Question

Pyridine (C₅H₅N) is a particularly foul-smelling substance used in manufacturing pesticides and plastic resins. Calculate the pH of a 0.145 M solution of pyridine (K_b = .00000000140).

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 = 0.00000000140

          c = concentration = 0.145 M

Plug the values we get α = 9.83x10^-5
So the concentration of [OH^-] = cα

                                          = 0.145 x9.83x10^-5
                                          = 1.42 x 10^-5 M

pOH = - log [OH^-]

        = - log  (1.42 x 10^-5)

       = 4.84

So pH = 14 - pOH

          = 14 - 4.84

          = 9.16