Buffer Capacity After Strong-Acid Addition
A student prepares a buffer by mixing 250.0 mL of 0.100 M weak acid HA (pKa = 4.76) with 150.0 mL of 0.100 M NaOH. A calibrated pH meter reads 4.94, close to the predicted value. The student then adds 10.0 mL of 1.00 M HCl and records a new pH. The stockroom asks whether the same buffer could absorb another 25.0 mL of the 1.00 M HCl without being exhausted, and whether replacing the weak acid with 0.100 M HCl at the same total volume would give a better buffer. The student has 400.0 mL of buffer before the HCl addition and must account for dilution. The final volume after HCl is 410.0 mL. The goal is to judge buffer capacity rather than simply plug numbers into the Henderson–Hasselbalch equation.
1. Calculate the pH after the 10.0 mL addition of 1.00 M HCl and show whether the buffer is still functioning.
Read the worked answer
Initial moles HA = 0.2500 L × 0.100 M = 0.0250 mol. Moles NaOH = 0.1500 L × 0.100 M = 0.0150 mol. After neutralization: HA = 0.0100 mol and A− = 0.0150 mol. pH = 4.76 + log(0.0150/0.0100) = 4.76 + 0.176 = 4.94. Added HCl = 0.0100 L × 1.00 M = 0.0100 mol. This converts A− to HA: A− = 0.0050 mol and HA = 0.0200 mol. New pH = 4.76 + log(0.0050/0.0200) = 4.76 − 0.602 = 4.16. The buffer is still functioning because both HA and A− remain, but the pH has moved 0.78 units, close to the practical capacity limit of about 1 pH unit. The volume is 410.0 mL, but the ratio makes volume unnecessary in the Henderson–Hasselbalch calculation.
2. Can the buffer absorb another 25.0 mL of 1.00 M HCl without being exhausted? Support with moles.
Read the worked answer
Another 25.0 mL × 1.00 M = 0.0250 mol H+. After the first HCl addition, A− = 0.0050 mol. The added H+ exceeds A− by 0.0200 mol, so all A− is consumed and 0.0200 mol strong acid remains in 435 mL. [H+] ≈ 0.0200 mol / 0.435 L = 0.0460 M, so pH ≈ 1.34. The buffer is exhausted well before 25.0 mL. The maximum additional HCl that can be absorbed without leaving excess strong acid is 0.0050 mol, or 5.0 mL of 1.00 M HCl. At that point HA = 0.0250 mol and A− is nearly zero, so the solution is essentially a weak acid solution, not a buffer.
3. The stockroom suggests replacing HA with 0.100 M HCl at the same total volume. Would that make a better buffer? Explain.
Read the worked answer
No. A buffer requires comparable concentrations of a weak acid and its conjugate base. HCl is a strong acid; it is fully dissociated and has no conjugate base reservoir to neutralize added OH− or added H+. A solution of HCl would have a very low pH and essentially no buffer capacity; adding base would cause a steep pH jump. The original HA/A− mixture is a genuine buffer because the weak acid neutralizes added base and A− neutralizes added acid. Replacing HA with HCl would remove the equilibrium that resists pH change.