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Determination of the partition coefficient of ethanoic acid between water and 2-methylpropan-1-ol

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Introduction

Chemistry TAS Report 1. Experiment Number : 11 2. Date : 29/01/2008 3. Title : Determination of the partition coefficient of ethanoic acid between water and 2-methylpropan-1-ol 4. Aims/Objective : To determine the partition coefficient of ethanoic acid between water and 2-methylpropan-1-ol. 5. Introduction / Theory: If a solute is added to two immiscible solvents, A and B, in contact with each other, it is distributed between the two solvents and an equilibrium is set up between the solute molecules in solvent A and B. The ratio of the concentration of the solute in the two solvents is K = concentration of solution in solvent A / concentration of solution in solvent B where K is known as the partition coefficient or distribution coefficient. 6. Relevant Equations/Chemical Reactions Involved : (1) At dynamic equilibrium, CH3COOH (alcohol) <=> CH3COOH (aq) (2) Titration CH3COOH(aq) + NaOH(aq) � CH3COONa(aq) + H2O(l) 7. Chemicals : 0.2 M ethanoic acid 120 cm3 0.1 M NaOH 100 cm3 2-Methylpropan-1-ol (density = 0.805 cm3) ...read more.

Middle

45 cm3 of aqueous ethanoic acid and 25 cm3 of 2-methylpropan-1-ol 10. Observations : The solution changed from colourless to red at the end point. 11. Data, Calculation and Results : Room temperature : 14.5oC Volume of 2-methylpropan-1-ol : 25 cm3 K = concentration of acid in aqueous layer / concentration of acid in alcohol layer Volume of 0.2 M ethanoic acid / cm3 Volume of 0.1 M NaOH titre for aqueous layer / cm3 Volume of 0.1 M NaOH titre for alcohol layer / cm3 Partition coefficient , K 25 10.70 12.15 1.14 35 11.85 13.90 1.17 45 12.75 19.50 1.53 Average value of partition coefficient K = (1.14 + 1.17 + 1.53) / 3 = 1.28 12. Conclusion : The partition coefficient of ethanoic acid between water and 2-methylpropan-1-ol was found to be 1.28. 13. Discussion : 1. Shaking is necessary in step (2) as equilibrium state can be attained more quickly. 2. The partition coefficient varies with temperature because equilibrium constant (K) ...read more.

Conclusion

This assumption is valid as the concentrations of the two solvents are low. 5. Two successive extractions by using 25cm3 portions of solvent make a better interaction between two kinds of solutions than only one extraction during shaking. A better equilibrium is then set up, resulting a higher extraction efficient. 6. (1) Solvent extraction : By partition law, the KD can be found experimentally. The amount of the solute that can be extracted using solvent extraction can be predicted, instead of using other complex method. (2) Chromatography : It is used to separate and analyse mixtures, based on the difference in solubilities of the solutes in the stationary and mobile phases. At a given temperature, the Rf value of a solute is a constant for a given solvent Rf = Distance travelled by spot / Distance travelled by solvent 7. Error analysis : There are some possible errors, (1) Wrong determination of end point and burette's reading in titration. (2) The 100 cm3 separating funnel, which contains the two solvents, are not shaken well and thus dynamic equilibrium may not be reached.. ...read more.

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