Question

The Ka for Formic acid (HCO2H) is 1.8 x 10^-4. What is the pH of a 0.20 M aqueous solution of sodium formate (NaHCO2)? Show all the work.

Answer 1

use:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/1.8*10^-4

Kb = 5.556*10^-11

HCO2- dissociates as

HCO2- + H2O -----> HCO2H + OH-

0.2 0 0

0.2-x x x

Kb = [HCO2H][OH-]/[HCO2-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((5.556*10^-11)*0.2) = 3.333*10^-6

since c is much greater than x, our assumption is correct

so, x = 3.333*10^-6 M

use:

pOH = -log [OH-]

= -log (3.333*10^-6)

= 5.4771

use:

PH = 14 - pOH

= 14 - 5.4771

= 8.5229

Answer: 8.52