Use the following solubility data to calculate a value of K_sp for each compound. (a) SrF₂: 1.03 x 10^-3 M (b) CuI: 1.05 x 10^-6 M

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Identify the dissolution equation for the compound. For SrF2, the dissolution equation is: SrF2(s) ⇌ Sr2+(aq) + 2F-(aq).

Determine the stoichiometric coefficients from the dissolution equation. Here, 1 mole of SrF2 produces 1 mole of Sr2+ and 2 moles of F-.

Use the given solubility to find the concentrations of the ions in solution. If the solubility of SrF2 is 1.03 x 10^-3 M, then [Sr2+] = 1.03 x 10^-3 M and [F-] = 2 x 1.03 x 10^-3 M.

Write the expression for the solubility product constant (Ksp) based on the dissolution equation. Ksp = [Sr2+][F-]^2.

Substitute the concentrations into the Ksp expression and calculate the value of Ksp.

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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 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. For example, for SrF2 dissolving into Sr²⁺ and 2F⁻, Ksp = [Sr²⁺][F⁻]².

Dissociation of Ionic Compounds

Ionic compounds dissociate into their constituent ions when they dissolve in water. For SrF2, it dissociates into one strontium ion (Sr²⁺) and two fluoride ions (F⁻). Understanding this dissociation is crucial for calculating Ksp, as the concentrations of the ions at equilibrium directly influence the Ksp value.

Equilibrium Concentrations

Equilibrium concentrations refer to the concentrations of reactants and products in a chemical reaction at equilibrium. In the context of Ksp, the equilibrium concentrations of the ions produced from the dissolution of the compound must be used to calculate the Ksp value. For SrF2, if the solubility is given as 1.03 x 10^-3 M, the concentration of Sr²⁺ will be 1.03 x 10^-3 M, while the concentration of F⁻ will be 2.06 x 10^-3 M due to the stoichiometry of the dissociation.

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