Question

Calculate \(\left[\mathrm{Cu}^{2+}\right]\) in a \(0.10 \mathrm{M} \mathrm{CuSO}_{4}(\) aq) solution that is also \(6.0 \mathrm{M}\) in free \(\mathrm{NH}_{3}\). \(\mathrm{Cu}^{2+}(\mathrm{aq})+4 \mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}(\mathrm{aq})\) \(K_{\mathrm{f}}=1.1 \times 10^{13}\)

Short answer

\([\mathrm{Cu}^{2+}]\) in the solution is approximately \(1.2 \times 10^{-19} M\).

Step-by-step solution

Initial Concentrations

Set up initial concentrations. From the initial molarity of \(\mathrm{CuSO}_{4}\), it is known that the initial concentration of \(\mathrm{Cu}^{2+}\) is 0.10 M. Also, the initial molar concentration of \(\mathrm{NH}_{3}\) is given as 6.0 M.

Establish change based on reaction

Discuss the change in concentrations that occur after the reaction. Based on the stoichiometry of the reaction, for each mole of \(\mathrm{Cu}^{2+}\) that reacts, four moles of \(\mathrm{NH}_{3}\) are consumed. Therefore, assuming all the \(\mathrm{Cu}^{2+}\) reacts, the change in concentration for \(\mathrm{NH}_{3}\) will be 4 times the initial concentration for \(\mathrm{Cu}^{2+}\). The \(\mathrm{NH}_{3}\) will decrease by (4)(0.10 M) = 0.40 M, giving a final concentration of 6.0 M - 0.40 M = 5.6 M.

Formation constant

Using the formation constant, set up an expression for the equilibrium. The right side of the equilibrium is predominant according to the high formation constant, so the concentration of the complex ion can be approximated by the initial concentration of \(\mathrm{Cu}^{2+}\), this is \(0.10 M\). Also, the concentration of \(\mathrm{NH}_{3}\) has already been calculated, so we presume \([\mathrm{Cu}^{2+}]\) is very small and can be considered as \(x\). Plugging all values into the equilibrium constant equation \(K_{f}=\frac{[\mathrm{Cu}(NH_{3})_{4}]^{2+}}{[\mathrm{Cu}^{2+}][NH_{3}]^{4}}\), this simplifies to \(1.1 \times 10^{13} = \frac{0.10}{x(5.6)^4}\).

Calculate concentration

Rearrange and solve for \([\mathrm{Cu}^{2+}]\). Solving the previous equation will find the concentration of \([\mathrm{Cu}^{2+}]\) in the solution. The answer is approximately \(1.2 \times 10^{-19} M\).

Key concepts

Complex Ion Formation

When certain metal ions react with ligands, new and more stable structures called complex ions form. In this exercise, copper (II) ions (\(\mathrm{Cu}^{2+}\)) react with ammonia molecules (\(\mathrm{NH}_3\)) to form the complex ion \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) through a reversible reaction.
This means the reaction can go forward or backward. The process of complex ion formation involves the association of metal ions with non-metal atoms having a pair of non-bonded electrons, which act as electron donors, called ligands.
When \(\mathrm{Cu}^{2+}\) and \(\mathrm{NH}_3\) interact, the ammonia ligands donate their lone pairs of electrons, forming a coordinate covalent bond with copper, resulting in a stable structure. In complex ions like this one, the metal center is surrounded by ligands that balance the charge and stabilize the ion. Such formations are crucial as they significantly lower the concentration of free metal ions like \(\mathrm{Cu}^{2+}\) in solution.

Equilibrium Constants

Equilibrium constants (\(K_f\)) provide us with insight into the extent of a chemical reaction at equilibrium. They are numerical representations that describe the balance between reactants and products in a reversible reaction. The magnitude of the equilibrium constant helps predict the direction and extent of the reaction. In this exercise, \(K_{f} = 1.1 \times 10^{13}\) indicates a strong tendency to form the complex ion, showing the equilibrium overwhelmingly favors the products.
The equilibrium constant is derived from the concentrations of the products and reactants raised to the power of their coefficients in the balanced equation:

• The equation for this reaction is: \[ K_{f} = \frac{[\mathrm{Cu}(NH_{3})_{4}]^{2+}}{[\mathrm{Cu}^{2+}][NH_{3}]^{4}} \]
• A high \(K_f\) implies that at equilibrium, almost all of the \(\mathrm{Cu}^{2+}\) ions are bound in the complex form, thus lowering their free concentration in the solution.
  

Understanding the equilibrium constant allows us to predict the behavior of complex ions in a solution under different conditions.

Coordination Compounds

Coordination compounds consist of a central metal atom or ion bonded to surrounding molecules or ions called ligands. These ligands are attached to the central metal through coordinate covalent bonds. In this exercise, the compound formed is \(\left[\mathrm{Cu}(\mathrm{NH}_3)_4\right]^{2+}\), featuring copper as the central metal ion and ammonia as the ligands.
The formation of coordination compounds changes several properties of the metal ions, such as their color, solubility, and reactivity. This change occurs because the ligands create a specific environment around the metal ion, thus stabilizing it and altering its original electron configuration.
Coordination compounds play important roles beyond simple chemical reactions. They are crucial in:

• Catalysis
• Biological systems, like hemoglobin
• Industrial applications, such as purification processes
  

Understanding these changes and their effects is essential in various scientific fields, including chemistry, biology, and environmental science.