Question

In: Chemistry

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

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

Solutions

Expert Solution

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, Kb = (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 Kb = cα2

==> α = √ ( Kb / c )

Given Kb = 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


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