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

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The pH of a buffer system created by mixing 1g of H2PO4- and 1g of HPO42- in 100 ml of water is calculated to be 7.21 using the Henderson-Hasselbalch equation. The relevant acid dissociation constants are Ka1 = 7.5x10^-3, Ka2 = 6.2x10^-8, and Ka3 = 4.8x10^-13. The discussion confirms that H2PO4- acts as a base and HPO42- can function as both an acid and a base, depending on the context of the reactions. The reasoning behind these roles is essential for understanding buffer systems in chemistry.

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yolo123
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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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