use:
Ka = Kw/Kb
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
Ka = (1.0*10^-14)/Kb
Ka = (1.0*10^-14)/6.3*10^-5
Ka = 1.587*10^-10
(CH3)3NH+ + H2O -----> (CH3)3N + H+
8.31*10^-3 0 0
8.31*10^-3-x x x
Ka = [H+][(CH3)3N]/[(CH3)3NH+]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((1.587*10^-10)*8.31*10^-3) = 1.148*10^-6
since c is much greater than x, our assumption is correct
so, x = 1.148*10^-6 M
So, [H+] = x = 1.148*10^-6 M
use:
pH = -log [H+]
= -log (1.148*10^-6)
= 5.9399
Answer: 5.94
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