Question

what is the pH of 0.124 M pyridine (C5H5N)?

Answer 1

Let α be the dissociation of the weak base,pyridine
                            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 = 1.7x10^-9

          c = concentration = 0.124 M

Plug the values we get α = 1.17x10^-4
So the concentration of [OH^-] = cα

                                           = 0.124 x1.17x10^-4
                                           = 1.45 x 10^-5 M

pOH = - log [OH^-]

        = - log  (1.45 x 10^-5)

        = 4.84

So pH = 14 - pOH

          = 14 - 4.84

          = 9.16