Triprotic Acids

A triprotic acid is an acid that has three ionizable H atoms. Phosphoric acid is one example:

$$\text{First ionizaiton: }H_3PO_4+H_2O(l)⇌H_3O^+(aq)+H_2PO_4^-(aq)\qquad K_{a1}=7.5×10^{-3}$$ $$\text{Second ionizaiton: }H_2PO_4^-+H_2O(l)⇌H_3O^+(aq)+HPO_4^{2-}(aq)\qquad K_{a2}=6.2×10^{-8}$$ $$\text{Third ionizaiton: }HPO_4^{2-}+H_2O(l)⇌H_3O^+(aq)+PO_4^{3-}(aq)\qquad K_{a3}=4.2×10^{-13}$$

As for the diprotic acid examples, each successive ionization reaction is less extensive than the former, reflected in decreasing values for the stepwise acid ionization constants. This is a general characteristic of polyprotic acids and successive ionization constants often differ by a factor of about 105 to 106.

This set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of H3PO4 complicated. However, because the successive ionization constants differ by a factor of 105 to 106, large differences exist in the small changes in concentration accompanying the ionization reactions. This allows the use of math-simplifying assumptions and processes, as demonstrated in the examples above.

Polyprotic bases are capable of accepting more than one hydrogen ion. The carbonate ion is an example of a diprotic base, because it can accept two protons, as shown below. Similar to the case for polyprotic acids, note the ionization constants decrease with ionization step. Likewise, equilibrium calculations involving polyprotic bases follow the same approaches as those for polyprotic acids.

$$H_2O(l)+CO_3^{2-}⇌HCO_3^-(aq)+OH^-(aq)\qquad K_{b1}=2.1×10^{-4}$$ $$H_2O(l)+HCO_3^-⇌H_2CO_3(aq)+OH^-(aq)\qquad K_{b2}=2.3×10^{-8}$$