A 0.20 mol/L solution of propionic acid has a pH of 2.79. What is the pKa of propionic acid?

The question is –

A 0.20 mol/L solution of propionic acid has a pH of 2.79. What is the pKa of propionic acid?

Answer:

⇒ The $\text{p}{K}_{\text{a}}$ of propanoic acid is 4.88.

Explanation:

Step 1: Understand the Dissociation of Propionic Acid

CH₃CH₂COOH is a weak acid that partially dissociates in water according to the following reaction:

⇒ CH₃CH₂COOH(aq) + H₂O(l) ⇌ CH₃CH₂COO⁻(aq) + H₃O⁺(aq)

Step 2: Understanding the Equilibrium Constant, K_a

The equilibrium constant for this reaction, K_a, is given by:

⇒ K_a = [CH₃CH₂COO⁻][H_₃O⁺] / [CH₃CH₂COOH]

Step 3: Understand the Concept of pH

The pH is a measure of the hydronium ion concentration, [H₃O⁺], and is given by:

⇒ pH = –log[H₃O⁺]

Step 4: Find [H₃O⁺]

Given that the pH of the propionic acid solution is 2.79, we can find the [H₃O⁺] by taking the antilog of -pH:

⇒ [H₃O⁺] = 10^-pH = 10^–2.79 = 1.62 × 10⁻³ M

Step 5: Calculation of K_a

In a 0.20 mol/L solution of propionic acid, the concentration of propionic acid [CH₃CH₂COOH] is approximately 0.20 M (since it’s a weak acid and doesn’t fully ionize).

The concentration of the propionate ion [CH₃CH₂COO⁻] is equal to the [H₃O⁺] at equilibrium, so we can substitute these values into the K_a expression:

⇒ K_a = [CH₃CH₂COO⁻][H₃O⁺] / [CH₃CH₂COOH]

⇒ K_a = (1.62 × 10⁻³ M) * (1.62 × 10⁻³ M) / 0.20 M

∴ K_a = 1.32 × 10⁻⁵

Step 6: Calculation of pK_a

The pK_a is the negative logarithm (base 10) of K_a:

⇒ pK_a = -log(K_a)

⇒ pK_a = -log(1.32 × 10⁻⁵)

∴ pK_a = 4.88

Therefore, the pK_a of propionic acid is 4.88.