Question

Calculate the pH of a 0.177 M aqueous solution of pyridine (C₅H₅N, K_b = 1.5×10^-9) and the equilibrium concentrations of the weak base and its conjugate acid.

pH	=
[C5H5N]equilibrium	=	M
[C5H5NH+]equilibrium	=	M

Answer 1

Let α be the dissociation of the weak base pyridine
                     C₅H₅N + H₂O <---> C₂H₅NH ⁺ + 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 = 1.5x10^-9

          c = concentration = 0.177 M

Plug the values we get α = 9.21x10^-5
So the concentration of [OH^-] = cα

                                          = 0.177 x9.21x10^-5
                                          = 1.63x 10^-5 M

pOH = - log [OH^-]

        = - log  (1.63x 10^-5 )

        = 4.78

So pH = 14 - pOH

          = 14- 4.78

          = 9.22

Therefore the pH of the solution is 9.22

Equilibrium concentration of C₂H₅NH ⁺ is cα

                                                            = 0.177 x 9.21x10^-5 M

                                                            = 1.63x 10^-5 M

Equilibrium concentration of C₂H₅NH is c(1-α)

                                                       = 0.177 x (1- 9.21x10^-5 M)

                                                       = 0.177M