CHROMATOGRAPHIC METHODS OF SEPARATIONS · distribution of a solute between two ... Common...

Chapter-1 CHROMATOGRAPHIC METHODS, Prepared by Dr. Khalid Shadid

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CHAPTER -1-----

CHROMATOGRAPHIC METHODS OF SEPARATIONS

Prepared By

Dr. Khalid Ahmad Shadid

Chemistry Department – Islamic University in Madinah 1

Islamic University in Madinah

Department of Chemistry

TRADITIONAL METHODS FOR SEPARATION AND PURIFICATION

Methods of Everyday Use

Crystallization

Sublimation

Solvent Extraction

Distillation

Criteria of Purity

Filtration and Evaporation

Centrifuging

Decantation

Chromatography2


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LIQUID-LIQUID EXTRACTION

• The process of transferring a dissolved

substance from one liquid phase to

another (immiscible) liquid phase.

• Solvent extraction involves the

distribution of a solute between two

immiscible liquid phases. This technique

is extremely useful for very rapid and

“clean” separations of both organic and

inorganic substances.

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CHOICE OF EXTRACTION SOLVENT

Although water is almost always one of the liquids in the liquid-liquid extraction

process, the choice of organic solvent is quite wide. A good extraction solvent

needs five essential features:

1. has high solubility for the organic compound.

2. be immiscible with the other solvent (usually water).

3. has a relatively low boiling point so as to be easily removed from the

compound after extraction.

4. extract little or none of the impurities and other compounds present in the

mixture.

5. be nontoxic, nonreactive, readily available, and inexpensive.

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Common Extraction Solvents

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• In some cases extraction can be controlled through pH of aqueous solution.

• Incase of extraction of phenol by using benzene solvent. Phenol can be

extracted only through acidic or neutral aqueous solvent.

• Phenol prefers benzene layer, in other hand, in a strong aqueous base phenol

will dissociate into anions and hence will return to water layer, so its

impossible to extract phenol in an aqueous base

C6H5OH C6H5O NaNaOH

HCl(water layer)(Benzene Layer)

CHOICE OF EXTRACTION SOLVENT

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• The partition of a solute between two immiscible phases is an equilibrium process

that is governed by the distribution law. If the solute species is allowed to distribute

itself between water and an organic phase to reach an equilibrium state. The ratio

concentrations of solute of each organic and water layer is constant and called

Concentration Distribution Ratio ( DC ). Or Distribution coefficient ( DC ).

• DC = 𝑨 𝑶

𝑨 𝑾

= 𝑴𝒐𝒍𝒂𝒓 𝒄𝒐𝒏𝒄𝒆𝒏𝒕𝒓𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝑺𝒐𝒍𝒖𝒕𝒆 𝒊𝒏 𝑶𝒓𝒈𝒂𝒏𝒊𝒄 𝒍𝒂𝒚𝒆𝒓

𝑴𝒐𝒍𝒂𝒓 𝒄𝒐𝒏𝒄𝒆𝒏𝒕𝒓𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝑺𝒐𝒍𝒖𝒕𝒆 𝒊𝒏 𝒘𝒂𝒕𝒆𝒓 𝒍𝒂𝒚𝒆𝒓

DISTRIBUTION COEFFICIENT

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• DC = 𝑨 𝑶

𝑨 𝑾

= 𝑴𝒐𝒍𝒂𝒓 𝒄𝒐𝒏𝒄𝒆𝒏𝒕𝒓𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝑺𝒐𝒍𝒖𝒕𝒆 𝒊𝒏 𝑶𝒓𝒈𝒂𝒏𝒊𝒄 𝒍𝒂𝒚𝒆𝒓

𝑴𝒐𝒍𝒂𝒓 𝒄𝒐𝒏𝒄𝒆𝒏𝒕𝒓𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝑺𝒐𝒍𝒖𝒕𝒆 𝒊𝒏 𝒘𝒂𝒕𝒆𝒓 𝒍𝒂𝒚𝒆𝒓

• Molar concentration A = number of millimole of solute in 1 mL of solvent.

