Problem 106 - A Complete Resource Book in Chemistry for JEE Main

Question

The equilibrium constant for the reaction \(\mathrm{H}_{3} \mathrm{PO}_{4} \rightleftharpoons \mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)is \(\mathrm{K}_{1}\) for reaction \(\mathrm{H}_{2} \mathrm{PO}_{4} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HPO}_{4}^{2-}\) is \(\mathrm{K}_{2}\) and for reaction \(\mathrm{HPO}_{4}^{2-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{PO}_{4}^{3-}\) is \(\mathrm{K}_{3}\) The equilibrium constant \((\mathrm{K})\) for \(\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow 3 \mathrm{H}^{+}+\mathrm{PO}_{4}^{3-}\) will be (a) \(\mathrm{K}_{1}, \times \mathrm{K}_{2} \times \mathrm{K}_{3}\) (b) \(\mathrm{K}_{1} / \mathrm{K}_{2} \mathrm{~K}_{3}\) (c) \(\mathrm{K}_{2} / \mathrm{K}_{1} \mathrm{~K}_{3}\) (d) \(\mathrm{K}_{1},+\mathrm{K}_{2}+\mathrm{K}_{3}\)

Step-by-Step Solution

Answer

The correct option is (a) \(\mathrm{K}_{1} \times \mathrm{K}_{2} \times \mathrm{K}_{3}\).

Step 1: Understand the Reactions

The problem involves several reactions where phosphoric acid (or its ions) donates protons. Each equilibrium constant represents a stage of ionization: \( \mathrm{K}_{1} \) for the first ionization, \( \mathrm{K}_{2} \) for the second, and \( \mathrm{K}_{3} \) for the third.

Step 2: Write the Overall Reaction

Combine the three given reactions to create the overall reaction: \( \mathrm{H}_{3} \mathrm{PO}_{4} \rightleftharpoons 3\mathrm{H}^{+} + \mathrm{PO}_{4}^{3-} \). This reaction involves transferring three protons from phosphoric acid to form phosphate and hydrogen ions.

Step 3: Determine the Equilibrium Expression

The equilibrium constant for the overall reaction is the product of the equilibrium constants for each step. This uses the rule that \( K_{overall} = K_{1} \times K_{2} \times K_{3} \) for combining sequential equilibria.

Step 4: Choose the Correct Option

Compare the obtained overall equilibrium expression with the given options. The expression \( K = \mathrm{K}_{1} \times \mathrm{K}_{2} \times \mathrm{K}_{3} \) matches option (a).

Key Concepts

Ionization of Phosphoric Acid

Phosphoric acid, denoted as \( \mathrm{H}_3\mathrm{PO}_4 \), undergoes ionization in a stepwise manner, transferring protons to form various ions. Each step involves the removal of one proton from the acid molecule, and each ionization stage has its own equilibrium constant.

• The first ionization: \( \mathrm{H}_3\mathrm{PO}_4 \longrightarrow \mathrm{H}^{+} + \mathrm{H}_2\mathrm{PO}_4^{-} \), with equilibrium constant \( K_1 \).
• The second ionization: \( \mathrm{H}_2\mathrm{PO}_4^{-} \longrightarrow \mathrm{H}^{+} + \mathrm{HPO}_4^{2-} \), with \( K_2 \).
• The third ionization: \( \mathrm{HPO}_4^{2-} \longrightarrow \mathrm{H}^{+} + \mathrm{PO}_4^{3-} \), with \( K_3 \).

Understanding these individual steps is essential to grasp the overall ionization process of the acid into the phosphate ion and hydrogen ions.

Phosphate Ion

The phosphate ion, \( \mathrm{PO}_4^{3-} \), is the final form obtained when phosphoric acid fully ionizes. This polyatomic ion is one of the major products of phosphoric acid ionization, occurring after the acid donates all three of its protons.Phosphate ions are significant in many chemical reactions, particularly in biochemistry, where they play key roles in energy transfer and are involved in various biological functions. The transition from \( \mathrm{H}_3\mathrm{PO}_4 \) to \( \mathrm{PO}_4^{3-} \) marks the completion of phosphoric acid's capacity to release protons, signifying its full conversion.

Equilibrium Expression

An equilibrium expression represents the mathematical relationship between the concentrations of reactants and products at equilibrium. For phosphoric acid's ionization into phosphate ion, the expression reflects the equilibrium constants of each ionization step.To express the overall process, where \( \mathrm{H}_3\mathrm{PO}_4 \) ionizes completely to form \( 3\mathrm{H}^{+} \) and \( \mathrm{PO}_4^{3-} \), we multiply the constants of each step: \[ K_{overall} = K_1 \times K_2 \times K_3 \]This multiplication stems from a rule applicable for sequential reactions, stressing how individual steps combine mathematically to provide the cumulative effect of the ionization.

Sequential Equilibria

Sequential equilibria refer to multiple reaction stages where each step proceeds through a distinct equilibrium. In the case of phosphoric acid ionization, the process is a set of sequential equilibria, with each step having its own equilibrium constant.In sequential equilibria, understanding each reaction stage builds a comprehensive picture of the overall process. As phosphoric acid transitions—from releasing the first proton to the third—the sequential nature allows for predicting the final ion distribution. By considering individual constants \( K_1, K_2, \) and \( K_3 \), and applying the multiplication rule for these steps, we can confidently ascertain the overall equilibrium constant. This approach highlights the cumulative influence of each ionization step in determining the state of the chemical mixture at equilibrium.