Question

A solution is formed by mixing \(50.0 \mathrm{~mL}\) of \(10.0 \mathrm{M} \mathrm{NaX}\) with \(50.0 \mathrm{~mL}\) of \(2.0 \times 10^{-3} \mathrm{M} \mathrm{CuNO}_{3} .\) Assume that \(\mathrm{Cu}(\mathrm{I})\) forms com- plex ions with \(\mathrm{X}^{-}\) as follows: $$\begin{aligned}\mathrm{Cu}^{+}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}(a q) & K_{1}=1.0 \times 10^{2} \\ \mathrm{CuX}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{2}^{-}(a q) & K_{2}=1.0 \times 10^{4} \\ \mathrm{CuX}_{2}^{-}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) & K_{3}=1.0 \times 10^{3} \end{aligned}$$ with an overall reaction $$\mathrm{Cu}^{+}(a q)+3 \mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \quad K=1.0 \times 10^{9}$$ Calculate the following concentrations at equilibrium. a. \(\mathrm{CuX}_{3}^{2-}\) b. \(\mathrm{CuX}_{2}\) c. \(\mathrm{Cu}^{+}\)

Short answer

a. \([\mathrm{CuX}_{3}^{2-}] \approx 9.99 \times 10^{-4} M\) b. \([\mathrm{CuX}_{2}^{-}] \approx 1.0 \times 10^{-8} M\) c. \([\mathrm{Cu}^{+}] \approx 1 \times 10^{-6} M\)

Step-by-step solution

Calculate the initial concentrations

First, we need to calculate the initial concentrations of \(\mathrm{Cu}^{+}\) and \(\mathrm{X}^{-}\). We can use the formula: Initial concentration = (amount of solute) / (final volume of solution) For the initial concentration of \(\mathrm{Cu}^{+}\) we have: Initial concentration of \(\mathrm{Cu}^{+}\) = (50.0 mL * 2.0 * 10^{-3} M)/((50.0+50.0) mL) = (1.0 * 10^{-4} mol) / (0.1 L) = 1.0 * 10^{-3} M For the initial concentration of \(\mathrm{X}^{-}\) we have: Initial concentration of \(\mathrm{X}^{-}\) = (50.0 mL * 10.0 M)/((50.0+50.0) mL) = (0.5 mol) / (0.1 L) = 5.0 M

Set up the reaction equation and ICE table

Now, we set up the reaction equation: \(\mathrm{Cu}^{+}(a q) + \mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}(a q) \rightarrow K_1\) \(\mathrm{CuX}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{2}^{-}(a q) \rightarrow K_2\) \(\mathrm{CuX}_{2}^{-}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \rightarrow K_3\) And set up an ICE table (Initial, Change, Equilibrium) for the concentrations of the chemical species: | | \(\mathrm{Cu}^{+}\) | \(\mathrm{X}^{-}\) | \(\mathrm{CuX}\) | \(\mathrm{CuX}_{2}^{-}\) | \(\mathrm{CuX}_{3}^{2-}\) | |-------------|---------------|-------------|-----------|----------------|----------------0| | Initial | 1.0 * 10^{-3} M | 5.0 M | 0 | 0 | 0 | | Change | -x | -3x | x | x | x | | Equilibrium | 1.0 * 10^{-3} - x | 5.0 - 3x | x | x | x | Where x is the concentration of \(\mathrm{CuX}_{3}^{2-}\) at equilibrium.

Use the equilibrium constant to find the equilibrium concentrations

Since the overall equilibrium constant \(K = 1.0 \times 10^{9}\), we can write the relationship between the equilibrium concentrations: \(K = \frac{[\mathrm{CuX}_{3}^{2-}]}{[\mathrm{Cu}^{+}][\mathrm{X}^{-}]^3}\) Substitute the equilibrium concentrations from the ICE table: \(1.0 \times 10^{9} = \frac{x}{(1.0 \times 10^{-3} - x)(5.0 - 3x)^3}\) Solving for x, we get the equilibrium concentration of \(\mathrm{CuX}_{3}^{2-}\) \(x \approx 9.99 \times 10^{-4} M \) Now we will use the other reactions to find \(\mathrm{CuX}_{2}^{-}\) and \(\mathrm{Cu}^{+}\): For \(\mathrm{CuX}_{2}^{-}\), we have: \(K_2 = \frac{[\mathrm{CuX}_{2}^{-}]}{[\mathrm{CuX}][\mathrm{X}^{-}]}\) Now substitute the known equilibrium concentrations: \(1.0 \times 10^4 = \frac{x}{[(5.0 - 3x)(9.99 \times 10^{-4})]}\) Solving for x, we get the equilibrium concentration of \(\mathrm{CuX}_{2}^{-}\): \(x \approx 1.0 \times 10^{-8} M\) Finally, for the concentration of \(\mathrm{Cu}^{+}\), we have: \( [\mathrm{Cu}^{+}] = 1.0 \times 10^{-3} - 9.99 \times 10^{-4}\) \( [\mathrm{Cu}^{+}] \approx 1 \times 10^{-6} M\)

