Question

Part The following data was collected from the titration curve. Identify the values for pka, pka , and paz. Volume (mL) 1.95 2.45 10.08 5.27 20.15 30.23 7.42 40.3 50.38 8.39 9.75 10.54 60.45

Answer 1

Considering the values of pH and the volumes, I guess that at 20.15 mL, 40.3 mL and 60.45 mL we have the three equivalence points, where the amount of moles of base added are equivalent to 1, 2 and 3 times the number of moles of the acid, respectively.

The rest of the values are the volumes halfway to the equivalence points (they are exactly half the volume of each equivalence point). At these points, half of each proton of the acid has been titrated and the other half hasn't, which means we have a mixture with the same amoun of protonated and deprotonated acid. If we calculate the pH with the Henderson-Hasselbach equation for, for example, the pH between the start of the titration and the first equivalence point, we have:

$pH=pKa_{1}+log\left ( \frac{[H_{2}A^{-}]}{[H_{3}A]} \right )$

Where H₂A^- is the acid which has lost its first proton and H₃A is the fully protonated acid. As we said, when the volume of base added is half that of the equivalence point, half the H₃A has reacted to become H₂A^-. This means that their concentrations will be the same, so their concentration ratio [H₂A^-]/[H₃A] = 1. Since log(1) = 0, we have pH = pKa, so the pH halfway to the equivalence point is equal to the pKa.

This means that:

pKa₁ = 2.45

pKa₂ = 7.42

pKa₃ =9.75