Question

Acetic acid (CH3COOH,CH3COOH, ?a=5.62×10−5)Ka=5.62×10−5) is a weak acid, so the salt sodium acetate (CH3COONa)CH3COONa) acts as a weak base. Calculate the pH of a 0.445 M0.445 M solution of sodium acetate.

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^-5

Kb = 5.556*10^-10

CH3COO- dissociates as

CH3COO- + H2O -----> CH3COOH + OH-

0.445 0 0

0.445-x x x

Kb = [CH3COOH][OH-]/[CH3COO-]

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^-10)*0.445) = 1.572*10^-5

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

so, x = 1.572*10^-5 M

use:

pOH = -log [OH-]

= -log (1.572*10^-5)

= 4.8035

use:

PH = 14 - pOH

= 14 - 4.8035

= 9.1965

Answer: 9.20