Question

For the complex ion equilibrium for the dissociation of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_{4}^{+}\), calculate (1) the ratio of \(\mathrm{Cu}^{2+}\) ion to \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_{4}^{+}\) in a solution \(1 \mathrm{M}\) in \(\mathrm{NH}_{3} ;\) (2) the equilibrium concentration of \(\mathrm{NH}_{3}\) needed to attain \(50 \%\) conversion of \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_{4}^{+}\) \(\mathrm{K}_{\mathrm{a}}\) for \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_{4}^{+}=2 \times 10^{-13}\)

Short answer

(1) The ratio of \(\mathrm{Cu}^{2+}\) ion to \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_4^{+}\) in a solution \(1\;\text{M}\) in \(\text{NH}_3\) is approximately \(2.00 \times 10^{-4}\). (2) The equilibrium concentration of \(\text{NH}_3\) needed to attain \(50\%\) conversion of \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_4^{+}\) is approximately \(2.15\;\text{M}\).

Step-by-step solution

Write the balanced dissociation reaction

The balanced dissociation reaction of the complex ion \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_4^{+}\) is: \[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_4^{+} \rightleftharpoons \mathrm{Cu}^{2+} + 4\text{NH}_3\]

Set up the equilibrium expression

Given the dissociation reaction written in Step 1, we can set up the equilibrium expression using the equilibrium constant \(K_a\) as: \[K_a = \frac{[\mathrm{Cu}^{2+}][\mathrm{NH}_3]^4}{[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_4^{+}]}\]

Calculate ratio of \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_4^{+}\) in a solution \(1\;\text{M}\) in \(\text{NH}_3\)

To calculate the ratio of \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_4^{+}\), we should express the concentration of each species in terms of one variable, let's say \(x\). Let the concentration of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_4^{+}\) decrease by \(x\) at equilibrium, then: \[ [\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_4^{+}] = 1 - x \] \[ [\mathrm{Cu^{2+}}] = x \] \[ [\mathrm{NH_3}] = 1 + 4x \] Now, let's substitute these equilibrium concentrations into the equilibrium expression: \[K_a = \frac{x(1 + 4x)^4}{1 - x}\] We are given that \(K_a = 2\times10^{-13}\). So, \[2 \times 10^{-13} = \frac{x(1 + 4x)^4}{1 - x}\]

Solve for variable x and find the ratio

To find the value of \(x\), we need to solve the above equation. Use numerical methods, or iteratively solve it: \[x \approx 2.00 \times 10^{-4}\] Now that we have the value of \(x\), we can find the ratio of \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_4^{+}\) as: \[\frac{[\mathrm{Cu}^{2+}]}{[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_4^{+}]} = \frac{x}{1 - x}\] Using the value of \(x\): \[\frac{[\mathrm{Cu}^{2+}]}{[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_4^{+}]} \approx 2.00 \times 10^{-4}\]

Calculate the equilibrium concentration of \(\text{NH}_3\) for \(50\%\) conversion

For \(50\%\) conversion of \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}_{4}^{+}\), we know that: \[[\mathrm{Cu}^{2+}] = [\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}_{4}^{+}]\] Let \(y\) be the equilibrium concentration of \(\mathrm{NH}_3\) required for this conversion, and we have: \[ [\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_4^{+}] = \frac{1}{2} \] \[ [\mathrm{Cu^{2+}}] = \frac{1}{2} \] \[ [\mathrm{NH_3}] = y - 4(1 - \frac{1}{2}) = y - 2 \] Plugging these concentrations into the equilibrium expression and solving for \(y\): \[K_a = \frac{\frac{1}{4}(y - 2)^4}{\frac{1}{2}}\] \[2 \times 10^{-13} = \frac{(y - 2)^4}{4}\] Solving for \(y\) numerically: \[y \approx 2.15\]

Final answers

(1) The ratio of \(\mathrm{Cu}^{2+}\) ion to \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_4^{+}\) in a solution \(1\;\text{M}\) in \(\text{NH}_3\) is approximately \(2.00 \times 10^{-4}\). (2) The equilibrium concentration of \(\text{NH}_3\) needed to attain \(50\%\) conversion of \(\mathrm{Cu}^{2+}\) to \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2}{ }_4^{+}\) is approximately \(2.15\;\text{M}\).

