Question

A solution is formed by mixing \(50.0 \mathrm{mL}\) of \(10.0 \mathrm{M}\) NaX with \(50.0 \mathrm{mL}\) of \(2.0 \times 10^{-3} \mathrm{M} \mathrm{CuNO}_{3} .\) Assume that \(\mathrm{Cu}^{+}\) forms complex ions with \(X^{-}\) as follows: $$\mathrm{Cu}^{+}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}(a q) \quad K_{1}=1.0 \times 10^{2}$$ $$\mathrm{CuX}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{2}^{-}(a q) \quad K_{2}=1.0 \times 10^{4}$$ $$\mathrm{CuX}_{2}^{-}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \quad K_{3}=1.0 \times 10^{3}$$ 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. [CuX3^2-]: \(\approx 1.35 \times 10^{-7} \mathrm{M}\) b. [CuX2^-]: \(\approx 1.26 × 10^{-6} \mathrm{M}\) c. [Cu^+]: \(\approx 9.68 × 10^{-4} \mathrm{M}\)

Step-by-step solution

Calculation of initial concentration of Cu+

\(Initial \: concentration \: of \: Cu^{+} = \frac{(50.0\: mL)(2.0 \times 10^{-3} M)}{50.0\: mL + 50.0\: mL}\) \(Initial \: concentration \: of \: Cu^{+} = 1.0 \times 10^{-3} \mathrm{M}\) Now, let's find the initial concentration of X-. Initial concentration of X- = (Volume of NaX solution × concentration of NaX) / total volume

Calculation of initial concentration of X-

\(Initial \: concentration \: of \: X^{-} = \frac{(50.0\: mL)(10.0 \: M)}{50.0\: mL + 50.0\: mL}\) \(Initial \: concentration \: of \: X^{-} = 5.0 \mathrm{M}\) Now we will set up the equilibrium expressions for each step in the reaction and solve them sequentially.

Equilibrium expression for the first reaction

\([CuX] = K_{1} [Cu^{+}][X^{-}]\)

Equilibrium expression for the second reaction

\([CuX_{2}^{-}] = K_{2} [CuX][X^{-}]\)

Equilibrium expression for the third reaction

\([CuX_{3}^{2-}] = K_{3} [CuX_{2}^{-}][X^{-}]\) Let's solve the equilibrium expressions in the order of the reactions. For the first reaction, let x represent the concentration of CuX at equilibrium

Solving for x

\(x = K_{1} [(1.0 \times 10^{-3}) - x][(5.0) - x]\) Next, we will solve for y, the equilibrium concentration of CuX2^- in the second reaction.

Solving for y

\(y = K_{2} [x][(5.0) - x - y]\) Now, we will solve for the concentration of CuX3^2-, which is the equilibrium concentration of the third reaction.

Solving for [CuX3^2-]

\([CuX_{3}^{2-}] = K_{3} [y][(5.0) - x - y]\) \([CuX_{3}^{2-}] = 1.0 \times 10^{3} [y][(5.0) - x - y]\) Since it's difficult to solve the set of equations directly, we can use the overall reaction.

Calculating the equilibrium concentration of CuX3^2-

\([CuX_{3}^{2-}] = K [(1.0 \times 10^{-3})][(5.0) - x]^{3}\) \([CuX_{3}^{2-}] = (1.0 \times 10^{9})[(1.0 \times 10^{-3})][(5.0) - x]^{3}\) Plug the value of x from step 6 into the equation. Using a calculator or software, we find that x = 3.16 x 10^(-5) and [CuX3^2-] ≈ 1.35 x 10^(-7)M.

Calculating the equilibrium concentration of CuX2^-

Now that we have the values of x and the concentration of CuX3^2-, we can calculate the equilibrium concentration of CuX2^- by plugging x into the equation from step 7: \([CuX_{2}^{-}] = 1.0\times 10^4 [x][(5.0) - x]\) \([CuX_{2}^{-}] = 1.0\times 10^4(3.16\times 10^{-5})(5.000 - 3.16\times 10^{-5})\) Using a calculator, we find that [CuX_{2}^{-}] ≈ 1.26 × 10^(-6)M.

