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:

    1. 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.
    2. Measure out 10mL of vinegar using a graduated cylinder. Transfer the vinegar into an Erlenmeyer flask. Add 2 drops of phenolphthalein to the flask.
    3. 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.
    4. Refill the buret if necessary and repeat the titration process at least 2 more times.
    5. Calculate the molarity of the NaOH solutions for each trial.
    6. Calculate the percent deviation between the trials. (high-low)/avg*100)
    7. 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:      CH3COOH+NaOH->CH3COONa+H2O

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.