DC = 𝒎𝒎𝒐𝒍𝒆𝒔 𝑨

𝑶/𝑽𝑶

𝒎𝒎𝒐𝒍𝒆𝒔 𝑨𝑾/𝑽𝑾

=𝒎𝒎𝒐𝒍𝒆𝒔 𝑨

𝑶𝒙 𝑽

𝑾

𝒎𝒎𝒐𝒍𝒆𝒔 𝑨𝑾𝒙 𝑽

𝑶

𝒎𝒎𝒐𝒍𝒆𝒔 𝑨 𝑶: A millimole number in organic layer

𝒎𝒎𝒐𝒍𝒆𝒔 𝑨 𝑾: A millimole number in water layer

𝑽𝑶 : volume of organic solvent in milliliter

𝑽𝑾 : volume of water in milliliter


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DC = 𝒎𝒎𝒐𝒍𝒆𝒔 𝑨 𝑶𝒙 𝑽𝑾

𝒎𝒎𝒐𝒍𝒆𝒔 𝑨 𝑾𝒙 𝑽𝑶

• If Dc increase it will afford into better or complete extraction

• Dc is constant. Hence the extraction amount of solute will increase by increasing volume of

organic solvent 𝑽𝑶.

• Mass Distribution ratio (Dm) is the ratio of amount of solute in organic solvent into solute

amount in water layer.

∵ Dm = 𝒎𝒎𝒐𝒍𝒆𝒔 𝑨 𝑶

𝒎𝒎𝒐𝒍𝒆𝒔 𝑨 𝑾

, ∴ DC = Dm

𝑽𝑾

𝑽𝑶


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DC = Dm

𝑽𝑾

𝑽𝑶

Although Vo/Vw increasing the ratio of extractions, but in most of

extractions Vo = Vw , hence, DC = Dm

To calculate unextracted part of solute in water layer ( F ):

F = 𝒎𝒎𝒐𝒍𝒆𝒔 𝑨

𝑾

𝒎𝒎𝒐𝒍𝒆𝒔 𝑨𝑶+ 𝒎𝒎𝒐𝒍𝒆𝒔 𝑨

𝑾

=𝟏

𝑫𝒎

+ 𝟏


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F = 𝒎𝒎𝒐𝒍𝒆𝒔 𝑨 𝑾

𝒎𝒎𝒐𝒍𝒆𝒔 𝑨 𝑶 + 𝒎𝒎𝒐𝒍𝒆𝒔 𝑨 𝑾

=𝟏

𝑫𝒎 + 𝟏

• In most of solvent extractions the extraction of solute in one time is not giving big Dc value,

hence we have to do it in multiple batch extraction.

∴ F = 𝟏

𝑫𝒎 + 𝟏 𝒏

% E = 100 [ 1-𝟏

𝑫𝒎

+ 𝟏 𝒏 ]


Percent extracted

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EXAMPLE

• Twenty milliliters of an aqueous solution of 0.10 M butyric acid is shaken with 10 mL

ether. After the layers are separated, it is determined by titration that 0.5 mmol butyric

acid remains in the aqueous layer. What is the distribution ratio, and what is the percent

extracted?

• Solution:

Calculate number of millimoles of (butyric acid A) in 20 ml of water before extraction,

therefore,

M (molar concentration) x V (in ml) mmoles A = 0.1 x 20 = 2.0 mmoles

∵ The remaining part = 0.5 mmole (mmoles A)w , ∴ extracted should be = 2.0 – 0.5 = 1.5

mmoles = (mmoles A)o

Dc = 𝟏.𝟓 𝒙 𝟐𝟎

𝟎.𝟓 𝒙 𝟏𝟎= 6.0

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• Dm = ? DC = Dm

𝑽𝑾

𝑽𝑶

Dm = Dc

𝑽𝒐

𝑽𝒘

= 6.0 𝟏𝟎

𝟐𝟎= 3.0

• To calculate percent extracted, since n = 1

%E = 100 [ 1-𝟏

𝑫𝒎

+ 𝟏 𝒏 ]

= 100 [ 1-𝟏

𝟑 + 𝟏 𝟏 ]

= 75 %

EXAMPLE 1

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• Assume that you have 100 ml of water solution contain of 1 gram of a solute.

