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.

Answer

`alpha = 6.53 xx 10^(-4), K_(a) = 2.35 xx 10^(-5)`

Explanation

`K_(b) = 4.27 xx 10^(-10)`
`c = 0.001 M`
`pH = ?`
`alpha = ?`
`k_(b) = calpha^(2)`
`4.27 xx 10^(-10) = 0.001 xx alpha^(2)`
`4270 xx 10^(-10) = alpha^(2)`
`65.34 xx 10^(-5) = alpha = 6.53 xx 10^(-4)`
Then, `["anion"] = calpha = .001 xx 65.34 xx 10^(-5)`
`= .065 xx 10^(-5)`
`pOH = - "log" (.065 xx 10^(-5))`
`= 6.187`
`pH = 7.813`
Now,
`K_(a) xx K_(b) = K_(w)`
`:. 4.27 xx 10^(-10) xx K_(a) = K_(w)`
`:. 4.27 xx 10^(-10) xx K_(a) = K_(w)`
`K_(a) = (10^(-14))/(4.27 xx 10^(-10))`
`= 2.34 xx 10^(-5)`
Thus, the ionization constant of the conjugate acid of aniline is `2.34 xx 10^(-5)`