What a half-neutralised acetate buffer can still absorb
A first-year student must prepare 100.0 mL of an acetate buffer of pH 4.76 and then test its capacity. She pipettes 50.0 mL of 0.100 M acetic acid (pKa = 4.76) into a 100 mL volumetric flask, adds 25.0 mL of 0.100 M NaOH and makes up to the mark with distilled water. The pH meter reads 4.76. She then adds 5.0 mL of 0.100 M HCl from a burette and records a small fall in pH. Her partner argues that the buffer was already half destroyed because half of the acid had been neutralised by the NaOH, and predicts that the pH will now collapse. Later, a second bottle labelled 0.100 M HCl is found to be 1.00 M, and the same 5.0 mL volume is added to a freshly prepared 100 mL portion of the buffer. Assume volumes are additive and activities equal concentrations.
1. Show why the initial pH equals pKa, and evaluate the partner's claim that a half-neutralised acid is a destroyed buffer.
Read the worked answer
n(CH3COOH) = 0.100 mol/L × 0.0500 L = 5.00 mmol. n(NaOH) = 0.100 × 0.0250 = 2.50 mmol.
The strong base converts acid to acetate quantitatively, so after mixing: CH3COO− = 2.50 mmol and CH3COOH = 5.00 − 2.50 = 2.50 mmol. Then
pH = pKa + log([A−]/[HA]) = 4.76 + log 1 = 4.76.
The partner's error is to treat the ratio as a measure of remaining usefulness. A 1:1 ratio is the best possible buffer composition, not a destroyed one; the reserve that matters is the absolute amount of each component, here 2.50 mmol of each. That reserve can absorb roughly 2.5 mmol of strong acid or base before the ratio leaves the useful 10:1 to 1:10 range. A buffer dilute in both components at 1:1 has the correct pH but poor capacity - the two ideas must not be conflated.
2. Calculate the pH after adding 5.0 mL of 0.100 M HCl, and compare the change with adding the same amount of HCl to 100 mL of pure water.
Read the worked answer
Added H+ = 0.100 mol/L × 0.0050 L = 0.500 mmol. This protonates acetate:
A− = 2.50 − 0.50 = 2.00 mmol; HA = 2.50 + 0.50 = 3.00 mmol.
pH = 4.76 + log(2.00/3.00) = 4.76 − 0.18 = 4.58.
In 105 mL of pure water the same 0.500 mmol gives [H+] ≈ 0.500/105 = 4.8 × 10−3 M, so pH ≈ 2.3. The buffer limits the change to 0.18 units instead of about 4.7 units. The pH does fall slightly, and it must: a buffer resists change, it does not abolish it. The small fall is the expected signature of a working buffer, not evidence of collapse.
3. Predict the pH when 5.0 mL of the 1.00 M acid is added to a fresh portion, and explain why Henderson–Hasselbalch can no longer be used.
Read the worked answer
Added H+ = 1.00 mol/L × 0.0050 L = 5.00 mmol, which is twice the acetate present (2.50 mmol).
- All 2.50 mmol of A− is converted to HA, leaving 2.50 mmol of strong acid unreacted in a total volume of about 105 mL.
- [H+] ≈ 2.50/105 = 0.0238 M, so pH ≈ 1.62.
The buffer is now exhausted: [A−] is essentially zero, so the ratio [A−]/[HA] is near zero and the Henderson–Hasselbalch expression no longer describes the system. The pH is set by the excess strong acid. The judgement being tested is that capacity is a mole-count limit: the same buffer composition would have coped easily with 2.5 mmol or less of strong acid, but not with a two-fold excess, and no ratio calculation can rescue it.