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Objective: 1)To practice the procedure for preparing a standard solution 2)To perform the standardization of an unknown hydrochloric acid solution 3)To determine the given sodium hydroxide solution 4)To estimate the ethanoic acid content in commercial

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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: 22nd September, 2010 Experiment 2: Acid base titration I Objective: 1) To practice the procedure for preparing a standard solution 2) To perform the standardization of an unknown hydrochloric acid solution 3) To determine the given sodium hydroxide solution 4) To estimate the ethanoic acid content in commercial vinegar solution Apparatus & equipments used: 1) Burette, 50mL capacity 2) Bulb pipette, 25mL capacity 3) Volumetric flask, 250mL capacity 4) Conical flask, 250mL capacity 5) Analytical balance Chemicals used: 1) Anhydrous sodium carbonate 2) ...read more.

Middle

HCl : Na2CO3 = 2 : 1 ? Number of mole of HCl = (0.0123)(2)(25/250) = 0.00246 Molarity of HCl = 0.00246 / (23.183 / 1000) = 0.1062 ~0.11M Titration II: Determination of the given sodium hydroxide solution Titrant (in burette): Hydrochloric acid Titrate (in Conical flask): 25.0cm3 of NaOH Indicator used: Methyl Orange Colour of indicator changed from: yellow to pink Titration No. 1(trial) 2 3 4 Final burette reading (cm3) 23.55 23.85 47.70 23.80 Initial burette reading (cm3) 0.00 0.00 23.85 0.00 Volume of titrant used (cm3) 23.55 23.85 23.85 23.8 Average volume: 23.83 cm3 Calculation: HCl(aq) + NaOH(aq) --> NaCl(aq) ...read more.

Conclusion

number of mole of CH3COOH(dil) = number of mole of NaOH = 0.00253 Molarity of CH3COOH(dil) = 0.00253 / (16.98 / 1000) = 0.1489999 Molarity of CH3COOH = (0.1489999)(250 / 25) = 1.489999 Number of mole of CH3COOH in commercial vinegar = (1.489999)(250/1000) = 0.3725 Mass of CH3COOH in commercial vinegar = (0.3725)[(2)(12.0107)+(4)(1.00794)+(2)(15.9994)] = 22.369 The percentage of CH3COOH in commercial vinegar = (22.369 / 250)(100%) = 8.948 ~ 8.95% Conclusion: We prepared standard solution 0.05M NaOH. Through the titration I, we know that the molarity of HCl was 0.11M. In the titration II, we determinate NaOH solution is 0.1012M. At the titration III, we found out the molarity of diluted commercial vinegar is ~0.15M and also calculation out the original vinegar was ~1.49M. In calculation, we found out there are 8.95% ethanoic acid in commercial vinegar. ...read more.

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