Write chemical equations and corresponding equilibrium expressions for each of the three ionization steps of phosphoric acid.

Understanding chemical equations is essential in the study of chemistry as they describe the substances involved in a chemical reaction, their relative quantities, and the direction of the reaction. In the context of phosphoric acid ionization, we see this through a series of steps where phosphoric acid loses protons (hydrogen ions, H⁺) one at a time.

For the first ionization step, the chemical equation is represented as:
\[ H_3PO_4 \rightarrow H^+ + H_2PO_4^- \]
This depicts the loss of one proton from phosphoric acid to form dihydrogen phosphate ion. In subsequent steps, the dihydrogen phosphate ion and then the hydrogen phosphate ion each lose a proton, resulting in the formation of hydrogen phosphate and then phosphate ions respectively:
\[ H_2PO_4^- \rightarrow H^+ + HPO_4^{2-} \]
\[ HPO_4^{2-} \rightarrow H^+ + PO_4^{3-} \]
Each equation conserves mass and charge, a fundamental principle in chemical reactions. For students, it's crucial to recognize the progressive nature of phosphoric acid ionization, where each step involves the formation of a new anion with one less hydrogen ion.

To grasp the concept of equilibrium expressions, think of it as a mathematical representation of a chemical reaction at equilibrium. It quantifies the ratio of the concentrations of products to reactants. For ionization reactions, these expressions involve the acid dissociation constant, designated as K_a.

In our exercise, the first ionization equilibrium expression looks like this:
\[ K_{a1} = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]} \]
This expression indicates that at equilibrium, the product of the concentrations of the hydrogen ions and dihydrogen phosphate ions, divided by the concentration of undissociated phosphoric acid, remains constant at a given temperature. The expressions for the second and third ionization steps follow the same pattern:
\[ K_{a2} = \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]} \]
\[ K_{a3} = \frac{[H^+][PO_4^{3-}]}{[HPO_4^{2-}]} \]
Equilibrium expressions are important as they allow chemists to predict the direction in which a reaction will proceed and to calculate the concentrations of substances in a reaction mixture at equilibrium.

The acid dissociation constant (K_a) is a pivotal concept in understanding the strength of an acid. It indicates how completely an acid dissociates into its ions in water.

For phosphoric acid, we have different K_a values for each ionization step due to the different affinities the acid has for its protons. The first ionization step usually has the highest K_a value as the first proton is the easiest to lose:
\[ K_{a1} > K_{a2} > K_{a3} \]
Each successive K_a is smaller since it's progressively more difficult for the molecule to lose additional protons. This is vital knowledge for chemists when predicting the behavior of the acid in different chemical environments.

For education purposes, it's beneficial to think of the K_a as a measure of acid strength — the larger the K_a, the stronger the acid. In practice, these constants allow us to calculate the pH of a solution and are crucial in many areas such as pharmaceuticals, agriculture, and food science.