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.
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
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