Question

Calculate the pH of a 0.59 M CH3COONa solution. (Ka for acetic acid = 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-

0.59 0 0

0.59-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)*0.59) = 1.81*10^-5

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

so, x = 1.81*10^-5 M

we have below equation to be used:

pOH = -log [OH-]

= -log (1.81*10^-5)

= 4.74

we have below equation to be used:

PH = 14 - pOH

= 14 - 4.74

= 9.26

Answer: 9.26