In: Chemistry
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)
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