Question

Calculate the pH of a solution of .250 M sodium acetate

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-

0.25 0 0

0.25-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)*0.25) = 1.179*10^-5

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

so, x = 1.179*10^-5 M

we have below equation to be used:

pOH = -log [OH-]

= -log (1.179*10^-5)

= 4.9287

we have below equation to be used:

PH = 14 - pOH

= 14 - 4.9287

= 9.0713

Answer: 9.07