Henderson-Hasselbalch & phosphate buffers

chops369

This is an example calculation about the phosphate buffer system from my Biochemistry textbook.

1. The problem statement, all variables and given/known data
If the total cellular concentration of phosphate is 20 mM (millimolar) and the pH is 7.4, the distribution of the major phosphate species is given by

pH = pKa + log₁₀ [HPO₄^2-] / [H₂PO₄^-]

7.4 = 7.20 + log₁₀ [HPO₄^2-] / [H₂PO₄^-]

[HPO₄^2-] / [H₂PO₄^-] = 1.58

Thus, if [HPO₄^2-] + [H₂PO₄^-] = 20 mM, then

[HPO₄^2-] = 12.25 mM and [H₂PO₄^-] = 7.75 mM


2. Relevant equations
pH = pKa + log₁₀ [A^-] / [HA]

pH = -log₁₀ [H⁺]


3. The attempt at a solution
I understand everything up until they provide the concentrations of each phosphate species. Since their ratio as shown in the equation is 1.58, one can clearly assume that [HPO₄^2-] > [H₂PO₄^-]. But the fact that no explanation is provided for arriving at their specific concentrations is driving me insane.

The Henderson-Hasselbalch equation shows that, when [HPO₄^2-] / [H₂PO₄^-] = 1, pH = pKa. But since we are at pH = 7.4, they obviously can't be equal. I think the solution must involve taking the 0.2 difference into account somehow.

Borek

These are two simultaneous equations in two unknowns - just solve.

chops369

Wow. I stared at that problem for 2 hours...I can't believe the answer was right there. Thanks for clearing that up.