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?

To determine the concentration of the phenolate ion in a 0.05 M solution of phenol, we first need to understand the ionization process of phenol in water. Phenol (C6H5OH) can ionize to form phenolate ions (C6H5O⁻) and hydrogen ions (H⁺). The ionization constant (Ka) for phenol is given as 1.0 × 10⁻¹⁰, which helps us calculate the degree of ionization in the solution.

Setting Up the Ionization Equation

The ionization of phenol can be represented by the following equilibrium equation:

C6H5OH ⇌ C6H5O⁻ + H⁺

Using the Ionization Constant

The ionization constant (Ka) is defined as:

Ka = [C6H5O⁻][H⁺] / [C6H5OH]

In our case, let's denote the degree of ionization as α. Initially, we have a concentration of phenol, [C6H5OH], equal to 0.05 M. At equilibrium, the concentrations will be:

• [C6H5OH] = 0.05 - α
• [C6H5O⁻] = α
• [H⁺] = α

Substituting into the Ka Expression

Now, substituting these values into the Ka expression gives:

1.0 × 10⁻¹⁰ = (α)(α) / (0.05 - α)

Assuming α is small compared to 0.05, we can simplify this to:

1.0 × 10⁻¹⁰ ≈ (α²) / 0.05

Solving for α

Rearranging the equation to solve for α yields:

α² = 1.0 × 10⁻¹⁰ × 0.05

α² = 5.0 × 10⁻¹²

Taking the square root gives:

α = √(5.0 × 10⁻¹²) ≈ 7.07 × 10⁻⁶

Calculating the Concentration of Phenolate Ion

Since α represents the concentration of the phenolate ion at equilibrium, we find:

[C6H5O⁻] = α ≈ 7.07 × 10⁻⁶ M

Considering the Presence of Sodium Phenolate

Now, let's analyze the situation when the solution also contains 0.01 M sodium phenolate. Sodium phenolate dissociates completely in solution to give phenolate ions:

NaC6H5O → Na⁺ + C6H5O⁻

This means we start with an initial concentration of phenolate ions of 0.01 M from sodium phenolate. The total concentration of phenolate ions in the solution will be:

[C6H5O⁻] = 0.01 M + α ≈ 0.01 M + 7.07 × 10⁻⁶ M ≈ 0.01 M (since α is negligible compared to 0.01 M)

Degree of Ionization with Sodium Phenolate

In this case, the degree of ionization can be calculated again using the same Ka expression, but now we have a much higher concentration of phenolate ions:

Ka = [C6H5O⁻][H⁺] / [C6H5OH]

Let’s denote the new degree of ionization as β. The equilibrium concentrations will be:

• [C6H5OH] = 0.05 - β
• [C6H5O⁻] = 0.01 + β
• [H⁺] = β

Substituting into the Ka expression gives:

1.0 × 10⁻¹⁰ = (0.01 + β)(β) / (0.05 - β)

Assuming β is small compared to 0.01, we can simplify this to:

1.0 × 10⁻¹⁰ ≈ (0.01)(β) / 0.05

Rearranging gives:

β = (1.0 × 10⁻¹⁰ × 0.05) / 0.01 = 5.0 × 10⁻¹⁰

Final Thoughts

The degree of ionization in the presence of 0.01 M sodium phenolate is quite small, approximately 5.0 × 10⁻¹⁰, indicating that the addition of sodium phenolate significantly suppresses the ionization of phenol due to the common ion effect. This is a classic example of how the presence of a common ion can affect the equilibrium of a weak acid or base in solution.