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

[yolo123]

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


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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.

 

[Qube]

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 HPO₄^2-. This is both an acid and a base.

 

[yolo123]

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.

 

[Qube]

Your work looks fine.