How to Determine the pH of a Buffer Made with K2HPO4 and KH2PO4?

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SUMMARY

The discussion focuses on calculating the pH of a buffer solution made with 10.0 g of K2HPO4 and 5.0 g of KH2PO4 in 1.0 L of water. The key equation used is the Henderson-Hasselbalch equation: pH = -log Ka + log (base/acid). Participants clarified that KH2PO4 acts as the acid while K2HPO4 serves as the conjugate base. The calculated pH value of 7.4, derived using a Ka of 6.2 x 10^-8, was confirmed as correct.

PREREQUISITES
  • Understanding of the Henderson-Hasselbalch equation
  • Knowledge of acid-base chemistry, specifically regarding conjugate acids and bases
  • Familiarity with the dissociation of salts in aqueous solutions
  • Basic skills in converting grams to moles and calculating molarity
NEXT STEPS
  • Study the dissociation reactions of multiprotic acids, particularly phosphoric acid
  • Learn about the properties and applications of buffer solutions in chemistry
  • Explore the significance of Ka and Kb values in acid-base equilibria
  • Investigate the role of pH in biochemical systems and laboratory settings
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Chemistry students, educators, and laboratory technicians who are involved in buffer solution preparation and pH calculations.

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Homework Statement



Calculate the pH of a buffer made with 10.0 g of K2HPO4 and 5.0 g of KH2PO4 in 1.0L of water.

Homework Equations



-log Ka + log (base/acid) = pH

The Attempt at a Solution



So...mathwise, I know how to do this problem. My only question is, aren't K2HPO4 and KH2PO4 both salts? How would they break up? I don't know which would be the acid or the base, and I couldn't find a Ka/Kb value to help me.

Basically I just can't figure out how they break up and what their conjugates are... :p
 
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Ignore cation. Write stepwise dissociation reactions for multiprotic acid.
 


So my teacher said that the KH2PO4 would be considered the acid, and that the K2HPO4 would form a strong conjugate base. I converted from grams to mols, and then to molarity. Using the given Ka value of 6.2 x 10^-8, I just plugged everything into the Henderson-Hasselbalch equation and got a pH of 7.4. Does this sound right? :/
 


Yes.

Do you understand why H2PO4- is an acid?
 

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