The fourth equilibrium does not directly consume or produce H3O + or OH -. However, its components-HA, H2A + , and A --participate in the first and second equilibria, which adjust the concentrations of H3O + and OH -, respectively. At the pI, the relationship in equation ( ) holds, ensuring that the H2A + and A -concentrations are equal, thereby influencing the overall acid-base balance. We can now use equation ( ) to determine how the autoprotolysis of HA influences H3O + in the first equilibrium or OH -in the second equilibrium, and thus the pH of the solution at the isoelectric point.