Question

Calculate the pH of a 0.16 M solution of NaCOOH (sodium formate). Ka for HCOOH (formic acid) is 1.8x10-4

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

HCOO- dissociates as

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

0.16 0 0

0.16-x x x

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

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)*0.16) = 2.981*10^-6

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

so, x = 2.981*10^-6 M

use:

pOH = -log [OH-]

= -log (2.981*10^-6)

= 5.5256

use:

PH = 14 - pOH

= 14 - 5.5256

= 8.4744

Answer: 8.47