Question

What is the pH of a 0.30 M solution of sodium propionate, NaC ₃H ₅O ₂, at 25°C? (propionic acid, HC ₃H ₅O ₂, is monoprotic and has a K_a = 1.3 × 10 ^–5 at 25°C.. K _w = 1.01 × 10 ^-14 )

Answer 1

use:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.01*10^-14 at 25 oC

Kb = (1.01*10^-14)/Ka

Kb = (1.01*10^-14)/1.3*10^-5

Kb = 7.769*10^-10

C3H5O2- dissociates as

C3H5O2- + H2O -----> HC3H5O2 + OH-

0.3 0 0

0.3-x x x

Kb = [HC3H5O2][OH-]/[C3H5O2-]

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 ((7.769*10^-10)*0.3) = 1.527*10^-5

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

so, x = 1.527*10^-5 M

use:

pOH = -log [OH-]

= -log (1.527*10^-5)

= 4.8163

use:

PH = 14 - pOH

= 14 - 4.8163

= 9.1837

Answer: 9.18