Question

Calculate the pH at 25°C of a 0.0019 M solution of a weak base with a K_b of 3.2 × 10⁻⁹.

Answer 1

Answer:

Let the weak base be represented by B-OH

Given the concentration = 0.0019 M and the equillibrium constant K_b = 3.2 x 10^-9

The dissociation equation is given by

               B-OH    B⁺    +           OH^-

The initial concentration       0.0019                         0                            0

                                                 Let x moles dissociated at equillibrium.

equillibrium concentration      0.0019 -x                      x                            x

                  K_b = [B⁺] [OH^-] / [B-OH]

                        = x * x / [0/0019 -x]

                        = x² / 0.0019          since x is small compared to 0.0019, it is neglected in the denominator.

but k_b =   3.2 x 10^-9           (given)

Therefore         3.2 x 10^-9 = x² / 0.0019

                           x² = 3.2 * 10^-9 * 0.0019

                                = 6.08 * 10^-12

Therefore x = (6.08 x 10^-12 )

                 = 2.466 x 10^-6

Concentration of [OH] = 2.466 x 10^-6

P^OH = - log [OH^-] = - log [2.466 x 10^-6]

                             = 6 - log 2.466

                             = 6 - 0.4 = 5.6

We know that P^H + P^OH = 14

                    P^H = 14 - P^OH

                         = 14 - 5.6

                         = 8.4

Thus

The P^H of the given weak base solution is 8.4