What is the pH of a 0.363 M propanoic acid solution and the concentrations of C2H5COOH and C2H5COO- at equilibrium?

To calculate the pH of a 0.363 M propanoic acid solution (C₂H₅COOH), we first need to understand that propanoic acid is a weak acid, and it will partially dissociate in water according to the equilibrium:

C₂H₅COOH ⇌ H⁺ + C₂H₅COO^–

The acid dissociation constant (K_a) for propanoic acid is provided as 1.34 x 10^-5. This will help us set up the equilibrium expression:

K_a =    [H⁺][C₂H₅COO^–]/[C₂H₅COOH]

Let’s assume that at equilibrium, the change in concentration due to dissociation is ‘x’. Therefore, at equilibrium we have:

• [C₂H₅COOH] = 0.363 – x
• [H⁺] = x
• [C₂H₅COO^–] = x

Substituting these into the Ka expression:

1.34 x 10^-5 = (x)(x) / (0.363 – x)

If we assume that x is small compared to 0.363 M, we can simplify the denominator to 0.363:

1.34 x 10^-5 = (x²) / 0.363

Solving for x gives:

x² = 1.34 x 10^-5 * 0.363

x² ≈ 4.85 x 10^-6

x ≈ √(4.85 x 10^-6) ≈ 0.00220

Thus, at equilibrium:

• [C₂H₅COOH] = 0.363 – 0.00220 ≈ 0.3608 M
• [C₂H₅COO^–] = 0.00220 M
• [H⁺] = 0.00220 M

To find the pH:

pH = – log[H⁺] = – log(0.00220) ≈ 2.65

In summary, for a 0.363 M propanoic acid solution:

• The pH is approximately 2.65.
• The concentration of C₂H₅COOH at equilibrium is about 0.3608 M.
• The concentration of C₂H₅COO^– at equilibrium is approximately 0.00220 M.