8. Vinegar Titration Lab

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11/12/24 
 8. Vinegar Titration Lab 
 Purpose: 
 Determine the concentration of acetic acid in vinegar. 
 Procedure: 
 
 
 
 Using a clean funnel, fill a buret with NaOH solution, which was determined to have a molarity of 1.00M. Record the initial volume of NaOH in the buret. 
 Measure out 10mL of vinegar using a graduated cylinder. Transfer the vinegar into an Erlenmeyer flask. Add 2 drops of phenolphthalein to the flask. 
 Slowly add the 1.00M NaOH to the vinegar in 1-2mL increments. As the endpoint nears, the color change of the indicator slows. If you feel that you need less than a drop of NaOH, wash the tip of the buret and walls of the flask with a very small amount of distilled water. Continue to add NaOH until the endpoint is reached. The endpoint should be a very faint pink color. Record the final volume of NaOH in the buret. 
 Refill the buret if necessary and repeat the titration process at least 2 more times. 
 Calculate the molarity of the NaOH solutions for each trial. 
 Calculate the percent deviation between the trials. (high-low)/avg*100) 
 If the % deviation is greater than 2%, repeat the titration process.

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Data: 
 
 
 
 
 Trial 1 
 Trial 2 
 Trial 3 
 Trial 4 
 Trial 5 
 Trial 6 
 
 
 
 
 Amount of vinegar 
 10mL 
 10mL 
 10mL 
 10mL 
 10mL 
 10mL 
 
 
 Initial NaOH 
 1mL 
 9.9mL 
 18.0mL 
 26.2mL 
 34.4mL 
 36.65mL 
 
 
 Final NaOH 
 9.9mL 
 18.0mL 
 26.2mL 
 34.4mL 
 42.65mL 
 44.6mL 
 
 
 
 Observations: 
 First trial turned dark pink, too much NaOH. Accidentally let out a stream of NaOH near the endpoint. Trial 2 was a light pink, better than trial 1. Trial 3 was terrible. I accidentally added a stream of NaOH. Trial 4 was clear. half a drop turned it pink. Trial 5 was also a drop over from being clear and pink. Trial 6, you can barely tell that it is pink. If you look really close, you can see a very faint pink. It is really close to perfect. With some distilled water. 
   
 Analysis:

 CH 3 COOH+NaOH->CH 3 COONa+H 2 O 
 Trial 1: 
 9.90mL-1.00mL= 8.90mL NaOH used 
 1.00M*8.90mL/1000=0.0089 mol NaOH 
 0.0089 mol/10mL/1000= 0.890M Acetic Acid 
 Trial 2: 
 18.0mL-9.9mL= 8.1mL 
 (8.1/1000)/(10mL/1000)= 0.81M 
 Trial 3: 
 26.2mL-18.0mL= 8.2mL 
 (8.2mL/1000)/(10mL/1000)= 0.82M

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Trial 4: 
 34.4mL-26.2mL= 8.2mL 
 (8.2/1000)/(10mL/1000)= 0.82M 
 Trial 5: 
 42.65mL-34.4mL= 8.25M 
 (8.25mL/1000)/(10mL/1000)= 0.825M 
 Trial 6: 
 44.6mL-36.65mL= 7.95mL 
 (7.95mL/1000)/(10mL/1000)= 0.795M 
 Average Molarity: 
 (0.89M+0.81M+0.82M+0.82M+0.825M+0.795M)6= 0.842M 
 % Deviation: 
 (0.825-0.810)/0.842*100= 1.78%

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Conclusion: 
 The purpose of the lab was to determine the concentration of acetic acid in vinegar. I achieved this by combining NaOH with vinegar until the indicator turned pink. The average concentration across the trials was determined to be 0.842M. The percent deviation was 1.78%, taking the 3 closest trials. This shows that I could be more consistent between trials. It shows that I am not great, but not terrible at doing titrations. Some possible errors include the calculations for trial 1(in notebook), improperly read measurements, and contamination during trial 6. For those reasons, I would take trial 1 and 6 with a grain of salt, as they are most definitely inaccurate. If I were to do this lab again, I would be more patient. I would spend more tme taking measurements and mixing the reactants.