Islamic University in Madinah Department of Chemistry CHAPTER - ----- CHROMATOGRAPHIC METHODS OF SEPARATIONS Prepared By Dr. Khalid Ahmad Shadid Chemistry Department Islamic University in Madinah TRADITIONAL METHODS FOR SEPARATION AND PURIFICATION Methods of Everyday Use Crystallization Sublimation Solvent Extraction Distillation Criteria of Purity Filtration and Evaporation Centrifuging Decantation Chromatography 2
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. 3 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:. 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. 4 2
Common Extraction Solvents 5 CHOICE OF EXTRACTION SOLVENT 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 C 6 H 5 OH (Benzene Layer) NaOH HCl C 6 H 5 O Na (water layer) 6 3
DISTRIBUTION COEFFICIENT 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 ( D C ). Or Distribution coefficient ( D C ). D C = A O Molar concentration of Solute in Organic layer = A W Molar concentration of Solute inwater layer 7 DISTRIBUTION COEFFICIENT D C = A O Molar concentration of Solute in Organic layer = A Molar concentration of Solute in water layer W Molar concentration A = number of millimole of solute in ml of solvent. D C = mmoles A O /VO mmoles A W /VW = mmoles A O x V W mmoles A W x V O mmoles A O : mmoles A W : V O : V W : A millimole number in organic layer A millimole number in water layer volume of organic solvent in milliliter volume of water in milliliter 8 4
DISTRIBUTION COEFFICIENT mmoles A D O x V W C = mmoles A W x V O 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 V O. Mass Distribution ratio (Dm) is the ratio of amount of solute in organic solvent into solute amount in water layer. D m = mmoles A O mmoles A W, D C = D m V W V O 9 DISTRIBUTION COEFFICIENT D C = D m Although V o /V w increasing the ratio of extractions, but in most of extractions V o = V w, hence, D C = D m To calculate unextracted part of solute in water layer ( F ): V W V O F = mmoles A W mmoles A O + mmoles A W = D m + 0 5
DISTRIBUTION COEFFICIENT F = mmoles A W mmoles A O + mmoles A W = 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 = D m D m + n + Percent extracted % E = 00 [ - D m + n ] EXAMPLE Twenty milliliters of an aqueous solution of 0.0 M butyric acid is shaken with 0 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. x 20 = 2.0 mmoles The remaining part = 0.5 mmole (mmoles A) w, extracted should be = 2.0 0.5 =.5 mmoles = (mmoles A) o.5 x 20 Dc = = 6.0 0.5 x 0 2 6
EXAMPLE D m =? D C = D m V W V O D m = D c V o = 6.0 0 = 3.0 V w 20 To calculate percent extracted, since n = % E = 00 [ - = 00 [ - = 75 % D m + n ] 3 + ] 3 EXAMPLE Assume that you have 00 ml of water solution contain of gram of a solute. Calculate the remaining part of solute in water layer and calculate percent extracted in the following cases: ) 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 = 0) Solution: D m = D c F = V o = 0 90 = 9 V w 00 D m + n = 9 + n = 0. g, % E = 00 [ - = 00 [ - = 90 % D m + n ] 9 + ] 4 7
EXAMPLE 2) after one time extraction using 30 ml of organic solvent. D m = D c V o = 0 30 = 3 V w 00 F = D m + n = 3 + n = 0.25 g, % E = 00 [ - 3 + ] = 75 % 5 EXAMPLE 3) after three successive times of extraction using 30 ml (in each time) of organic solvent. (Dc = 0) D m = D c V o = 0 30 = 3 V w 00 F = D m + n = 3 + 3 = 0.056 g, % E = 00 [ - = 98.4 % 3 + 3 ] 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. 6 8
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 7 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 and organic phase 2. Suppose that the conjugate acid, BH +, is soluble only in aqueous phase. Let s denote its acid dissociation constant as K a. The distribution coefficient, D, is defined as distribution coefficient: D = Total concentration in phase 2 Total concentration in phase For a weak base (B) where BH + only exists in phase : D = B 2 B + BH+ Substituting K B = [B] 2 /[B] and K a = [H + ][B]/[BH + ] into above leads to Distribution of base between two phases: D = K B Ka K a + H+ D is directly related to [H + ] 8 9
ph EFFECTS A similar expression can be written for a weak acid (HA) D K K HA a [ H [ H ] ] where: K HA [ HA] [ HA] 2 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) 9 ph EFFECTS 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 +. Effect of ph on the distribution coefficient for the extraction of a base into an organic solvent. In this example, K = 3.0 and pk a for BH + is 9.00. 20 0
