February 24 - Sour Titration Lab

After learning about titrations, we had to put our knowledge to the test. We did so using by performing a titration ourselves. Our titrant, or the substance with a known concentration was NaOH, which had a molarity of 0.8. Next, we were given acetic acid and asked to find its concentration.

To recall, the steps to a titrations include making a table:

Trial	1	2	3	4
Final Reading
Initial Reading
Volume Used

We did this to find the volume of titrant used. We then take the average to compensate for potential errors. With this number, we can find the number of moles of NaOH that we used.

Remember, we pour enough titrant in so that the indicator, phenolphthalein, in this case, just turns pink.

With this number, we use simple stoichiometry to find the number of moles of acetic acid used. This requires a the balanced equation, which is: NaOH + CH₃COOH à NaCH₃COO + H₂O

With the found number of moles, we divide it by the volume of acetic acid that was originally present in the Erlenmeyer flask, which was 10mL, and we have our concentration.

Now let’s go over the same thing, utilising numbers this time.

Trial	1 (disregard)	2	3	4
Final Reading mL	14.6	25.5	36.9	46.9
Initial Reading mL	1.6	14.4	25.5	35.9
Volume Used mL	13.0	11.1	11.4	11.0

We determined trial one was done inaccurately, since it was far away from the other three. Therefore, we will disregard that in our calculations. From these results, we obtained 11.3 mL to be our average. Now time for some stoichiometry.

11.2 mL (0.8 mol NaOH/ 1000 mL) (1 / 1) = 0.00904 mol CH₃COOH (1 / 0.01L) = 0.896 [CH₃COOH]

And there you have it! We completed the titration. The actual value for acetic acids concentration is 0.85M. Therefore, our percent error is: 5.4%. Not bad, but we could have done better.

And that’s all! So, next time you’ll be using vinegar, you’ll know it was 0.8 moles of solute per litre of solution!

Posted by Andrew.