Question

Calculate the pH of a 0.070 M C5H5N solution. The Kb of C5H5N is 1.7 x 10-9.

Answer 1

C5H5N dissociates as:

C5H5N +H2O -----> C5H5NH+ + OH-

7*10^-2 0 0

7*10^-2-x x x

Kb = [C5H5NH+][OH-]/[C5H5N]

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 ((1.7*10^-9)*7*10^-2) = 1.091*10^-5

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

so, x = 1.091*10^-5 M

so.[OH-] = x = 1.091*10^-5 M

use:

pOH = -log [OH-]

= -log (1.091*10^-5)

= 4.96

use:

PH = 14 - pOH

= 14 - 4.96

= 9.04

Answer: 9.04