Attempt 2 A buffer with a pH of 4.77 contains 0.37 M of sodium benzoate and 0.10 M of benzoic acid. What is the concentration of [ H 3 O ] in the solution after the addition of 0.056 mol HCl to a final volume of 1.3 L

[H₃O⁺] = 2.75x10⁻⁵M

Explanation:

The benzoic acid (C₆H₅COOH) dissociates in water producing benzoate ion (C₆H₅COO⁻) and H₃O⁺ ion:

C₆H₅COOH + H₂O ⇄ C₆H₅COO⁻ + H₃O⁺

As pH of the buffer is 4.77, H-H equation gives:

pH = pKa + log₁₀ [Benzoate] / [Benzoic acid]

4.77 = pKa + log₁₀ [0.37M] / [0.10M]

4.20 = pKa

Moles of benzoate and benzoic acid before reaction are:

benzoate: 1.3L × 0.37M = 0.481 moles benzoate

benzoic acid: 1.3L × 0.10M = 0.130 moles benzoic acid

0.056 mol HCl reacts with benzoate ion, producing benzoic acid thus:

C₆H₅COO⁻ + HCl → C₆H₅COOH + Cl⁻

That means, after reaction, moles of benzoate ion and benzoic acid are:

benzoate: 0.481mol - 0.056mol = 0.425 moles benzoate

benzoic acid: 0.130mol + 0.056mol = 0.186 moles benzoic acid

Concentration is:

0.425 moles benzoate / 1.3L = 0.327M

0.186 moles benzoic acid / 1.3L = 0.143M

Replacing in H-H equation, pH after addition of HCl is:

pH = 4.20 + log₁₀ [0.327M] / [0.143M]

pH = 4.56

As pH = -log [H₃O⁺]

[H₃O⁺] = = 2.75x10⁻⁵M

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