Polyprotic Solutions

Some common acids have more than one proton to donate to solution. A diprotic acid can donate two protons to solution. A triprotic acid can donate three protons to solution. These acids do not lose all protons at once. Stepwise ionization is a process by which a diprotic acid or a triprotic acid ionizes by losing one proton at a time. Each ionization has its own equilibrium constant, the constant that relates the amount of reactants and products in an equilibrium, and its own contribution of protons to the solution. This process can be illustrated with the dissolution of triprotic phosphoric acid, H₃PO₄.

Step 1: Step 2: Step 3: For nearly all polyprotic acids, the equilibrium constants for the second and third ionizations are less than the previous K_a by several orders of magnitude. In other words, after the first ionization, the concentration of products from the second (and third, if applicable) ionizations is relatively small, and the vast majority of protons donated to solution have come from the first ionization. Therefore, the pH of a polyprotic solution can be approximated by using only K_a for the first ionization: Ionized acid molecules can also be thought of as conjugate bases that can accept one, two, or three protons and would accept them in a similar, stepwise manner. Consider what happens when trisodium phosphate (Na₃PO₄) is dissolved in water. Sodium ions (Na⁺) and phosphate ions (PO₄^3–) are released into solution. Sodium does not produce either protons or hydroxide ions in solution. Phosphate (PO₄^3–), however, is a conjugate base of a weak acid (HPO₄^2–) and can accept one or more protons. The phosphate (PO₄^3–) ions would serve as a polyprotic base. The ionization of water, however, produces far fewer protons than the ionization of phosphoric acid. So, the K_b value for the three conjugate bases is very small.