Phosphoric Acid Equilibrium: Understanding pH and pKa

What is the equilibrium equation for phosphoric acid at a pH of 12?

How does the pH being equal to the pKa3 of phosphoric acid affect the equilibrium?

Equilibrium Equation and pH Effect

The equilibrium equation for phosphoric acid at a pH of 12 involves the deprotonation of the dihydrogen phosphate ion (H2PO4−) to form the hydrogen phosphate ion (HPO4−2).

The pH being equal to the pKa3 of phosphoric acid means that the concentration of hydroxide ions ([OH−]) is relatively high, shifting the equilibrium towards the production of more HPO4−2 ions through deprotonation.

Phosphoric acid (H3PO4) is a polyprotic acid with three dissociation steps. One of these steps, corresponding to the pKa3 value, occurs at a pH of 12. At this pH, the equilibrium favors the deprotonation of the dihydrogen phosphate ion (H2PO4−) to form the hydrogen phosphate ion (HPO4−2).

The equilibrium equation for this deprotonation reaction is: H2PO4−(aq) + H2O(l) ⇌ HPO4−2(aq) + H3O+(aq)

Since the pH is equal to the pKa3 value, it indicates that the concentration of hydroxide ions ([OH−]) is relatively high in the solution. This high concentration of hydroxide ions drives the equilibrium towards the formation of more HPO4−2 ions through deprotonation of H2PO4−.

This shift in equilibrium due to the pH being equal to pKa3 highlights the importance of understanding the interplay between pH, pKa values, and equilibrium concentrations in acid-base reactions.