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?
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} $
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