The ionization constant of phenol is $1.0 \times 10^{-10}.$ What is the concentration of phenolate ion in $0.05 \ M$ solution of phenol? What will be its degree of ionization if the solution is also $0.01 \ M$ in sodium phenolate?

(N/A) Ionization of phenol: $C_6H_5OH + H_2O \leftrightarrow C_6H_5O^{-} + H_3O^{+}$
Initial concentration: $0.05 \ M$,$0$,$0$
At equilibrium: $0.05 - x$,$x$,$x$
$K_a = \frac{[C_6H_5O^{-}][H_3O^{+}]}{[C_6H_5OH]} = \frac{x^2}{0.05 - x}$
Since $K_a$ is very small,$x$ is negligible compared to $0.05$. So,$0.05 - x \approx 0.05$.
$x = \sqrt{K_a \times C} = \sqrt{1.0 \times 10^{-10} \times 0.05} = \sqrt{5.0 \times 10^{-12}} = 2.236 \times 10^{-6} \ M$.
Thus,$[C_6H_5O^{-}] = 2.236 \times 10^{-6} \ M$.
Now,in the presence of $0.01 \ M$ sodium phenolate $(C_6H_5ONa)$:
$C_6H_5ONa \rightarrow C_6H_5O^{-} + Na^{+}$
$[C_6H_5O^{-}] = 0.01 \ M$ (due to common ion effect).
Let $\alpha$ be the degree of ionization of phenol.
$[C_6H_5OH] = 0.05 \ M$,$[H_3O^{+}] = 0.05 \alpha$,$[C_6H_5O^{-}] = 0.01 \ M$.
$K_a = \frac{[C_6H_5O^{-}][H_3O^{+}]}{[C_6H_5OH]} \Rightarrow 1.0 \times 10^{-10} = \frac{0.01 \times 0.05 \alpha}{0.05}$.
$1.0 \times 10^{-10} = 0.01 \alpha \Rightarrow \alpha = 1.0 \times 10^{-8}$.