• Calculate the remaining part of solute in water layer and calculate percent extracted in the following cases:

• 1) after one time extraction using 90 ml of organic solvent.

• 2) after one time extraction using 30 ml of organic solvent.

• 3) after three successive times of extraction using 30 ml (in each time) of organic solvent. (Dc = 10)

• Solution:

• Dm = Dc

𝑽𝒐

𝑽𝒘

= 10 𝟗𝟎

𝟏𝟎𝟎= 9

• F = 𝟏

𝑫𝒎 + 𝟏 𝒏 = 𝟏

𝟗 + 𝟏 𝒏 = 0.1 g,

EXAMPLE 11

%E = 100 [ 1-𝟏

𝑫𝒎

+ 𝟏 𝒏 ]

= 100 [ 1-𝟏

𝟗 + 𝟏 𝟏 ]

= 90 % 14


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2) after one time extraction using 30 ml of organic solvent.

Dm = Dc

𝑽𝒐

𝑽𝒘

= 10 𝟑𝟎

𝟏𝟎𝟎= 3

F = 𝟏

𝑫𝒎 + 𝟏 𝒏 = 𝟏

𝟑 + 𝟏 𝒏 = 0.25 g,

% E = 100 [ 1-𝟏

𝟑 + 𝟏 𝟏 ]

= 75 %

EXAMPLE 11

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Dm = Dc

𝑽𝒐

𝑽𝒘

= 10 𝟑𝟎

𝟏𝟎𝟎= 3

F = 𝟏

𝑫𝒎 + 𝟏 𝒏 = 𝟏

𝟑 + 𝟏 𝟑 = 0.0156 g,

% E = 100 [ 1-𝟏

𝟑 + 𝟏 𝟑 ]

= 98.4 %

3) after three successive times of extraction using 30 ml (in each time) of organic solvent. (Dc = 10)

From this example its better to use a small volume of organic

solvent in multiple times rather than using in one time a large

volume of organic solvent.

EXAMPLE 11

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pH EFFECTS

For weak acids (HA) and Bases (B)

- Protonated and non-protonated forms usually have different partition coefficients (K)

- Charged form (A- or BH+) will not be extracted

- Neutral form (HA or B) will be extracted

Partitioning is Described in Terms of the Total Amount of a Substance

- Individual concentrations of B & BH+ or HA & A- are more difficult to determine

- Partitioning is regardless of the form in both phases

- Described by the distribution coefficient (D)

Distribution coefficient: D = 𝑻𝒐𝒕𝒂𝒍 𝒄𝒐𝒏𝒄𝒆𝒏𝒕𝒓𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝒑𝒉𝒂𝒔𝒆 𝟐

𝑻𝒐𝒕𝒂𝒍 𝒄𝒐𝒏𝒄𝒆𝒏𝒕𝒓𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝒑𝒉𝒂𝒔𝒆 𝟏 17

pH EFFECTS

• If a solute is an acid or base, its charge changes as the pH is changed. Usually, a neutral species is

more soluble in an organic solvent and a charged species is more soluble in aqueous solution.

• Consider a basic amine whose neutral form, B, has partition coefficient K between aqueous phase 1

and organic phase 2.

• Suppose that the conjugate acid, BH+, is soluble only in aqueous phase 1. Let’s denote its acid

dissociation constant as Ka . The distribution coefficient, D, is defined as

distribution coefficient: D = 𝑻𝒐𝒕𝒂𝒍 𝒄𝒐𝒏𝒄𝒆𝒏𝒕𝒓𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝒑𝒉𝒂𝒔𝒆 𝟐

𝑻𝒐𝒕𝒂𝒍 𝒄𝒐𝒏𝒄𝒆𝒏𝒕𝒓𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝒑𝒉𝒂𝒔𝒆 𝟏

D = 𝑩 𝟐

𝑩 𝟏 + 𝐁𝐇⁺ 𝟏

Substituting KB = [B]2/[B]1 and Ka = [H+][B]/[BH+] into above leads to

Distribution of base between two phases: D = 𝑲𝑩𝑲𝒂

𝑲𝒂+ 𝐇⁺

For a weak base (B) where BH+ only exists in phase 1:

D is directly related to [H+]

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pH EFFECTS

A similar expression can be written for a weak acid (HA)

The ability to change the distribution ratio of a weak acid or weak base

with pH is useful in selecting conditions that will extract some

compounds but not others.

