Question

Calculate the pH and concentrations of CH3NH2 and CH3NH3 in a 0.0323 M methylamine (CH3NH2) solution. The Kb of CH3NH2 = 4.47 × 10-4.

Answer 1

Let α be the dissociation of the weak base
                       CH₃NH₂ + H₂O <---> CH₃NH₃⁺ + OH^-

initial conc.            c                 0         0

change               -cα              +cα      +cα

Equb. conc.         c(1-α)       cα      cα

Dissociation constant, K_b = (cα x cα) / ( c(1-α)               

                                          = c α² / (1-α)

In the case of weak bases α is very small so 1-α is taken as 1

So K_b = cα²

==> α = √ ( K_b / c )

Given K_b = 4.47x10^-4

          c = concentration = 0.0323 M

Plug the values we get α = 0.118
So the concentration of [OH^-] = cα

                                            = 0.0323 x 0.118
                                            = 3.70 x 10^-3 M

pOH = - log [OH^-]

        = - log  (3.70x10^-3)

        = 2.42

So pH = 14 - pOH = 14-2.42 =11.58

[CH₃NH₂ ] = c(1-α) = 0.0285 M

[CH₃NH₃⁺ ] = cα = 3.81x10^-3 M