Designing a pH 7.40 phosphate buffer and the tempting pH 6.00 wash
A technician must prepare 250 cm³ of a phosphate buffer of pH 7.40 for an enzyme assay run at 25 °C, using solid sodium dihydrogen phosphate (NaH2PO4, Mr = 120.0) and disodium hydrogen phosphate (Na2HPO4, Mr = 142.0). For H2PO4⁻ ⇌ HPO4²⁻ + H⁺, pKa = 7.20 at 25 °C. The technician has 5.00 g of each solid and deionised water. One colleague suggests mixing equal masses of the two salts 'because a 50:50 buffer is the best buffer'. A second colleague wants a pH 6.00 wash made by adding hydrochloric acid to a portion of the same phosphate mixture. The assay is sensitive to ionic strength above 0.20 mol dm⁻³, and the technician proposes a total phosphate concentration of 0.100 mol dm⁻³.
1. Calculate the masses of the two salts needed to make 250 cm³ of pH 7.40 buffer at a total phosphate concentration of 0.100 mol dm⁻³.
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Ratio of base to acid Henderson–Hasselbalch: pH = pKa + log([HPO4²⁻]/[H2PO4⁻]) 7.40 = 7.20 + log r, so log r = 0.20 and r = 10^0.20 = 1.58
Total phosphate n(total) = 0.100 mol dm⁻³ × 0.250 dm³ = 0.0250 mol Let n(acid) = x and n(base) = 1.58x, then x + 1.58x = 0.0250 x = 0.0250 ÷ 2.58 = 9.69 × 10⁻³ mol H2PO4⁻ n(HPO4²⁻) = 0.0250 − 0.00969 = 1.53 × 10⁻² mol
Masses m(NaH2PO4) = 9.69 × 10⁻³ × 120.0 = 1.16 g m(Na2HPO4) = 1.53 × 10⁻² × 142.0 = 2.17 g
Practical check Both masses are below the 5.00 g available. Dissolve the salts in about 200 cm³ of water, check the pH with a calibrated meter (the calculated ratio uses concentrations, whereas the meter responds to activity), then make up to 250 cm³. The ionic strength of this mixture is roughly 0.06 mol dm⁻³, comfortably below the assay's 0.20 limit.
2. Assess the first colleague's 'equal masses is best' suggestion.
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What equal masses actually give Equal masses do not mean equal moles, because the two salts have different Mr values. For 5.00 g of each in 250 cm³: n(NaH2PO4) = 5.00 ÷ 120.0 = 0.0417 mol n(Na2HPO4) = 5.00 ÷ 142.0 = 0.0352 mol r = 0.0352 ÷ 0.0417 = 0.845 pH = 7.20 + log(0.845) = 7.20 − 0.073 = 7.13
So the mixture would be about 0.27 pH units too acidic for the assay - enough to matter for an enzyme with a sharp optimum.
'50:50 is best' A 1:1 mole ratio gives pH = pKa = 7.20, and maximum buffer capacity does indeed occur at pH = pKa. But a buffer works well within about ±1 unit of its pKa, and at pH 7.40 (0.20 units above pKa) the mole ratio 1.58 still gives close to the maximum capacity. The colleague's rule of thumb is therefore doubly wrong here: equal masses are not equal moles, and the required pH is not the pKa anyway. The correct approach is to calculate the ratio from the required pH, not to guess a 'balanced' recipe.
3. Evaluate the second colleague's proposal for a pH 6.00 wash made by adding HCl to a portion of the phosphate mixture.
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What the pH 6.00 wash would require At pH 6.00: log r = 6.00 − 7.20 = −1.20, so r = [HPO4²⁻]/[H2PO4⁻] = 10⁻¹²⁰ = 0.063. Only about 6% of the phosphate would be present as HPO4²⁻, with ~94% as H2PO4⁻. The mixture is 1.2 pH units below the pKa, outside the ±1 working range, so its buffer capacity is poor: adding a small amount of acid or base would change the pH markedly.
Practical objections Making it by adding HCl to the pH 7.40 buffer consumes the HPO4²⁻ and adds Cl⁻, increasing ionic strength - the assay already has a 0.20 mol dm⁻³ ceiling. The volume also changes as acid is added, so the final concentration is no longer the designed 0.100 mol dm⁻³.
Judgement The proposal is chemically possible but technically poor. A buffer whose pKa lies near 6.0–6.4 (for example MES, pKa ≈ 6.1, or the third dissociation of citric acid, pKa ≈ 6.40) would give the required pH at a near 1:1 ratio, with far better capacity and a defined, low ionic strength. The pH 7.40 buffer should be kept for the assay and a separate buffer prepared for the wash.