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

Let c M be the initial concentration of $\left[C_6{H}_{5}N{H}_3^{+}\right]$ and x be the degree of ionization.
$\underset{c(1-x)}{PhN{H}_2}+H_{2}O\rightleftharpoons \underset{cx}{PhN{H}_3^{+}}+\underset{cx}{O{H}^{-}}$
$K_b=\frac{\left[PhN{H}_3^{+}\right]\left[O{H}^{-}\right]}{\left[PhN{H}_2\right]}=\frac{\left(cx\right)\left(cx\right)}{c(1-x)}=\frac{c{x}^2}{1-x}$
But x is very small. Hence, $1-x\approx 1$.
Hence, $K_b=c{x}^2$.
The degree of ionization of aniline, $x=\sqrt{\frac{K_b}{c}}=\sqrt{\frac{4.3\times {10}^{-10}}{0.001}}=6.56\times {10}^{-4}$
$\left[O{H}^{-}\right]=cx=0.001\times 6.56\times {10}^{-4}=6.56\times {10}^{-7}M$
$\left[H^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\frac{{10}^{-14}}{6.56\times {10}^{-7}}=1.52\times {10}^{-8}$
$pH=-log\left[H^{+}\right]=-log1.52\times {10}^{-8}=7.818$
$K_a$ (conjugate base of aniline) $=\frac{K_w}{K_b}=\frac{{10}^{-14}}{4.3\times {10}^4}=2.32\times {10}^{-5}$