Present the results

So, the equilibrium concentrations are: a. \([\mathrm{CuX}_{3}^{2-}] \approx 9.99 \times 10^{-4} M\) b. \([\mathrm{CuX}_{2}^{-}] \approx 1.0 \times 10^{-8} M\) c. \([\mathrm{Cu}^{+}] \approx 1 \times 10^{-6} M\)

Key concepts

Equilibrium Constant

Understanding the equilibrium constant (K) is essential for mastering chemical reactions that do not go to completion but instead reach a state where the reactants and products coexist. The equilibrium constant is a numerical value that expresses the ratio of the concentrations of products to reactants, each raised to the power of their coefficients in the balanced chemical equation.

For the given reaction \[\mathrm{Cu}^{+}(a q) + 3 \mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q)\] the reaction quotient, or K, is expressed as \[K = \frac{[\mathrm{CuX}_{3}^{2-}]}{[\mathrm{Cu}^{+}][\mathrm{X}^{-}]^3}\] where the square brackets denote the concentration of each species at equilibrium.

The larger the value of K, the farther to the right the equilibrium lies, indicating a higher concentration of products. In our exercise, K is exceptionally high (\(1.0 \times 10^{9}\)), signifying that the formation of \(\mathrm{CuX}_{3}^{2-}\) is very favorable under the given conditions. By establishing the initial concentrations, an ICE table can be utilized to evaluate how these concentrations shift to reach equilibrium based on K.

ICE Table

When dealing with equilibrium problems, an ICE table – standing for Initial, Change, and Equilibrium – is an invaluable tool that helps visualize and calculate the shifts in concentrations over the course of a reaction. It sets the stage for understanding how the system reaches equilibrium.

The initial step involves listing the initial concentrations of reactants and products before any reaction occurs. The 'Change' row denotes the change in concentration as the system moves toward equilibrium, often represented by a variable like \(x\). Lastly, the 'Equilibrium' row reflects the initial concentrations adjusted by the change. For our complex ion formation problem, the ICE table shows that the initial concentration of \(\mathrm{CuX}_{3}^{2-}\) is zero, with changes leading to an increase by \(x\), the unknown equilibrium concentration of \(\mathrm{CuX}_{3}^{2-}\).

By solving the adjusted ICE table equation with the given equilibrium constant, students can find the value of \(x\), thus determining the concentration of all species at equilibrium. This step-wise approach is key to enhancing comprehension and provides a clear method for tackling diverse chemical equilibrium problems.

Complex Ion Formation

Complex ions are species formed when a central metal ion binds with one or more ligands – molecules or ions that can donate a pair of electrons. In our case, the ligand, \(\mathrm{X}^{-}\), coordinates with the copper ion forming a series of complex ions: \(\mathrm{CuX}^{+}\), \(\mathrm{CuX}_{2}^{-}\), and \(\mathrm{CuX}_{3}^{2-}\).

These stepwise reactions are guided by individual equilibrium constants, \(K_1\), \(K_2\), and \(K_3\), where each constant represents the affinity between the metal ion and the ligands at each stage. As the reaction proceeds, additional ligands may add to the complex ion in a sequential manner, creating a chain of equilibria that can be analyzed separately or as an overall reaction with an overall equilibrium constant, \(K\).

In our exercise, given the high values of the constants, we infer a strong preference for complex ion formation, with an overwhelming majority of \(\mathrm{Cu}^{+}\) ions ending up in the fully complexed form of \(\mathrm{CuX}_{3}^{2-}\). This concept is critical in areas such as coordination chemistry and is frequently applied in understanding biochemical interactions and industrial processes involving metal complexes.