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Determining an equilibrium constant

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DATA COLLECTION: PART A: Table 1.1- Components of tube 1 (a) Mass of empty tube/ g 19.076 � 0.001 Volume of HCl (aq) added/ cm 3 5.00 � 0.01 Mass of tube after addition/ g 20.201 � 0.001 Mass of HCl (aq) added/ g 5.125 � 0.001 Table 1.2 - Components of tube 1 (b) (A1: measurements in order) Mass of empty tube/ g 15.231 � 0.001 Volume of HCl (aq) added/ cm 3 5.00 � 0.01 Mass of tube after addition/ g 20.372 � 0.001 Mass of HCl (aq) added/ g 5.141 � 0.001 Table 1.3 - Components of tube 2 (A1: measurements in order) Mass of empty tube/ g 14.604 � 0.001 Volume of HCl (aq) added/ cm 3 5.00 � 0.01 Mass of tube after addition/ g 19.749 � 0.001 Volume of ethy ethanoate added/ cm3 5.0� 0.1 Mass of tube after addition/ g 24.060 � 0.001 Mass of ethylethanoate added/ g 4.311 � 0.002 Mass of HCl (aq) added 5.145 � 0.002 Table 1.4- Components of tube 3 (A1: measurements in order) Mass of empty tube/ g 15.030 � 0.001 Volume of HCl (aq) added/ cm 3 5.00 � 0.01 Mass of tube after addition/ g 20.137 � 0.001 Volume of ethy ethanoate added/ cm3 4.0 � 0.1 Mass of tube after addition/ g 23.546 � 0.001 Volume of water added/cm3 1.0 � 0.1 Mass of tube after addition/ g 24.380 � 0.001 Mass of ethylethanoate added/ g 3.409 � 0.002 Mass of HCl (aq) ...read more.


and 1(b)) = 0.009615 dm3 � 0.0001 Total amount of acid at equilibrium = 0.02651 � 0.48% Eqm amount of ethanoic acid/ mol = (0.02651 � 0.0001) - (0.009615 � 0.0001) = 0.016895 � 1.18% Eqm amount of ethanol/ mol = 0.016895 � 1.18% Initial amount of ethylethanoate = = = 0.019995 mol � 0.011% Eqm amount of ethylethanoate = Initial amount of ethylethanoate - Eqm amount of ethanoic acid = (0.019995 � 0.00002) - (0.016895� 0.0002) = 0.031 � 6.51% Mass of pure HCl/g = amount x molar mass = (0.009615 dm3 � 0.0001) x 36.5 = 0.3506 � 1.14% Mass of water in the aqueous HCl added/g = mass of HCl (aq) - mass of HCl = (5.105 � 0.002) - (0.35056 � 0.004) = 4.75444 � 0.13% Initial amount of water/mol = = = 0.427633 � 0.10% Eqm amount of water = initial amount of water - Eqm amount of ethanoic acid = (0.427633 � 0.10%) - (0.016895 � 1.18%) = 0.410735 � 0.15% K c = [CH3COOH] [CH3OH] = (0.016895� 1.18%) (0.016895� 1.18%) = 0.224341 � 9.02% [CH3COOCH3] [H2O] (0.410735 � 0.15)(0.031 � 6.51) CONCLUSION: QUALITIVE: Many esterifications are slow at room the temperature. In the following experiment, the equilibrium reaction is as follows: CH3CH2OH(l) + CH3COOH(l) CH3COOCH2CH3(l) + H2O(l) The formation of ethylethanoate is well- suited for the determination of an equilibrium mixture. The reaction described above is slow enough at room temperature so that the order of mixing, temperature fluctuations over the 2 days of the reaction time, and the final titration with the NaOH solution have little effect on the reaction. ...read more.


In addition, the transition pH of phenolphthalein indicator is 8 - 9.6. This is above the optimal range for an acid-base reaction where such small concentrations are involved. Despite the fact that it is a weak acid - strong base reaction the end point of the reaction is does not increase infinitely when it reaches its turning point. Repeated trials should be performed to account for errors and mistakes. Errors are natural variations (e.g., small difference in drop size). Mistakes include forgetting to count a drop of base, or adding two when you think you only added one. HOAc + NaOH --> NaOAc + H2O (A4)During titration, NaOH in contact with carbon dioxide, in the presence of minimal moisture, will react with NaOH forming sodium bicarbonate as shown below: NaOH(s) Na + (l) + OH-(1) + CO2 HCO3- + Na+ NaHCO3 This results in a lower concentration of NaOH, so the observed volume value is greater than the expected volume value, hence a discrepancy in the final Ka value. To prevent air contact, NaOH solutions can be stored in isolated delivery systems, utilising soda lime tubes. One could also standardize NaOH solution using a primary standard of KHP. (A4) The volumetric pipette and burette only give correct readings at r.t.p. The factors of r.t.p were controlled except the temperature of the room was at 25�C instead of 20�C. (Improvement: difficult to determine) (A5) The presence of acidic impurities in the burette used for the addition of NaOH that could of neutralized some of the NaOH and decreased its concentration, hence leading to a greater value of titre, resulting in a discrepancy in the Kc value. ...read more.

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