Key concepts

Dissociation of Complex Ions

Understanding the dissociation of complex ions is critical for grasping chemical equilibrium concepts in solutions. A complex ion comprises a central metal ion surrounded by molecules or anions, known as ligands. The complex ion in our exercise, \(\mathrm{Cu}(\mathrm{NH}_{3})^{2}_{4}^{+}\), dissociates into \(\mathrm{Cu}^{2+}\) ions and \(\mathrm{NH}_3\) ligands.

Dissociation can be influenced by the concentration of the ligands in the solution. For instance, a high concentration of \(\mathrm{NH}_3\) ligand shifts the equilibrium towards the formation of the complex ion, as per Le Chatelier's Principle. Conversely, lowering the ligand concentration can lead to an increased dissociation of the complex ion. This process is reversible and reaches a state of dynamic equilibrium where the rates of formation and dissociation of the complex ion occur at the same rate, leading to constant concentrations of the species involved in the reaction.

To visualize the process and calculate equilibrium concentrations, it's helpful to set up a reaction table that details the changes in concentration for all species involved from the initial state to the equilibrium state, as shown in the provided exercise solution.

Equilibrium Constant

The equilibrium constant, \(K_a\), is a numeric value that represents the ratio of the concentration of the products to the reactants at equilibrium for a reversible chemical reaction, raised to the power of their stoichiometric coefficients. It's a fundamental concept when dealing with dissociation of complex ions:

For the Reaction

\[\mathrm{Cu}(\mathrm{NH}_{3})^{2}_{4}^{+} \rightleftharpoons \mathrm{Cu}^{2+} + 4\text{NH}_3\]
the equilibrium constant is expressed as:

\[K_a = \frac{[\mathrm{Cu}^{2+}][\mathrm{NH}_3]^4}{[\mathrm{Cu}(\mathrm{NH}_{3})^{2}_{4}^{+}]}\]

This expression allows us to understand how the concentrations of each species will affect the position of the equilibrium. A larger equilibrium constant indicates a greater extent of complex ion dissociation, while a smaller constant suggests the reverse. The value of \(K_a\) is temperature-dependent and remains constant for a given temperature, regardless of the initial concentrations of the reactants and products.

The students should note that the equilibrium constant provided in the exercise is incredibly small (\(2 \times 10^{-13}\)), signifying that at equilibrium, the concentration of the complex ion is much greater than the concentration of the \(\mathrm{Cu}^{2+}\) ions and \(\mathrm{NH}_3\) molecules that have dissociated.

Chemical Equilibrium Concentration

Chemical equilibrium concentration refers to the stable concentration of reactants and products in a reversible chemical reaction when the rate of the forward reaction equals the rate of the reverse reaction. At this point, no net change is observed in the concentration of either reactants or products, although the reaction may still proceed at the molecular level.

For our exercise involving \(\mathrm{Cu}(\mathrm{NH}_{3})^{2}_{4}^{+}\), reaching equilibrium means that the ratio of dissociated \(\mathrm{Cu}^{2+}\) ions to the undissociated complex ion becomes fixed. By manipulating the solution's \(\mathrm{NH}_3\) concentration or the pressure and temperature conditions, we can influence the equilibrium concentrations and, consequently, the position of the equilibrium. Calculating these equilibrium concentrations, as in the second part of the exercise, provides valuable information about the extent of the reaction and aids in designing conditions that favor either the formation or dissociation of the complex ion.

By understanding these basic principles of chemical equilibrium, students will be able to apply this knowledge to various equilibrium systems, making predictions about the behavior of reactions under different conditions and interpreting experimental data effectively.