Choosing a pH 7.40 buffer compatible with calcium and with added acid
An enzyme assay requires 1.00 L of buffer at pH 7.40 ± 0.10 and must tolerate addition of 5.0 mmol of strong acid from the substrate stock. The assay also requires 2.0 mmol L−1 CaCl2 because the enzyme is calcium-dependent. The laboratory has 0.100 M acetic acid/sodium acetate (pKa 4.76), 0.100 M NaH2PO4/Na2HPO4 (pKa2 7.21), 0.100 M Tris/Tris-HCl (pKa 8.06) and 0.100 M boric acid/borate (pKa 9.24), plus 0.100 M HCl and NaOH. A student proposes acetate buffer because pH 7.4 is close to neutral and plans to adjust with NaOH. A second student prefers phosphate because its pKa is nearest 7.4. The supervisor is concerned about calcium phosphate precipitation and about buffer capacity after acid addition. Ka3 of phosphoric acid is 4.8×10−13; Ksp of Ca3(PO4)2 is 2.1×10−33.
1. Evaluate the acetate proposal quantitatively. Why is acetate unsuitable despite pH 7.4 being near neutral?
Read the worked answer
Acetate is a poor choice. pH − pKa = 7.40 − 4.76 = 2.64, so [A−]/[HA] = 10^2.64 ≈ 4.4×102. In 0.100 M total acetate, [A−] ≈ 0.0998 M and [HA] ≈ 0.00023 M. There is almost no acid form to consume added H+, so buffer capacity is negligible.
Adding 5.0 mmol HCl to 1.00 L would convert nearly all remaining acetate: [HA] ≈ 0.00523 M, [A−] ≈ 0.0948 M, ratio ≈ 18.1, pH = 4.76 + log 18.1 = 6.02. That is far outside 7.40 ± 0.10. Adjusting the initial pH with NaOH does not create capacity at 7.40; it simply makes the same tiny [HA] from the other direction. Acetate should be rejected on both pKa and capacity grounds.
2. For a phosphate buffer of total phosphate 0.100 M, calculate the volumes of 0.100 M NaH2PO4 and Na2HPO4 needed for pH 7.40. Then calculate the pH after adding 5.0 mmol HCl and judge whether the capacity requirement is met.
Read the worked answer
Use pH = pKa2 + log([HPO4^2−]/[H2PO4−]): 7.40 = 7.21 + log r, so r = 10^0.19 = 1.55.
With total phosphate 0.100 M: [HPO4^2−] = 0.100 × 1.55/2.55 = 0.0608 M [H2PO4−] = 0.0392 M.
From 0.100 M stocks, mix 608 mL Na2HPO4 with 392 mL NaH2PO4.
After adding 5.0 mmol HCl: acid = 39.2 + 5.0 = 44.2 mmol base = 60.8 − 5.0 = 55.8 mmol r = 55.8/44.2 = 1.26 pH = 7.21 + log 1.26 = 7.31.
This is just inside the lower limit 7.30, with only 0.01 pH unit to spare. The buffer is technically acceptable on capacity but too marginal for routine use; a higher total phosphate or a buffer with pKa closer to 7.40 would be safer.
3. Show whether calcium phosphate will precipitate at pH 7.40 with 2.0 mM Ca2+. Choose an alternative buffer from the available list and calculate its acid/base ratio.
Read the worked answer
At pH 7.40, [H+] = 3.98×10−8 M. Ka3 = [H+][PO4^3−]/[HPO4^2−], so: [PO4^3−] = 4.8×10−13 × 0.0608 / 3.98×10−8 = 7.3×10−7 M.
Q = [Ca2+]3[PO4^3−]2 = (2.0×10−3)3 (7.3×10−7)2 = 8.0×10−9 × 5.3×10−13 = 4.3×10−21.
Ksp = 2.1×10−33, so Q ≫ Ksp: calcium phosphate will precipitate and the assay will lose both phosphate buffer and free Ca2+. Phosphate is incompatible with the 2 mM CaCl2 requirement.
Tris is the better available choice. At pH 7.40, ratio base/acid = 10^(7.40 − 8.06) = 10−0.66 = 0.219. With 0.100 M total, [acid] = 0.082 M and [base] = 0.018 M. Tris does not precipitate calcium, but its pH is temperature-sensitive, so the pH must be adjusted at the assay temperature.