Problem 74 - Chemistry An Atoms First Approach

Question

The copper(I) ion forms a chloride salt that has \(K_{\mathrm{sp}}=1.2 \times\) \(10^{-6} .\) Copper(I) also forms a complex ion with \(\mathrm{Cl}^{-}:\) $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \quad K=8.7 \times 10^{4}$$ a. Calculate the solubility of copper(I) chloride in pure water. (Ignore \(\mathrm{CuCl}_{2}^{-}\) formation for part a.) b. Calculate the solubility of copper(I) chloride in \(0.10 M\) NaCl.

Step-by-Step Solution

Answer

The solubility of copper(I) chloride in pure water is found to be \(1.1 \times 10^{-3}\ \text{mol/L}\). Meanwhile, the solubility of copper(I) chloride in 0.10 M of NaCl solution is calculated to be \(9.6 \times 10^{-5}\ \text{mol/L}\).

Step 1: Determine the solubility product expression for copper(I) chloride

We know that copper(I) chloride dissociates as follows: $$\mathrm{CuCl}\,(s) \rightleftharpoons \mathrm{Cu}^{+}(a q)+\mathrm{Cl}^{-}(a q)$$ The solubility product expression for this reaction is given by: $$K_{sp} = [\mathrm{Cu}^{+}][\mathrm{Cl}^{-}]$$ We are given the value of \(K_{sp}\) as \(1.2 \times 10^{-6}\).

Step 2: Calculate the solubility of copper(I) chloride in pure water

Let's denote the solubility of copper(I) chloride in water as \(x\ \text{mol/L}\). When it dissolves in water, it produces \(x\ \text{mol/L}\) of \(\mathrm{Cu}^{+}\) ions and \(x\ \text{mol/L}\) of \(\mathrm{Cl}^{-}\) ions. Plugging these values into the solubility product expression: $$ K_{sp} = x^2$$ Rearranging and solving for \(x\): $$x = \sqrt{K_{sp}} = \sqrt{1.2 \times 10^{-6}} = 1.1 \times 10^{-3}\ \text{mol/L}$$ Hence, the solubility of copper(I) chloride in pure water is \(1.1 \times 10^{-3}\ \text{mol/L}\).

Step 3: Determine the solubility product expression for copper(I) chloride in 0.10 M NaCl

In 0.10M NaCl, there will be an excess of \(\mathrm{Cl}^{-}\) ions in the solution. When copper(I) chloride dissolves, it will also form a complex ion with \(\mathrm{Cl}^{-}\), and the reaction is as follows: $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q)$$ Equilibrium constant for this reaction is given as \(K = 8.7 \times 10^{4}\).

Step 4: Set up solubility equilibria expressions for copper(I) chloride in 0.10M NaCl

Let the solubility of CuCl in 0.10M NaCl also be \(x\ \text{mol/L}\). When CuCl dissolves, we get \(x\ \text{mol/L}\) of \(\mathrm{Cu}^{+}\) ions and (0.10+x) \(\text{mol/L}\) of \(\mathrm{Cl}^{-}\) ions. The solubility product expression for CuCl dissociation in 0.10M NaCl is: $$K_{sp} = [\mathrm{Cu}^{+}][\mathrm{Cl}^{-}] = x(0.10 + x)$$ Since, \(x \ll 0.10\), we can approximate \(0.10 + x \approx 0.10\): $$K_{sp} = x(0.10)$$ Now, the expression for the formation of \(\mathrm{CuCl}_{2}^{-}\) ion is: $$K = \frac{[\mathrm{CuCl}_{2}^{-}]}{[\mathrm{Cu}^{+}][\mathrm{Cl}^{-}]^2} = \frac{x}{[x][(0.10 + x)]^2}$$ Approximating here as well (as in denominator, \(x \ll 0.10\)): $$K = \frac{x}{[x][(0.10)^2]} = \frac{1}{(0.10)^2}$$

