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Calculate the pH of a buffer made

  • Thread starter kk727
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
Borek
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Ignore cation. Write stepwise dissociation reactions for multiprotic acid.
 
  • #3
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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? :/
 
  • #4
Borek
Mentor
28,398
2,800


Yes.

Do you understand why H2PO4- is an acid?
 

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