solvent, two solvents are used in this case.
 The extracting solvent (e.g. a non-aqueous solvent), which is
lighter than the sample solution (e.g. an aqueous solution), is
placed in a container (A).
 It is connected to a container (B) holding the Sample solution to
half its capacity.
 A glass tube (C), having a funnel shaped opening at one end
and a glass bulb with holes at another end, is placed inside
container B. This body is then connected to a condenser (D),
which is attached at the upper end of the container B.
29

i. The extracting solvent being
lighter than sample solution
 when extracting solvent in the container (A) is heated.
 Vapor of the extracting solvent passes to the condenser
where it get condense.
 The droplets of extracting liquids enter the glass tube C
and then escape through the holes in the glass bulb .
 it passes through the sample solution and extract the
solute and accumulated on the top of the sample solution.
 When sufficient quantity of extracting solution get
collected in container B, it overflows in to the flask (A)
from the side arm (S) the process is continued till the
extraction is complete.
30

ii) Extracting solvent heavier than the sample
solution
 Apparatus used for this type of extraction is shown in fig
 The extracting solvent placed in RBF (A) and is heated.
 The sample solution to be extracted is present in a container (B)
 Vapor of the extracting solvent passes into the condenser and
falls as droplets into the funnel type glass tube (C), and pass
through the sample solution.
 In this process it extracts the solute and get accumulated at the
bottom of the container B. When volume of extractant rises to
sufficient height it overflows in to the flask (A) from the side arm
(S) and the process is continued till the extraction is complete.
 Continuous extraction method is used when the material to be
extracted has low partition coefficient for a pair of solvents.
31

Continuous Extractor
i) Extracting solvent lighter ii) Extracting solvent heavier than the
than the sample solution sample solution
32

C. Countercurrent Distribution
 Generally for simple extraction separating funnel
method is used but if sample solution contains 2 or
more substances having similar distribution coefficients
then their separation using single batch extraction is
very poor. So by a technique of succesive extraction
with fresh solvent, complex mixture with similar
distribution coefficient may be separated.
 Eg counter currentextraction technique
 CCD is basic liquid liquid extraction that permits
separation of sub with very similar partiton coefficient.
33

Working of CCD
 Term countercurrent indicates that 2 phases move in
apposite directions. In this one phase is stable and
other phase is in moving state. The countercurrent
distribution process involve train of tubes each
containing two separate chamber within which the
individual equilibrium occur.
 Chamber A in all tubes filled with solvent A (which is
denser than extracting solvent 2)
 Tube one contain sample and extracting solvent (2) is
introduced into first tube through inlet B
34

Working of CCD
 After shaking back and forth and then allow the phase
to separate. Then tube is tilted to 900
.
 The less dense solvent flow through tube C into
chamber D then assembly is rotated to original position
and solvent flows out through tube E into next tube
leaving solvent 1 in chamber A of first tube.
 Fresh solvent 2(lighter solvent extractant) is added to
first tube and process is repeated.
35

 Solute originally extracted into solvent 2 (extractant or less
denser solvent)has been transferred to second tube
containing solvent 1 (no sample) where it is redistributed.
This procedure is repeated several times depending on
separation efficiency or no of tubes.
 In this way transfer of solute in the direction of motion of
upper phase (solvent 2) takes place.
 If sample contains 2 solutes with different partition
coefficient they will pass through tube with different rate
i.e. sub with larger partition coefficient will travel faster.
 Counter current is very efficient process for separation of
closely related compounds.
Working of CCD
36

Diagram of CCD tube
B
A
E
D
C
1st
position
B
A
E
C
D
2nd
position
CCD tube is
rotated to 900
37

CCD
 The greater the difference of the distribution ratio
of various substances, the better the separation
between each other. A much larger number of
tubes is required to separate mixtures of
substances with almost similar distribution ratios.
38

Extraction of drug in CCD
39

Solid liquid extraction
Soxhlet apparatus:
 Continuous hot percolation process or
Soxhlet extraction Apparatus.
 The apparatus consists of three parts-
A flask: For boiling of solvent.
A Soxhlet extractor: Having the drug in
thimble, a side tube & siphon tube
A reflux condenser: For condensation of
vapors
40

Soxhlet extraction
 Typically soxhlet extraction is only required where the
desired compound has limited solubility in a solvent.
Extraction carried out by distillation and condensation of
the solvent .
 Apparatus allows multiple extractions to be done
repeatedly using same volume of solvent.
 Soxhlet extraction has been a standard method of
extraction of solids from crude material
 Extraction of solid is imp step in preparation of many
pharmaceuticals.
41

