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

Ionization of phenol:

                      ${C_6}{H_3}OH\,\quad  + \quad {H_2}O \leftrightarrow {C_6}{H_5}{O^ - }\quad  + \quad {H_3}{O^ + }$

Initial conc           $0.05$                                            $0$                          $0$

At equilibrium   $0.05-x$                                     $x$                           $x$ 

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

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

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

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

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

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

Since $\left[ H _{3} O ^{+}\right]=\left[ C _{6} H _{5} O ^{-}\right],$ $\left[ C _{6} H _{5} O ^{-}\right]=2.2 \times 10^{-6} \,M$

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

      $C _{6} H _{5} ONa \longrightarrow C _{6} H _{5} O ^{-}+ Na ^{+}$

Conc.                                                    $0.01$

Also,

          $C _{6} H _{5} OH + H _{2} O \longleftrightarrow C _{6} H _{5} O ^{-}+ H _{3} O ^{+}$

Conc.   $0.05-0.05 \alpha$                      $0.05 \alpha$         $0.05 \alpha$

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

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

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

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

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

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

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