Calculate the pH of 4.55 M aniline, C6H5NH2 ( Kb = 4.0 x 10^-10)

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Expert Solution

for simplicity lets write weak base as BOH

C6H5NH2 + H2O -----> C6H5NH3+   +   OH-
4.55                    0                      0
4.55-x                x                       x

Kb = [C6H5NH3+][OH-]/[C6H5NH2 ]
Kb = x*x/(c-x)
since kb is small, x will be small and it can be ignored as compared to c
So, above expression becomes

Kb = x*x/(c)
so, x = sqrt (Kb*c)

x = sqrt ((4.0E-10)*4.55) = 4.27E-5

pOH = -log [OH-] = -log (4.27E-5) = 4.37
PH = 14 - pOH = 14 - 4.37 = 9.63

Answer: 9.63

queen_honey_blossom answered 1 year ago

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