1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

H2PO42- HPO4- buffer system

  1. May 13, 2014 #1
    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.
     

    Attached Files:

  2. jcsd
  3. May 13, 2014 #2

    Qube

    User Avatar
    Gold Member

    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.
     
  4. May 13, 2014 #3
    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.
     
  5. May 13, 2014 #4

    Qube

    User Avatar
    Gold Member

    Your work looks fine.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted