The pKa for the dissociation of H3PO4 is 2.15. What is the concentration of H2PO4-1 (in M) at pH 3.21 if the original concentration of the phosphate was 2.37 M? My Attempt: Key Information: pH final: 3.21 Initial Molarity of phosphate: 2.37 M pKa: 2.15 I started with the first dissociation of the weak acid into its proton and conjugate base. The stoicheometry is as follows: H3PO4 <-----> H3O+ + H2PO4- I am only looking at the first dissociation constant K1. pKa = -log Ka 2.15 = -log Ka = 10^-2.15 = 7.08e-3 -----This is where I am stuck------ When the question refers to the "original concentration of the phosphate", is it referring to the conjugate base or the original acid? When I try to set up an "ICE table" plugging the given molarity into the "Initial" conjugate base column, I end up with too many unknowns and can not solve quadratically. Yet when I plug the Molarity into the [HA] column I get values that do not make sense. I understand that: Ka= [H+][A-]/[HA] If I gloss over this middle step I can continue to find the concentration of H2PO4- ions in the final solution. Since the first dissociation is 1:1, the concentration of H+ ions will be the same as the concentration of H2PO4- ions. Therefore: pH = 3.21 = 10^-3.21 = 6.17e-4 M I feel like I am missing a crucial part of this problem (obviously, since my attempts have yielded wrong answers). I have solved for the original pH of the solution at 2.37 M (0.89), I just dont know how to pull all the pieces together to solve the problem. Any help would be greatly appreciated. Thank you in advance.