Titration equivlance point

Question

Calculate the pH at the equivalence point for the titration of 4.938 M HA (a weak acid with pKa = 3.55) with 4.938 M NaOH. 

Answer 1

When a weak acid is titrated with a strong base, as in this case, the pH at the equivalence point will be >7 because the weak acid is converted to a "stronger" conjugate base which hydrolyzes.

HA + OH^- ==> H₂O + A^-

A- hydrolyzes as follows: A- + H₂O ==> HA + OH-

Here A- is acting as a base and we can find the Kb for this base from the pKa of the weak acid.

All of HA is converted to A- and volume doubles because [NaOH] = [HA]. Final [A-] = 4.938/2 = 2.469 M

pKa + pKb = 14

pKb = 14 - 3.55 = 10.45

Kb = 1x10^-10.45 = 3.55x10^-11 = Kb for A-

Kb = [HA][OH-]/[A-]

3.55x10^-11 = (x)(x)/2.469

x² = 8.76x10^-11

x = 2.96x10^-6 M = [OH-]

pOH = -log 2.96x10^-6 = 5.53

pH = 14 - pOH

pH = 8.47 (> 7 as predicted)