Question

The copper(I) ion forms a chloride salt that has \(K_{\text {sp }}=1.2 \times\) \(10^{-6}\). 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 \(0.10 M\)

Short answer

The solubility of copper(I) chloride in pure water is approximately \(1.1 \times 10^{-3}\, M\), while its solubility in a 0.10 M chloride solution is approximately \(1.9 \times 10^{-4}\, M\).

Step-by-step solution

Part a: Solubility of copper(I) chloride in pure water

First, we write the balanced chemical equation for the dissolution of CuCl in water: \[ \mathrm{CuCl}(s) \rightleftharpoons \mathrm{Cu}^{+}(a q) + \mathrm{Cl}^{-}(a q) \] Now, we can write the expression for the solubility product, \(K_{sp}\), for this reaction: \[ K_{sp} = [\mathrm{Cu}^{+}][\mathrm{Cl}^{-}] \] Since the stoichiometry of the reaction is 1:1, we can let the solubility of CuCl in water be \(s\) and then we have: \[ K_{sp} = (s)(s) = s^2 \] Using the given value for \(K_{sp}\), we can solve for the solubility of CuCl: \[ s^2 = 1.2 \times 10^{-6} \] \[ s = \sqrt{1.2 \times 10^{-6}} \]

Calculation of Solubility

To find the solubility of copper(I) chloride in pure water, we just need to find the value of \(s\): \[ s = \sqrt{1.2 \times 10^{-6}} \approx 1.1 \times 10^{-3}\, M \] So the solubility of copper(I) chloride in pure water is approximately \(1.1 \times 10^{-3}\, M\).

Part b: Solubility of copper(I) chloride in 0.10 M chloride

Now, we need to consider the complex ion formation as well as the dissolution of CuCl in water. The balanced chemical equation for the complex ion formation is: \[ \mathrm{Cu}^{+}(a q) + 2\, \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \] We are given that the concentration of chloride ions, [\(\mathrm{Cl}^{-}\)], is 0.10 M. Let \(x\) be the solubility of CuCl in 0.10 M chloride solution. Since the complex formation involves 1 Cu+ ion and 2 Cl- ions, we can write the concentrations of these ions as: \[[\mathrm{Cu}^{+}] = x \] \[[\mathrm{Cl}^{-}] = 0.10 + 2x \] Now, we can write the expression for the equilibrium constant, \(K\) for the complex ion formation: \[ K = \frac{[\mathrm{CuCl}_{2}^{-}]}{[\mathrm{Cu}^{+}][\mathrm{Cl}^{-}]^2 } \] Since the stoichiometry is 1:1, we can let the concentration of CuCl2- be \(x\), and then we get: \[ K = \frac{x}{x(0.10 + 2x)^2 } \]

Calculation of Solubility

Using the given value of \(K\), we can solve the equation and get the value of \(x\), which represents the solubility of CuCl in 0.10 M chloride solution: \[\frac{x}{x(0.10 + 2x)^2 } = 8.7 \times 10^{4} \] \[ 0.10 + 2x = \sqrt[3]{\frac{1}{8.7 \times 10^{4}}} \] \[ x = \frac{1}{2}\left (\sqrt[3]{\frac{1}{8.7 \times 10^{4}}} - 0.10\right) \approx 1.9 \times 10^{-4}\, M \] So the solubility of copper(I) chloride in 0.10 M chloride solution is approximately \(1.9 \times 10^{-4}\, M\).

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, often abbreviated as \( K_{sp} \), is crucial in the study of ionic compounds in water. It's a specific type of equilibrium constant used for sparingly soluble salts. Simply put, \( K_{sp} \) helps indicate how much of a salt can dissolve in water to achieve saturation. The smaller the \( K_{sp} \) value, the less soluble the compound is.
For any dissolution process like that of copper(I) chloride, \( \text{CuCl} \), the equilibrium can be represented by the equation \( \text{CuCl}(s) \rightleftharpoons \text{Cu}^{+}(aq) + \text{Cl}^{-}(aq) \). At equilibrium, the rate of dissolution equals the rate of precipitation, and the \( K_{sp} \) expression can be defined as \( [\text{Cu}^{+}][\text{Cl}^{-}] \).
In the context of copper(I) chloride, knowing the \( K_{sp} \) enables the calculation of its solubility, even if it's extremely low. The calculation involves solving \( s^2 = K_{sp} \), where \( s \) represents the molar solubility.

Complex Ion Formation

Complex ion formation significantly affects the solubility of ionic compounds. A complex ion is an assembly consisting of a central metal ion bonded to molecules or anions, known as ligands. This process can further dissolve compounds that might otherwise be sparingly soluble.
With copper(I) chloride, the formation of the complex ion \( \text{CuCl}_2^{-} \) happens when \( \text{Cu}^{+} \) combines with two \( \text{Cl}^{-} \) ions. This reaction is described by the equation: \( \text{Cu}^{+}(aq) + 2 \text{Cl}^{-}(aq) \rightleftharpoons \text{CuCl}_2^{-}(aq) \).
The constant \( K \) for this reaction helps in understanding the extent of complex ion formation. When this happens, the presence of extra chloride ions drives the equilibrium to form more complex ions, significantly impacting the solubility as seen in quantitative calculations.

Copper(I) Chloride

Copper(I) chloride, denoted as \( \text{CuCl} \), is an interesting salt due to its relatively low solubility in water. In water, it dissociates slightly into copper(I) ions \( \text{Cu}^{+} \) and chloride ions \( \text{Cl}^{-} \). However, in the presence of excess chloride ions, its solubility behavior alters due to complex ion formation.
Despite its limited solubility, copper(I) chloride is useful in various chemical applications. Its behavior in different concentrations of \( \text{Cl}^{-} \) solutions demonstrates fundamental chemical principles, particularly those concerning dynamic equilibria and solubility alterations under changing conditions.

Chemical Equilibrium

Chemical equilibrium is a state where the forward and reverse reactions occur at the same rate, resulting in no net change in the concentration of reactants and products. It is a balance between the dissolution and precipitation processes for sparingly soluble salts, like copper(I) chloride.
Writing equilibrium expressions involves using the concentrations of the ions at equilibrium. For the dissolution of \( \text{CuCl} \), the equilibrium constant is \( K_{sp} = [\text{Cu}^{+}][\text{Cl}^{-}] \). This illustrates how equilibrium can be affected by factors such as concentration changes.
The addition of \( \text{Cl}^{-} \) ions shifts the equilibrium, encouraging the formation of the complex ion \( \text{CuCl}_2^{-} \), thereby reducing the concentration of free \( \text{Cu}^{+} \). This dynamic nature showcases how equilibria respond to different conditions, a foundational concept in chemistry.