- Use low pH to extract HA but not BH+ (weak acid extractions)

- Use high pH to extract B but not A- (weak base extractions)

]H[K

]H[KD

a

HA

1

2

]HA[

]HA[KHA where:

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• The distribution coefficient D is used in place of the partition coefficient K when dealing with a species that has

more than one chemical form, such as B and BH+.

pH EFFECTS

Effect of pH on the distribution coefficient for the

extraction of a base into an organic solvent. In this

example, K = 3.0 and pka for BH+ is 9.00.

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• Suppose that the partition coefficient for an amine, B, is K = 3.0 and the acid dissociation constant

of BH+ is = 1.0 X 10-9. If 50 mL of 0.010 M aqueous amine is extracted with 100 mL of solvent, what

will be the formal concentration remaining in the aqueous phase (a) at pH 10.00? (b) at pH 8.00?

• SOLUTION:

(a) At pH 10.00, D = 𝑲 𝑲𝒂

𝑲𝐚+ 𝐇⁺=

(3.0)(1.0 X 10−9)(1.0 X 10−9 + 1.0 X 10−10 )

= 2.73 Using D in place of K

Fraction remaining in the aqueous after extraction can be calculated from q = 𝑽𝒘

𝐕𝐰+ 𝑫 𝑽𝑶

q = 𝟓𝟎

𝟓𝟎 + (𝟐.𝟕)(𝟏𝟎𝟎)= 0.15 => 15% left in water

The concentration of amine in the aqueous phase is 15% of 0.010 M = 0.0015 M.

pH EFFECTS

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• (b) At pH 8.00, D = 𝑲 𝑲𝒂

𝑲𝐚+ 𝐇⁺=

(3.0)(1.0 X 10−9)(1.0 X 10−9 + 1.0 X 10−8 )

= 0.273

Therefore,

q = 𝑽𝒘

𝐕𝐰+ 𝑫 𝑽𝑶

, q = 𝟓𝟎

𝟓𝟎 + (𝟎.𝟐𝟕𝟑)(𝟏𝟎𝟎)= 0.65 => 65% left in water

, The concentration in the aqueous phase is 0.0065 M.

At pH 10, the base is predominantly in the form B and is extracted into the organic solvent.

At pH 8, it is in the form BH+ and remains in the water.

pH EFFECTS

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EXTRACTION WITH A METAL CHELATOR

• Usually neutral complexes can be extracted into organic solvents. Charged complexes (e.g. MEDTA2-) i.e. Fe(EDTA)¯ or

Fe(l,10-phenanthroline)3+are not very soluble in organic solvents. Commonly used: dithizone, 8-hydroquinoline, and cupferron.

dithizone

+ Hg+2 + 2H+

Crown ethers can extract alkali metal ions and can bring them into non-polar solvents Cu+2 is completely extracted at pH 5

while Zn2+ remains in aqueous phase

pH selectivity of dithizone metal ion extraction

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SOLID PHASE EXTRACTION

• The principle of SPE is similar to that of liquid-liquid extraction, involving a partitioning of

solutes between two phases. However, instead of two immiscible liquid phases. SPE involves

partitioning between a liquid (sample matrix or solvent with analytes) and a solid

(sorbent) phase.

• This sample treatment technique enables the concentration and purification of analytes

from solution by sorption on a solid sorbent and purification of extract after extraction.

Analyte either adsorps or absorbs to a solid support.

•

• Limitations of Liquid-Liquid Extraction, Extraction solvents are limited, formation of emulsions, and

waste disposal problems

Advantages:

• Large concentrations possible

• Limited use of solvents

• Very effective clean-up technique

Disadvantages

• Time consuming

• Precision can be difficult

• Expensive24


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• SPE is based on selective adsorption and desorption of analyte on a bonded phases material, where matrix compounds are

washed before eluting the analyte. Selectivity is controlled by the type of stationary phase (C8, C18, OH, CN, etc).