Calculating the equilibrium concentration of Cu+

The change in the concentration of Cu+ will be equal to the change in CuX, since one Cu+ ion reacts with one X- ion in the first step: \([Cu^{+}] = [Initial \; concentration \; of \; Cu^{+}] - x\) \([Cu^{+}] = (1.0 \times 10^{-3}) - (3.16 \times 10^{-5})\) Using a calculator, we find that [Cu^+] ≈ 9.68 × 10^(-4)M. So, the equilibrium concentrations are: a. [CuX3^2-]: \(\approx 1.35 \times 10^{-7} \mathrm{M}\) b. [CuX2^-]: \(\approx 1.26 × 10^{-6} \mathrm{M}\) c. [Cu^+]: \(\approx 9.68 × 10^{-4} \mathrm{M}\)

Key concepts

Complex Ion Formation

Complex ion formation occurs when a central metal ion binds to surrounding ions or molecules called ligands. In this exercise, the metal ion is \({\text{Cu}}^{+}\) and the ligand is \({\text{X}}^{-}\). They combine to form different complex ions such as \({\text{CuX}}\), \({\text{CuX}}_{2}^{-}\), and \({\text{CuX}}_{3}^{2-}\).
The formation of a complex ion is a stepwise process, often described by multiple equilibrium equations. Each reaction represents the addition of one ligand to the growing complex. This sequence involves:

• \({\text{Cu}}^{+}(aq) + {\text{X}}^{-}(aq) \rightleftharpoons {\text{CuX}}(aq)\)
• \({\text{CuX}}(aq) + {\text{X}}^{-}(aq) \rightleftharpoons {\text{CuX}}_{2}^{-}(aq)\)
• \({\text{CuX}}_{2}^{-}(aq) + {\text{X}}^{-}(aq) \rightleftharpoons {\text{CuX}}_{3}^{2-}(aq)\)

Complex ion formation often significantly changes the chemical properties and solubility of the metal ions.

Equilibrium Constant

The equilibrium constant, denoted as \(K\), quantifies the balance between the reactants and products in a reversible chemical reaction at equilibrium. It is specific to a particular reaction at a given temperature.
For this problem, the stepwise equilibrium constants are:

• \(K_{1} = 1.0 \times 10^{2}\)
• \(K_{2} = 1.0 \times 10^{4}\)
• \(K_{3} = 1.0 \times 10^{3}\)

The overall equilibrium constant for the combined reaction is \(K = 1.0 \times 10^{9}\).

Equilibrium constants allow us to understand how far a reaction will proceed and are crucial for calculating concentrations of substances in equilibrium. Large values of \(K\) imply the reaction strongly favors product formation.

Molar Concentration

Molar concentration measures the amount of a solute (like \(\text{Cu}^{+}\) or \(\text{X}^{-}\)) present in a given volume of solution. It is expressed in moles per liter (M).
In this exercise, we mix two solutions, requiring calculations of their initial concentrations in the combined volume:

• Initial \(\text{Cu}^{+}\) concentration: \(\frac{(50.0\, \text{mL})(2.0 \times 10^{-3} \, \text{M})}{100.0\, \text{mL}} = 1.0 \times 10^{-3} \, \text{M}\)
• Initial \(\text{X}^{-}\) concentration: \(\frac{(50.0\, \text{mL})(10.0 \, \text{M})}{100.0\, \text{mL}} = 5.0 \, \text{M}\)

Molar concentration is essential in setting up the initial conditions for calculating equilibrium concentrations.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It provides a way to calculate the amounts of substances consumed and produced.
In the formation of complex ions, stoichiometry helps us write balanced chemical equations at each step. For every mole of \(\text{Cu}^{+}\) reacting with three moles of \(\text{X}^{-}\), one mole of \(\text{CuX}_{3}^{2-}\) is formed.
By understanding stoichiometry, we can calculate equilibrium concentrations of products and reactants efficiently. We used this concept to derive expressions for each step in the reaction, guiding us in determining the equilibrium concentrations:

• For \(\text{CuX}\), \(\text{CuX}_{2}^{-}\), and \(\text{CuX}_{3}^{2-}\), we followed balanced equations and stoichiometric coefficients.

This ensures that conservation of mass and charge is maintained throughout the reaction process.