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Finding the Equilibrium Constant for the Ester Formation of Ethyl ethanoate

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Introduction

Finding the Equilibrium Constant for the Ester Formation of Ethyl ethanoate Michael Zuber Introduction: In this experiment 6 different mixtures of known amount of acid, alcohol and/or ester were mixed with dilute hydrochloric acid in a stoppered bottle. The mixture is then set aside for a week to reach equilibrium at room temperature. The mixture is then titrated against a solution of 1 mol.dm-3 NaOH to determine the total amount of acid present in the equilibrium. From this the amount of ethanoic acid can be found and therefore the amount of the other reagents can be found. These values can then be used to find the equilibrium constant for the reaction. The equation for the reaction is: Raw Data: Below is a table of the different volumes of the chemicals initially added in each solution: Solution HCl (cm3) Water (cm3) Ethyl ethanoate (cm3) Ethanoic acid (cm3) Ethanol (cm3) 1 5 0 5 0 0 2 5 1 4 0 0 3 5 3 2 0 0 4 5 0 4 1 0 5 5 0 0 1 4 6 5 0 0 2 3 Below is a table of the volume of NaOH needed to neutralize the different solutions at equilibrium: Solution NaOH (cm3) used to neutralise soltuion (1) NaOH (cm3) used to neutralize soltuion (2) NaOH (cm3) used to neutralise soltuion (3) 1 44.2 43.7 42.9 2 44.5 45.2 3 32.4 32.1 31.1 4 52.9 50.9 51.8 5 22.8 23 6 37.5 36.4 From these values we can find an average amount of 1 mol.dm-3 NaOH needed to neutralize each solution can be found. ...read more.

Middle

If this is taken away from the mass of the HCl(aq) than the mass of water can be found. Mass of water = 5.28 - 0.5475 = 4.73g Converting this to moles: Moles water = 4.73/18 = 0.263 moles Since water has a density of 1 g.cm-3 the amount of cm3 of water added to the initial mixture equals the mass of water added to the initial mixture. Therefore this value must be added to the mass of water and a new mole value will be found. The amount of moles of water added to each solution initially is shown in the below table. Solution Initial Moles water 1 0.263 2 0.318 3 0.429 4 0.263 5 0.263 6 0.263 To find the moles of ethyl ethanoate, a similar approach must be used. In the solution where no ethy; ethanoate was introduced initially, the amount of moles of ethyl ethanoate produced will equal to the amount of moles of ethanoic acid reacted (due to the 1 : 1 mole ratio of the reaction equation). However, for the solutions where ethyl ethanoate was put into the mixture the amount of moles of ethyl ethanoate will equal the amount of moles initially put in subtracted by the amount of moles of ethanoic acid. The mass of 5cm3 ethyl ethanoate was measured out to be 4..37g. Therefore for 5cm3 of ethyl ethanoate there is 4.37/88 = 0.0497 moles. For 4 cm3 the amount of moles will equal 4/5 x 0.0497 = 0.0397 For 2 cm3: moles = 0.0199 These values are then subtracted by the moles of ethanoic acid in their respective solutions. ...read more.

Conclusion

Labeling of Liquids: An error in this practical was that not all of the liquids were properly labeled. Students were told where each liquid was in the room but not all of the prepared biurettes were properly labeled with the liquids inside of them students could have easily gotten confused. For example, since both the distilled water and the aqueous hydrochloric acid have the same colorless appearance, a student could easily get confused between the two and mix them up therefore leading to inaccurate mixtures therefore affecting our results. Improvements: Accuracy of Apparatus: As explained above, the use of the biurettes is a fairly accurate method of measuring out the liquids. However, more accurate apparatus could be used. For example, electronic flow meters would give even more accurate results. Time Management: In order to allow students to get there own sets of results more time should have been allotted to this experiment so that each individual student could get there own results. Sharing of Data: The sharing of data is very useful to get a wider range of values. However, when there is a very limited amount of results it can be more problematic than helpful. Therefore, if the above proposal is used data could still be shared because if one set of results was wrong it would have less of an effect on the overall results and it would stick out more. Labelling of Liquids: In order to avoid confusion, all biurettes should have been properly labeled with the liquids that they contained. ...read more.

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