• Sample capacity is governed by amount of stationary phase.


The general procedure is to load a solution onto the SPE solid phase, wash away

undesired components, and then wash off the desired analytes with another solvent

into a collection tube:

The Four Basic Steps of SPE

There are four basic steps to every SPE method. These steps

are:

1. Condition

2. Load

3. Wash

4. Elute25

• Conditioning

Conditioning is critical to the success of your SPE method. Conditioning is a

two step process. For the material in the cartridge to be active and effective,

it must be exposed to common solvents. The SPE material must first be

conditioned with an organic solvent. Next, equilibrate with an aqueous

solution.

• Loading

For loading there are two important basics to remember:

• Be sure to understand the type of matrix associated with your compounds

of interest

• Make sure you know the physical characteristics associated with your

analyte.

Make sure to allow the sample to interact with the SPE sorbent as long as

possible by employing proper loading flowrates.

Si

C18 tails


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• Washing

This step refers to using a solvent or solution to wash interferences from the SPE

material without removing the analytes of interest.

• Elution

Elution refers to the solvent required to elute the compounds of interest from

the SPE material. It is not uncommon to employ the use of stronger or larger

volumes of solvent.


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• The solid-phase extraction of 10 ppb (10 ng/mL) of

steroids from urine?

• First a syringe containing 1 mL, of C18-silica is conditioned with 2

mL of methanol to remove adsorbed organic material.

• Then the column is washed with 2 mL of water. When the 10-mL

urine sample is applied, nonpolar components adhere to the C18

silica, and polar components pass through.

• The column is then washed with 4 mL of 25 mM borate buffer at

pH 8 to remove polar substances.

• The column is then rinses with 4 mL of 40% methanol/60%

water and 4 mL 20% acetonitrile/80% water to remove less

polar substances. Finally, elution with two 0.5-mL aliquots 80%

methanol/20% water washes the steroids from the column.HPLC of anabolic steroids preconcentrated

from urine by solid phase extraction.


Sample soluble in water or

organic solvent

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COUNTER CURRENT EXTRACTION

•Craig method: used to separate two species by solvent extraction, but their Distribution

coefficient are not sufficiently different. Then carry out a series of extractions

Counter current extraction

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After the first transfer (n = 1), the fraction of solute remaining in tube 0 is

1/Dm + 1, since the two immiscible liquids forming the two phases have equal

volumes. Upon equilibration, the second tube (tube 1; r = 1) contains:

(1 /Dm + 1)(Dc /Dm + 1) = Dc /(Dm + 1)2 in the bottom phase and

(Dc /Dm + 1) )(Dc /Dm + 1) = (D2c /Dm + 1)2 in the top phase.

In the second transfer, the top phase is moved to tube 2. Tube 1 then contains

Dc / (Dm + 1)2 transferred from tube 0 and an equal amount remaining after

the second transfer, to give a total of 2Dc /(Dm + 1)2

The distribution of solute during three extractions is shown in next slide.

COUNTER-CURRENT DISTRIBUTION

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http://www.chem.uoa.gr/Applets/AppletCraig/Appl_Craig2.html

http://www.chem.uoa.gr/Applets/AppletCraig/Appl_Craig2.html


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counter-current distribution

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CRAIG APPARATUS

• (1) Counter-current extraction are useful in that they improve both the recovery and

purification yield of A. However, the technique is time-consuming and tedious to perform.

(2) To overcome these difficulties L. C. Craig developed a device in 1994 to automate this

method. Known as the Craig Apparatus, this device uses a series of “separatory funnels” to

perform a counter-current extraction.

• The patern formed by the movement of a solute through the system is known as a counter-

current distribution.

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COUNTER-CURRENT DISTRIBUTION

• (4) The result of this process is that solutes partition between the phases in each

tube, but eventually all travel to the right and off of the apparatus, where they are

collected.

