Question

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

Answer 1

The balanced chemical equation for the ionization of phenol is:
$PhOH+H_{2}O\rightleftharpoons Ph{O}^{-}+H_3{O}^{+}$.

The initial concentrations of phenol, phenoxide ion and hydronium ion are 0.05, 0 and 0 respectively.
Corresponding equilibrium concentrations are $0.05-x,$ x and x respectively.

$K_a=\frac{\left[Ph{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[PhOH\right]}$

$1\times {10}^{-10}=\frac{x\times x}{0.05-x}$
x is very small (as the ionization constant is very small) . Hence, x can be ignored in the denominator.

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

$x=2.2\times {10}^{-6}M=\left[H_3{O}^{+}\right]=\left[Ph{O}^{-}\right]$
Let $\alpha$ be the degree of dissociation of phenol in presence of 0.01 M phenoxide.

The dissociation of phenoxide is as represented:
$PhONa\to Ph{O}^{-}+N{a}^{+}$
Also,
$PhOH+H_{2}O\rightleftharpoons Ph{O}^{-}+H_3{O}^{+}$
$\left[PhOH\right]=0.05-0.05\alpha \simeq 0.05$ M (as the degree of dissociation is very small.)
$\left[Ph{O}^{-}\right]=0.01+0.05\alpha \simeq 0.01$ M
$\left[H_3{O}^{+}\right]=0.05\alpha$

$K_a=\frac{\left[Ph{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[PhOH\right]}$
$1.0\times {10}^{-10}=\frac{0.01\times 0.05\alpha }{0.05}$
$\alpha =1\times {10}^{-8}$