Question

calculate the pH of a liter of 0.10 M ammonium chlorate solution. (kb=1.7x10^-5)

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Answer 1

Let α be the dissociation of the weak base
                            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^-5

          c = concentration = 0.10 M

Plug the values we get α = 0.013
So the concentration of [OH^-] = cα

                                           = 0.10 x0.013

                                           = 1.304x10^-3 M

pOH = - log [OH^-]

        = - log  (1.304x10^-3 )

        = 2.88

So pH = 14 - pOH

          = 14 - 2.88

         = 11.12