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

Concept: Weak Acid Ionization – Phenol ($\ce{C6H5OH}$) is a weak acid; its ionization constant $K_a$ governs the equilibrium $\ce{C6H5OH <=> H+ + C6H5O-}$.

Step 1 – Phenolate ion in pure 0.05 M phenol

For a weak acid, $[\ce{C6H5O-}] \approx \sqrt{K_a \cdot C}$ when $K_a$ is very small.

$[\ce{C6H5O-}] = \sqrt{(1.0 \times 10^{-10})(0.05)} = \sqrt{5.0 \times 10^{-12}} = 2.24 \times 10^{-6} \text{ M}$.

Step 2 – Degree of ionization in presence of common ion

With 0.01 M sodium phenolate (common ion $\ce{C6H5O-}$), use the common-ion effect. Let $\alpha$ be the degree of ionization of phenol.

$[\ce{H+}] = 0.05\alpha$, $[\ce{C6H5O-}] \approx 0.01$ (since added salt dominates). …

This problem is a classic weak acid ionization, but with a twist: the second part introduces the common ion effect. Let’s start with the core concept.

Phenol ($C_6H_5OH$) is a weak acid. It partially dissociates in water:

The ionization constant $K_a$ is given as $1.0 \times 10^{-10}$. This tiny number tells us the equilibrium lies far to the left — very few molecules actually ionize. The phenolate ion ($C_6H_5O^-$) is the conjugate base.

For the first part, we need the concentration of phenolate ion in a pure 0.05 M solution. For the second part, we add sodium phenolate, a salt that fully dissociates to give $C_6H_5O^-$ ions. This common ion (phenolate) shifts the equilibrium left, suppressing further ionization — that’s the common ion effect. The degree of ionization ($\alpha$) will drop dramatically.

Let’s work through it step by step.

1. Set up the equilibrium for pure phenol. Let initial concentration of phenol be $c = 0.05$ M. Let $x$ be the concentration of $C_6H_5O^-$ (and also $H^+$) at equilibrium.

Initial: $0.05$ M, $0$, $0$

Change: $-x$, $+x$, $+x$

Equilibrium: $0.05 - x$, $x$, $x$

The $K_a$ expression is:

2. Solve for $x$ using the weak acid approximation. Since $K_a$ is very small, $x$ will be tiny compared to 0.05. So we can approximate $0.05 - x \approx 0.05$. This is valid if $x < 5\%$ of 0.05 — we’ll check later.

So the concentration of phenolate ion in pure 0.05 M phenol is $[C_6H_5O^-] = x = 2.236 \times 10^{-6}$ M. Since the ionization is 1:1, $[C_6H_5O^-] = [H^+] = \sqrt{K_a c} = \sqrt{5 \times 10^{-12}} = 2.24 \times 10^{-6}$ M — the answer to the first part.

3. Now the second part: solution also 0.01 M in sodium phenolate. Sodium phenolate ($C_6H_5ONa$) is a strong electrolyte — it dissociates completely:

So initially, we have 0.01 M phenolate ions from the salt, plus the phenol at 0.05 M. Let’s set up the new equilibrium. Let $y$ be the concentration of phenol that ionizes (i.e., the additional $C_6H_5O^-$ and $H^+$ produced). But note: the initial phenolate from salt is 0.01 M, so at equilibrium:

The $K_a$ expression:

4. Apply the common ion effect approximation. …