Preparing and testing a phosphate buffer for an enzyme assay
A technician must prepare 250.0 cm³ of a phosphate buffer of pH 7.40 for an enzyme assay. Stock solutions are 0.100 mol dm⁻³ sodium dihydrogen phosphate, NaH2PO4, and 0.100 mol dm⁻³ disodium hydrogen phosphate, Na2HPO4. For H2PO4⁻, pKa2 = 7.20 at 25 °C. The assay is sensitive to pH: the enzyme loses half its activity if pH falls below 7.1 or rises above 7.7. The technician mixes 125.0 cm³ of each stock solution, checks the pH and finds it is 7.20, then adds 5.00 cm³ of 0.100 mol dm⁻³ hydrochloric acid to 100.0 cm³ of the buffer and asks whether the preparation is still fit for the assay. A colleague suggests using an ethanoic acid/sodium ethanoate buffer instead because 'it is a weak acid and will resist pH change better'.
1. Calculate the volumes of the two 0.100 mol dm⁻³ stock solutions needed to make 250.0 cm³ of pH 7.40 buffer. Show whether the technician's 125.0 cm³ + 125.0 cm³ mixture is suitable.
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Use the Henderson–Hasselbalch equation: pH = pKa + log([HPO4²⁻]/[H2PO4⁻]) 7.40 = 7.20 + log r, where r is the base/acid ratio. log r = 0.20, so r = 10^0.20 = 1.585.
Let the volume of NaH2PO4 solution be x cm³. Then the volume of Na2HPO4 solution is 1.585x cm³, and: x + 1.585x = 250.0 2.585x = 250.0 x = 96.7 cm³.
Volume of acid stock = 96.7 cm³; volume of base stock = 250.0 – 96.7 = 153.3 cm³.
Check: 153.3/96.7 = 1.585, so pH = 7.20 + log 1.585 = 7.40.
The technician's equal-volume mixture has [base] = [acid], so pH = pKa = 7.20. That is not the requested 7.40. Although 7.20 is still within the enzyme's 7.1–7.7 range, the buffer is not at its intended pH and has less base reserve than the correct buffer; it should be remade with 96.7 cm³ acid stock and 153.3 cm³ base stock.
2. 5.00 cm³ of 0.100 mol dm⁻³ HCl is added to 100.0 cm³ of the correct buffer. Calculate the new pH and decide whether the buffer remains fit for the assay.
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Amounts in the full 250.0 cm³ correct buffer: n(acid) = 0.100 × 0.0967 = 9.67 × 10⁻³ mol n(base) = 0.100 × 0.1533 = 1.533 × 10⁻² mol
In 100.0 cm³, the fraction is 100.0/250.0 = 0.400: n(acid) = 3.868 × 10⁻³ mol n(base) = 6.132 × 10⁻³ mol
HCl added: n = 0.100 × 0.00500 = 5.00 × 10⁻⁴ mol. The added H⁺ converts base to acid: new n(acid) = 3.868 × 10⁻³ + 5.00 × 10⁻⁴ = 4.368 × 10⁻³ mol new n(base) = 6.132 × 10⁻³ – 5.00 × 10⁻⁴ = 5.632 × 10⁻³ mol
Ratio = 5.632/4.368 = 1.289. pH = 7.20 + log(1.289) = 7.20 + 0.110 = 7.31.
The pH has fallen by about 0.09 units, from 7.40 to 7.31. This is still within the 7.1–7.7 range, so the buffer remains fit for the assay after this addition. The calculation assumes the volumes are additive and activity effects are negligible, which is reasonable for a school treatment.
3. Evaluate the colleague's suggestion to use an ethanoic acid/sodium ethanoate buffer instead, given pKa(ethanoic acid) = 4.76.
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A buffer is most effective when the required pH is close to the pKa of its acid, usually within about one pH unit. At pH 7.40 the ethanoate system would have: ratio [CH3COO⁻]/[CH3COOH] = 10^(7.40 – 4.76) = 10^2.64 ≈ 4.4 × 10².
This means that almost all the acid is present as ethanoate and there is very little undissociated ethanoic acid available to neutralise added acid. The buffer would therefore have a very low capacity towards acid, even though it might resist added base somewhat. The required pH is also 2.64 units above the pKa, far outside the useful buffering range.
Phosphate is a better choice because its pKa2 is 7.20, close to 7.40, giving a ratio of 1.585 and substantial amounts of both acid and base forms. The colleague's statement that a weak acid automatically resists pH change better is too general: the acid must have a pKa near the target pH. The ethanoate suggestion should be rejected for this assay.