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Ch.17 - Additional Aspects of Aqueous Equilibria
Chapter 17, Problem 58a

Calculate the solubility of LaF3 in grams per liter in (a) pure water.

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Identify the chemical formula and dissociation of LaF3 in water. LaF3 dissociates into La^{3+} and F^{-} ions as: LaF3(s) \rightarrow La^{3+}(aq) + 3F^{-}(aq).
Write the expression for the solubility product constant (K_{sp}) of LaF3. The expression is K_{sp} = [La^{3+}][F^{-}]^3.
Let the solubility of LaF3 be 's' mol/L. This means [La^{3+}] = s and [F^{-}] = 3s.
Substitute the values of [La^{3+}] and [F^{-}] into the K_{sp} expression and solve for 's'.
Convert the solubility from moles per liter to grams per liter by multiplying the molar solubility 's' by the molar mass of LaF3.

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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 represents the product of the molar concentrations of the ions, each raised to the power of their coefficients in the balanced dissolution equation. For LaF3, the Ksp expression would involve the concentrations of La^3+ and F^- ions, allowing us to calculate its solubility in water.
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Dissociation of Ionic Compounds

Ionic compounds, such as LaF3, dissociate into their constituent ions when dissolved in water. For LaF3, this dissociation can be represented as LaF3(s) ⇌ La^3+(aq) + 3F^-(aq). Understanding this dissociation is crucial for calculating solubility, as it directly relates to the concentration of ions in solution and their contributions to the Ksp value.
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Concentration Units

Concentration is a measure of how much solute is present in a given volume of solvent, typically expressed in moles per liter (M) or grams per liter (g/L). When calculating the solubility of LaF3, it is important to convert the molar solubility (from Ksp) into grams per liter using the molar mass of LaF3. This conversion allows for a practical understanding of how much of the compound can dissolve in water.
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