Question

Consider the reaction \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)\) (a) Calculate \(K\) given that for \(\mathrm{Cu}(\mathrm{OH})_{2} K_{s p}=2 \times 10^{-19}\) and for \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+} K_{\mathrm{f}}=2 \times 10^{12}\) (b) Determine the solubility of \(\mathrm{Cu}(\mathrm{OH})_{2}\) (in \(\mathrm{mol} / \mathrm{L}\) ) in \(4.5 \mathrm{M} \mathrm{NH}_{2}\).

Short answer

Question: Calculate the equilibrium constant (K) for the reaction between Cu(OH)₂ and NH₃ given the solubility product (Ksp) of Cu(OH)₂ is 2 × 10^(-19) and the formation constant (Kf) of Cu(NH₃)₄²⁺ is 2 × 10^(12). Then, find the solubility of Cu(OH)₂ in a 4.5 M NH₃ solution. Answer: The equilibrium constant (K) for the reaction is 10^(-31), and the solubility of Cu(OH)₂ in a 4.5 M NH₃ solution is approximately 1.45 × 10^(-9) mol/L.

Step-by-step solution

Calculating the equilibrium constant (K)

Since we know the values of Ksp for Cu(OH)₂ and Kf for Cu(NH₃)₄²⁺, we can use them to find the equilibrium constant (K) for the given reaction. K = Ksp / Kf Now, plug in the given values of Ksp and Kf and find 'K': K = (2 × 10^(-19)) / (2 × 10^(12)) K = 10^(-31)

Setting up the equilibrium equation

Write the balanced reaction equation and the corresponding equilibrium equation: Cu(OH)₂(s) + 4 NH₃(aq) ⇌ Cu(NH₃)₄²⁺(aq) + 2 OH⁻(aq) K = [Cu(NH₃)₄²⁺][OH⁻]² / [Cu(OH)₂][NH₃]⁴ Since the initial concentrations of Cu(NH₃)₄²⁺ and OH⁻ are both 0, let 'x' mol/L be the change in their concentrations as Cu(OH)₂ dissolves: [Cu(NH₃)₄²⁺] = x [OH⁻] = 2x The concentration of NH₃ decreases by 4x as the reaction proceeds: [NH₃] = 4.5 - 4x Now, rewrite the equilibrium equation with these values: K = (x)(2x)² / (4.5 - 4x)⁴

Solving for 'x' (the concentration of Cu(OH)₂)

Since 'K' is extremely small (10^(-31)), we can assume that the change in the NH₃ concentration (4x) is negligible compared to the initial concentration (4.5 M). Thus, we can simplify the equilibrium equation: K ≈ x(2x)² / 4.5⁴ Now, plug in the value of 'K' we calculated earlier and solve for 'x': 10^(-31) ≈ x(4x²) / 4.5⁴ x ≈ 1.45 × 10^(-9)

Finding the solubility of Cu(OH)₂

The value of 'x' we found is the concentration of Cu(OH)₂ that dissolves in the 4.5 M NH₃ solution. Therefore, the solubility of Cu(OH)₂ in 4.5 M NH₃ is approximately 1.45 × 10^(-9) mol/L.

Key concepts

Understanding the Solubility Product Constant (Ksp)

The solubility product constant, often represented as Ksp, plays a crucial role in understanding the solubility of sparingly soluble ionic compounds. It is a special type of equilibrium constant that measures the extent to which a solid ionic compound can dissolve in water to form a saturated solution.

To comprehend Ksp, let's consider a generic soluble salt AB that dissociates in water: \[ AB(s) \rightleftharpoons A^+(aq) + B^-(aq) \]The Ksp expression for this dissociation would be: \[ K_{\text{sp}} = [A^+][B^-] \]It's important to note that Ksp does not include the concentration of the solid since it's a constant value. The Ksp is temperature-dependent and specific to each compound. Higher Ksp values indicate greater solubility. In our exercise, the Ksp for copper(II) hydroxide, \( Cu(OH)_2 \), is given, representing how much of this compound can dissolve to reach equilibrium.

When calculating equilibrium concentrations from Ksp, we usually begin with the assumption that the dissolution of the compound is the only process happening. However, in the presence of other chemicals that react with the dissolved ions, such as NH3 with Cu(OH)2, this assumption may not hold. This is where our understanding of complex ion formation and equilibrium calculations evolve.

Formation Constants and Complex Ions

The formation constant, denoted as Kf, is another equilibrium constant used to describe the stability of complex ions formed in solution. Complex ions are species formed from a central metal ion bonded to one or more ligands. The ligands are molecules or ions capable of donating electron pairs to the metal ion, forming coordinate covalent bonds.

In the illustrated reaction, ammonia is acting as a ligand to copper, forming the complex ion \( Cu(NH_3)_4^{2+} \). The reaction can be written as:\[ Cu^{2+}(aq) + 4NH_3(aq) \rightleftharpoons Cu(NH_3)_4^{2+}(aq) \]The formation constant for this reaction is expressed as: \[ K_{\text{f}} = [Cu(NH_3)_4^{2+}] / [Cu^{2+}][NH_3]^4 \]If Kf is significantly high, as in our exercise (\(2 \times 10^{12}\)), it implies a very stable complex ion. Understanding the Kf value is vital when we deal with equilibrium calculations since it indicates how much the metal-ligand complex will form from the reactants.

Mastering Equilibrium Calculations

Equilibrium calculations are integral to understanding how chemical reactions balance out over time. When a reaction reaches equilibrium, the rates of the forward and reverse reactions are equal, and the concentrations of reactants and products remain constant (though not necessarily equal). To predict the position of equilibrium or to find the concentrations of substances involved in a reaction at equilibrium, we use the equilibrium constant K.

Here's a step-by-step process to tackle equilibrium calculations:

• Write down the balanced chemical equation.
• Write the expression for the equilibrium constant, K.
• Determine the initial concentrations of all species involved.
• Calculate the changes in concentrations as the system moves towards equilibrium using a table (often called an ICE table for Initial, Change, and Equilibrium).
• Substitute these equilibrium concentrations into your K expression.
• Solve the resulting equation(s) for the unknown concentration(s).

In our example, by knowing the solubility product constant (Ksp) for \( Cu(OH)_2 \) and the formation constant (Kf) for \( Cu(NH_3)_4^{2+} \), we calculated a much smaller overall equilibrium constant K for our reaction, which implied a shift far toward the reactants in the presence of high concentrations of ammonia. Given that K is so small, the simplifying assumption that 4x is negligible in comparison to 4.5 M allowed us to calculate 'x', which gives the concentration of copper(II) hydroxide in a solution of 4.5 M ammonia—demonstrating the application of equilibrium principles in a complex system.