1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Henderson-Hasselbalch & phosphate buffers

  1. Sep 11, 2011 #1
    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 [HPO42-] / [H2PO4-]

    7.4 = 7.20 + log10 [HPO42-] / [H2PO4-]

    [HPO42-] / [H2PO4-] = 1.58

    Thus, if [HPO42-] + [H2PO4-] = 20 mM, then

    [HPO42-] = 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 [HPO42-] > [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 [HPO42-] / [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.
  2. jcsd
  3. Sep 11, 2011 #2


    User Avatar

    Staff: Mentor

    These are two simultaneous equations in two unknowns - just solve.
  4. Sep 11, 2011 #3
    Wow. I stared at that problem for 2 hours...I can't believe the answer was right there. Thanks for clearing that up.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook