Question

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

Answer 1

K_a = \(\frac{x\times x}{(0.05-x)}\)

or, 1.0 x 10^-10 = \(\frac{x^2}{0.05-x}\) = \(\frac{x^2}{0.05}\)

(\(x\) is very small because phenol is a very weak acid)

\(x\)² = 10 x 10^-10 x 0.05

\(x\) = (1.0 x 10^-10 x 0.05)^1/2

= 2.2 x 10^-6

Therefore, [C₆H₅O^-] = 2.2 x 10^-6 mol L^-1

and [H⁺] = 2.2 x 10^-6 mol L^-1

and pH = -log[H⁺]

= -log(2.2 x 10^-6) = 5.66

If α is the degree of ionization of phenol in a solution which was

[C₆H₅OH] = 0.05 mol L^-1

[C₆H₅O^-] = 0.01 mol L^-1

(Sodium phenate is fully dissociated into ions in solution)

The α is very small. Therefore

1.0 x 10^-10 = \(\frac{0.01\times 0.05\, \alpha}{0.05}\) = 0.01 α

or, α = \(\frac{1.0\times 10^{-10}}{0.01}\) = 1 x 10^-8

The degree of ionisation of 0.05 M phenol in the presence of 0.01 M sodium phenate is 1 x 10^-8.