A buffer that misses its target pH
A student must prepare 250.0 mL of a pH 5.00 buffer from 0.200 M acetic acid (pKa = 4.76) and 0.200 M sodium acetate. Using the Henderson–Hasselbalch ratio [A⁻]/[HA] = 10^(5.00 − 4.76) = 1.74, she mixes 60.0 mL of the acid with 105.0 mL of the acetate solution (measured with a graduated cylinder) and dilutes to the mark with deionized water. The pH meter, calibrated that morning at pH 4.00 and 7.00, reads 4.94. She repeats the preparation with volumetric pipettes and gets 4.95. Her calculation predicts 5.00. The ionic strength of the final solution is about 0.084 M. A demonstrator suggests the discrepancy is 'just a dirty electrode'.
1. Check the arithmetic behind the predicted pH, then decide whether the 0.05–0.06 unit gap is more likely a calculation error, random error, or a real systematic effect.
Read the worked answer
Amounts: n(HA) = 0.200 × 0.0600 = 12.0 mmol; n(A⁻) = 0.200 × 0.1050 = 21.0 mmol. Ratio = 21.0/12.0 = 1.75, log = 0.243, pH = 4.76 + 0.243 = 5.00. The arithmetic is correct, and the volume terms cancel because both stock solutions are 0.200 M.
Reproducibility with volumetric glassware (4.94 vs 4.95) rules out random error of that size. A systematic effect is expected: the student used concentrations in place of activities.
Davies equation at I = 0.084 M, z = 1 for acetate: log γ = −0.51(√I/(1+√I) − 0.3I) = −0.51(0.2247 − 0.0252) = −0.102, so γ(A⁻) ≈ 0.79 while γ(HA) ≈ 1 (neutral). pH = pKa + log([A⁻]/[HA]) + log(γ_A/γ_HA) = 5.00 − 0.10 = 4.90.
The activity correction has the right sign and a comparable magnitude, so the electrode is not the first suspect. The residual 0.03–0.04 unit is within the combined uncertainty of pKa, calibration buffers, and glassware. To actually reach pH 5.00, the ratio must be raised to 10^(0.24 + 0.102) ≈ 2.2, i.e. about 132 mL of acetate stock rather than 105 mL.
2. She adds 5.00 mL of 0.100 M HCl to 100.0 mL of the buffer. Calculate the new pH and compare it with adding the same acid to 100.0 mL of pure water.
Read the worked answer
In 100.0 mL of buffer: n(HA) = 12.0 × 0.400 = 4.80 mmol; n(A⁻) = 21.0 × 0.400 = 8.40 mmol. Initial pH = 4.76 + log(8.40/4.80) = 4.76 + 0.243 = 5.00.
HCl added = 0.100 × 5.00 = 0.500 mmol. Strong acid converts acetate to acetic acid: n(A⁻) = 8.40 − 0.50 = 7.90 mmol; n(HA) = 4.80 + 0.50 = 5.30 mmol. pH = 4.76 + log(7.90/5.30) = 4.76 + 0.174 = 4.93.
So 0.5 mmol of strong acid moves the buffer by only 0.07 unit. In 100 mL of water (final volume 105 mL), [H⁺] = 0.500/105 = 4.76 × 10⁻³ M, pH = 2.32 - a drop of about 4.7 units. The buffer is doing real work; the earlier 0.06-unit offset is a separate, systematic calibration/activity issue.
3. A later step requires 100.0 mL of this buffer to absorb 5.00 mmol of HCl with a pH change below 0.10 unit. Is the formulation adequate? Support your answer with numbers and state how to fix it.
Read the worked answer
No. With 5.00 mmol of HCl: n(A⁻) = 8.40 − 5.00 = 3.40 mmol; n(HA) = 4.80 + 5.00 = 9.80 mmol. pH = 4.76 + log(3.40/9.80) = 4.76 − 0.460 = 4.30. ΔpH = 0.70 unit, seven times the tolerance. The buffer is not exhausted (acetate remains), but the ratio has collapsed from 1.75 to 0.35.
Buffer capacity scales with total buffer concentration, not with the ratio alone. To keep ΔpH < 0.10, the added acid must be a small fraction of n(A⁻). Roughly, ΔpH ≈ 0.434 × Δn(A⁻)/n(A⁻); for ΔpH = 0.10, Δn/n(A⁻) ≈ 0.23, so n(A⁻) must be at least 5.00/0.23 ≈ 22 mmol in 100 mL. A total buffer concentration around 0.45–0.50 M (with the ratio still giving pH 5.00) would work, or the challenge volume/concentration must be reduced.