Question

In 0.100 M CH3NH2, only 6.4% of the base has undergone ionization. Calculate the pKb of methylamine.

The answer is 3.36 but I have no idea how to get it. Please show steps. Thanks!

Answer 1

Answer –

We are given, [CH₃NH₂] = 0.100 M , 6.47 % ionization, Kb = ?

We know, percent ionization = x / initial concentration * 100 %

So, x =   percent ionization * initial concentration / 100 %

         = 6.4 % * 0.100 M / 100 %

         = 0.0064 M

So at the equilibrium there is base dissociate only 0.0064 M

   Now we need put ICE table

    CH₃NH₂ + H₂O ------> CH₃NH₃⁺ + OH^-

I     0.100                           0                0

C      -x                            +x               +x

E   0.100-x                       +x               +x

We calculated , x = 0.0064 M

So at the equilibrium,

[CH₃NH₂] = 0.100-x

                  = 0.100 -0.0064 M

                 = 0.0936 M

[CH₃NH₃⁺] = x = 0.0064 M

[OH^-] = 0.0064 M

So, Kb = [CH₃NH₃⁺] [OH^-] / [CH₃NH₂]

            = 0.0064 * 0.0064 / 0.0936

            = 4.38*10^-4