ph EFFECTS Suppose that the partition coefficient for an amine, B, is K = 3.0 and the acid dissociation constant of BH + is =.0 X 0-9. If 50 ml of 0.00 M aqueous amine is extracted with 00 ml of solvent, what will be the formal concentration remaining in the aqueous phase (a) at ph 0.00? (b) at ph 8.00? SOLUTION: (a) At ph 0.00, D = K Ka K a + H+ = (3.0)(.0 X 0 9 ) (.0 X 0 9 +.0 X 0 0 ) = 2.73 Using D in place of K Fraction remaining in the aqueous after extraction can be calculated from q = q = 50 50 + (2.7)(00) = 0.5 => 5% left in water V w V w + D V O The concentration of amine in the aqueous phase is 5% of 0.00 M = 0.005 M. 2 ph EFFECTS (b) At ph 8.00, D = K Ka K a + H+ = (3.0)(.0 X 0 9 ) (.0 X 0 9 +.0 X 0 8 ) = 0.273 Therefore, V q = w, q = V w + D V O 50 50 + (0.273)(00), The concentration in the aqueous phase is 0.0065 M. = 0.65 => 65% left in water At ph 0, 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. 22
EXTRACTION WITH A METAL CHELATOR Usually neutral complexes can be extracted into organic solvents. Charged complexes (e.g. MEDTA 2- ) i.e. Fe(EDTA) or Fe(l,0-phenanthroline) 3+ are not very soluble in organic solvents. Commonly used: dithizone, 8-hydroquinoline, and cupferron. Crown ethers can extract alkali metal ions and can bring them into non-polar solvents Cu +2 is completely extracted at ph 5 while Zn 2+ remains in aqueous phase + Hg +2 + 2H + dithizone 23 ph selectivity of dithizone metal ion extraction SOLID PHASE EXTRACTION Limitations of Liquid-Liquid Extraction, Extraction solvents are limited, formation of emulsions, and waste disposal problems 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. Advantages: Large concentrations possible Limited use of solvents Very effective clean-up technique Disadvantages Time consuming Precision can be difficult Expensive 24 2
SOLID PHASE EXTRACTION 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 (C 8, C 8, 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:. Condition 2. Load 3. Wash 4. Elute 25 SOLID PHASE EXTRACTION C8 tails 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. Si 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. 26 3
SOLID PHASE EXTRACTION 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. 27 SOLID PHASE EXTRACTION Sample soluble in water or organic solvent The solid-phase extraction of 0 ppb (0 ng/ml) of steroids from urine? First a syringe containing ml, of C 8 -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 0-mL urine sample is applied, nonpolar components adhere to the C 8 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. 28 HPLC of anabolic steroids preconcentrated from urine by solid phase extraction. 4
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 29 COUNTER-CURRENT DISTRIBUTION After the first transfer (n = ), the fraction of solute remaining in tube 0 is /D m +, since the two immiscible liquids forming the two phases have equal volumes. Upon equilibration, the second tube (tube ; r = ) contains: ( /D m + )(D c /D m + ) = D c /(D m + ) 2 in the bottom phase and (D c /D m + ) )(D c /D m + ) = (D 2 c /D m + ) 2 in the top phase. In the second transfer, the top phase is moved to tube 2. Tube then contains D c / (D m + ) 2 transferred from tube 0 and an equal amount remaining after the second transfer, to give a total of 2D c /(D m + ) 2 30 The distribution of solute during three extractions is shown in next slide. 5
counter-current distribution 3 CRAIG APPARATUS () 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 994 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 countercurrent distribution. 32 6
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 34are going to be absorbed and are going to stop moving with the mobile phase. 7
CHROMATOGRAPHY 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. Paper chromatography Thin-layer chromatography 35 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 R t or t R, the time between injection and detection. In interrupted development systems like TLC the retention is measured as the retention factor R f, 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 8
WHAT IS THE RETENTION FACTOR, R F? The retention factor, R f, is a quantitative indication of how far a particular compound travels in a particular solvent. The R f value is a good indicator of whether an unknown compound and a known compound are similar, if not identical. If the R f value for the unknown compound is close or the same as the R f value for the known compound then the two compounds are most likely similar or identical. 37 38 9
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43 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. 44 22
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49 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. 50 25
ADSORPTION CHROMATOGRAPHY 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. 5 52 26
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GC 59 THANK YOU 60 30