Soxhlet extractor
 Soxhlet extractor is used only for thermostable compounds
 A finely ground drug is held in a porous bag or thimble
(made up of filter paper) is placed in chamber E
 Extracting solvent is placed in flask A and heated to boiling
vapors rise through side arm B are condensed in condenser
D.
 The condensed extractant drips into the thimble containing
crude drug extracting it by contact.
 When level of exctractant in chamber E rises to top of
siphon tube c, extractant come back to flask A and process
is repeated several times to achieve max extraction of
42

43

 Separation of pair of substances is much influence by their
distribution behavior.
 The feasibility of resoling two subs is explained in terms of
separability factor α
 α =K1/K2
 Where K is partition coefficient of two substances in LLE
 If value of α is unity means two subs can not separated by
extraction. So greater the deviation of α from unity more
feasible is the separation.
 In separation of two substances by LLE requires that one of
the solute has small k, so other sub is extractable from
Factors affecting LLEFactors affecting LLE
44

Factors affecting LLEFactors affecting LLE
 Number of Factors affect distribution coefficient of
solute which affect extraction and these can be
utilized for effective separation.
 K can be increase by considering following points:
 Choice of solvent
 pH effect
 Salting out effect
 Control of hydrophobicity
45

Conditions of a choice of solvent
which is used as extractant:
Criteria for Selection of solvent :
1. The density of extractant should be
difference from water density.
2. Low Boiling point
2. Should be selective.
3. High distribution coefficient
4. Should have the minimum viscosity.
6. Should be inexpensive and of low toxicity.
46

Choice of solvent
 The partition coefficient is influenced by chemical nature
of the solvent So choice of solvent is important factor to
achieve good recovery.
 Widely used extractant are diethylether, chloroform and
hydrocarbons.
 Ex: separation of sodium benzoate(SB) and caffeine(C)
 Mixture of SB and C is dissolved in water both are soluble.
 Extraction is carried out with chloroform(C is quite soluble
but SB is insoluble)
 Final result is total C is present in Chloroform layer and SB
remains in water layer.
 Separated constituents is then quantitatively analyzed
47

Control of ionic strength:
salting out effect
 If the salt concentration of an aqueous solution is made
very high, the solubility of nonelectrolyte will be decreased.
 The reduction of solubility by an increase in ionic strength
is called as salting out effect.
 This is because due to high salt conc. in water, availability
of water molecule to act as solvent for non-electrolyte get
reduced.
 The ions of the salt tying up much of water (through strong
ion dipole force) so solute due to unavailability of water
get extracted in to organic solvent. This helps to increase
extraction efficiency of solute having greater solubility in
water.
48

pH effect
 pH plays an important role in extraction process.
 Many comp encountered by pharmaceutical analyst
are weak acids or bases.
 Solubility of these subs depends upon their ionic
form.
 Generally ionized species are soluble in polar
solvents and nonionoised in nonpolar solvents.
 These form can be Inter-converted by adjusting pH
of medium. Hence pH control is most powerful
means for influencing the value of partition
coefficient.
49

Effect of pH
on separation of mixture of acid and base
 Acetylsalicylic acid (acid)and antihistamine(base)
 Both drugs are separated by dissolving in water. Then
acidify the solution using HCl and extracting the
Acetylsalicylic acid (unionized in HCl)completely with
ether .
 Then antihistamine which is in its salt form(due to reaction
of basis drug with HCL) in aqueous layer cab be extracted
by adding base to release the drug in free form.
 Finally using other organic solvent extract antihistamine
 Pka of weak neutral acid =5 i.e at pH 5 equal amount of
acid are present in the anion and neutral form
 At pH 6 about 10% remains in neutral form
 At pH 7 1% in neutral form and at pH8 0.1 % present as the acid.
50

Control of hydrophobicity
 The partition coefficient of an ion can be altered by
making the ion more hydrophobic.
 In two phase system (aq and org phase) ions are
normally expected to partition almost completely to
aq. Phase.
 If hydrophobic counter ions are added, so ion pair
may be hydrophobic enough to partition into org.
phase.
 Another way is use of macrocyclic sequestering
agent such as crown ether.
51

 Ex benzene water system, KMnO4 partitions
completely into aq layer and benzene layer is
colorless.
 In presence of crown ether, benzene layer
becomes colored indicating that some of
KMnO4 have distributed in benzene layer.
 This is because permanganate ion is
complexed within the cavity of the crown
ether, which thus masks the high polarity of
ion, presenting hydrophobic external surface.
52

 Emulsion : It may be defined as-a dispersed system
containing at least two immiscible liquid phases’.
 The effective and meaningful extraction of an analyte is
rendered almost impossible when one encounters an
emulsion formation during an extraction process thereby
the separation of the two phases becomes difficult.
Actually, it offers a frequent and serious problem when
dealing with the extraction of drugs from biological as well
as pharmaceutical formulations.
Emulsion formation problem
53

