Question

Copper(I) chloride has \(K_{\mathrm{sp}}=1.7 \times 10^{-7} .\) Calculate the molar solubility of copper(I) chloride in (a) pure water, (b) \(0.0200 \mathrm{M} \mathrm{HCl}\) solution, (c) \(0.200 \mathrm{M} \mathrm{HCl}\) solution, and (d) \(0.150 \mathrm{M} \mathrm{CaCl}_{2}\) solution.

Short answer

Molar solubility of copper(I) chloride (a) in pure water: s = square root of Ksp, (b) in 0.0200 M HCl: s = Ksp / 0.0200 M, (c) in 0.200 M HCl: s = Ksp / 0.200 M, (d) in 0.150 M CaCl2: s = Ksp / 0.300 M.

Step-by-step solution

Write Down the Dissociation Equation for Copper(I) Chloride

Copper(I) chloride dissolves in water according to the following equation: ز A reversible reaction where solid copper(I) chloride (CuCl) dissociates into its ions. The equation is: CuCl(s) ⇌ Cu^+(aq) + Cl^-(aq).

Write the Expression for the Solubility Product Constant (Ksp)

Based on the dissociation equation, we can write the expression for the solubility product constant as: Ksp = [Cu^+][Cl^-], since the stoichiometric coefficients of the ions in the balanced equation are both 1.

Calculate Molar Solubility in Pure Water

In pure water, let the molar solubility of copper(I) chloride be 's'. This means that [Cu^+] = 's' and [Cl^-] = 's'. Substituting these into the Ksp expression, we get Ksp = s^2. By solving this equation with the given Ksp value (1.7 x 10^-7), we find the molar solubility 's' in pure water.

Calculate Molar Solubility in 0.0200 M HCl Solution

In a 0.0200 M HCl solution, the concentration of Cl^- ions is significantly increased. The additional Cl^- from HCl shifts the equilibrium to the left, thereby decreasing the solubility of CuCl. Assuming 's' to be the additional solubility of CuCl, we then have [Cu^+] = 's' and [Cl^-] = 0.0200 + 's'. Substituting these into the Ksp expression, we get Ksp = s(0.0200 + s). However, because 's' will be much smaller than 0.0200 M, we can approximate [Cl^-] as 0.0200 M to simplify the calculation.

Calculate Molar Solubility in 0.200 M HCl Solution

The process for calculating molar solubility in a 0.200 M HCl solution is similar to the 0.0200 M HCl solution step. Here also [Cu^+] = 's' and [Cl^-] = 0.200 + 's', with the approximation [Cl^-] ≈ 0.200 M.

Calculate Molar Solubility in 0.150 M CaCl2 Solution

In a 0.150 M calcium chloride (CaCl2) solution, we consider that each mole of CaCl2 contributes two moles of Cl^- ions. Therefore, the initial concentration of Cl^- ions due to CaCl2 will be 2 × 0.150 M. When calculating the solubility 's' in this case, [Cu^+] = 's' and [Cl^-] = 0.300 + 's', with the approximation [Cl^-] ≈ 0.300 M.

Solve for Molar Solubility 's' in Each Case

Using the approximations and the Ksp value, solve for 's' in each case: (a) Ksp = s^2 in pure water, (b) Ksp = s(0.0200) in 0.0200 M HCl solution, (c) Ksp = s(0.200) in 0.200 M HCl solution, and (d) Ksp = s(0.300) in 0.150 M CaCl2 solution. This will yield the molar solubility of copper(I) chloride in each scenario.

Key concepts

Solubility Product Constant

The solubility product constant, commonly denoted as K_sp, is a crucial term in the study of solubility and precipitation in chemistry. It is used to describe the equilibrium between a solid salt and its ions in a saturated solution. Specifically, K_sp is the product of the molar concentrations of the ions produced from the salt, each raised to the power of its stoichiometric coefficient in the dissociation equation.

For example, the dissociation of copper(I) chloride into copper ions (Cu⁺) and chloride ions (Cl^-) in water can be represented by the equation CuCl(s) ⇌ Cu⁺(aq) + Cl^-(aq). The K_sp expression for this reaction is K_sp = [Cu⁺][Cl^-], which indicates that the product of the concentrations of Cu⁺ and Cl^- at equilibrium equals the solubility product constant. It is important to note that solids and pure liquids are not included in the K_sp expression as their concentrations do not change.

Common Ion Effect

The common ion effect is a phenomenon where the solubility of a salt is reduced in a solution that already contains one of the ions present in the salt. This effect is based on Le Chatelier's principle, which states that a dynamic equilibrium will adjust to counteract any change in conditions.

When additional chloride ions (Cl^-) are introduced into the solution, for instance from HCl or CaCl₂, the equilibrium of the dissociation of CuCl is disturbed. The system responds by shifting the equilibrium to favor the formation of the undissociated solid, CuCl, resulting in a decrease in molar solubility of the salt. This means in a situation where you are determining molar solubility in a solution containing a common ion, the presence of the common ion (Cl^- in this case) must be included in the solubility calculations, which will yield lower solubility values than in pure water.

Equilibrium Calculation

Equilibrium calculations are mathematical approaches to quantify the concentrations of reactants and products at equilibrium. These calculations incorporate the principles of chemical equilibrium and the stoichiometry of the reactions involved.

Starting with the K_sp expression and knowing the solubility product constant, we can solve for the molar solubility 's'. This is done by substituting in the known values and solving the resulting equation for 's'. In cases where simplifications are possible, such as when the concentration of common ions is much higher than the molar solubility, we can approximate the common ion concentration as constant, which simplifies the computation significantly. Various scenarios, such as different ionic strengths in the solution, as seen in steps 4 through 6 of the original solution, show the utility of equilibrium calculations in determining how molar solubility changes under different conditions.