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

Step 1: Understand the Dissociation of Propionic Acid

CH_₃CH_₂COOH, is a weak acid that partially dissociates in water according to the following reaction:

⇒ CH_₃CH_₂COOH(aq) + H_₂O(l) ⇌ CH_₃CH_₂COO⁻(aq) + H_₃O⁺(aq)

Step 2: Understanding the Equilibrium Constant, Ka

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

⇒ K_a = [CH_₃CH_₂COO⁻][H_₃O⁺] / [CH_₃CH_₂COOH]

Step 3: Understand the Concept of pH

The pH is a measure of the hydronium ion concentration, [H_₃O⁺], and is given by:

⇒ pH = –log[H_₃O⁺]

Step 4: Find [H₃O⁺]

Given that the pH of the propionic acid solution is 2.79, we can find the [H_₃O⁺] by taking the antilog of -pH:

⇒ [H_₃O⁺] = 10^-pH = 10^–2.79 = 1.62 × 10⁻³ M

Step 5: Calculation of Ka

In a 0.20 mol/L solution of propionic acid, the concentration of propionic acid [CH_₃CH_₂COOH] is approximately 0.20 M (since it’s a weak acid and doesn’t fully ionize). The concentration of the propionate ion [CH_₃CH_₂COO⁻] is equal to the [H_₃O⁺] at equilibrium, so we can substitute these values into the Ka expression:

⇒ K_a = [CH_₃CH_₂COO⁻][H_₃O⁺] / [CH_₃CH_₂COOH]

⇒ 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 pKa of propionic acid is 4.88.