Question

(7b)

The hydronium ion concentration of an aqueous solution of 0.432 M pyridine (a weak base with the formula C₅H₅N) is ...

[H₃O⁺] = _____M.

Answer 1

Kb of Pyridine = 1.7*10^-9
                                C5H5N + H2O ----> C5H5NH⁺ +   OH^-
initial                        0.432                        0                0
at equilibrium:        0.432-x                         x                x

Kb = [C5H5NH⁺] [OH^-] /[C5H5N]

1.7*10^-9 = x*x / (0.432-x)

since Kb is very small, x will be small and it can be ignored as compared to 0.432
so, 0.432-x is approximately equal to 0.432
Above equation becomes:
1.7*10^-9 = x*x / 0.432
x= 2.71*10^-5
since [OH-]= x, [OH-] = 2.71*10^-5 M

To get [H+], use:
[H+][OH-]=10^-14
[H+] = 10^-14 / [OH-]
    = 10^-14 / (2.71*10^-5)
   =3.7*10^-10 M

Answer: 3.7*10^-10 M