Problem 54 - Chemistry The Molecular Nature of Matter

Question

Mercury(I) chloride has \(K_{\mathrm{sp}}=1.4 \times 10^{-18} .\) Calculate the molar solubility of mercury(I) chloride in (a) pure water, (b) \(0.010 M \mathrm{HCl}\) solution, (c) \(0.010 \mathrm{M} \mathrm{MgCl}_{2}\) solution, and (d) \(0.010 \mathrm{M} \mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}\) solution.

Step-by-Step Solution

Answer

The molar solubility of mercury(I) chloride in pure water is \(1.4 \times 10^{-18} / 4)^{1/3} \) mol/L. In 0.010 M HCl, the molar solubility is \(1.4 \times 10^{-18} / (0.010)^2) \) mol/L of Hg2Cl2. In 0.010 M MgCl2, it's \(1.4 \times 10^{-18} / (0.020)^2) \) mol/L of Hg2Cl2. In 0.010 M Hg2(NO3)2, the molar solubility is determined by solving for [Cl-] in the equation 1.4 x 10^-18 = (0.010)[Cl-]^2.

Step 1: Write the Dissociation Reaction and Expression for Ksp

Mercury(I) chloride, Hg2Cl2, dissociates into one Hg2^2+ ion and two Cl- ions. The dissociation reaction is: Hg2Cl2(s) ⇌ Hg2^2+(aq) + 2Cl-(aq).The expression for the solubility product constant, Ksp, is: Ksp = [Hg2^2+][Cl-]^2.

Step 2: Calculate Molar Solubility in Pure Water

Let the molar solubility of Hg2Cl2 in water be s mol/L. Therefore, [Hg2^2+] = s and [Cl-] = 2s. The Ksp expression becomes:1.4 x 10^-18 = (s)(2s)^2 = 4s^3.Solve for s to find the molar solubility in pure water.

Step 3: Calculate Molar Solubility in 0.010 M HCl Solution

In a 0.010 M HCl solution, the [Cl-] is already present at 0.010 M due to the HCl dissociation. Since Cl- is a common ion, the presence of Cl- ions from HCl will affect the solubility of Hg2Cl2. The expression now becomes:Ksp = [Hg2^2+](0.010)^2.Solve for [Hg2^2+] to find the molar solubility in the presence of HCl.

Step 4: Calculate Molar Solubility in 0.010 M MgCl2 Solution

MgCl2 dissociates to give Mg2+ ions and double the concentration of Cl- ions. The initial concentration of Cl- due to MgCl2 is 0.020 M. Applying the common ion effect:Ksp = [Hg2^2+](0.020)^2.Solve for [Hg2^2+] to find the molar solubility in the presence of MgCl2.

Step 5: Calculate Molar Solubility in 0.010 M Hg2(NO3)2 Solution

Hg2(NO3)2 gives Hg2^2+ ions and NO3- ions in solution. Since there is 0.010 M of Hg2^2+ already present because of Hg2(NO3)2, the solubility of Hg2Cl2 will be further suppressed. Apply the common ion effect with the Hg2^2+ concentration fixed at 0.010 M, and solve for the [Cl-] to find the molar solubility.

Step 6: Solve for Molar Solubility in Pure Water

1.4 x 10^-18 = 4s^3, s^3 = 1.4 x 10^-18 / 4. Take the cube root of both sides to solve for s: s = (1.4 x 10^-18 / 4)^(1/3).

Step 7: Solve for Molar Solubility in 0.010 M HCl Solution

1.4 x 10^-18 = [Hg2^2+](0.010)^2, [Hg2^2+] = 1.4 x 10^-18 / (0.010)^2. Calculate [Hg2^2+] to find molar solubility, which is equal to [Hg2^2+] in this case.

Step 8: Solve for Molar Solubility in 0.010 M MgCl2 Solution

1.4 x 10^-18 = [Hg2^2+](0.020)^2, [Hg2^2+] = 1.4 x 10^-18 / (0.020)^2. Calculate [Hg2^2+] to find the molar solubility, which is equal to [Hg2^2+] in this case.

Step 9: Solve for Molar Solubility in 0.010 M Hg2(NO3)2 Solution

Since the [Hg2^2+] is fixed at 0.010 M due to the presence of Hg2(NO3)2, the Ksp equation becomes 1.4 x 10^-18 = (0.010)[Cl-]^2. Solve for [Cl-] to find the molar solubility as the chlorine concentration would be twice the molarity of dissolved Hg2Cl2.

Key Concepts

Solubility Product Constant (Ksp)

Understanding the solubility product constant, often denoted as Ksp, is crucial for predicting whether a salt will dissolve in a solution, and to what extent. It represents the maximum product of the concentrations of the ions involved in the dissolution when the solution is saturated. A low Ksp value, such as that of mercury(I) chloride (Hg₂Cl₂), indicates very low solubility.
In a solubility equation, if Hg₂Cl₂ dissociates into Hg₂²⁺ and Cl^-, the Ksp expression would be defined as Ksp = [Hg₂²⁺][Cl^-]². The key to solving molar solubility problems is setting up and solving this equilibrium expression for the ionic concentrations in a saturated solution.
To establish the initial molar solubility in pure water, the concentrations of the ions are expressed in terms of a variable, such as s, which represents the molarity of Hg₂Cl₂ that dissolves. With the Ksp known, one can then calculate the value of s. It's a straightforward application of equilibrium principles that entails substituting into the equilibrium expression and solving for the unknown.

Common Ion Effect

The common ion effect refers to the decrease in solubility of an ionic compound when a common ion is added to the solution. In this case, the presence of Cl^- or Hg₂²⁺ ions from other sources influences the solubility of Hg₂Cl₂.
For example, when HCl is added to the solution, it completely dissociates into H⁺ and Cl^-. Since Cl^- is already in the solution, the system shifts towards the left (the solid form of Hg₂Cl₂) to re-establish equilibrium, as per Le Chatelier's Principle. This results in a lower molar solubility of Hg₂Cl₂ in the solution with added HCl. The decreased solubility is calculated by adjusting the Ksp expression to account for the increased concentration of Cl^- ions.
Understanding and calculating the effects of common ions can be challenging, but it becomes manageable by carefully setting up the equilibrium expressions to reflect the real conditions in the solution. The lessons from such exercises are widely applicable, including understanding biochemical pathways in living organisms and predicting the behavior of environmental pollutants.

Dissociation Reaction

The dissociation reaction of an ionic compound describes how it breaks apart into its constituent ions when dissolved in water. For mercury(I) chloride, the reaction is as follows: Hg₂Cl₂(s) ⇌ Hg₂²⁺(aq) + 2Cl^-(aq). This reversible process is central to understanding the concept of Ksp.
During dissolution, the solid compound (Hg₂Cl₂) becomes associated with its ions in solution (Hg₂²⁺ and Cl^-). The reaction reaches a dynamic equilibrium when the rate at which the compound dissolves equals the rate at which the ions combine to form the solid compound. This equilibrium doesn't mean the reaction has stopped; both processes continue but at the same rate, maintaining a constant concentration of ions in solution.
Quantitative analysis of dissociation reactions are vital in many fields, such as pharmacy and environmental science. By understanding how compounds dissociate, we can predict how they will interact with their environment, how to precipitate them out of solution for purification, or how best to formulate them for absorption in biological systems.