The Ka for the H2PO4–/HPO42– buffer is given by the equation:
Ka= ([HPO4^-2]gamma [H^+])/[H2PO4^-]=10^-7.20
What is the pH of a 2:3 mixture of H2PO4– to HPO42– at 0.10 M ionic strength? (See the table for the activity coefficients.)
I just am not sure how to figure out the concentrations.. I know I have to solve for H+ but I am not sure what to put for HPO4^-2 and H2PO4- Thank you!
1 answer
I must confess that I, too, am confused but mostly because I don't know what gamma is in your equation. I THINK, but I'm not positive, that you look in the table, find the activity coefficient for H2PO4^- and HPO4^2-. Then I would be inclined to use 2 for H2PO4^- and 3 for HPO4^2- and multiply each by its activity coefficient. Since it's the ratio in Ka it makes little difference what REAL concn it is because 2M/3M is the same as 0.2/0.3 or 0.002/0.003. And since you are given the ionic strength that will correct for whatever difference you have because the activity coefficient will not be the same for each. I think this takes care of the 2:3 mixture of 0.1 M ionic strength part of the problem if you know what to do with gamma.
2 answers