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

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To determine the pH of a buffer made with K2HPO4 and KH2PO4, the relevant equation is the Henderson-Hasselbalch equation, which incorporates the acid dissociation constant (Ka). K2HPO4 acts as the conjugate base, while KH2PO4 is the acid in this buffer system. The dissociation of these salts in water leads to the formation of H2PO4- and HPO4^2- ions. Using the provided Ka value of 6.2 x 10^-8 and the calculated molarities from the given weights, a pH of 7.4 was determined, which is confirmed as accurate. Understanding the roles of these ions is crucial for calculating the pH of the buffer solution.
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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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