Calculate the molar solubility of barium fluoride in each liquid or solution. a. pure water

Verified step by step guidance

1. Write the dissolution equation for barium fluoride: \( \text{BaF}_2 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + 2\text{F}^- (aq) \).
2. Write the expression for the solubility product constant \( K_{sp} \) for barium fluoride: \( K_{sp} = [\text{Ba}^{2+}][\text{F}^-]^2 \).
3. Let the molar solubility of \( \text{BaF}_2 \) be \( s \). Then, at equilibrium, \([\text{Ba}^{2+}] = s\) and \([\text{F}^-] = 2s\).
4. Substitute these expressions into the \( K_{sp} \) expression: \( K_{sp} = (s)(2s)^2 = 4s^3 \).
5. Solve for \( s \) by setting \( 4s^3 = K_{sp} \) and taking the cube root of both sides to find the molar solubility of \( \text{BaF}_2 \) in pure water.

Key Concepts

Molar Solubility

Molar solubility refers to the maximum amount of a solute that can dissolve in a given volume of solvent at equilibrium, expressed in moles per liter (mol/L). It is a crucial concept in understanding how substances interact in solution and is influenced by factors such as temperature and the presence of other ions.

Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is an equilibrium constant that applies to the dissolution of sparingly soluble ionic compounds. It quantifies the extent to which a compound can dissolve in water, represented by the product of the molar concentrations of its ions, each raised to the power of their coefficients in the balanced dissolution equation.

Dissociation of Ionic Compounds

Dissociation of ionic compounds occurs when an ionic solid separates into its constituent ions in solution. For barium fluoride (BaF2), this process can be represented by the equation BaF2(s) ⇌ Ba²⁺(aq) + 2F⁻(aq). Understanding this dissociation is essential for calculating molar solubility, as it directly relates to the concentrations of the ions in solution.