Emulsion formation problem
 Emulsion formation enhances the area of the interface
between the two immiscible solvents and as a result also
enhances the ‘free energy’ of the system, which may be
designated by the following expression :
 Free energy = γ × ΔA
 where γ = Interfacial tension, and A = change in surface
area resulting from emulsification. Obviously the ‘lowest
free energy’ is given by the most stable state for a
system at constant pressure and, therefore, in due
course an emulsion shall ‘break’ spontaneously to the
two-layered system.
54

 It has been observed that once an emulsion is formed it is very
difficult to break it. Therefore, it is absolutely necessary to
adhere to the following guidelines, as far as possible, in order to
avoid forming emulsions in the course of an extraction process :
 (1) Always affect very cautious and gentle agitation besides
employing a sufficiently large liquid liquid interface to obtain a
reasonably good extraction. Especially when the two-liquid
layers have a large contact surface in an extraction process,
vigorous or thorough shaking of the two phases is not required
at all,
 (2) The removal of any finely divided insoluble material(s) in a
liquid phase must be done by filtration before carrying out the
extraction process.
Ways to minimize Emulsion formation problem
55

 (3) Always prefer and use such solvent pairs that have a large
density difference and a high interfacial tension, for instance :
water and hexane, as they are less prone to emulsion problems. In
contrast, such solvent pairs as water and benzene should not be
used in the extraction process.
 (4) When performing extraction from water always ensure not to
work at pH extremes and particularly at high pH ranges to avoid
emulsification.
 (5) In cases, of acute emulsion-problems substances like-anion
exchangers alumina or silica gel are used specifically to resolve the
problem by adsorption of the emulsifying agents. In fact, it would
be advisable to employ the technique of column chromatography
for the effective separation of the analyte as compared to an
extraction process.
Ways to minimize Emulsion formation problem
56

Applications of extraction in the drug
analysis
1. Separation of elements
2. Concentrating impurities
3. Clearings of the basic component from impurities in the
process of synthesis of substances of drugs
4. Definition of the basic component from impurities in the
process of synthesis of substances of drugs
5. For identification and quantitative definition of chemical
agent or substances-markers in the process of the analysis
of phytogenesis drugs
6. Increase of sensitivity and selectivity of reactions
57

5. Studying of formation constant of complexs
6. Studying of substance condition in a solution (a
charge, polymerisation degree)
 Separation – controlled by pH which controls
ionization and complex formation
 Clean up before analysis
 Preconcentration: Extract from a large aqueous
volume into a much smaller organic volume.
 treatment of extracts, tinctures, antibiotics,
preparations from a different biological material.
Applications of extraction in the drug
analysis
58

Difference between continuous extraction and
Multiple extraction
 Ex. Soxhlet apparatus
 Multiple extraction is one phase
system i.e.Single solvent is used
 Not suitable for thermolabile
compound
 Multiple extraction method can
separate only one constituent at a
time
 It is solid liquid extraction technique
 Technique is used when K value is
quite differ.
 Same extracting solvent is used
repeatedly for extraction
 Ex. Counter current distribution
 CCD is two phase system i.e. Two
solvents are used
 suitable for thermolabile
compound
 CCD method can separate two
constituent at a time
 It is Liquid liquid extraction
technique
 Technique is used when K values
of 2 drugs in solvents are almost
similar.
 Fresh extracting solvent is used
every time for extraction.
59

Difference between multiple extraction
and single extraction
 Single extraction (Batch)
 Used when partition coefficient of
solute between two solvents is
extremely large.
 Maximum recovery is not possible in
single extraction.
 Extraction efficiency is less as
compared to multiple extraction
 Suitable for thermolabile compound
 Total volume of extractant required
in single extraction technique is more
as compared to multiple extraction
 Ex. Separating funnel
 Multiple extraction
 Used when partition coefficient of
solute between two solvents is
almost same.
 Maximum recovery is possible in
multiple extraction
 Extraction efficiency is more as
compared to single extraction
 Not suitable for thermolabile comps
 Total volume of extractant required
in multiple extraction technique is
less as compared to single extraction
 Ex. Soxhlet apparatus
60

1. Practical Pharmaceutical Chemistry by Beckett, A H &
Stenlake, J B , 2005, 4th edition, Part I and II, CBS
Publishers and Distributors, India.
2. A Textbook of Pharmaceutical Analysis by Kenneth A
Connors, 2002, 3rd edition, John Wiley and Sons,
Canada.
4 Fundamentals of Analytical Chemistry by Douglas A.
Skoog, Donald M. West, F. James Holler, 1991, 7th
edition, Saunders College Publishing, USA.
5. Analytical Chemistry by Gary D. Christian, 6th edition,
John Wiley & Sons, Singapore.
References
61 Dr. Archana Naik, Saraswathi Vidya Bhavans College of Pharmacy

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