Textbook Question
Calculate the molar solubility of barium fluoride in each liquid or solution. c. 0.15 M NaF
Ch.17 - Aqueous Ionic Equilibrium
Chapter 17, Problem 96c
Calculate the molar solubility of MX (Ksp = 1.27⨉10-36) in each liquid or solution. c. 0.20 M Na2X
Verified step by step guidance1
Identify the dissociation of MX in water: MX(s) ⇌ M⁺(aq) + X⁻(aq).
Recognize that the presence of Na₂X introduces additional X⁻ ions into the solution, which affects the solubility of MX due to the common ion effect.
Write the expression for the solubility product constant (Ksp) for MX: Ksp = [M⁺][X⁻].
Set up the equation considering the initial concentration of X⁻ from Na₂X (0.40 M, since each Na₂X dissociates into 2 X⁻ ions) and the additional X⁻ produced from the dissociation of MX.
Solve for the molar solubility of MX by substituting the known values into the Ksp expression and solving for [M⁺], which equals the molar solubility of MX in the 0.20 M Na₂X solution.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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 (M). It is a critical concept in understanding how substances interact in solution and is influenced by factors such as temperature and the presence of other ions.
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Molar Solubility Example
Solubility Product Constant (Ksp)
The solubility product constant (Ksp) is an equilibrium constant that quantifies the solubility of a sparingly soluble ionic compound. It is defined as the product of the molar concentrations of the ions, each raised to the power of their coefficients in the balanced dissolution equation. A lower Ksp value indicates lower solubility.
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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, the presence of Na2X introduces the ion X, which shifts the equilibrium of the dissolution reaction, reducing the molar solubility of MX due to Le Chatelier's principle.
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Related Practice
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