Group3 - Liquid-Liquid Extraction_print

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<p>Reporters: Jonathan Jamito &amp; Paula Tayo Group 3, AB2</p> <p>`</p> <p>It is a method of separating a component from a liquid solution by mixing it with a liquid solvent.</p> <p>`</p> <p>`</p> <p>The added solvent must be more volatile than the desired component. It must also specifically dissolve the desired component.</p> <p>`</p> <p>The component must have a greater tendency to dissolve in the added solvent than in the solution.</p> <p>KD= CO = SO = CH20 SH20Ratio of concentration = Ratio of solubility</p> <p>`</p> <p>`</p> <p>`</p> <p>The added solvent must be immiscible with the solution. The added solvent and the solution must be of different densities. If possible, the added solvent must not be toxic.</p> <p>`</p> <p>`</p> <p>` `</p> <p>`</p> <p>To understand the theory and to develop the skills needed in liquid-liquid extraction. To identify the characteristics of an ideal solvent in such a process. To study the different types of solvents. To determine the best possible solvent based on the outcomes of both solubility and miscibility tests. To learn how to compute for the percent recovery [%rec] for each extraction.</p> <p>Figure 1.1 The actual solubility test. L-R: Acetone+benzoic acid, Ether+benzoic acid, Hexane+benzoic acid</p> <p>Figure 1.2 The idealized solubility test. L-R: Acetone+benzoic acid, Ether+benzoic acid, Hexane+benzoic acid</p> <p>Figure 1.1 The actual miscibility test. L-R: Acetone+water, Ether+water, Hexane+water</p> <p>Figure 1.2 The idealized miscibility test. L-R: Acetone+water, Ether+water, Hexane+water</p> <p>Solvent Acetone Ether Hexane</p> <p>Water Miscible Immiscible Immiscible</p> <p>Benzoic Acid Soluble Soluble Insoluble</p> <p>` ` `</p> <p>`</p> <p>The best extracting solvent: Ether Benzoic acid in saturated solution: 0.47 g Amount of undissolved benzoic acid: 0.03 g %rec : 10.64%</p> <p>In choosing the ideal solvent for the extraction, there are parameters measured in the solubility and miscibility tests. 1. The component must have a greater tendency to dissolve in the added solvent than in the solution.`</p> <p>Measured by the solubility test[Why? To allow solute transfer].</p> <p>Greater solubility means greater rate of recovery.</p> <p>2.</p> <p>The added solvent must be immiscible with the solution.Measured by the miscibility test [Why? Layer formation].</p> <p>Figure 2.1 A diagramatic representation of a separatory funnel showing layer formation.</p> <p>3.</p> <p>The added solvent and the solution must be of different densities.Why? Layer separation The less dense ether layer forms above the aqueous layer.</p> <p>Figure 2.2 Layer separation showing ether forming above the denser aqueous layer.</p> <p>4.</p> <p>The added solvent must be more volatile than the desired component.Why? Benzoic acid is retrieved from the ether layer by evaporation.</p> <p>Figure 2.3 The ether component is separated from the benzoic acid by evaporation.</p> <p>5.</p> <p>The added solvent must specifically dissolve the desired component and nothing else from the solution.Why? To prevent the extraction of undesired and unnecessary solutes.</p> <p>6.</p> <p>If possible, the added solvent should not be toxic.Why? For reasons of good health.</p> <p>`</p> <p>The result of both miscibility and solubility tests can be deduced from the polarity of the substances:Benzoic acid is polar and its IMF forms two Hbonds.</p> <p>Hexane is non-polar and its IMF comprises of London Dispersion Forces.</p> <p>Ether is slightly polar and its IMF forms a dipoledipole bond.</p> <p>Acetone is polar and its IMF forms an H-bond.</p> <p>Water is polar is polar and its IMF forms H-bonds</p> <p>`</p> <p>The principle behind the solubility test?Like dissolves likeBenzoic acid [polar] + Acetone [polar] = soluble Benzoic acid [polar] + Ether [slightly polar] = soluble Benzoic acid [polar] + Hexane [non-polar] = insoluble Benzoic acid [polar] + Water [polar] = slightly soluble Why? Because water is too polar to completely dissolve benzoic acid.</p> <p>`</p> <p>On the miscibility testo o o</p> <p>Water [polar] + Acetone [polar] = miscible Water [polar] + Ether [slightly polar] = immiscible Water [polar] + Hexane [non-polar] = immiscible</p> <p>Solvent Acetone Ether Hexane`</p> <p>Water Miscible Immiscible Immiscible</p> <p>Benzoic Acid Soluble Soluble Insoluble</p> <p>This based on the results of the two tests, the best extraction solvent is ether.o o o o</p> <p>It can dissolve the desired solute [benzoic acid] It is immiscible in the solute [water] of the solution It is of a different density than the solute [water] of the solution It is more volatile than the desired solute [benzoic acid]</p> <p>` 1.</p> <p>So solving for the %recovery: Determine the amount of benzoic acid dissolved in the solution. 0.5g initial amount 0.03g amount recovered from filter paper = 0.47gOnly 20mL of the 100mL was used in the solution.</p> <p>2.