Question

Calculate the pH of a 0.22 M CH3NH3Br solution. Kb(CH3NH2) = 4.4 ×10–4

Answer 1

use:

Ka = Kw/Kb

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Ka = (1.0*10^-14)/Kb

Ka = (1.0*10^-14)/4.4*10^-4

Ka = 2.273*10^-11

CH3NH3+ + H2O -----> CH3NH2 + H+

0.22 0 0

0.22-x x x

Ka = [H+][CH3NH2]/[CH3NH3+]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((2.273*10^-11)*0.22) = 2.236*10^-6

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

so, x = 2.236*10^-6 M

So, [H+] = x = 2.236*10^-6 M

use:

pH = -log [H+]

= -log (2.236*10^-6)

= 5.6505

Answer: 5.65