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

`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)][H_(2)O]]=[[OH^(-)]^(2)]/[[C_(6)H_(5)NH_(2)]]`
`[OH^(-)] =[K_(a) xx(C_(6)H_(5)NH_(2))]^(1//2) =[(4.27 xx 10^(-10)) xx (10^(-3))]^(1//2)`
`=6.534 xx 10^(-7) M`
`pOH =- log (6.534 xx 10^(-7)) =(7- log 6.534) =(7-0.8152) = 6.18`
`pH =14 -6.18 =7.82`
(ii).
`K_(b)=(Calpha.Calpha)/(C(1-alpha))=(Calpha^(2))/(1-alpha)=Calpha^(2) " or " alpha =(K_(b)//C)^(1//2)`
`alpha =(4.27 xx10^(-10)//10^(-3))^(1//2) =6.53 xx 10^(-4)`
`(iii) . pK_(a) + pK_(b) =14 ("for conjugate acid , base pair")`
`pK_(b) =- log K_(b)= - log (4.27 xx 10^(-10)) =(10 - log 4.27) =10- 0.62 = 9.38`
`pK_(a) = 14 - pK_(b) =14 - 9.38 =4.62 , pK_(a) =- log K_(a) =4.62`
`" or "" "log K_(a)= - 4.62 = bar(5) .38 , K_(a) = " Antilog "(bar(5) .38) =2. 399 xx 10^(-5) =2.4 xx10^(-5)`