What Is the pH of a Buffer System with H2PO4- and HPO42-?

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The discussion centers on calculating the pH of a buffer system created by mixing 1g of H2PO4- and 1g of HPO42- in 100 ml of water. The participant used the Henderson-Hasselbalch equation to determine a pH of 7.21, considering the relevant Ka and Kb values for the two species. There is a debate about whether H2PO4- acts as a base and HPO42- as an acid, with emphasis on the importance of their respective acid-base properties in the buffer system. Clarification is sought on the reasoning behind these assumptions and their appropriateness for exam responses. The conversation highlights the complexities of buffer chemistry and the interpretation of equilibrium constants.
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Hello Forum!

1g of H2PO4- and 1 g of HPO42- are put together into 100 ml of H2O. What is the pH of the buffer created.

Ka1= 7.5x10^-3
Ka2=6.2x10^-8
Ka3=4.8x10^-13


______________________
Okay. I uploaded my solutions. (Please disregard the part 1.184 g/mol and the 34% on my answers sheet. That was part of another problem.) Basically, I decided to put down all the reactions and say that H2PO4- and HPO4- will be the important reactions according to Ka/b values.
Then, I used Henderson to get 7.21.
Are my steps of thinking that H2PO4- will not act as base because of Kb1, and that HPO4- will not act as acid because of Ka3 necessary and correct? Are they part of the reasoning that I should put on, say an exam?

Many thanks.
 

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Your handwriting is hard to read. And of course H2PO4- will be acting as a base in the system; the question is just to what extent. Same with HPO42-. This is both an acid and a base.
 
Yes, I looked at it from the point of view of the values of the Ka/b values. Does it seem to make sense? Sorry about my handwriting. This is actually my third "clean" version I made just for the forum.
 
Your work looks fine.
 

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