Question

CH3NH2 is a weak base (K = 5.0 * 10 ) and so the salt, CH3NH2NO3, acts as a weak acid. What is the pH of a solution that is 0.0540 M in CH3NH2NO3 at 25°C? Number pH-

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)/5*10^-4

Ka = 2*10^-11

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

5.4*10^-2 0 0

5.4*10^-2-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*10^-11)*5.4*10^-2) = 1.039*10^-6

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

so, x = 1.039*10^-6 M

So, [H+] = x = 1.039*10^-6 M

use:

pH = -log [H+]

= -log (1.039*10^-6)

= 5.9833

Answer: 5.98