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 \(\mathrm{H}_{3} \mathrm{PO}_{4} \rightleftharpoons 3 \mathrm{H}^{+}+\mathrm{PO}_{4}^{3-}\) 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) \(K_{1},+K_{2}^{1}+K_{3}\)

Step-by-Step Solution

Answer

The correct answer is (a) \( K_1 \times K_2 \times K_3 \).

Step 1: Understand the Given Reactions

We are given three equilibrium reactions:1. \( \mathrm{H}_{3} \mathrm{PO}_{4} \rightleftharpoons \mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{PO}_{4}^{-} \) with equilibrium constant \( K_1 \).2. \( \mathrm{H}_{2} \mathrm{PO}_{4} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HPO}_{4}^{2-} \) with equilibrium constant \( K_2 \).3. \( \mathrm{HPO}_{4}^{2-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{PO}_{4}^{3-} \) with equilibrium constant \( K_3 \).

Key Concepts

Acid-Base Equilibrium

Acid-base equilibrium plays a crucial role in many chemical processes. When acids and bases are mixed, they undergo a reversible reaction to form water and salts. In equilibrium chemistry, this process is represented by double arrows indicating that the reaction occurs in both directions.
In the context of phosphoric acid, each dissociation step leading to different ions demonstrates an acid-base equilibrium. Each reaction has a unique equilibrium constant, symbolized as:

• \( K_1 \) for the reaction where phosphoric acid (\( \mathrm{H}_3\mathrm{PO}_4 \)) releases its first proton to become \( \mathrm{H}_2\mathrm{PO}_4^- \)
• \( K_2 \) when \( \mathrm{H}_2\mathrm{PO}_4^- \) discharges another proton forming \( \mathrm{HPO}_4^{2-} \)
• \( K_3 \) for the final deprotonation step resulting in \( \mathrm{PO}_4^{3-} \).

Understanding acid-base equilibrium helps predict product formation and directions under various conditions, which is essential for understanding chemical reactions on a larger scale.

Phosphoric Acid

Phosphoric acid is a triprotic acid, meaning it can lose three protons successively. The chemical formula for phosphoric acid is \( \mathrm{H}_3 \mathrm{PO}_4 \). It is widely used in fertilizers, food flavoring, and industrial cleaning. Various equilibrium reactions reflect each proton being dissociated step-by-step until nothing is left but phosphate ion \( \mathrm{PO}_4^{3-} \).

• The first dissociation: \( \mathrm{H}_3 \mathrm{PO}_4 \rightleftharpoons \mathrm{H}^+ + \mathrm{H}_2\mathrm{PO}_4^- \)
• The second dissociation: \( \mathrm{H}_2 \mathrm{PO}_4^- \rightleftharpoons \mathrm{H}^+ + \mathrm{HPO}_4^{2-} \)
• The third dissociation: \( \mathrm{HPO}_4^{2-} \rightleftharpoons \mathrm{H}^+ + \mathrm{PO}_4^{3-} \)

The stepwise dissociation is important for various applications where specific pH levels are needed, as each step has a different equilibrium constant. It’s a practical and measurable way to control acid's behavior in solution.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the backward reaction, so there is no net change in the concentrations of the reactants and products.
A balance is achieved and maintained in this dynamic state, which is crucial for controlling chemical processes. The concept is central when studying phosphoric acid because different dissociation stages each reach their own equilibrium. The constants \( K_1, K_2, \) and \( K_3 \) give insights into the stability of each ionic form.
To find the overall equilibrium constant for the complete dissociation of phosphoric acid, the constants from each step are multiplied together: \[ K = K_1 \times K_2 \times K_3 \]This principle underscores the balanced state that natural systems strive to achieve, providing a powerful tool for chemists in both theoretical calculations and practical applications.

Ionic Equilibria

Ionic equilibria relate to the balance established between various ionic species in a solution during a chemical reaction. With phosphoric acid, ionic equilibria involve the different ions produced when the acid dissociates.
It integrates both the concentrations of the ions and the pH of the solution. Each dissociation of phosphoric acid has its equilibrium, dictated by its own constant (\( K_1, K_2, \) and \( K_3 \)):

• \( \mathrm{H}_3\mathrm{PO}_4 \rightleftharpoons \mathrm{H}^+ + \mathrm{H}_2\mathrm{PO}_4^- \)
• \( \mathrm{H}_2\mathrm{PO}_4^- \rightleftharpoons \mathrm{H}^+ + \mathrm{HPO}_4^{2-} \)
• \( \mathrm{HPO}_4^{2-} \rightleftharpoons \mathrm{H}^+ + \mathrm{PO}_4^{3-} \)

Each equilibrium constant reflects the affinity of the acid to hold onto its protons compared to losing them during the reactions. Ionic equilibria are vital for understanding and predicting the behavior of solutions under various conditions, like changes in pH and dilution, helping chemists design experiments and products effectively.