epenguin

Would you care to have another look at that Borek? I may not have much time next days.

I think you can simplify by ignoring the first and third phosphoric acid dissociations. Even if you couldn't you would worry if you couldn't solve the same problem for some other substance with just one pK about 2 units below that of NH₄⁺, so how to do that?

I think you can think you have your 2/3 mole of phosphate buffer at pH = pK = 7.21 and to it are adding 1/3 mole of H₂PO₄^- and 1/3 mole NH₄⁺.

This added H₂PO₄^- stays in that form on the basis its dissociable proton 'has nowhere to go'. The ammonia is already protonated and if it protonates a HPO₄^2- the result is no change!

So the ratio [H₂PO₄^-]/[HPO₄^2-] is (1/3 + 1/3)/(1/3) = 2, which corresponds to a pH close to 6.9 which sounds to me about right - the ammonium biphosphate solution is somewhat acid (I think pH is mean of that of NH₄⁺ and the first pK of H₃PO₄ -> about 5.5); you are adding this weakly buffered solution to the maximally buffered K phosphate.


When I do a calculation with all the equations I do get a result close to that. I have a vague inkling but not got to the bottom of the fact I get a quadratic and its other root is positive but unphysical.
Edit: I suppose that's OK. 'Unphysical' meant there was a concentration that was higher than it can be, but that just means by conservation that another one is negative. You must have just one physical solution but if equation degree is even you must get another real root which should be negative to make it unphysical. Maybe sometimes (often? always? negative real solutions with odd degree equations also).

It is all quite hard to think about.