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Lab.6. Extraction

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Lab.6. Extraction Lab.4. Extraction Key words: Solubility, immiscible solvent, partition solutes in two phase system, distribution constant, partition neutral compound, partition acids and basis, factors (pH, temperature), partition in octanol/water system, QSAR, QSSR. Literature: D.A. Skoog, F.J. Holler, T.A. Nieman: Principles of Instrumental Analysis; Part I, pp. 17 – 224, Part VI pp. 875 – 996, Part VII, pp. 1021 -1041, http:// chemistry.brookscole.com/skoogfac/. J. Crowe. T. Bradshaw, P. Monk, Chemistry for the Biosciences. The essential concepts., Oxford University Press, 2006; Chapter 17, pp. 515 - 549. Medicinal Chemistry Theoretical background Liquid – liquid extraction is a useful method to separate components (compounds) of a mixture. The success of this method depends on the difference in solubility of a compound in two solvents, which form extraction system. In the practical use, usually one phase is water (or water – based aqueous solution) and the other is an organic solvent which is immiscible with water. It is known that water does not mix with many solvents, eg. chloroform, diethyl ether, benzene, diethyl ketone et c. You well know that water does not mix with gasoline. If one try to mix eg. chloroform and water, they’ll obtain two phase system. water Upper phase Bottom phase chloroform Fig.1. Example of two phases system Such two phases system – composed of two immiscible solvents – is used during extraction process. If the solute A is present (dissolved) in e.g. water phase, it can be distributed between water and an organic solvent. Some amounts of A will remain in the aqueous phase and some will have been transferred to the organic phase. Liquid – liquid extraction is based on the transfer of a solute substance from one liquid phase e.g. water into another liquid phase (e.g. diethyl ketone), according to the solubility. 2 The ratio of molar concentrations for sample A in the two phases (organic and aqueous) will be constant and independent of the total quantity of A, that is, at any temperature: Aorg P = (1) Aaq where [A]org and [A]aq are molar concentrations of A in organic and aqueous phase respectively. The equilibrium constant P is known as the distribution constant or partition coefficient. The equation (1) applies for solutes which don’t dissociate or associate. These two processes strongly influence on the molecule amount (concentration) in both phases. Ionisable solutes Solutes such as weak acids or bases dissociate in water depending on the pH of the solution. The solute total concentration in water phase will be equal sum of the concentration of non- dissociated and dissociated form whereas the concentration in organic phase is equal only concentration of non-dissociated form. So in the equilibrium the equation (1) has the following form: A D = org For acids c − (2) Aaq+[A ]aq Borg For bases Dc = + (3) Baq + B aq where Dc is the extraction coefficient; [A]org, [B]org are molar concentration of acid (A) and base (B) in organic phase, [A]aq [B]aq are molar concentration of acid (A) and base (B) in - + water phase and [A ]aq and [B ]aq are molar concentration of ionized acid or base in water phase. The mathematic formulas connect the distribution constant (P), extraction coefficient (Dc) and acid or bases dissociation constant (KA or KB) and they are presented below: For acids P D = acid c K (4) 1+ A H + 3 where [H+] is the molar concentration of hydrogen ions, for the remaining abbreviations see lines above. So according to this formula the nominative is close to 1.0 when the concentration of hydrogen ion is high (e.g. pH 0-4) thus in such solution the acid dissociation is small and DC is equal P of acid. For bases: P D = base c K (5) 1+ B OH − where [OH-] is the molar concentration of hydroxyl ions, for the remaining abbreviations see lines above. So according to this formula the nominative is close to 1.0 when the concentration of hydroxyl