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

Explanation

`pH = 3.44 , i.e., log [H^(+)]=-3.44 = bar(4).56 :. [H^(+)]=3.63xx10^(_4)M`
`C_(5)H_(5)N.HCl+aq hArr C_(5)H_(5)overset(+)NCl^(-)+H^(+)`
`K_(a)=([C_(5)H_(5)overset(+)NCl^(-)][H^(+)])/([C_(5)H_(5)N.HCl])=((3.63xx10^(-4))(3.63xx10^(-4)))/(2xx10^(-2))=6.588xx10^(-6)`
`pK_(a)=-log(6.588xx10^(-6))=6-0.8187=5.18`
`pK_(a)+pK_(b)=14 :. pK_(b)=14-5.18 = 8.82`
`-log K_(b) = 8.82 or log K_(b) = - 8.82 = bar(-).18 :. K_(b)=1.514xx10^(-9)`
Alternatively, Pyridine hydrochloride is a salt of weak base and strong acid . Hence,
`pH=-(1)/(2)[log K_(w)-logK_(b)+logc],i.e., 3.44 = - (1)/(2) [-14-logK_(b)+log(2xx10^(-2))]`
or `6.88=14+log K_(b) + 1.70 or log K_(b) = - 8.82 = bar(9).18 or K_(b)=1.5xx10^(-9)`.