Diagnosing and Repairing an Acetate Buffer Batch
A teaching laboratory provides 0.100 M acetic acid (pKa 4.76) and 0.100 M sodium acetate. A student must prepare 250.0 mL of pH 4.50 buffer with a total analytical acetate concentration of 0.0800 M. Instead, she mixes 66.0 mL of the acetic acid stock with 134.0 mL of the sodium acetate stock and dilutes to 250.0 mL with deionized water. The pH meter, calibrated at pH 4 and 7, reads 5.06 and is stable. She has 0.100 M HCl and 0.100 M NaOH available, but no more concentrated acid. The instructor asks her to diagnose the ratio error, test the buffer capacity with a 25.0 mL aliquot, and decide whether the batch can be repaired in place without a large dilution. Assume 25°C, activity coefficients near 1, and negligible volume contraction.
1. Calculate the acid-to-base ratio and the stock volumes needed for pH 4.50 at total acetate 0.0800 M in 250.0 mL. What ratio did the student actually use?
Read the worked answer
HH: pH = pKa + log([A−]/[HA]). 4.50 = 4.76 + log r, so log r = −0.26 and r = 10^−0.26 = 0.550. Total acetate = 0.0800 × 0.2500 = 0.0200 mol. Let HA = x and A− = 0.550x; 1.550x = 0.0200, so x = 0.0129 mol HA and A− = 0.00710 mol. At 0.100 M, volumes are 129 mL acetic acid and 71.0 mL sodium acetate, diluted to 250.0 mL. The student used 66.0 mL HA = 0.00660 mol and 134.0 mL A− = 0.0134 mol, so r = 2.03 and pH = 4.76 + log(2.03) = 5.07. She reversed the required proportions: at pH below pKa, HA must exceed A−.
2. She adds 10.0 mL of 0.100 M HCl to a 25.0 mL aliquot of the incorrect buffer. Calculate the approximate new pH and assess whether the buffer has useful capacity near pH 4.50.
Read the worked answer
A 25.0 mL aliquot contains 0.0200 mol × (25.0/250.0) = 0.00200 mol total acetate. From r = 2.03, A− = 0.00134 mol and HA = 0.000660 mol. HCl added = 0.0100 L × 0.100 M = 0.00100 mol, which converts A− to HA: A− = 0.00034 mol, HA = 0.00166 mol. New r = 0.00034/0.00166 = 0.205; pH = 4.76 + log(0.205) = 4.07. The pH falls about 1.0 unit, so the buffer is nearly exhausted by that acid challenge. This is expected because the starting buffer was at pH 5.06, not near its target; at pH 4.50 the ratio 0.55 is closer to pKa and capacity is better, but capacity also depends on total concentration.
3. Can she repair the 250.0 mL batch in place with the available 0.100 M HCl without significantly lowering total acetate concentration? Show the calculation and recommend a practical fix.
Read the worked answer
Target r = 0.550. Let x = mol HCl added: (0.0134 − x)/(0.00660 + x) = 0.550. Solving: 0.0134 − x = 0.00363 + 0.550x; 0.00977 = 1.550x; x = 0.00630 mol. With 0.100 M HCl, volume = 63.0 mL. That raises total volume to 313 mL and lowers total acetate to 0.0200/0.313 = 0.0639 M, a 20% decrease, so the buffer capacity is materially reduced. With only 0.100 M acid, in-place repair is not satisfactory. The practical fix is to discard and remake using 129 mL HA + 71.0 mL A− and dilute to 250.0 mL. If 1.0 M HCl were available, only 6.3 mL would be needed and dilution would be acceptable.