A `0.02 M` solution of pyridinium hydrochloride has `pH=3.44`. Calculate the ionization constant of pyridine.

Explanation

Pyridinium hydrochloride `(C_(6)H_(5)N^(+)HCI^(-))` is salt of strong acid and weak base. The pH of the solution is given as:
`pH =- 1/2 [log K_(w) - log K_(b) + log C]`
From the available date,
`pH = 3.44 , K_(w) =10^(-14) , C = 0.02 M`
`:. " "3.44 =- 1/2 [log (10^(-14)) - log K_(b) +log (2xx 10^(-2))]`
`3.44 =-1/2 [-14 -log K_(b) + (0.3010 -2)]`
`3.44 =- 1/2 [-14 - log K_(b) -1.70]`
`6.88 =14 + log K_(b) +1.70`
`log K_(b) =6.88 -14 -1.70 =- 8.82`
`K_(b) = " Antilog " (-8.82) = " Antilog " =(bar(9).18) = 1.51 xx 10^(-9)`