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What is the pH of 0.020 M aniline (Kb = 4.3*10^-10)?

What is the pH of 0.020 M aniline (Kb = 4.3*10^-10)?

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Answer #1

Let us assume aniline as A.(C6H5NH2)
A + H2O ↔ AH+ + OH-

0.02             0         0

0.02-x          x          x

Kb = x^2/(A)

Kb = 4.3*10^-10

4.3*10^-10 = x^2/[A]

4.3*10^-10 = x^2/(0.02-x)
x = 2.93*10^-6

pOH = -log(2.93*10^-6)

        = 5.53

pH = 14-5.53

     = 8.47

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