Question

The ionization constant of phenol is 1.0 × 10–10. What is the concentration of phenolate ion in a 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate?

Answer 1

Ionization of phenol: C6H5OH + H2O $\leftrightarrow$ C6H5O- + H3O+
Initial conc. 0.05 0 0
At equilibrium 0.05 - x x x

Ka = $\frac{[C_6H_5O^-][H_3O^+]}{ [C_6H_5OH]}$
Ka =$\frac{x\times x}{0.05-x}$
As the value of the ionization constant is much less, x will be very small. Thus, we can ignore x in the denominator.
∴ x = $\sqrt{1 × 10 - 10 × 0.05 }$ = $\sqrt{5 × 10 ^{- 12}}$ = 2.2 × 10-6 M = [H3O+]
Since [H3O+] = [C6H5O-], [C6H5O-] = 2.2 × 10-6 M.
Now, the degree of ionization of phenol in the presence of 0.01 M C6H5ONa.
C6H5ONa $\rightarrow$C6H5O- + Na+
Conc. 0.01
Also, C6H5OH + H2O $\leftrightarrow$ C6H5O- + H3O+
[C6H5OH] = 0.05 - 0.05a ; 0.05M
[C6H5OH] = 0.01 + 0.05a ; 0.01M
[H3O+] = 0.05a
Ka = $\frac{[C_6H_5O^-][H_3O^+]}{ [C_6H_5OH]}$
Ka = $\frac{(0.01)(0.05a)}{0.05}$
1.0 × 1.0 - 10 = .01a
a = 1 × 10-8