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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


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    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.
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