Question

In a solution that is \(0.0500 \mathrm{M}\) in \(\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{3-}\) and \(0.80 \mathrm{M}\) in free \(\mathrm{CN}^{-}\), the concentration of \(\mathrm{Cu}^{+}\) is \(6.1 \times 10^{-32} \mathrm{M}\) Calculate \(K_{\mathrm{f}}\) of \(\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{3-}\). \(\mathrm{Cu}^{+}(\mathrm{aq})+4 \mathrm{CN}^{-}(\mathrm{aq}) \rightleftharpoons\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{3-}(\mathrm{aq}) \quad K_{\mathrm{f}}=?\)

Short answer

The formation constant \(K_f\) of \([Cu(CN)_4]^{3-}\) is approximately \(6.41 \times 10^{36}\).

Step-by-step solution

Understand the equilibrium concept

The equilibrium constant \(K\) defines the ratio of the concentrations of the products to the reactants at equilibrium. In the reaction \(Cu^+ + 4CN^- \rightarrow [Cu(CN)_4]^{3-}\), the formation constant \(K_f\) is thus given by: \[K_f = \frac{([Cu(CN)_4]^{3-})}{[Cu^+][CN^-]^4}\]

Substitute given concentration values

Substitute the given concentration values of \([Cu(CN)_4]^{3-}\) = \(0.0500 M\), \([Cu^+]\) = \(6.1 \times 10^{-32} M\), and \([CN^-]\) = \(0.80 M\) into the equation for \(K_f\). So: \[K_f = \frac{(0.0500)}{(6.1 \times 10^{-32})(0.80)^4}\]

Solve for \(K_f\)

Carrying out the calculation we get: \[K_f \approx 6.41 \times 10^{36}\]

Key concepts

Solution Chemistry

Solution chemistry is all about the behavior and characteristics of substances when dissolved in a solvent, usually water. These solutions can be electrolytic or nonelectrolytic based on whether the solute conducts electricity when dissolved.
In chemical terms, a solute is the substance being dissolved, while the solvent is the medium that does the dissolving. For example, in our exercise, \([Cu(CN)_4]^{3-}\), a complex ion, is formed in a solution containing free \(\text{CN}^–\) ions and \(\text{Cu}^+\) ions.
This dissolution process is crucial in facilitating various chemical reactions, including those involving complex ion formations and equilibria.Understanding the concentration of solutes is essential. This is often expressed in molarity (M), which tells us how many moles of solute are present in one liter of solution:

• Molarity (M) = moles of solute / liters of solution

Knowing these concentrations allows chemists to predict the behavior of the solution and its reactivity.

Complex Ion Formation

Complex ion formation is a key concept in chemistry, involving the combination of simple ions to form a more complex entity. A complex ion usually consists of a central metal atom or ion surrounded by molecules or anions, known as ligands.
In the given reaction, \([Cu^+]\) interacts with \(\text{CN}^-\) ions to form the complex ion \([Cu(CN)_4]^{3-}\).
Ligands, like \(\text{CN}^-\) in this example, donate electron pairs to form coordinate covalent bonds with the metal ion. This process can influence the properties of both the metal and the ligand.
Such formations typically occur in specific geometries, affecting the color, solubility, and reactivity of the complex ion. These characteristics make understanding complex ions essential for fields like coordination chemistry and material science.

Equilibrium Concepts

Chemical equilibrium refers to the state in which the concentrations of all reactants and products remain constant over time. In a dynamic equilibrium, the forward and reverse reactions proceed at the same rate, creating a constant ratio between products and reactants.
The reaction in our exercise, \(\text{Cu}^+ + 4\text{CN}^- ightleftharpoons [\text{Cu}(\text{CN})_4]^{3-}\), demonstrates equilibrium in complex ion formation. Here, the equilibrium constant, \(K_f\), quantifies the tendency of the complex ion to form in the solution state.Key points on equilibrium:

• Equilibrium is dynamic, rather than static
• Forward and reverse reaction rates are equal
• Concentrations do not change over time

Determining equilibrium constants like \(K_f\) provides insight into the position of equilibrium and the stability of complex ions in solution.

Formation Constant Calculation

The formation constant, often denoted as \(K_f\), is a specific type of equilibrium constant that applies to the formation of complex ions. It expresses the ratio of concentration of a complex ion to the concentrations of its individual components at equilibrium.
In our problem, the formation constant \(K_f\) is calculated as follows:\(K_f = \frac{\left[Cu(CN)_4]^{3-}\right}{\left[Cu^+\right]\left(CN^-\right)^4}\)This formula shows how \(K_f\) is dependent on the concentration of the resultant complex ion compared to the original reactants.
Calculating \(K_f\) involves substituting the concentration values given in the exercise:

• \([Cu(CN)_4]^{3-} = 0.0500 M\)
• \([Cu^+] = 6.1 \times 10^{-32} M\)
• \([CN^-] = 0.80 M\)

The calculated \(K_f\) value of \(6.41 \times 10^{36}\) demonstrates the strength and stability of the complex ion under the given conditions.