(5) Since this system involves both rate and phase separation processes (i.e.,

distribution of solutions between two phases affecting their rate of travel through

the system), The Craig countercurrent distribution is often as a simple model to

describe chromatography. In fact, anther term often used for countercurrent

distribution is countercurrent chromatography (CCC).

• We can get the distribution of solute among Craig tubes (chromatographic

column)33

CHROMATOGRAPHY

• Chromatography: is a separation method that exploits the differences in partitioning

behavior between a mobile phase and a stationary phase to separate the components in a

mixture. Components of a mixture may be interacting with the stationary phase based on

charge, relative solubility or adsorption.

• In all chromatography there is a mobile phase and a stationary phase. The stationary phase

is the phase that doesn't move and the mobile phase is the phase that does move. The mobile

phase moves through the stationary phase picking up the compounds to be tested. As the

mobile phase continues to travel through the stationary phase it takes the compounds with it.

At different points in the stationary phase the different components of the compound are

going to be absorbed and are going to stop moving with the mobile phase. 34


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• In Paper and Thin-layer chromatography the mobile phase is the solvent.

• The stationary phase in Paper chromatography is the strip or piece of paper that is placed

in the solvent.

• In Thin-layer chromatography the stationary phase is the thin-layer cell. Both these kinds

of chromatography use capillary action to move the solvent through the stationary phase.

CHROMATOGRAPHY

Paper chromatography Thin-layer chromatography

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RETENTION

• The retention is a measure of the speed at which a substance moves in a chromatographic

system.

• In continuous development systems like HPLC or GC, where the compounds are eluted with

the Eluent, the retention is usually measured as the retention time Rt or tR, the time

between injection and detection.

• In interrupted development systems like TLC the retention is measured as the retention

factor Rf, the run length of the compound divided by the run length of the eluent front:

It is important to compare the retention of the test compound to that of one or more

standard compounds under absolutely identical conditions. 36


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WHAT IS THE RETENTION FACTOR, RF ?

• The retention factor, Rf, is a quantitative indication of how far a particular compound travels in

a particular solvent.

• The Rf value is a good indicator of whether an unknown compound and a known compound

are similar, if not identical. If the Rf value for the unknown compound is close or the same as

the Rf value for the known compound then the two compounds are most likely similar or

identical.

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PARTITION CHROMATOGRAPHY

• Molecular species separate because they differ in their distribution between Mobile and

stationary phases.

• The stationary phase is the sorbent. If the sorbent is a liquid held stationary by a solid, the

solid is called the support or matrix.

• The mobile phases is the solvent or developer and the components in the mixture to be

separated constitute then called solute.

• If two phases are in contact with one another and if

one or both phases contain a solute, the solute will

distribute itself between the two phases.

This is called partitioning and is described by the

Partition coefficient (K), the ratio of the

concentrations of the solute in the two phases.

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ADSORPTION CHROMATOGRAPHY

• Adsorption chromatography is one of the oldest types of chromatography around. It utilizes a

mobile liquid or gaseous phase that is adsorbed onto the surface of a stationary solid phase.

• The equilibration between the mobile and stationary phase accounts for the separation of

different solutes.

• Consider a solid surface containing a wide varity of binding sites – for example, regions that

are electron – rich (negatively charged), electron – poor (positively charged), nonpolar and so

forth, and a liquid containing solute in contact with the surface.

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• If binding is reversible, the number of molecules bound to the surface will depend on the solute

concentration. This dependency is shown in figure. Curves of this sort are called adsorption

isotherms. The most common is the convex curve – that is, binding sites with high affinity are

filled first so that additional amounts of solute are bound less tightly. The binding isotherm is a

characteristic of a particular molecule and sorbent.

The rate at which the substance moves is related to

the strength of binding – that is, the tighter the binding,

the slower the movement. Clearly then, molecules can

be separated if they have different adsorption

isotherms.

ADSORPTION CHROMATOGRAPHY

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HPLC

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GC

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THANK YOU

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CHROMATOGRAPHIC METHODS OF SEPARATIONS · distribution of a solute between two ... Common...

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