The ionization constant of phenol is 1.0 × 10–10.

Solution:

Ionization of phenol:

$K_{a \pi}=\frac{\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right]}$

$K_{a}=\frac{x \times x}{0.05-x}$

As the value of the ionization constant is very less, $x$ will be very small. Thus, we can ignore $x$ in the denominator.

$\therefore x=\sqrt{1 \times 10^{-10} \times 0.05}$

$=\sqrt{5 \times 10^{-12}}$

$=2.2 \times 10^{-6} \mathrm{M}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$

Since $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}\right]$

$\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}\right]=2.2 \times 10^{-6} \mathrm{M}$

Now, let $\propto$ be the degree of ionization of phenol in the presence of $0.01 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{ONa}$.

$\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right]=0.05-0.05 \alpha ; 0.05 \mathrm{M}$

$\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}\right]=0.01+0.05 \alpha ; 0.01 \mathrm{M}$

$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=0.05 \alpha$

$K_{o}=\frac{\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}^{-}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right]}$

$K_{a}=\frac{(0.01)(0.05 \alpha)}{0.05}$

$1.0 \times 10^{-10}=.01 \alpha$

$\alpha=1 \times 10^{-8}$