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Objective: 1)To standardize the given unknown sodium hydroxide solution 2)To analyze the commercial aspirin tablet by back titration method

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Introduction

Course Code: 13554/Y1 Student Name: Chow Man Chung Lam Kwok Kei Lee Long Sing Chan Ka Chun Date performed: 28th September, 2010 Experiment 3: Acid base titration II Objective: 1) To standardize the given unknown sodium hydroxide solution 2) To analyze the commercial aspirin tablet by "back titration" method Apparatus & equipments used: 1) Burette, 50mL capacity 2) Bulb pipette, 25mL capacity 3) Volumetric flask, 250mL capacity 4) Mortar and pestle Chemicals used: 1) Standard Potassium Hydrogen Phthalate Solution, 0.1M 2) Sodium Hydroxide Solution, 1M 3) Commercial Aspirin Tablet 4) Phenol Red 5) Phenolphthalein Indicator Solution Results and Data Treatment: Weighting Data: Mass of vial: 6.0134g Mass of vial and Aspirin: 7.5142g Mass of ...read more.

Middle

0.00 0.00 25.00 0.00 Volume of titrant used (cm3) 35.20 24.85 24.80 24.90 Average volume: 24.85 cm3 Calculation: HK(C8H4O4)(aq) + NaOH(aq) --> NaK(C8H4O4)(aq) + H2O(l) Number of mole of HK(C8H4O4) = (0.1005)(24.85/1000) = 0.002497 ? HK(C8H4O4) : NaOH = 1 : 1 ? Number of mole of NaOH = Number of mole of HK(C8H4O4) = 0.002497 Molarity of NaOH(dil) = (0.002497) / (25/1000) = 0.09988M ~ 0.10M Molarity of NaOH = (0.10)(250/25) = 1.00M Titration II: Determination of the commercial Aspirin tablet Titrant (in burette): Potassium Hydrogen Phthalate Titrate (in Conical flask): 25.0cm3 of Aspirin and Sodium Hydroxide Mixture Solution Indicator used: Phenol Red Colour of indicator changed from: Purple to Yellow Titration No. 1(trial) ...read more.

Conclusion

= (0.000999975)(250/25) = 0.00999975 Number of mole of NaOH(original) = (1)(25/1000) = 0.025 Number of mole of NaOH(reaction) = 0.025-0.00999975 = 0.015 ? CH3COOC6H4COOH : NaOH = 1 : 2 ? Number of mole of CH3COOC6H4COOH = (Number of mole of NaOH) / 2 = 0.015 / 2 = 0.0075 Mass of CH3COOC6H4COOH = [(9)(12.0107)+(8)(1.00794)+(4)(15.9994)](0.0075) = (180.15742)(0.0075) = 1.3512 Percentage of CH3COOC6H4COOH in a Aspirin tablet = (1.3512 / 1.5008)(100%) = 90.03% Conclusion: In titration I, we found out given sodium hydroxide solution was ~1.00M and is fine with the given molarity. In titration II, we use the determined sodium hydroxide solution to react the 2-acetoxybenzoic acid which contain in 1.5g Aspirin tablet powder and found out it has 90.03% of 2-acetoxybenzoic acid in a Aspirin tablet by calculation. ...read more.

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