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

What is the pHof 0.001 M anilinesolution ? The ionization constant of

1.	What is the pHof 0.001 M anilinesolution ? The ionization constant of aniline is 4.27xx10^(-10). Calculate degree of ionization of aniline in the solution . Also calculate the ionization constant of the conjugate acid of aniline.
Answer» SOLUTION :(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)`