Question

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties. a. \(\mathrm{Ag}_{3} \mathrm{PO}_{4}, K_{\mathrm{sp}}=1.8 \times 10^{-18}\) b. \(\mathrm{CaCO}_{3}, K_{\mathrm{sp}}=8.7 \times 10^{-9}\) c. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}, K_{\mathrm{sp}}=1.1 \times 10^{-18}\left(\mathrm{Hg}_{2}^{2+}\right.\) is the cation in solution.)

Short answer

The solubilities of the given compounds are as follows: a. Ag3PO4: ≈ 3.18 x 10⁻⁶ moles per liter b. CaCO3: ≈ 2.95 x 10⁻⁴ moles per liter c. Hg2Cl2: ≈ 7.32 x 10⁻⁷ moles per liter

Step-by-step solution

Write the dissociation equation

For this compound, the dissociation equation is: Ag3PO4 (s) <=> 3 Ag⁺ (aq) + PO₄³⁻ (aq)

Express the solubility in terms of x

Let the solubility of Ag3PO4 be x moles per liter. Then, the concentration of Ag⁺ will be 3x and the concentration of PO₄³⁻ will be x.

Write the expression for Ksp

According to the dissociation equation, the Ksp expression is: Ksp = [Ag⁺]³ * [PO₄³⁻] Substitute the solubility expressions from step 2: Ksp = (3x)³ * x

Solve for x

Given Ksp = 1.8 x 10⁻¹⁸, we have: 1.8 x 10⁻¹⁸ = (3x)³ * x 1.8 x 10⁻¹⁸ = 27x⁴ x = \( \sqrt[4]{\frac{1.8 \times 10^{-18}}{27}} \) x ≈ 3.18 x 10⁻⁶

Express the solubility

The solubility of Ag3PO4 is approximately 3.18 x 10⁻⁶ moles per liter. #b. CaCO3, Ksp = 8.7 x 10^-9#

Write the dissociation equation

For this compound, the dissociation equation is: CaCO₃ (s) <=> Ca²⁺ (aq) + CO₃²⁻ (aq)

Express the solubility in terms of x

Let the solubility of CaCO₃ be x moles per liter. Then, the concentration of Ca²⁺ will be x and the concentration of CO₃²⁻ will be x.

Write the expression for Ksp

The Ksp expression is: Ksp = [Ca²⁺] * [CO₃²⁻] Substitute the solubility expressions from step 2: Ksp = x * x = x²

Solve for x

Given Ksp = 8.7 x 10⁻⁹, we have: 8.7 x 10⁻⁹ = x² x = \( \sqrt{8.7 \times 10^{-9}} \) x ≈ 2.95 x 10⁻⁴

Express the solubility

The solubility of CaCO3 is approximately 2.95 x 10⁻⁴ moles per liter. #c. Hg2Cl2, Ksp = 1.1 x 10^-18 (Hg2²⁺ is the cation)#

Write the dissociation equation

For this compound, the dissociation equation is: Hg2Cl2(s) <=> Hg2²⁺(aq) + 2 Cl⁻(aq)

Express the solubility in terms of x

Let the solubility of Hg2Cl2 be x moles per liter. Then, the concentration of Hg2²⁺ will be x and the concentration of Cl⁻ will be 2x.

Write the expression for Ksp

The Ksp expression is: Ksp = [Hg2²⁺] * [Cl⁻]² Substitute the solubility expressions from step 2: Ksp = x * (2x)²

Solve for x

Given Ksp = 1.1 x 10⁻¹⁸, we have: 1.1 x 10⁻¹⁸ = x * (2x)² 1.1 x 10⁻¹⁸ = x * 4x² x = \( \sqrt[3]{\frac{1.1 \times 10^{-18}}{4}} \) x ≈ 7.32 x 10⁻⁷

Express the solubility

The solubility of Hg2Cl2 is approximately 7.32 x 10⁻⁷ moles per liter.

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, or Ksp, is a fundamental concept in chemistry that describes the extent to which a compound will dissolve in water. It's a special type of equilibrium constant that applies to the dissolution of sparingly soluble salts. These salts reach a point in dissolution where the process is in dynamic equilibrium, with the rates of the forward reaction (dissolution) and the reverse reaction (precipitation) being equal.
For example, the dissociation of silver phosphate, Ag_{3}PO_{4}, is represented in the equilibrium expression as Ksp = [Ag^+]^3 * [PO_4^{3-}]. Calculating the Ksp involves determining the concentrations of the ions at this equilibrium state. Essentially, Ksp gives us insight into the solubility of a compound: the higher the Ksp value, the more soluble the compound is.

Molar Solubility

Molar solubility is a measurement that indicates how many moles of a substance can dissolve in a liter of solution to reach saturation. It's usually denoted as 'x' in solubility calculations, representing the number of moles that dissolve in one liter of water. For compounds that dissociate into multiple ions, the molar solubility becomes quite important when calculating the concentration of each individual ion. For instance, the molar solubility of calcium carbonate, CaCO_{3}, is (x), then both calcium ions, Ca^{2+}, and carbonate ions, CO_3^{2-}, have a concentration of (x) in the solution. This concept allows us to calculate the ionic concentrations needed to plug into the Ksp expression.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction in a chemical process, resulting in no net change in the concentration of reactants and products over time. At equilibrium, the system may appear static, but it's actually dynamic—reactions continue to occur, but the concentrations stay constant. It's important to recognize that equilibrium does not mean that the reactants and products are present in equal quantities, but rather that their concentrations have stabilized in a fixed ratio. This state of balance is modeled by the equilibrium constant, such as the Ksp for solubility, which quantifies the proportions of reactants and products at this point of dynamic balance.

Dissociation Equations

Dissociation equations are crucial for understanding how ionic compounds break apart in solution. They show the separation of a compound into its constituent ions. For solubility problems, the dissociation equation provides the framework for setting up the expression for the solubility product constant, Ksp.

Example of a Dissociation Equation

For mercury(I) chloride, Hg_{2}Cl_{2}, the dissociation equation is: Hg_{2}Cl_{2}(s) <=> Hg_{2}^{2+}(aq) + 2 Cl^{-}(aq). The stoichiometry of the ions as they appear in this equation is critical when calculating Ksp and molar solubility. Specifically, it indicates that for each mole of Hg_{2}Cl_{2} that dissolves, one mole of Hg_{2}^{2+} and two moles of Cl^{-} are produced in the solution.