Question

If the Kb of a weak base is 4.1 × 10-6, what is the pH of a 0.50 M solution of this base?

Answer 1

Let α be the dissociation of the weak base
                            BOH <---> B ⁺ + 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 = 4.1x10^-6

          c = concentration = 0.50 M

Plug the values we get α = 2.86x10^-3
So the concentration of [OH^-] = cα

                                           = 0.50 x2.86x10^-3
                                           = 1.43x 10^-3 M

pOH = - log [OH^-]

        = - log  (1.43x 10^-3 )

        = 2.84

So pH = 14 - pOH

          = 14 - 2.84

          = 11.16

Therefore the pH of the base is 11.16