Question

The base protonation constant K, of azetidine (C2H NH) is 1.5 x 10”. Calculate the pH of a 1.4 M solution of azetidine at 25 °C. Round your answer to 1 decimal place. pH = 0 . I ?

Answer 1

C3H6NH dissociates as:

C3H6NH +H2O -----> C3H6NH2+ + OH-

1.4 0 0

1.4-x x x

Kb = [C3H6NH2+][OH-]/[C3H6NH]

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 ((1.5*10^-8)*1.4) = 1.449*10^-4

since c is much greater than x, our assumption is correct

so, x = 1.449*10^-4 M

So, [OH-] = x = 1.449*10^-4 M

use:

pOH = -log [OH-]

= -log (1.449*10^-4)

= 3.8389

use:

PH = 14 - pOH

= 14 - 3.8389

= 10.1611

Answer: 10.2