Solving Biochem pH Problem for Na2HPO4 Solution

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SUMMARY

The pH of a solution formed by dissolving solid Na2HPO4 in water is approximately 9. This conclusion is derived from the equilibrium of phosphate species, specifically H3PO4, H2PO4-, HPO42-, and PO43-, with relative pKa values of ~2, ~7, and ~12. The calculation for the pH utilizes the average of the pKa values for the relevant species, leading to the determination that the resulting pH is 1/2 (7 + 12) = ~9.

PREREQUISITES
  • Understanding of acid-base equilibria
  • Familiarity with pKa values and their significance
  • Knowledge of phosphate species and their protonation states
  • Basic concepts of pH calculation
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  • Research the properties of amphiprotic substances in solution
  • Study the Henderson-Hasselbalch equation for pH calculations
  • Explore the behavior of phosphate buffers in biochemical applications
  • Learn about the role of pKa in determining the protonation state of weak acids
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Homework Statement



Solid Na2HPO4 is dissolved in water. What is the resulting pH of the solution? Relative pka's of phosphate derivaties are ~2, ~7, and ~12.

Homework Equations



No equation needed, just look at the species which phosphate would develop.


The Attempt at a Solution



I believe the specified compound would be in equilibrium between one of these species;
H3PO4, H2PO4 (-), HPO4 (-2) and PO4 (-3). The choices are 12, 7, 2, 9, or there is not enough information.
 
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Thank you very much! If I understand correctly then the answer would be 1/2 (7 +12) which would be ~ 9 ph?

That is the answer I chose. Thank you.
 

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