What is the `pH` of `0.001 M` aniline solution? The ionization constant of aniline `4.27xx10^(-10)`. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Explanation

(i) `C_(6)H_(5)NH_(2)+H_(2)O hArr C_(6)H_(5)NH_(3)^(+)+OH^(-)`
`K_(a)=([C_(6)H_(5)NH_(3)^(+)][OH^(-)])/([C_(6)H_(5)NH_(2)])=([OH^(-)]^(2))/([C_(6)H_(5)NH_(2)])`
`[OH^(-)]=sqrt(K_(a)[C_(6)H_(5)NH_(2)])=sqrt((4.27xx10^(-10))(10^(-3)))=6.534xx10^(-7)M`
`pOH=-log (6.534xx10^(7))=7-0.8152=6.18 :. pH = 14 - 6.18 = 7.82`
`{:("(ii)Also",C_(6)H_(5)NH_(2)+H_(2)O ,hArr,C_(6)H_(5)NH_(3)^(+),+,OH^(-),,),("Initial conc.",C,,,,,,),("At. eqm.",C- C alpha,,C alpha,,C alpha ,,):}`
`K_(b)=(C alpha . C alpha)/(C (1-alpha))=(C alpha^(2))/(1-alpha)=C alpha^(2) :. alpha=sqrt(K_(b)//C)=sqrt(4.27xx10^(-10)//10^(-3))=6.53xx10^(-4)`
(iii) `pK_(a)+pK_(b) = 14` (for a pair of conjugate acid and base)
`pK_(b)=-log(4.27xx10^(-10))=10-0.62=9.38 :. pK_(a) = 14-9.38=4.62`
i.e., `-log K_(a)=4.62 or log K_(a) = - 4.62 = bar(5).38 or K_(a) = ` Antilog `bar(5).38 = 2.399xx10^(-5)~=2.4xx10^(-5)`