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) \(\mathrm{K}_{1},+\mathrm{K}_{2}+\mathrm{K}_{3}\)

Short answer

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

Step-by-step solution

Identify Individual Reactions and Constants

The three given reactions are dissociation reactions of phosphoric acid. The respective equilibrium constants for these reactions are denoted as \(K_1, K_2,\) and \(K_3\). They represent the dissociation of: 1) \( \text{H}_3\text{PO}_4 \rightleftharpoons \text{H}^+ + \text{H}_2\text{PO}_4^- \) (\(K_1\))2) \( \text{H}_2\text{PO}_4^- \rightleftharpoons \text{H}^+ + \text{HPO}_4^{2-} \) (\(K_2\))3) \( \text{HPO}_4^{2-} \rightleftharpoons \text{H}^+ + \text{PO}_4^{3-} \) (\(K_3\))

Combine the Reactions to Form Overall Reaction

The overall reaction is the combination of the three individual steps. When summed, the reactions yield:\( \text{H}_3\text{PO}_4 \rightleftharpoons 3\text{H}^+ + \text{PO}_4^{3-} \). This is the overall dissociation of \(\text{H}_3\text{PO}_4\).

Calculate the Equilibrium Constant for the Overall Reaction

The equilibrium constant for the overall reaction is the product of the equilibrium constants of the individual steps. This is given by:\( K = K_1 \times K_2 \times K_3 \).The reasoning is that equilibrium constants multiply when individual reactions are added to form an overall reaction.

Choose the Correct Option

Based on the calculation in Step 3, the equilibrium constant for the overall reaction is the product \( K_1 \times K_2 \times K_3 \). Thus, the correct answer is (a) \( K_1 \times K_2 \times K_3 \).

Key concepts

Equilibrium Constant

In a chemical reaction, the equilibrium constant, denoted as \( K \), is a crucial aspect that characterizes the balance between reactants and products. This value gives insights into the extent to which a reaction has progressed. The equilibrium constant is determined using the concentrations of the reactants and products at equilibrium conditions. For a generic reaction \( aA + bB \leftrightharpoons cC + dD \), it is expressed as:

• \( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)

Here, square brackets represent the concentration of each species, and the lowercase letters are stoichiometric coefficients from the balanced chemical equation.
This constant is fundamental in predicting the direction a reaction will favor under specific conditions. A large \( K \) value indicates products are favored, while a small \( K \) emphasizes reactants.

Dissociation Reactions

Dissociation reactions are processes where compounds break down into two or more constituents, often ionic species. For example, water dissociating into hydrogen and hydroxide ions is well known. In the context of phosphoric acid, dissociation involves several stages. Each stage involves the loss of a proton, leading to unique ions in solution. Key equations include:

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

These equations show phosphoric acid losing protons step-by-step, characteristic of polyprotic acids. Studying dissociation reactions helps in understanding the acidity and buffering capabilities of such systems.

Phosphoric Acid

Phosphoric acid is a triprotic acid, meaning it releases three protons \((\text{H}^+)\) sequentially. It plays a vital role in various industrial and biological contexts, including as a food additive and in dental products. Each dissociation step has its specific equilibrium constant \((K_1, K_2, K_3)\), indicating the strength of each ionization:

• First dissociation produces \( \text{H}_2\text{PO}_4^- \).
• Second results in \( \text{HPO}_4^{2-} \).
• Third forms \( \text{PO}_4^{3-} \).

The dissociation constants are crucial in applications requiring precise control over pH. Phosphoric acid's ability to donate multiple protons makes it a versatile component in chemical reactions.

Overall Reaction

An overall reaction is the summation of individual steps contributing to the final chemical change. In phosphoric acid’s dissociation, combining the three stages gives the overall equation: \( \text{H}_3\text{PO}_4 \leftrightharpoons 3\text{H}^+ + \text{PO}_4^{3-} \).
This represents the complete transformation from the acid to its ionized base form. To determine its equilibrium constant \( K \), one multiplies the individual constants \( (K_1 \times K_2 \times K_3) \), showcasing how each step contributes to the final state. Understanding the overall reaction is vital in fields like biochemistry, where complex reactions proceed through multiple stages.

Equilibrium Problems

Equilibrium problems involve solving for unknowns in reactions at equilibrium, usually focusing on concentrations or equilibrium constants. These problems require a good grasp of both the reaction stoichiometry and the relationships between species concentrations. Techniques involve setting up the balanced chemical equation and using the equilibrium constant expression to solve for unknown variables.
For students, mastering such problems involves careful step-by-step analysis:

• Write the balanced reaction and determine the \( K \).
• Identify initial concentrations and changes during the reaction.
• Apply these values into the equilibrium constant expression.

Such practice highlights the real-world applications of equilibrium concepts, including pharmaceuticals and environmental chemistry.