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

Explanation

`{:(PyH^(+),+H_(2)OhArr,PyH^(+)OH^(-)+H^(+)),(C(1-h),Ch,Ch):}`
`[H^(+)]=Ch=Csqrt((K_(H))/(C ))=sqrt((C.K_(w))/(K_(b))`
`3.63xx10^(-4)`
`(.: pH=3.44 :. [H^(+)]=3.63xx10^(-4))`
`:. K_(b)=(0.02xx10^(-14))/((3.63xx10^(-4))^(2))=1.52xx10^(-9)`