Acetic Acid–Acetate Buffer Near Capacity
A first-year pharmacy student must prepare 100.0 mL of an acetic acid–acetate buffer for a spectrophotometric assay. The stock solutions are 0.100 M CH3COOH and 0.100 M CH3COONa; the student mixes 50.0 mL of the acid with 40.0 mL of the salt and dilutes to the mark with distilled water. The laboratory procedure requires the buffer to resist a pH change of no more than 0.20 unit when 5.0 mL of 0.100 M HCl is added. The pKa of acetic acid is 4.74 at 25 °C, and the student may treat activities as concentrations. A supervisor warns that the buffer may be too close to its capacity limit and asks for a quantitative check before the assay. No solid sodium acetate is available, and the total volume after HCl addition is 105.0 mL.
1. Calculate the pH of the prepared buffer before any HCl is added.
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Moles of CH3COOH = 0.100 mol/L × 0.0500 L = 5.00×10⁻³ mol. Moles of CH3COO⁻ = 0.100 × 0.0400 = 4.00×10⁻³ mol. In the buffer, volume cancels in the ratio, so pH = pKa + log([A⁻]/[HA]) = 4.74 + log(4.00×10⁻³ / 5.00×10⁻³) = 4.74 + log(0.800) = 4.74 − 0.097 = 4.64. The prepared buffer is pH 4.64, not pH 5.0; it is acidic as expected for an acetic acid–acetate buffer.
2. Calculate the pH after adding 5.0 mL of 0.100 M HCl and decide whether the buffer meets the laboratory criterion of a pH change no greater than 0.20 unit.
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HCl added = 0.100 mol/L × 0.0050 L = 5.00×10⁻⁴ mol. It converts acetate to acetic acid: new HA = 5.00×10⁻³ + 0.500×10⁻³ = 5.50×10⁻³ mol; new A⁻ = 4.00×10⁻³ − 0.500×10⁻³ = 3.50×10⁻³ mol. Ratio = 3.50/5.50 = 0.6364. pH = 4.74 + log(0.6364) = 4.74 − 0.196 = 4.54. Change = 4.64 − 4.54 = 0.10 unit, which is less than the allowed 0.20 unit. The buffer passes this test, but the margin is only about a factor of two; the result is near the limit because the initial ratio was already 0.80 rather than 1.0.
3. Suppose the assay instead requires adding 50.0 mL of the same HCl. Predict quantitatively what happens to the pH and state whether the buffer is suitable.
Read the worked answer
50.0 mL of 0.100 M HCl = 5.00×10⁻³ mol, exactly equal to the acetate present. The added H⁺ consumes essentially all A⁻: HA ≈ 5.00×10⁻³ + 5.00×10⁻³ = 1.00×10⁻² mol; A⁻ ≈ 0. Total volume = 100.0 + 50.0 = 150.0 mL, so [HA] ≈ 1.00×10⁻²/0.150 = 0.0667 M. The solution is now essentially a weak acid solution: pH ≈ ½(pKa − log[HA]) = ½(4.74 − log 0.0667) = ½(4.74 + 1.176) = 2.96. The drop from 4.64 is about 1.68 units, far exceeding 0.20. The buffer is not suitable for that procedure; its capacity is exhausted at about 5.00×10⁻³ mol HCl, and a more concentrated buffer or a larger acetate reserve would be required.