</p> <p>Determine the amount of benzoic acid recovered from the ether layer.Amount = 0.01g</p> <p>3.</p> <p>Determine the %recovery .%recovery = Xpure x 100% = 0.01g x 100% = 10.64% benzoic acid Ximpure 0.094 g benzoic acid</p> <p>* 20mL x 0.47g = 0.094 g benzoic acid 100mL</p> <p>1.</p> <p>Suppose 20mL of the saturated benzoic acid solution was treated with 10% NaOH, which solvent [acetone, ether, hexane] could extract most of the benzoic acid from the aqueous solution?The presence of sodium hydroxide makes the benzoic acid more polar as the hydroxide ions ionize the benzoic acid, forming an organic salt. Despite the addition of sodium hydroxide, ether is still the best extracting solvent because it dissolves more sodium benzoate than hexane and unlike acetone, it is immiscible with the sodium hydroxide solution.</p> <p>2.</p> <p>What effect does partial miscibility of the two solvents have on the efficiency of extraction?Partial miscibility of the two solvents makes the extraction less efficient. This makes the two layers more difficult to separate as compared to complete immiscibility between the two solvents. Partial miscibility also implies that the relative affinity of the solute to the extracting solvent is not as great as to efficiently extract the solute from the solution.</p> <p>3.</p> <p>A substance X can be isolated from its plant source by solvent extraction. However, a minor component Y has an appreciable solubility in the solvents that may be used. Given below are the solubilities of X and Y in different solvents:Solvent T, C Solubility in 100g solvent at 28C X Ethyl methyl ketone Cyclohexane Benzene CCl4 Water 80 81 80 78 100 6 8 5 8.75 2 Y 5 2 1.8 1.25 1</p> <p>a.</p> <p>Which is the best extracting solvent?CCl4 is the best extracting solvent it dissolves only a little amount of Y relative to the X component needed. It also dissolves the greatest amount of X among the choices. It is immiscible with water, which happens to be the solvent in the solution.</p> <p>Mathematically speaking, the choice relies on the KD and the best result is the one with the highest KD. The KD values with respect to water as calculated are the following:SolventEthyl methyl ketone Cyclohexane Benzene CCl4</p> <p>KD3 4 2.5 4.375</p> <p>From this table, it is evident that CCl4 is the yields the highest ratio, ergo the best possible choice as the extracting solvent.</p> <p>b.</p> <p>Given a saturated aqueous solution of X and Y and using 100mL of solvent in (1), determine the percent recovery of X in a single extraction.</p> <p>KD = SCCl4 = 8.75g/100g = 4.375 SH20 2.00g/100g KD = CCCl4 = Xg/100ml = 4.375 CH20 (2.00g-Xg)/100ml X = 1.63g %rec = Xpure x 100% = 1.63g x 100% = 81.4% Ximpure 2.00g</p> <p>c.</p> <p>Repeat (b) using 50mL of solvent in each of the two successive extractions. Determine the percent recovery and compare this with (b).</p> <p>KD = SCCl4 = 8.75g/100g = 4.375 SH20 2.00g/100g First extraction: KD = CCCl4 = Xg/50ml = 4.375 CH20 (2.00g-Xg)/100ml X = 1.37g</p> <p>Second extraction: KD = CCCl4 = Xg/50ml = 4.375 CH20 (0.63g-Xg)/100ml X = 0.43g %rec = Xpure x 100%= (1.37+0.43)g x 100%= 90.0% Ximpure 2.00g In the double extraction, the percent recovery is comparably higher, considering the same amount of the solvent added.</p> <p>d.</p> <p>What is the percent recovery of the minor component in a single extraction using 100mL of the solvent in (a)?</p> <p>KD = SCCl4 = 1.25g/100g = 1.25 SH20 1.00g/100g Yg/100g = 1.25 KD = CCCl4 = CH20 (1.00g-Yg)/100g Y = 0.56 %rec = Ypure x 100% = 0.56g x 100% = 56% Yimpure 1.00g</p> <p>`</p> <p>1.</p> <p>2.</p> <p>3.</p> <p>4.</p> <p>That these are the characteristics of an ideal solvent for a liquidliquid extraction: It can dissolve the solute to be extracted better than the original solvent of the solution. It is immiscible with the solvent of the solution like how ether is immiscible with water. It is of a different density with the solvent of the solution in order for layering to occur. It can also be separated from the solute by evaporation like how ether is more volatile than its solute benzoic acid.</p> <p>The aforementioned criteria can be used in determining the best extracting solvent which is of utmost importance in the liquid-liquid extraction. Solubility and miscibility tests are effective means by which one can determine the best extracting solvent.</p> <p>To avoid experimental errors, ` laboratory apparatuses should be in good shape, ` Laboratory techniques should be properly executed, ` The filter papers used should be thoroughly dried and the organic layer of the ether-benzoic acid solution, completely evaporated.</p> <p>In extracting benzoic acid from an aqueous solution, ether proved to be the best extracting solvent:Benzoic acid is insoluble in hexane and Acetone is miscible in water making both solvents inefficient in extracting benzoic acid from water.`</p> <p>So the best choice for the extraction: ether. The more appropriate the solvent used, and the more extractions done, the more solute can be recovered from the solution.</p>