What is the pH of a 0.500 M solution of sodium acetate?
Ka of acetic acid (CH3COOH) is 1.8*10^-5
lets find Kb for CH3COO- (acetate ion)
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-
0.5 0 0
0.5-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)*0.5) = 1.667*10^-5
since c is much greater than x, our assumption is correct
so, x = 1.667*10^-5 M
we have below equation to be used:
pOH = -log [OH-]
= -log (1.667*10^-5)
= 4.7782
we have below equation to be used:
PH = 14 - pOH
= 14 - 4.7782
= 9.2218
Answer: 9.22
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