What is the pH of 0.001 (M) aniline solution? The ionisation constant of aniline can be taken from standard table. Calculate the degree of ionisation of aniline in the solution. Also calculate the ionisation constant of the conjugate acid of aniline.

Explanation

`C_(6)H_(5)NH_(2)(aq)+H_(2)O(l)hArr C_(6)H_(5)overset(+)(N)H_(3)(aq)+OH^(-)(aq)`
`K_(b)` for `C_(6)H_(5)NH_(2)=4.27xx10^(-10)` for a solution of weak base like `C_(6)H_(5)NH_(2)`.
`pOH=(1)/(2) pK_(b)-(1)/(2)logc`
`therefore pOH=-(1)/(2)log_(10)(4.27xx10^(-10))-(1)/(2)log(0.001)`
`=4.68+1.5=6.18`
If the degree of ionisation of `C_(6)H_(5)NH_(2)` in its solution be `alpha`, then 0.001 `alpha=[OH^(-)]=10^(-pOH)=10^(-6.18)=6.6xx10^(-7)(M)`
`therefore alpha=6.6xx10^(-4)`
we know, `K_(a)xxK_(b)=10^(-14)`
Therefore, `K_(a)(C_(6)H_(5)overset(+)(N)H_(3))xxK_(b)(C_(6)H_(5)NH_(2))=10^(-14)`
`therefore K_(a)(C_(6)H_(5)overset(+)(N)H_(3))=(10^(-14))/(4.27xx10^(-10))=2.34xx10^(-5)`