Question

In an excess of \(\mathrm{NH}_{3}(a q), \mathrm{Cu}^{2+}\) ion forms a deep blue complex ion, \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\), which has a formation constant \(K_{\mathrm{f}}=5.6 \times 10^{11} .\) Calculate the concentration of \(\mathrm{Cu}^{2+}\) in a solution prepared by adding \(5.0 \times 10^{-3} \mathrm{~mol}\) of \(\mathrm{CuSO}_{4}\) to \(0.500 \mathrm{~L}\) of \(0.40 \mathrm{M} \mathrm{NH}_{3}\).

Short answer

The concentration of uncomplexed \(\mathrm{Cu}^{2+}\) is approximately zero.

Step-by-step solution

Write the Balanced Equation for Complex Formation

The balanced chemical equation for the formation of the complex ion is: \[ \mathrm{Cu}^{2+} + 4 \mathrm{NH}_3 \rightleftharpoons \mathrm{Cu}(\mathrm{NH}_3)_4^{2+} \] This equation shows that each \(\mathrm{Cu}^{2+}\) ion reacts with 4 molecules of \(\mathrm{NH}_3\) to form the complex ion \(\mathrm{Cu}(\mathrm{NH}_3)_4^{2+}\).

Calculate Initial Concentrations

The initial molarity of \(\mathrm{Cu}^{2+}\) from \(\mathrm{CuSO}_4\) is calculated by dividing the moles by the volume in liters: \[ \frac{5.0 \times 10^{-3} \text{ mol}}{0.500 \text{ L}} = 0.010 \text{ M} \]The initial concentration of \(\mathrm{NH}_3\) is given as \(0.40 \text{ M}\).

Define the Equilibrium Expression

Using the formation constant \(K_f = 5.6 \times 10^{11}\), the equilibrium expression for the complex formation is: \[ K_f = \frac{[\mathrm{Cu}(\mathrm{NH}_3)_4^{2+}]}{[\mathrm{Cu}^{2+}][\mathrm{NH}_3]^4} \] This equation relates the concentrations of the complex ion and reactants at equilibrium.

Set Up Equilibrium Concentrations

Let \(x\) be the change in concentration of \(\mathrm{Cu}^{2+}\) that forms the complex. At equilibrium:- \([\mathrm{Cu}^{2+}] = 0.010 - x\)- \([\mathrm{NH}_3] = 0.40 - 4x\)- \([\mathrm{Cu}(\mathrm{NH}_3)_4^{2+}] = x\)

Simplify Assumptions and Solve for x

Given \(K_f\) is very large, it is reasonable to assume that nearly all \(\mathrm{Cu}^{2+}\) is complexed, therefore:- Assume \(4x\) is small compared to initial \(0.40 \text{ M}\), so \([\mathrm{NH}_3] \approx 0.40\).- Substitute into \(K_f\) expression:\[ 5.6 \times 10^{11} = \frac{x}{(0.010 - x)(0.40)^4} \]- Solve for \(x\) assuming \(0.010 - x \approx 0.010\). This simplifies to:\[ x = 5.6 \times 10^{11} \times 0.010 \times (0.40)^4 \approx [\mathrm{Cu}(\mathrm{NH}_3)_4^{2+}] \]- Calculate and solve to find \(x\), the concentration of complexed \(\mathrm{Cu}^{2+}\).

Calculate Remaining \(\mathrm{Cu}^{2+}\) Concentration

Since nearly all \(\mathrm{Cu}^{2+}\) is converted to the complex, \[ [\mathrm{Cu}^{2+}] \approx 0.010 - x \approx 0 \]Using the calculated \(x\) from the previous step shows that the uncomplexed \(\mathrm{Cu}^{2+}\) concentration is negligible.

Key concepts

Formation Constant

The formation constant, denoted as \( K_{f} \), plays a crucial role in describing the extent to which a complex ion will form. In the given problem, the formation constant for the complex ion \( \mathrm{Cu(NH}_3)_4^{2+} \) is a very large value, \( 5.6 \times 10^{11} \). This indicates that the reaction strongly favors the formation of the complex ion from \( \mathrm{Cu}^{2+} \) and ammonia molecules.
The formation constant is defined by the ratio of the concentration of products to reactants at equilibrium for a given reaction. In this complex formation, the equation can be written as:

• The product's concentration: \([\mathrm{Cu(NH}_3)_4^{2+}]\)
• The reactants' concentrations: \([\mathrm{Cu}^{2+}][\mathrm{NH}_3]^4 \)

This large \( K_{f} \) value implies that the equilibrium position is significantly shifted towards the products, meaning that most of the copper ions will be in the form of the complexed \( \mathrm{Cu(NH}_3)_4^{2+} \). Therefore, when calculating the equilibrium composition, it's usually safe to assume that a vast majority of \( \mathrm{Cu}^{2+} \) will be complexed, simplifying calculations.

Equilibrium Expression

Writing the equilibrium expression is an essential first step in solving equilibrium problems related to complex ion formation. For the reaction \[ \mathrm{Cu}^{2+} + 4 \mathrm{NH}_3 \rightleftharpoons \mathrm{Cu(NH}_3)_4^{2+} \] we use the provided formation constant, \( K_{f} \), to express:

• \( K_{f} = \frac{[\mathrm{Cu(NH}_3)_4^{2+}]}{[\mathrm{Cu}^{2+}][\mathrm{NH}_3]^4} \)

This equation is a mathematical representation of chemical equilibrium and helps in determining how concentrations change when the system reaches equilibrium.
The equation expresses how the concentration of the product (the complex ion) is related to the concentrations of the reactants (\( \mathrm{Cu}^{2+} \) and \( \mathrm{NH}_3 \)).
In practical terms, the equilibrium expression allows us to solve for unknown concentrations, given a set of initial conditions and the equilibrium constant. It’s important to set up this equation correctly, as it forms the basis for calculating the concentration of remaining species at equilibrium.

Initial Concentration

Understanding initial concentration is key to determining the system's behavior as it approaches equilibrium. In our problem, we calculate the initial concentration of the reactants involved in the complex ion formation step.
First, we find the initial concentration of \( \mathrm{Cu}^{2+} \) by dividing the amount of \( \mathrm{CuSO}_4 \) (given as moles) by the volume of the solution:

• \(\text{Initial concentration of } [\mathrm{Cu}^{2+}]: \frac{5.0 \times 10^{-3} \text{ mol}}{0.500 \text{ L}} = 0.010 \text{ M}\)

Similarly, the initial concentration of \( \mathrm{NH}_3 \) is directly provided as

• \(0.40 \text{ M}\)

Initiating our calculations with these concentrations allows us to apply them to our equilibrium expression. Changes in concentration can then be related back to these initial values.
By understanding how the initial concentrations interact through the equilibrium expression, we can predict the behavior of the system and how it will shift in order to reach equilibrium.