# 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. 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:

<table border="1" id="bkmrk-trial-1-trial-2-tria" style="border-collapse: collapse; width: 100%; height: 116.4px;"><colgroup><col style="width: 14.2857%;"></col><col style="width: 14.2857%;"></col><col style="width: 14.2857%;"></col><col style="width: 14.2857%;"></col><col style="width: 14.2857%;"></col><col style="width: 14.2857%;"></col><col style="width: 14.2857%;"></col></colgroup><thead><tr style="height: 27.6px;"><td style="height: 27.6px;">  
</td><td style="height: 27.6px;">Trial 1  
</td><td style="height: 27.6px;">Trial 2  
</td><td style="height: 27.6px;">Trial 3  
</td><td style="height: 27.6px;">Trial 4  
</td><td style="height: 27.6px;">Trial 5  
</td><td style="height: 27.6px;">Trial 6  
</td></tr></thead><tbody><tr style="height: 29.6px;"><td style="height: 29.6px;">Amount of vinegar  
</td><td style="height: 29.6px;">10mL  
</td><td style="height: 29.6px;">10mL  
</td><td style="height: 29.6px;">10mL  
</td><td style="height: 29.6px;">10mL  
</td><td style="height: 29.6px;">10mL  
</td><td style="height: 29.6px;">10mL  
</td></tr><tr style="height: 29.6px;"><td style="height: 29.6px;">Initial NaOH  
</td><td style="height: 29.6px;">1mL  
</td><td style="height: 29.6px;">9.9mL  
</td><td style="height: 29.6px;">18.0mL  
</td><td style="height: 29.6px;">26.2mL  
</td><td style="height: 29.6px;">34.4mL  
</td><td style="height: 29.6px;">36.65mL  
</td></tr><tr style="height: 29.6px;"><td style="height: 29.6px;">Final NaOH  
</td><td style="height: 29.6px;">9.9mL  
</td><td style="height: 29.6px;">18.0mL  
</td><td style="height: 29.6px;">26.2mL  
</td><td style="height: 29.6px;">34.4mL  
</td><td style="height: 29.6px;">42.65mL  
</td><td style="height: 29.6px;">44.6mL  
</td></tr></tbody></table>

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<sub>3</sub>COOH+NaOH-&gt;CH<sub>3</sub>COONa+H<sub>2</sub>O

Trial 1:

9.90mL-1.00mL=<span style="text-decoration: underline;">8.90mL NaOH used</span>

1.00M\*8.90mL/1000=0.0089 mol NaOH

0.0089 mol/10mL/1000=<span style="text-decoration: underline;">0.890M Acetic Acid</span>

Trial 2:

18.0mL-9.9mL=<span style="text-decoration: underline;">8.1mL</span>

(8.1/1000)/(10mL/1000)=<span style="text-decoration: underline;">0.81M</span>

Trial 3:

26.2mL-18.0mL=<span style="text-decoration: underline;">8.2mL</span>

(8.2mL/1000)/(10mL/1000)=<span style="text-decoration: underline;">0.82M</span>

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Trial 4:

34.4mL-26.2mL=<span style="text-decoration: underline;">8.2mL</span>

(8.2/1000)/(10mL/1000)=<span style="text-decoration: underline;">0.82M</span>

Trial 5:

42.65mL-34.4mL=<span style="text-decoration: underline;">8.25M</span>

(8.25mL/1000)/(10mL/1000)=<span style="text-decoration: underline;">0.825M</span>

Trial 6:

44.6mL-36.65mL=<span style="text-decoration: underline;">7.95mL</span>

(7.95mL/1000)/(10mL/1000)=<span style="text-decoration: underline;">0.795M</span>

Average Molarity:

(0.89M+0.81M+0.82M+0.82M+0.825M+0.795M)6=<span style="text-decoration: underline;">0.842M</span>

% Deviation:

(0.825-0.810)/0.842\*100=<span style="text-decoration: underline;">1.78%</span>

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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.