Checking a Phosphate Buffer Recipe and Its Capacity
A teaching lab needs 500.0 mL of 0.100 M sodium phosphate buffer at pH 7.40 for an enzyme assay. The stockroom has solid anhydrous NaH2PO4 (MW 119.98 g/mol) and Na2HPO4 (MW 141.96 g/mol), both assayed at ≥99%. The relevant pKa is 7.20 at 25 °C, and the assay runs at 25 °C. The lab has a calibrated pH meter, a 500.0 mL volumetric flask, and 1.00 M HCl and 1.00 M NaOH. A student weighs the two salts, dissolves them in about 400 mL of deionized water, adjusts the pH with HCl, and brings the volume to 500.0 mL. Later, a postdoc adds 5.0 mL of 0.100 M HCl to 100.0 mL of the finished buffer to test its capacity. The student's notebook lists 4.36 g Na2HPO4 and 2.32 g NaH2PO4, which the student claims should give pH 7.40 'because the ratio is about 1.6.'
1. Check the student's masses and show whether they produce pH 7.40 in 500.0 mL.
Read the worked answer
Use the Henderson–Hasselbalch equation: pH = pKa + log([HPO4^2−]/[H2PO4^−]). 7.40 = 7.20 + log(r), so log(r) = 0.20 and r = 10^0.20 = 1.585.
Total phosphate = 0.500 L × 0.100 M = 0.0500 mol. Let acid = x and base = 1.585x. Then 2.585x = 0.0500, so x = 0.01934 mol acid and base = 0.03066 mol.
Mass NaH2PO4 = 0.01934 × 119.98 = 2.32 g. Mass Na2HPO4 = 0.03066 × 141.96 = 4.35 g. The student's 4.36 g and 2.32 g are within rounding. The actual ratio is 0.03071/0.01934 = 1.588, giving pH = 7.20 + log(1.588) = 7.401. The masses are acceptable, but the pH must still be checked and adjusted because weighing error and activity coefficients shift the true pH.
2. Calculate the pH after adding 5.0 mL of 0.100 M HCl to 100.0 mL of the buffer. Has the buffer exceeded its capacity?
Read the worked answer
In 100.0 mL there is 0.0100 mol total phosphate: acid = 0.003868 mol and base = 0.006132 mol. Added HCl = 0.0050 L × 0.100 M = 0.000500 mol.
HCl converts base to acid: new base = 0.006132 − 0.000500 = 0.005632 mol; new acid = 0.003868 + 0.000500 = 0.004368 mol. Ratio = 0.005632/0.004368 = 1.289.
pH = 7.20 + log(1.289) = 7.20 + 0.110 = 7.31. The pH changes by only about 0.09 unit, so the buffer absorbs the acid and is not exhausted. The capacity is limited but adequate for this small addition.
3. How much 1.00 M HCl would be needed to consume all the base in 100.0 mL of the buffer, and what does that show about buffer capacity?
Read the worked answer
The base in 100.0 mL is 0.006132 mol. At 1.00 M, HCl volume = 0.006132 L = 6.13 mL. At that point the buffer is essentially gone, and further acid causes a steep pH drop.
Buffer capacity is therefore roughly 0.0061 mol per 100 mL, or 0.061 mol/L, not unlimited. The postdoc's 5.0 mL addition used 0.000500 mol, about 8% of the capacity. For the assay, avoid adding strong acid directly to the stock; dilute the buffer first or use dialysis, and check pH after any addition.