Question

Determine the pH of a solution that is 0.025 M C2H8N2 where Kb(C2H8N2) = 8.3 * 10−5?

Answer 1

Let α be the dissociation of the weak base,C2H8N2 simply treat it as BOH
                            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 = 8.3x10^-5

          c = concentration = 0.025 M

Plug the values we get α = 0.0576
So the concentration of [OH^-] = cα

                                           = 0.025 x0.0576
                                           = 1.44x 10^-3 M

pOH = - log [OH^-]

        = - log  (1.44 x 10^-3)

        = 2.84

So pH = 14 - 2.84 = 11.16