Question

What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Answer 1

$K_b=4.27\times {10}^{-10}c=0.001M$

$pH=?\alpha =?K_b=c{\alpha }^{2}4.27\times {10}^{-10}=0.001\times {\alpha }^{2}4270\times {10}^{-10}={\alpha }^{2}6.534\times {10}^{-4}=\alpha$

Find the pOH and pH

$\left[O{H}^{-}\right]=0.001\times 65.34\times {10}^{-5}\left[O{H}^{-}\right]=0.065\times {10}^{-5}pOH=-log(.065\times {10}^{-5})=6.187pH=7.813$
Now, find the Ka
$K_a\times K_b=K_w\therefore 4.27\times {10}^{-10}\times K_a=w{K}_a=\frac{{10}^{-14}}{4.27\times {10}^{-10}}=2.34\times {10}^{-5}$
Thus, the ionization constant of the conjugate acid of aniline is $2.34\times {10}^{-5}$.