Designing a pH 7.50 Buffer for a Zinc-Dependent Enzyme
A biochemistry group needs 500.0 mL of a 0.100 M buffer at pH 7.50 for an assay. The enzyme being studied requires a tightly bound Zn²⁺ ion as a cofactor, and the assay must run for 90 minutes at 25 °C. The stockroom holds solid NaH₂PO₄ (M = 120.0 g mol⁻¹) and Na₂HPO₄ (M = 142.0 g mol⁻¹) (phosphoric acid pKa₁ = 2.15, pKa₂ = 7.20, pKa₃ = 12.35), solid HEPES in its free-acid form (M = 238.3 g mol⁻¹, pKa = 7.55), 1.00 M HCl and 1.00 M NaOH. As a quality check, 100.0 mL of the prepared buffer will be challenge-tested with 2.00 mL of 1.00 M HCl. Assume activity coefficients are unity and that volumes are additive.
1. Calculate the masses of NaH₂PO₄ and Na₂HPO₄ needed for the phosphate buffer at pH 7.50, showing the equilibrium reasoning.
Read the worked answer
Use the Henderson–Hasselbalch expression for the H₂PO₄⁻/HPO₄²⁻ couple, whose pKa is pKa₂ = 7.20 (the only pKa near 7.50; pKa₁ and pKa₃ are irrelevant at this pH).
pH = pKa₂ + log([HPO₄²⁻]/[H₂PO₄⁻]) 7.50 − 7.20 = 0.30 = log r, so r = 10⁰·³⁰ = 2.00.
The chosen buffer has a base:acid ratio of 2:1 - close enough to 1:1 that the buffer still has useful capacity in both directions. Total phosphate = 0.500 L × 0.100 M = 0.0500 mol, split as n(HPO₄²⁻) = 0.0500 × 2/3 = 0.0333 mol; n(H₂PO₄⁻) = 0.0167 mol.
Mass Na₂HPO₄ = 0.0333 × 142.0 = 4.73 g Mass NaH₂PO₄ = 0.0167 × 120.0 = 2.00 g
Dissolve both in about 400 mL of water, check the pH with a calibrated electrode, and adjust with 1.00 M NaOH or HCl before making up to 500.0 mL - the calculation gives the target, not a guaranteed result.
2. The challenge test adds 2.00 mL of 1.00 M HCl to 100.0 mL of the buffer. Calculate the new pH and comment on whether the buffer is adequate for a 90-minute assay.
Read the worked answer
Moles in the 100.0 mL aliquot: acid H₂PO₄⁻ = 0.00333 mol, base HPO₄²⁻ = 0.00667 mol. HCl added = 0.00200 L × 1.00 = 0.00200 mol. It converts base into acid: base = 0.00667 − 0.00200 = 0.00467 mol; acid = 0.00333 + 0.00200 = 0.00533 mol.
pH = 7.20 + log(0.00467/0.00533) = 7.20 + log(0.876) = 7.20 − 0.057 = 7.14
The pH falls by 0.36 units - a change of about 5 % in [H⁺]. Strong acid added was 0.0020 mol against a total buffer pool of 0.0100 mol, i.e. 20 % of the pool, which is near the useful limit of a buffer (roughly 10 % of the pool per pH unit). For an enzyme assay with a pH optimum near 7.5, a 0.36-unit drift during 90 minutes is a real risk; a 0.200 M buffer, or a larger volume, would be the safer choice. Note also that the buffer capacity is symmetric around pH = pKa = 7.20, not around the working pH of 7.50.
3. State the chemical problem with using phosphate for this particular assay, and show, with a calculation, how you would reformulate using HEPES instead.
Read the worked answer
Problem. Phosphate precipitates Zn²⁺ as sparingly soluble Zn₃(PO₄)₂, stripping the cofactor out of the enzyme and giving non-reproducible, time-dependent inactivation over the 90-minute run. A buffer must not sequester the metal ion it is meant to leave available.
Reformulation with HEPES (pKa 7.55): 7.50 = 7.55 + log([A⁻]/[HA]) → log r = −0.05 → r = 0.891. So the base form is 0.891/(1 + 0.891) = 47.1 % of total buffer. Total = 0.500 L × 0.100 M = 0.0500 mol → n(A⁻) = 0.0236 mol, n(HA) = 0.0264 mol. Mass of HEPES free acid = 0.0500 × 238.3 = 11.9 g.
Procedure. Suspend 11.9 g of HEPES free acid in ~400 mL of water and titrate with 1.00 M NaOH until pH 7.50 (about 23.6 mL, i.e. 0.0236 mol OH⁻ to convert that fraction to the sodium salt), then dilute to 500.0 mL and re-check the pH. HEPES is chosen because it buffers in the right range and does not precipitate the zinc.