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Titration lab report - determine the percentage of acetic acid present in vinegar

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Introduction

´╗┐LAB REPORT PERCENTAGE OF ACETIC ACID IN VINEGAR Experiment #20 OBJECTIVE: Propose of this experiment is to determine the percentage of acetic acid present in vinegar. PROCEDURE: 1. Carefully clean a 50 mL buret and rinse it with water and then 2 times with 5 mL portions of NaOH solution. 2. Fill the buret and open the pinch clamp to fill the tip. 3. If the exact concentration of the base is given, record the value and move on. If not, standardize the base solution with primary standard H2C2O4.2H2O (oxalic acid): 1. Weigh approximately 0.15 g (down to 0.0002 g) of oxalic acid in a 125 mL Erlenmeyer flask. 2. Dissolve the solid in about 25 mL of distilled water. ...read more.

Middle

7. Deliver the 5 mL aliquot into a 125 mL Erlenmeyer flask. Touch off the last drop on the inside wall. 8. Fill the buret with the standardized NaOH solution 9. Record the initial buret reading. 10. Add 2 drops of phenolphthalein indicator to the vinegar solution. 11. Titrate to a faint pink end point. 12. Record final buret reading. 13. Repeat the titration at least once more until the volume of NaOH solution used in 2 different titrations agrees within 0.2 mL. Chemical equations: H2C2O4.2H2O+2NaOH?Na2C2O4.2H2O+2H2O CH3COOH+NaOH?CH3COONa+H2O DATA and CALCULATION: Standardization of NaOH solution: Titration 1 Titration 2 Titration 3 Weight of Erlenmeyer flask + solid H2C2O4.2H2O (g) 79.2380 82.9368 81.0613 Weight of empty Erlenmeyer flask (g) ...read more.

Conclusion

5.155% 5.168% 5.168% CH3COOH+NaOH→CH3COONa+H2O (2) From the equation (2): mol of CH3COOH=mol of NaOH=VNaOH (used)×MNaOH (average) Mvinegar=mol of CH3COOHVvinegar % of CH3COOH in vinegar=mass of CH3COOHmass of vinegar×100%=mol of CH3COOH×molar mass of CH3COOHdensity of vineger×Vvineger×100%=mol of CH3COOH×60.052 g/mol1 g/mL×5mL×100%=mol of CH3COOH×12.0104 mol-1×100% Titration 1: mol of CH3COOH=mol of NaOH=42.00×10-3L×0.1022molL=4.292×10-3 mol Mvinegar=4.292×10-3 mol5×10-3L=0.8584 M % of CH3COOH in vinegar=4.292×10-3 mol×12.0104 mol-1×100%=5.155% Titration 2 and 3: mol of CH3COOH=mol of NaOH=42.10×10-3L×0.1022molL=4.303×10-3 mol Mvinegar=4.303×10-3 mol5×10-3L=0.8605 M % of CH3COOH in vinegar=4.303×10-3 mol×12.0104 mol-1×100%=5.168% CONCLUSION: We did the experiment to standardize the NaOH solution in order to get the exact concentration of NaOH solution to be 0.1022 M that agrees with the given concentration of NaOH solution, 0.1 M. We did the titrations successfully in order to get the percentages of acetic acid in vinegar to be approximately to be 5.2%. The value is acceptable for the table vinegar that was given in the lab. ...read more.

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