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)/1.3*10^-5
Kb = 7.692*10^-10
C3H3O2- dissociates as
C3H3O2- + H2O -----> HC3H3O2 + OH-
0.36 0 0
0.36-x x x
Kb = [HC3H3O2][OH-]/[C3H3O2-]
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 ((7.692*10^-10)*0.36) = 1.664*10^-5
since c is much greater than x, our assumption is correct
so, x = 1.664*10^-5 M
use:
pOH = -log [OH-]
= -log (1.664*10^-5)
= 4.7788
use:
PH = 14 - pOH
= 14 - 4.7788
= 9.2212
Answer: a
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