The ionization constant of propanoic acid is 1.32 × 10–5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

To tackle the problem of calculating the degree of ionization of propanoic acid in a 0.05 M solution, as well as its pH, we need to start with the ionization constant (Ka) and apply some principles of equilibrium chemistry. Let's break this down step by step.

Understanding Ionization and Equilibrium

Propanoic acid (C2H5COOH) is a weak acid, which means it does not completely dissociate in solution. The ionization constant (Ka) gives us a measure of how much the acid ionizes in water. The formula for the ionization of propanoic acid can be represented as:

C2H5COOH ⇌ C2H5COO⁻ + H⁺

The equilibrium expression for this reaction is:

Ka = [C2H5COO⁻][H⁺] / [C2H5COOH]

Calculating Degree of Ionization in 0.05 M Solution

Let’s denote the degree of ionization as α. In a 0.05 M solution of propanoic acid, we can express the concentrations at equilibrium as follows:

• Initial concentration of C2H5COOH = 0.05 M
• Change in concentration due to ionization = -α for C2H5COOH and +α for both C2H5COO⁻ and H⁺
• At equilibrium: [C2H5COOH] = 0.05 - α
• [C2H5COO⁻] = α
• [H⁺] = α

Substituting these into the Ka expression gives:

1.32 × 10⁻⁵ = (α)(α) / (0.05 - α)

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

1.32 × 10⁻⁵ ≈ (α²) / 0.05

Now, solving for α:

α² = 1.32 × 10⁻⁵ × 0.05

α² = 6.6 × 10⁻⁷

α = √(6.6 × 10⁻⁷) ≈ 0.00081

To find the degree of ionization as a percentage, we multiply by 100:

Degree of ionization = (0.00081 / 0.05) × 100 ≈ 1.62%

Calculating pH of the Solution

The pH of the solution can be calculated using the concentration of H⁺ ions at equilibrium:

pH = -log[H⁺] = -log(α) = -log(0.00081) ≈ 3.09

Effect of Adding HCl

Now, let’s consider the scenario where we have a 0.01 M solution of HCl added to the propanoic acid solution. HCl is a strong acid and will completely dissociate, providing additional H⁺ ions to the solution. The total concentration of H⁺ ions will now be:

[H⁺] = 0.01 M (from HCl) + α (from propanoic acid)

Since α is very small compared to 0.01 M, we can approximate:

[H⁺] ≈ 0.01 M

In this case, the presence of H⁺ ions from HCl will suppress the ionization of propanoic acid due to the common ion effect. Thus, we can assume that the degree of ionization of propanoic acid will be negligible, leading to:

Degree of ionization ≈ 0%

Summary of Findings

In summary, for a 0.05 M solution of propanoic acid, the degree of ionization is approximately 1.62%, and the pH is around 3.09. When 0.01 M HCl is added, the degree of ionization of propanoic acid effectively becomes negligible due to the common ion effect, resulting in a degree of ionization close to 0%.