Henderson-Hasselbalch & phosphate buffers


by chops369
Tags: buffers, hendersonhasselbalch, phosphate
chops369
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#1
Sep11-11, 12:27 PM
P: 57
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.
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Borek
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Sep11-11, 12:52 PM
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Quote Quote by chops369 View Post
[HPO42-] / [H2PO4-] = 1.58

Thus, if [HPO42-] + [H2PO4-] = 20 mM, then
These are two simultaneous equations in two unknowns - just solve.
chops369
chops369 is offline
#3
Sep11-11, 02:15 PM
P: 57
Quote Quote by Borek View Post
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.


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