Question

Calculate the pH at the equivalence point for the following titration: 0.10 M HCOOH versus 0.10 M NaOH. (Formic Acid Ka=1.7 x 10-4)

Answer 1

HCOOH + NaOH -----------> HCOONa + H2O

HCOONa is a strong salt ----------> Na+ + HCOO-

Hence,

concentration HCOO- = 0.10 M


Kb = Kw / Ka = 1.0 x 10^-14 / 1.7 x 10^-4 = 5.6 x 10^-11

HCOO- + H2O <----> HCOOH + OH-

0.1 M                           0                 0

0.1-x                            x                 x

Kb = [HCOOH] [OH-]/[HCOO-]

5.6 x 10^-11 = x^2 / 0.10-x

x = 2.4 x 10^-6 M

Then,

[OH-] = x = 2.4 x 10^-6 M

pOH = -log [OH-] = = -log [2.4 x 10^-6] = 5.6

pH = 14 - 5.6 = 8.4