Step 5: Calculate the solubility of copper(I) chloride in 0.10M NaCl

Substitute the given values of \(K_{sp}\) and \(K\) and solve for \(x\): $$K_{sp} = x(0.10) \Rightarrow x = \frac{K_{sp}}{0.10}$$ $$K = \frac{1}{(0.10)^2} \Rightarrow x = \frac{[(0.10)^2]}{K}$$ Now equate both values of \(x\) and solve for \(K_{sp}\): $$\frac{K_{sp}}{0.10} = \frac{[(0.10)^2]}{K}$$ $$K_{sp} = 0.10 \cdot [(0.10)^2/K]$$ $$K_{sp} = \frac{(0.10)^3}{K}$$ Solving for x, $$x = \frac{(0.10)^3}{K_{sp} \cdot K} = \frac{(0.10)^3}{(1.2 \times 10^{-6})(8.7 \times 10^{4})} = 9.6 \times 10^{-5}\ \text{mol/L}$$ Hence, the solubility of copper(I) chloride in \(0.10\ M\) NaCl is \(9.6 \times 10^{-5}\ \text{mol/L}\).

Key Concepts

Copper(I) chloride

Copper(I) chloride is a chemical compound composed of copper and chloride ions. It appears as a white solid under normal conditions and is known for its limited solubility in water.
When dissolved, it dissociates into copper ions, Cu⁺, and chloride ions, Cl^-. Despite being a chloride salt, it does not dissolve as readily as other chloride compounds due to its specific ionic properties.
Understanding the solubility of copper(I) chloride is crucial, especially when considering dissolution in various solutions like pure water or salty water containing additional chloride ions.

Solubility product

The solubility product (\(K_{sp}\)) is an essential concept in chemistry that represents the equilibrium constant for the dissolution of a sparingly soluble ionic compound. For copper(I) chloride, the solubility product is calculated from the equation:\[K_{sp} = [\text{Cu}^+][\text{Cl}^-]\]
This expression indicates the product of the molar concentrations of copper and chloride ions at equilibrium.

• For copper(I) chloride, \(K_{sp}\) is \(1.2 \times 10^{-6}\), indicating low solubility.

By knowing \(K_{sp}\), one can calculate the solubility of a compound in a given solution. This involves determining the concentration of ions that can exist in equilibrium without forming more of the solid precipitate.

Complex ions

Complex ions involve a central metal ion bonded to one or more molecules or ions, increasing the possible variations in chemical equilibrium. With copper(I) and chloride ions, a significant complex ion is formed:\[\text{Cu}^+(aq) + 2\text{Cl}^-(aq) \rightleftharpoons \text{CuCl}_2^-(aq)\]
The formation of these complex ions from copper(I) and excess chloride ions impacts the overall solubility of the compound. This is represented by an equilibrium constant, \(K = 8.7 \times 10^{4}\).

• Complex ion formation can increase the solubility of metal salts by stabilizing metal ions in the solution.

Understanding complex ions is vital for predicting how dissolved ions will react under various conditions and how they are manipulated in different chemical processes.

Equilibrium

Equilibrium in a chemical reaction refers to the state where the rate of the forward reaction equals the rate of the reverse reaction. For copper(I) chloride, this involves its dissolution and the formation of complex ions.
At equilibrium, the concentrations of all species remain constant over time. In our case, even if reactions proceed to form other species like \(\text{CuCl}_2^-\), the amounts of each ion do not change, and this balance is described by the equilibrium constant.

• The equilibrium constant for the dissolution of copper(I) chloride is characterized by its \(K_{sp}\).
• The equilibrium constant for complex ion formation is denoted as \(K\).

Recognizing equilibrium allows chemists to forecast how conditions shift with altered concentrations or environmental factors, critical for calculations of solubility in varied media.

Common ion effect

The common ion effect is a phenomenon that influences the solubility of compounds. It occurs when a compound dissolves in a solution already containing one of its ions.
For copper(I) chloride, when dissolved in a sodium chloride solution (which has chloride ions), its solubility decreases compared to pure water because of this effect.

• The presence of additional Cl^- ions shifts the dissolution equilibrium of CuCl.
• This results in a significant decrease in its solubility, from \(1.1 \times 10^{-3} \ ext{mol/L}\) in pure water to \(9.6 \times 10^{-5} \ ext{mol/L}\) in 0.1M NaCl solution.

The common ion effect is crucial in processes like precipitation reactions and in controlling concentrations of ions in chemical and industrial applications.