Buffer Choice for a Calcium-containing Enzyme Assay
A molecular-biology laboratory needs 1.0 L of 0.10 M buffer at pH 7.40 for an enzyme assay that also contains 1.0 mM CaCl2 and 1.0 mM MgCl2. Available stock solutions, each 0.10 M, are sodium acetate (pKa 4.76), NaH2PO4/Na2HPO4 (pKa2 7.20), Tris base/Tris-HCl (pKa 8.06) and NH4Cl/NH3 (pKa 9.25). A junior researcher proposes phosphate because its pKa is closest to 7.40 and it is the 'standard' biological buffer. The assay is sensitive to phosphate because calcium phosphate and magnesium phosphate are insoluble; Ksp(Ca3(PO4)2) = 2.07×10⁻³³ and Ka3(H3PO4) = 4.5×10⁻¹³. The buffer must also resist addition of 5.0 mmol HCl from the assay quench, and the laboratory can measure pH at 25 °C but not at 37 °C.
1. Select the buffer system for pH 7.40 and calculate the volumes of the 0.10 M stock solutions needed for 1.0 L. Justify the choice against the Ca²⁺/Mg²⁺ constraint.
Read the worked answer
Phosphate has the best pKa match: with pKa2 7.20,
[\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]} = 10^{7.40-7.20} = 1.585.]
For 0.10 M total, [HPO4²⁻] = 0.0613 M and [H2PO4⁻] = 0.0387 M, i.e. 613 mL of 0.10 M Na2HPO4 plus 387 mL of 0.10 M NaH2PO4. However, phosphate is chemically incompatible with the assay: Ca²⁺ and Mg²⁺ form very insoluble phosphates, so the buffer would precipitate the metal ions the enzyme needs. Tris has pKa 8.06, giving
[\frac{[\text{Tris base}]}{[\text{Tris-H}^+]} = 10^{7.40-8.06} = 0.219.]
For 0.10 M total, [Tris base] = 0.0180 M and [Tris-H⁺] = 0.0820 M: 180 mL of 0.10 M Tris base plus 820 mL of 0.10 M Tris-HCl. Acetate and NH4⁺ are too far from pH 7.40 to give useful capacity. Tris does not chelate Ca²⁺ or Mg²⁺, so it is the only listed choice that is chemically compatible, even though its capacity is lower and its pKa is strongly temperature-dependent. The pH should be verified at the assay temperature if possible, because a buffer set at 25 °C will not read 7.40 at 37 °C.
2. Calculate the pH after adding 5.0 mmol HCl to 1.0 L of each candidate buffer (phosphate and Tris) and compare their capacities.
Read the worked answer
Phosphate: initially 61.3 mmol HPO4²⁻ and 38.7 mmol H2PO4⁻. The 5.0 mmol H⁺ converts 5.0 mmol HPO4²⁻ into H2PO4⁻, leaving 56.3 mmol and 43.7 mmol:
[\text{pH} = 7.20 + \log\frac{56.3}{43.7} = 7.20 + 0.110 = 7.31.]
The change is −0.09 pH unit. Tris: initially 18.0 mmol base and 82.0 mmol acid; 5.0 mmol H⁺ consumes base, leaving 13.0 mmol and 87.0 mmol:
[\text{pH} = 8.06 + \log\frac{13.0}{87.0} = 8.06 - 0.826 = 7.23.]
The change is −0.17 pH unit. Phosphate therefore has roughly twice the capacity at pH 7.40, but it is excluded by the Ca²⁺/Mg²⁺ precipitation. With 0.10 M Tris, a 5.0 mmol acid challenge shifts the pH outside a typical ±0.05 tolerance. Practical options are to increase the Tris concentration if the assay tolerates it, reduce the quench volume, or switch to a zwitterionic buffer such as HEPES (pKa about 7.55) that does not precipitate divalent cations.
3. Using Ksp and Ka3, decide whether 1.0 mM Ca²⁺ will precipitate in the 0.10 M phosphate buffer at pH 7.40.
Read the worked answer
At pH 7.40, [H⁺] = 3.98×10⁻⁸ M. For the HPO4²⁻/PO4³⁻ equilibrium,
[K_{a3} = \frac{[\text{H}^+][\text{PO}_4^{3-}]}{[\text{HPO}_4^{2-}]},]
so
[[\text{PO}_4^{3-}] = \frac{4.5\times10^{-13}\times0.0613}{3.98\times10^{-8}} = 6.9\times10^{-7}\ \text{M}.]
The ion product for Ca3(PO4)2 is
[Q = [\text{Ca}^{2+}]^3[\text{PO}_4^{3-}]^2 = (1.0\times10^{-3})^3(6.9\times10^{-7})^2 = 4.8\times10^{-22}.]
Since (K_{sp} = 2.07\times10^{-33}), (Q/K_{sp} \approx 2.3\times10^{11}). Precipitation is strongly favoured even though only a tiny fraction of the phosphate is present as PO4³⁻ at pH 7.40. This confirms that phosphate cannot be used in the presence of 1.0 mM Ca²⁺; the buffer choice must be changed rather than relying on the low PO4³⁻ concentration.