Question

The titration of 0.02500 L of a diprotic acid solution with 0.1000 M NaOH requires 34.72 mL of titrant to reach the second equivalence point. The pH is 3.95 at the first equivalence point and 9.27 at the second equivalence point.

What is the pKa1 and pKa2 of the acid?

Thank You

Answer 1

Lets consider dissociation of a diprotic acid H₂X

H₂X <---------> HX^- + H⁺

K_a1 = [HX^- ][H⁺] / [H₂X]

At half-equivalence point half of H₂X is neutralised,

Hence [H₂X] = [HX^- ]

So K_a1 = [H⁺]

Or LogK_a1 = Log[H⁺]

pKa1 = pH

Similarly it can be shown pKa2 = pH₂

pH₂ = pH at second half-equivalence point

Therefore pKa of an acid is equal to the pH at half-equivalence point.

In our problem pH at first half equivalence point = 3.95

Hence pH at first half-equivalence point = 3.95 / 2 = 1.975

Hence pKa1 = 1.975

Similarly pKa2 = 9.27 / 2 = 4.635