Use of Acid/Base Distributions in pH Problems

Once the fractional amounts of all species are known, we can put these expression to work in complex equilibria problems involving acids and bases at known pH's.

First, let's determine the relative amounts of all species in the phosphoric acid equilibrium. If we were to control the pH, and to somehow measure the relative amounts of H3PO4, H2PO4-, HPO42-, and PO43-, what would we expect to observe? Phosphoric acid has three equilibria

The corresponding a are

where FA is the formal concentration of phosphoric acid in solution. One expects to find more H3PO4 at lower pH since a0 is proportional to [H3O+]3 (low pH corresponds to high [H3O+]). Similarly, at very high pH (thus low [H3O+]) [H3PO4] will tend to zero while [PO43-] would be maximum. It is easy to calculate the fractional concentrations if we know the equilibrium constants. The equilibrium constants for phosphoric acid are; Ka1=7.11x10-3; Ka2=6.32x10-8; and Ka3=7.1x10-13. Using the a one could calculate the relative concentrations at any pH.

It is instructive to examine a plot of the a as a function of pH. Several interesting 'features' are apparent by examining trends illustrated in the plot with the equations describing the concentrations. First, notice that the pH where two species concentrations are the same is around the pKa for that equilibrium. In fact, for polyprotic acids with pKa's that differ by over 3 to 4 units, the pH is equal to the pKa. Take for example the point where [H3PO4]=[H2PO4-]. The equilibrium equation relating these two species is

If we take the -log10, or "p", of this equation

Since [H3PO4]=[H2PO4-], and log10(1) = 0, pH=pKa1.

Second, you might notice that the concentrations of the conjugate bases are maximum half-way between the pKa points. For example, the point where [H2PO4-] is a maximum lies half-way between between pKa1 and pKa2. Since H2PO4- is the major species present in solution, the major equilibrium is the disproportionation reaction

This equilibrium cannot be used to solve for pH because [H3O+] doesn't occur in the equilibrium equation. We solve the pH problem adding the first two equilibria equations

Note that when we add chemical equilibria, we take the product of the equilibrium equations. Taking the -log10 of the last equation

Since the disproportionation reaction predicts [H3PO4]=[HPO42-]