Question

Calculate the pH of a liter of 0.125 M ammonium perchlorate solution. (kb= 1.7 x 10^-5)

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Answer 1

Let α be the dissociation of the weak base,ammonium perchlorate.
                            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 = 0.125 M

Plug the values we get α = 0.01166
So the concentration of [OH^-] = cα

                                               = 0.125 x0.01166
                                               = 1.46x10^-3 M

pOH = - log [OH^-]

        = - log (1.46x10^-3 )

       = 2.84

So pH = 14 - 2.84 = 11.16

Therefore the pH of the solution is 11.16