Problem 67 - Chemistry

Question

In the presence of \(\mathrm{CN}^{-}, \mathrm{Fe}^{3+}\) forms the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-} .\) The equilibrium concentrations of \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) are \(8.5 \times 10^{-40} \mathrm{M}\) and \(1.5 \times 10^{-3} M,\) respectively, in a \(0.11-M\) KCN solution. Calculate the value for the overall formation constant of \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) . $$\mathrm{Fe}^{3+}(a q)+6 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \qquad K_{\text { overall }}=?$$

Step-by-Step Solution

Answer

The overall formation constant for the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) is \(2.97 \times 10^{41}\).

Step 1: Write the equilibrium reaction equation.

The equilibrium reaction equation for the formation of the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) involves \(\mathrm{Fe}^{3+}\) and \(\mathrm{CN}^{-}\) as reactants, forming the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\): $$\mathrm{Fe}^{3+}(a q)+6 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q)$$

Step 2: Write the equilibrium constant expression for the reaction.

The equilibrium constant expression, K_overall, would be the ratio of the concentrations of the products to the concentrations of the reactants, with each concentration raised to its stoichiometric coefficient in the balanced reaction equation: $$K_\text{overall} =\frac{[\mathrm{Fe}(\mathrm{CN})_{6}^{3-}]}{[\mathrm{Fe}^{3+}][\mathrm{CN}^{-}]^6}$$

Step 3: Plug in the given equilibrium concentrations.

We are given the equilibrium concentrations of \(\mathrm{Fe}^{3+}\), \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\), and \(\mathrm{CN}^{-}\) as \(8.5 \times 10^{-40} \mathrm{M}\), \(1.5 \times 10^{-3} \mathrm{M}\), and \(0.11 \mathrm{M}\), respectively. Plug these values into the equilibrium constant expression: $$K_\text{overall} =\frac{(1.5 \times 10^{-3})}{(8.5\times 10^{-40})(0.11)^6}$$

Step 4: Calculate the overall formation constant, K_overall.

We can now solve for K_overall: $$K_\text{overall} =\frac{(1.5 \times 10^{-3})}{(8.5\times 10^{-40})(0.11)^6} \approx 2.97 \times 10^{41}$$ So, the overall formation constant for the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) is \(2.97 \times 10^{41}\).

Key Concepts

Complex Ions

Complex ions form when a central metal ion binds with molecules or anions, known as ligands. In the context of the problem, the complex ion is \( \mathrm{Fe} (\mathrm{CN})_{6}^{3-} \), where iron (\( \mathrm{Fe}^{3+} \)) is surrounded by six cyanide ions (\( \mathrm{CN}^- \)).

This assembly forms a stable, charged particle. Complex ions are integral in many chemical processes. They influence solubility and reactivity and are central to coordination chemistry.

• **Central Atom or Ion:** Usually a metal that can accept electron pairs.
• **Ligands:** These donate electron pairs to form coordinate bonds with the metal ion.
• **Coordination Number:** Refers to the number of ligand atoms bonded to the metal center.

Understanding complex ions helps explain behaviors in both natural and industrial chemical systems.

Formation Constant

The formation constant, often denoted as \( K_f \), quantifies the stability of a complex ion in solution. It’s a measure of how well reactants convert into products at equilibrium.

In the provided example, the formation constant \( K_{\text{overall}} \) for \( \mathrm{Fe} (\mathrm{CN})_{6}^{3-} \) is very large \((2.97 \times 10^{41})\). This indicates a very stable complex, meaning that once \( \mathrm{Fe}^{3+} \) and \( \mathrm{CN}^- \) combine, they remain mostly in the complex form in the solution.

Significance of a Large \( K_f \)

• A large \( K_f \) implies that the equilibrium significantly favors the products.
• It signifies the efficiency of ligand binding and the preference for complex formation.
• Useful for predicting reaction direction and understanding chemical equilibria.

Formation constants are crucial for anticipating the behavior of metal ion systems in different environments.

Chemical Equilibrium

Chemical equilibrium occurs when the forward and reverse reactions happen at the same rate, resulting in constant concentrations of reactants and products.

For complex ions like \( \mathrm{Fe} (\mathrm{CN})_{6}^{3-} \), equilibrium involves the interaction between metal ions and ligands until a stable configuration is achieved.

• **Dynamic Nature:** Though concentrations remain constant, the process is continuous and dynamic.
• **Equilibrium Expression:** Expresses the ratio of product and reactant concentrations, each raised to the power of their stoichiometric coefficients.
• **Effect of Concentration Changes:** Shifting the concentrations can disturb the equilibrium, leading to shifts to re-establish balance (Le Châtelier's Principle).

Understanding equilibrium helps predict how different conditions alter reactions and the formation of complexes.

Coordination Compounds

Coordination compounds are formed when central atoms or ions bind with surrounding ligands to create structured entities. These compounds are pivotal in various fields, including catalysis and biochemistry.

In the exercise, \( \mathrm{Fe} (\mathrm{CN})_{6}^{3-} \) is a classic coordination compound comprised of the central metal ion \( \mathrm{Fe}^{3+} \) and cyanide ligands.

Properties of Coordination Compounds

• **Stability:** Influenced by ligand types and the central metal.
• **Color and Magnetism:** Dependent on the electron configuration and ligand field.
• **Applications:** From catalysis in industrial processes to oxygen transport in biological systems (e.g., hemoglobin).

Grasping the characteristics of coordination compounds is essential for innovations in materials science and pharmaceuticals.