Question

What is the pH of a 0.108 M aqueous solution of potassium acetate, KCH₃COO at 25 °C? (K_a for CH₃COOH = 1.8×10^-5)

Answer 1

we have below equation to be used:

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-

0.108 0 0

0.108-x x x

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

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that 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)*0.108) = 7.746*10^-6

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

so, x = 7.746*10^-6 M

we have below equation to be used:

pOH = -log [OH-]

= -log (7.746*10^-6)

= 5.1109

we have below equation to be used:

PH = 14 - pOH

= 14 - 5.1109

= 8.89

Answer: 8.89