Question

Determine the pH of a 0.20 M KCHO2 (potassium formate) aqueous solution. [Ka for HCHO2 = 1.8 x 10-4]

 

Answer 1

 

find Kb for KCHO2

use:

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

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

Kb = 5.556*10^-11

CHO2- dissociates as

CHO2- + H2O -----> HCHO2 + OH-

0.2 0 0

0.2-x x x

Kb = [HCHO2][OH-]/[CHO2-]

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

So, [OH-] = 3.333*10^-6 M

use:

pOH = -log [OH-]

= -log (3.333*10^-6)

= 5.48

use:

PH = 14 - pOH

= 14 - 5.48

= 8.52

Answer: 8.52