ion is high (e.g. pH 10-14) thus in such solution the base dissociation is small and DC is equal P of base. Example The distribution constant K of Vitamin A (non-dissociation solute), between an organic solvent and water is 4. Find the concentration of the Vitamin A remaining in the aqueous phase after extraction of 50.0 mL of 1.0·10-3 M L-1 Vitamin A, with 50 mL of organic solvent. The situation before (I) and after (II) extraction is presented in Fig. 2. I II organic solvent no Vit. A X moles of Vit. A water 1.0·10-3 M L-1 Vit. A 1.0·10-3 M L-1– X moles of Vit. A Fig. 2. Extraction of Vitamin A. System I- before extraction process, system II in equilibrium. We don’t know how many moles will be transferred to the organic solvent, therefore assume that it was X moles. In such situation, in water remain 1.0·10-3 - X moles. Substitution of these quantities into Eq. 1 gives (note that K = 4): 4 X K = = 4 (6) 1.0 10-3 - X 4 (1.0·10-3 - X) = X 4.0·10-3 – 4X = X 4.0·10-3 = X + 4X 4.0·10-3 = 5X X = 8.0·10-4 Substitution of this result into Eq. 2 gives: 0.0008 0.0008 4 = org = org 0.001 − 0.0008aq 0.0002aq Conclusion: 4 times more Vit. A was transferred to the organic phase than remained in water phase. EXPERIMENTAL Procedure: Step 1 From the containers with stock solutions of acetic acid (concentrations 1.00; 0.75; 0.50 and 0.25 M L-1) pour 25 mL of solution to each of the 4 special Erlenmeyer flasks with plugs. Step 2 Add 25 mL of ether to each flask (use the graduated cylinder) stem the flask with a plug and shake (all operations make in the fume-hood). Leave flasks for few minutes until the two phase system is achieved. Step 3 From the containers with diluted acetic acid (concentrations 1.00; 0.75; 0.50 and 0.25 M L-1) pour of solution to separate Erlenmeyer flasks 5 mL (use pipette) to each flask. Step 4 Titrate the acid solution with sodium hydroxide solution (0.1 M L-1) using solution of phenolphthalein as the indicator. The pink color of the solution determines the end of the titration. 5 Repeat this step twice. The results report in the Table.1. Table. 1. The result of the acetic acid titration with NaOH solution. Acetic acid Volume of Volume of Mean volume of concentration NaOH solution NaOH solution NaOH solution [M L-1] used in titration used in titration [mL] 1 2 (V1) [mL] [mL] 1.0 0.75 0.50 0.25 Step 5 Transfer 5 mL of solution (use pipette) from the bottom layer (water layer) of the special Erlenmayer flask, after extraction process, into titration flask. Remember that the solution contain ether so the end of the pipette should be covered with your finger when you dip the pipette. Additionally all procedure should be made in fume-hood. Step 6 Repeat the procedure from step 4 using solutions after extraction. The results report in the Table 2. Table. 2. The results of the titration of solution after extraction process. Acetic acid Volume of Volume of Mean volume of concentration NaOH solution NaOH solution NaOH solution [M L-1] used in titration used in titration [mL] 1 2 (V2) {mL] {mL] 1.0 0.75 0.50 0.25 6 Step 7 Fill the table 3 with the results from Tables 1 and 2 in Table 3. Table. 3. Determination of extraction coefficient. Acetic acid Before After extraction The hypothetical concentration extraction process volume of [M L-1] process NaOH used for Vet Mean volume of Mean volume of titration of acid Dc = NaOH NaOH in ether phase V1 (V1) (V2) Vet = V1-V2 1.0 0.75 0.50 0.25 Calculate mean DC using formula below: D1 + D2 + D3 + D4 D = : DC = = .................. C 4 Step 8 Draw the graph: Vet = f(V2). 7 Graph 1. Volume of acid in organic (ether) phase vs. volume of acid in water phase at equilibrium. Answer the questions: Does the concentration of acid in water phase influences on the concentration of acid in organic phase? Why we can use volume of base used in titration processes instead the acid concentration? Step 9 Draw the graph: DC = f(V2). 8 Graph 2. Extraction coefficient vs. volume of acid in water phase at equilibrium .
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