Buffer Capacity at the Bench: How Much Acid Before the Wheels Come Off
A teaching lab supplies 0.100 M acetic acid (pKa = 4.76) and 0.100 M sodium acetate. A student mixes 50.0 mL of each, checks the pH with a freshly calibrated meter, then adds 1.00 mL of 1.00 M HCl with stirring and re-reads the pH. Before touching another pipette, the instructor asks the student to predict two things: how the same 1.00 mL of 1.00 M HCl would change the pH of 101 mL of pure water, and how many millimoles of strong acid the buffer can absorb before its pH behavior becomes qualitatively different. Assume volumes are additive, activity coefficients are unity, and temperature is 25 degrees C.
1. Calculate the pH of the initial mixture and the pH after adding 1.00 mL of 1.00 M HCl.
Read the worked answer
Initial buffer
- n(HA) = 0.100 M x 0.0500 L = 5.00 mmol
- n(A-) = 0.100 M x 0.0500 L = 5.00 mmol
- pH = pKa + log([A-]/[HA]) = 4.76 + log(1) = 4.76
After 1.00 mL of 1.00 M HCl (1.00 mmol H+) The strong acid is consumed stoichiometrically by acetate, not by dilution:
- A- : 5.00 - 1.00 = 4.00 mmol
- HA : 5.00 + 1.00 = 6.00 mmol
- Volume change (101 mL total) cancels in the ratio, so pH = 4.76 + log(4.00/6.00) = 4.76 - 0.18 = 4.58
Net change is only -0.18 pH units. The key move is converting everything to millimoles and tracking the conjugate pair, not plugging concentrations into a strong-acid formula.
2. Compare the buffer result with the pH of 101 mL of pure water receiving the same 1.00 mL of 1.00 M HCl, and explain the difference mechanistically.
Read the worked answer
Pure water case
- 1.00 mmol H+ in 0.101 L = 9.9 x 10^-3 M
- pH = -log(9.9 x 10^-3) = 2.00
Why the buffer barely moves
- In water, essentially every added proton remains free; pH tracks the total amount added divided by volume.
- In the buffer, the added proton is consumed: A- + H+ -> HA. Free [H+] is not set by how much acid was added but by the ratio of the conjugate pair through Ka.
- Because the ratio went from 1.0 to 0.67, the pH shifts only logarithmically (0.18 units) even though the absolute amount of strong acid is the same.
Judgment point: a buffer's protection comes from reservoir size (moles of A-) plus a favorable ratio. Water has no reservoir, so the same aliquot produces a ~2.8-unit collapse.
3. Estimate the buffer's capacity for strong acid and predict how the pH behaves at and just past that limit.
Read the worked answer
Capacity The only proton acceptor present is acetate: 5.00 mmol. That corresponds to 5.00 mL of 1.00 M HCl. At that point [A-] approaches zero and the Henderson-Hasselbalch ratio goes to zero, so the equation stops being a useful description - the pH plunges with each further drop.
At the limit The solution becomes ~10.0 mmol acetic acid in ~106 mL, i.e. about 0.094 M HA:
- pH = 0.5(pKa - log C) = 0.5(4.76 + 1.03) = ~2.9
- Any additional HCl after this point appears essentially as free strong acid, so pH falls steeply (approaching the water-like behavior).
Practical reading Buffering is best near a 1:1 ratio; the useful window is roughly one pH unit on either side of the pKa. Here the student used half of the total capacity in a single 1.00 mL addition (1.00 of 5.00 mmol), which is why the ratio moved appreciably. A weaker acid (lower pKa relative to the target) or a larger acetate reservoir would resist better.