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The objective of this experiment is to calculate the % purity of aspirin, by first neutralizing the acetylsalicylic acid with excess sodium hydroxide, and then back titrate the excess

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Introduction

The titration of aspirin Objective : The objective of this experiment is to calculate the % purity of aspirin, by first neutralizing the acetylsalicylic acid with excess sodium hydroxide, and then back titrate the excess base with standardised hydrochloric acid. Apparatus : burette, pipette, volumetric flask, funnel. Procedure : 1) About 1.5 gm of aspirin was accurately weighed, the mass was recorded. 2) The weighed amount of aspirin was transferred into a beaker , 25 cm3 of 1 molar NaOH was then transferred into the beaker. The beaker was then gently heated for 10 minutes. 3) After heating, the solution was then transferred into a volumetric flask via a funnel. More distilled water was then added to the flask to make a final volume of 250 a cm3. ...read more.

Middle

7) The above titration is repeated 3 times, and the result is recorded. Result and Calculation: Mass of aspirin = 1.504 gm Concentration of NaOH = 1 mole/L Concentration of HCl = 0.1 mole/L Reaction after adding of NaOH is : 2H2O 1 mole of aspirin will react with 2 moles of NaOH Reaction for the titration is : NaOH + HCl ==> NaCl + H2O 1 mole of NaOH will react with 1 mole of HCl The result of the titration is tabulated as follows: Burret Reading Titration 1 Titration2 Titration Final Reading 9.2 16 25 Initioal reading 0 7.0 16 Volume used 9.2 9 9 % purity 94.55% 95.74% 95.74% Calculation : Sample calculation using result of titration 2 # moles of HCl use = Molarity * Volume in litres ...read more.

Conclusion

Of the above only 0.09 moles was left behind, so the aspirin has reacted with 0.025 - 0.009 = 0.016 moles of NaOH. 1 mole of aspirin will react with 2 moles of NaOH So the number of moles of aspirin must be 0.016/2 = 0.008 moles Molecular mass of aspirin (C9H8O4) = 180, so mass of acetylsalicylic acid = 0.008 * 180 = 1.44 g So % purity of aspirin = (1.44/1.504) * 100 = 95.74% The other titration results was calculated the same way. Average purity = 1/3 (94.55 + 95.74 + 95.74) = 95.34% Conclusion and discussion: The error in this lab is in the judgement of the end-point, in which the solution will completely becomes colourless. The result was quite consistent, the average % purity is 95.34% ...read more.

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