Calculate the molar solubility of barium fluoride in each liquid or solution. c. 0.15 M NaF

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1. Identify the dissolution reaction of 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. Recognize that the presence of 0.15 M NaF provides a common ion, \( \text{F}^- \), which affects the solubility of \( \text{BaF}_2 \).
4. Let the molar solubility of \( \text{BaF}_2 \) be \( s \). Then, \([\text{Ba}^{2+}] = s\) and \([\text{F}^-] = 0.15 + 2s\).
5. Substitute these concentrations into the \( K_{sp} \) expression and solve for \( s \), considering that \( s \) is small compared to 0.15 M.

Key Concepts

Molar Solubility

Molar solubility refers to the maximum amount of a solute that can dissolve in a given volume of solvent at a specific temperature, expressed in moles per liter (M). It is a crucial concept in understanding how ionic compounds dissociate in solution and is often determined through equilibrium expressions.

Common Ion Effect

The common ion effect describes the decrease in solubility of an ionic compound when a common ion is added to the solution. In this case, adding NaF introduces fluoride ions (F-) into the solution, which shifts the equilibrium of barium fluoride dissolution, reducing its molar solubility due to Le Chatelier's principle.

Equilibrium Constant (Ksp)

The solubility product constant (Ksp) is an equilibrium constant that applies to the dissolution of sparingly soluble ionic compounds. It quantifies the product of the molar concentrations of the ions in a saturated solution, raised to the power of their coefficients in the balanced dissolution equation, and is essential for calculating molar solubility in different conditions.