Henderson-Hasselbalch & phosphate buffers

[chops369]

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

Homework Statement

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

Homework Equations

pH = pKa + log₁₀ [A^-] / [HA]

pH = -log₁₀ [H⁺]

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]

[HPO₄^2-] / [H₂PO₄^-] = 1.58

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

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

 

[chops369]

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

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