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An unknown triprotic acid (H3A) is titrated with NaOH. After the titration Ka1 is determined to be 0.0013 and Ka2 is determined to be 6.5e-07. The pH at the second equivalence point of the titration is measured as 8.181. What is Ka3 for this triprotic acid

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manuelacastro87

Answer:

6.68 X 10^-11

Explanation:

From the second Ka, you can calculate pKa = -log (Ka2) = 6.187

The pH at the second equivalence point (8.181) will be the average of pKa2  and pKa3. So,

8.181 = (6.187 + pKa3) / 2

Solving gives pKa3 = 10.175, and Ka3 = 10^-pKa3 = 6.68 X 10^-11

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