Buffer Capacity and pH After Adding Strong Acid
A student prepares 1.00 L of buffer by mixing 500 mL of 0.10 M acetic acid (Ka = 1.8×10^-5) with 500 mL of 0.10 M sodium acetate. The measured pH is 4.74, matching pKa. The student then adds 10.0 mL of 1.0 M HCl, measures pH, and finds it close to the predicted value. Encouraged, the student proposes to add another 100 mL of 1.0 M HCl to the same buffer and expects only a small pH change because "buffers resist pH change." The instructor asks the student to calculate the pH after each addition, determine the buffer capacity, and explain why the second addition cannot be treated the same way. Assume volumes are additive and temperature is 25°C.
1. Calculate the pH after the first addition of 10.0 mL of 1.0 M HCl. Show the mole balance and the Henderson–Hasselbalch calculation.
Read the worked answer
Initial moles: Acetic acid = 0.500 L × 0.10 mol/L = 0.050 mol. Acetate = 0.500 L × 0.10 mol/L = 0.050 mol. HCl added = 0.0100 L × 1.0 mol/L = 0.010 mol.
HCl converts acetate to acetic acid: Acetate remaining = 0.050 − 0.010 = 0.040 mol. Acetic acid formed = 0.050 + 0.010 = 0.060 mol.
pH = pKa + log([acetate]/[acetic acid]) = 4.74 + log(0.040/0.060) = 4.74 + log(0.6667) = 4.74 − 0.176 ≈ 4.56.
The volume changes to 1.010 L, but the ratio is unaffected because both species are in the same volume. The small pH drop is consistent with buffer action.
2. Calculate the pH after the second addition, using a total of 110.0 mL of 1.0 M HCl added to the original 1.00 L buffer. Explain why the Henderson–Hasselbalch equation no longer applies directly.
Read the worked answer
Total HCl added = 0.110 L × 1.0 mol/L = 0.110 mol.
The original buffer contains only 0.050 mol of acetate. The first 0.050 mol of HCl consumes all the acetate; the remaining HCl is excess strong acid: Excess HCl = 0.110 − 0.050 = 0.060 mol. Total volume = 1.000 + 0.110 = 1.110 L. [H+] from strong acid = 0.060 / 1.110 ≈ 0.0541 M. pH ≈ −log(0.0541) ≈ 1.27.
Henderson–Hasselbalch no longer applies because the conjugate base has been exhausted. The solution is no longer a buffer; it is essentially a dilute strong-acid solution. The pH falls by about 3.5 units, not a small amount.
3. Define buffer capacity in this context, estimate it for the original buffer, and state one change that would increase it without changing the initial pH.
Read the worked answer
Buffer capacity is the amount of strong acid or base that a given volume of buffer can absorb with only a small pH change. It is limited by the number of moles of the conjugate base or acid available.
Here the buffer contains 0.050 mol of acetate in 1.00 L, so it can neutralize roughly 0.050 mol of strong acid before the buffer is destroyed. That is the practical capacity for added H+.
To increase capacity without changing the initial pH, increase the concentrations of both acetic acid and sodium acetate while keeping their ratio at 1:1. For example, 0.50 M acetic acid with 0.50 M sodium acetate has the same pKa and initial pH but about ten times the capacity. Simply adding more acetate would raise the pH and change the buffer.