20 mL of 0.1 M acetic acid (aq) was placed into a beaker, and 0.2 M NaOH (aq) so

Question

20 mL of 0.1 M acetic acid(aq) was placed into a beaker, and 0.2 M NaOH (aq) solution was added in 1 mL aliquots until either 25 mL of NaOH was added or the pH of the HCl and NaOH mixture was 12. I now have to derive the pKa value from the graph, but I am unsure how to.

This is the data that was used for the graph:

mL of 0.2 M NaOH Added pH Reading 0 3.2 1 3.7 2 4.1 3 4.3 4 4.5 5 4.7 6 4.8 7 5.0 8 5.2 9 5.4 10 6.0 11 10.9 12 11.9 13 12.2 Titration of 0.1M Acetic Acid 14 13 12 10 10 12 13 Amount of NaOH Added (mL)

Explanation / Answer

Answer:-

[CH3COOH] = 0.10 M

[NaOH] = 0.20

Volume of CH3COOH = 29 mL

So, the pH at which Half of CH3COOh has been titrated is its pKa.

So, to reach half equivalence point , volume of NaOH used = 1/2 x 0.1 x 20/0.2

= 5 mL

So, pH at 5 ml of NaOH will be pKa value of CH3COOh

so, pKa = 4.7

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