Buffer Capacity After an Acid Addition
A laboratory assistant prepares 500 mL of an acetic acid–acetate buffer for a spectrophotometry experiment. The solution is made by mixing 250 mL of 0.100 M acetic acid with 250 mL of 0.100 M sodium acetate. The label on the acetic acid bottle gives Ka = 1.8 × 10⁻⁵ at 25 °C. The experiment requires the pH to remain between 4.60 and 4.80 after small acid additions. A student accidentally adds 10.0 mL of 0.100 M HCl and measures the pH with a calibrated meter, obtaining 4.70. The assistant must decide whether the buffer is still usable, calculate the expected pH after the addition, and explain why the observed pH is close to the calculated value even though the solution is no longer exactly equimolar. A second concern is how much more 0.100 M HCl could be added before the lower tolerance limit is crossed.
1. Calculate the initial pH and the pH after adding 10.0 mL of 0.100 M HCl. Is the buffer still within the required range? Show the mole-table reasoning.
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Initial moles: HA = 0.250 L × 0.100 M = 0.0250 mol; A⁻ = 0.0250 mol. pH = pKa + log(A⁻/HA). pKa = -log(1.8 × 10⁻⁵) = 4.745; log(1) = 0, so initial pH = 4.745. Added H⁺ = 0.0100 L × 0.100 M = 0.00100 mol. H⁺ converts A⁻ to HA: HA = 0.0260 mol, A⁻ = 0.0240 mol. New pH = 4.745 + log(0.0240/0.0260) = 4.745 + log(0.9231) = 4.745 - 0.0348 = 4.710. The buffer is still within the required 4.60–4.80 range. The measured 4.70 is close to the calculated value; small differences can come from activity coefficients, meter calibration, temperature, and dissolved CO₂. The ratio, not the total volume, controls the Henderson–Hasselbalch calculation, so the added volume does not change the ratio term.
2. Determine the additional volume of 0.100 M HCl that can be added before pH falls to 4.60. Use the buffer ratio at that pH and account for the HCl already added.
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At pH 4.60, 4.60 = 4.745 + log(A⁻/HA), so log(A⁻/HA) = -0.145 and A⁻/HA = 10⁻⁰·¹⁴⁵ ≈ 0.716. Let n be total moles of HCl added from the start. Then HA = 0.0250 + n and A⁻ = 0.0250 - n. Solve (0.0250 - n)/(0.0250 + n) = 0.716: 0.0250 - n = 0.0179 + 0.716n; 0.0071 = 1.716n; n = 0.00414 mol. Since 0.00100 mol has already been added, additional HCl = 0.00314 mol. Volume = 0.00314/0.100 = 0.0314 L = 31.4 mL. Thus roughly 31 mL more of 0.100 M HCl can be added before the pH reaches the lower tolerance limit. This is an estimate because activity coefficients and CO₂ absorption shift measured pH, and the buffer capacity falls as the ratio moves away from 1.