Question

Benzoic acid is a weak monoprotic acid (Ka = 6.3 x 10-5). Calculate the pH of the solution at the equivalence point when 25.0 mL of a 0.100 M solution of benzoic acid is titrated against 0.050 M

Answer 1

The neutralization reaction can be represented as

HA + BOH ----------> BA + H2O

25x0.1 Vx0.05 0 0 initial mmoles

at equivalence point mmoles of acid = mmoles of base

25x0.1 = V x0.05

Thus volume of base used = 50 ml

At equivalence

   HA + BOH ----------> BA + H2O

25x0.1 50x0.05 0 0 initial mmoles

0 0 2.5 - mmoles at equivalent point

0 0 2.5/75 - concentration at equivalent point

Thus the pH of the solution is decided by the [salt], as the salt undergoes anionic hydrolysis and hence the solution will be basic in aqueous solution.

thus the pH of the salt of weak acid and strong base is given by

pH = 1/2[pkw + pka + log C]

= 1/2 [14 + ( 5 - log6.3) + log (2.5/75)

= 8.361