The solubility of CaCO3 is pH dependent. (a) Calculate the molar solubility of CaCO3 given Ksp = 4.5 * 10^-92, neglecting the acid-base character of the carbonate ion. (c) If we assume that the only sources of Ca2+, HCO3-, and OH- ions are from the dissolution of CaCO3, what is the molar solubility of CaCO3 using the equilibrium expression from part (b)?

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Identify the dissolution reaction of calcium carbonate (CaCO3) in water: CaCO3(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq).

Write the expression for the solubility product constant (Ksp) for the dissolution of CaCO3: Ksp = [Ca²⁺][CO₃²⁻].

Assume the molar solubility of CaCO3 is 's'. At equilibrium, [Ca²⁺] = s and [CO₃²⁻] = s. Substitute these into the Ksp expression: Ksp = s * s = s².

Solve for 's' by taking the square root of both sides: s = √(Ksp). Substitute the given Ksp value (4.5 * 10^-92) into the equation to find the molar solubility.

For part (c), consider the equilibrium expression from part (b) and the assumption that the only sources of ions are from CaCO3. Set up the equilibrium expressions for the dissolution and any additional equilibria involving HCO3⁻ and OH⁻, and solve for the molar solubility using these expressions.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is an equilibrium constant that applies to the solubility of sparingly soluble ionic compounds. It is defined as the product of the molar concentrations of the ions, each raised to the power of their coefficients in the balanced equation. For calcium carbonate (CaCO3), Ksp = [Ca2+][CO3^2-]. Understanding Ksp is crucial for calculating the molar solubility of CaCO3 in different pH conditions.

pH and its Effect on Solubility

pH is a measure of the acidity or basicity of a solution, which can significantly influence the solubility of certain compounds. In the case of CaCO3, an increase in pH (more basic conditions) can lead to a decrease in the concentration of carbonate ions (CO3^2-), thus affecting its solubility. This relationship is essential for understanding how the dissolution of CaCO3 varies with changes in pH.

Equilibrium Expressions

Equilibrium expressions describe the relationship between the concentrations of reactants and products at equilibrium. For the dissolution of CaCO3, the equilibrium expression can be derived from its dissociation into Ca2+ and CO3^2- ions. This expression is vital for calculating molar solubility, as it allows for the determination of ion concentrations based on the Ksp value and the stoichiometry of the dissolution reaction.

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For each pair of compounds, use Ksp values to determine which has the greater molar solubility: (a) CdS or CuS (b) PbCO₃ or BaCrO₄ (c) Ni(OH)₂ or NiCO₃ (d) AgI or Ag₂SO₄.

Textbook Question

The solubility of CaCO₃ is pH dependent. (b) Use the K_b expression for the CO₃^2- ion to determine the equilibrium constant for the reaction CaCO₃(s) + H₂O(l) ⇌ Ca²⁺(aq) + HCO₃^-(aq) + OH^-(aq)

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Tooth enamel is composed of hydroxyapatite, whose simplest formula is Ca51PO423OH, and whose corresponding Ksp = 6.8 * 10-27. As discussed in the Chemistry and Life box on page 746, fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, Ca51PO423F, whose Ksp = 1.0 * 10-60. (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite.