Acetate buffer: two errors that a pH meter will not catch
A student must prepare 250.0 mL of a pH 5.00 acetate buffer of total acetate concentration 0.100 mol L⁻¹, starting from 0.200 mol L⁻¹ acetic acid (pKa = 4.76) and 0.200 mol L⁻¹ NaOH. The student's notebook plan reads: 'Pipette 25.0 mL of the acetic acid into a 250 mL volumetric flask, add 12.5 mL of the NaOH, make up to the mark with deionised water, check with the pH meter.' A demonstrator glances at the plan and says the volumes are wrong in two independent ways: one error affects the concentration and one affects the pH. She also warns that adding NaOH dropwise at the end to 'hit 5.00' would be guesswork rather than buffer design. The stock bottle contains 500 mL of the acetic acid.
1. Identify both errors in the plan and calculate the volumes the student should actually use.
Read the worked answer
Error 1 - concentration. 25.0 mL of 0.200 mol L⁻¹ acid contains 5.00 mmol of acetate; made up to 250.0 mL that is 0.0200 mol L⁻¹, one fifth of the required 0.100 mol L⁻¹.
Error 2 - pH. 12.5 mL of 0.200 mol L⁻¹ NaOH is 2.50 mmol, which neutralises exactly half of the 5.00 mmol of acid, so [HA] = [A⁻] and pH = pKa = 4.76, not 5.00.
Correct recipe. Total acetate needed = 0.100 × 0.2500 = 25.0 mmol, so acid volume = 25.0 / 0.200 = 125.0 mL.
Ratio from Henderson–Hasselbalch: [A⁻]/[HA] = 10^(5.00 − 4.76) = 10^0.24 = 1.74. So [A⁻] = 25.0 × 1.74/2.74 = 15.9 mmol and [HA] = 25.0 − 15.9 = 9.1 mmol.
Converting 15.9 mmol of HA into A⁻ requires 15.9 mmol NaOH = 15.9/0.200 = 79.4 mL.
Method: mix 125.0 mL acid with 79.4 mL NaOH, allow to cool to room temperature, then dilute to 250.0 mL. The total volume before dilution (204.4 mL) is not the final volume - diluting to the mark does not change the ratio, so the pH is set by the mole ratio, not by the volume.
2. Does making up to the mark with water change the pH? Should the student use the pH meter to fine-tune with NaOH after dilution?
Read the worked answer
Dilution does not change the ratio [A⁻]/[HA], so Henderson–Hasselbalch predicts no pH change at all to first order. The second-order effect is on activity coefficients: diluting lowers the ionic strength, which changes the measured (practical) pH by only a few hundredths at these concentrations - well inside the ±0.02 typically achievable with a well-calibrated meter. So the buffer should not need re-adjustment after dilution.
Fine-tuning after dilution is poor practice for three reasons. First, adding NaOH after the volume is set changes the total acetate concentration and therefore the buffer capacity you were asked to deliver, and the volume is no longer 250.0 mL. Second, it is uncontrolled: each drop changes the ratio non-linearly near the ends of the buffer range and you cannot easily undo an overshoot without adding acid and chasing the reading. Third, it masks a recipe error rather than fixing it - if the meter reads 4.94 you should suspect the pipetting, the stock concentration or the calibration, not titrate blindly.
The defensible workflow is: calibrate the meter against fresh pH 4.01 and 7.00 buffers, adjust the pH before final dilution if it is grossly wrong, make up to the mark, then verify and record. A reading near pKa = 4.76 is a strong warning sign that the student has made the exact error in the plan, since that is the value you get whenever [HA] = [A⁻].
3. Suppose only 50.0 mL of the 0.200 mol L⁻¹ acetic acid remains. Can the specified buffer still be made? Give two options and say how you would choose between them.
Read the worked answer
50.0 mL × 0.200 = 10.0 mmol of acetate - 40% of the 25.0 mmol needed for 250 mL at 0.100 mol L⁻¹, so the specified buffer cannot be made from that stock alone. Two options:
Option A - keep the concentration, reduce the volume. Make 100.0 mL at 0.100 mol L⁻¹, which needs exactly 10.0 mmol. Required A⁻ = 10.0 × 1.74/2.74 = 6.35 mmol and HA = 3.65 mmol. Use all 50.0 mL of acid (10.0 mmol) and neutralise 6.35 mmol of it with 6.35/0.200 = 31.7 mL of NaOH, then dilute to 100.0 mL. Same pH, same concentration, a quarter of the volume.
Option B - keep the volume, reduce the concentration. Dilute the same 10.0 mmol to 250.0 mL, giving 0.040 mol L⁻¹, with 6.35 mmol A⁻ and 3.65 mmol HA. The pH is identical but the buffer capacity is 40% of specification.
Choose A if the method only needs a small working volume and the assay is judged on pH and molarity; choose B only if the method does not state a minimum buffer capacity. Because the capacity requirement is usually in the method, A is normally correct. Either way, do the mole arithmetic before picking up the pipette - the mistake students make is to start from the written volumes and then dilute the error.