Question

5. Calculate pH of 0.25M aniline (C₆H₅NH₂) solution.

Answer 1

C6H5NH2 dissociates as:

C6H5NH2 +H2O -----> C6H5NH3+ + OH-

0.25 0 0

0.25-x x x

Kb = [C6H5NH3+][OH-]/[C6H5NH2]

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 ((4.3*10^-10)*0.25) = 1.037*10^-5

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

so, x = 1.037*10^-5 M

So, [OH-] = x = 1.037*10^-5 M

use:

pOH = -log [OH-]

= -log (1.037*10^-5)

= 4.9843

use:

PH = 14 - pOH

= 14 - 4.9843

= 9.0157

Answer: 9.02