Question

Calculate the molar solubility of \(\operatorname{SrF}_{2}\) in: (a) \(0.010 \mathrm{M} \mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}\) (b) \(0.010 \mathrm{M} \mathrm{NaF}\)

Short answer

(a) 7.9 x 10^-4 M; (b) 2.5 x 10^-5 M.

Step-by-step solution

Write the solubility equilibrium

The solubility equilibrium for strontium fluoride (\( \operatorname{SrF}_{2} \)) can be written as: \[ \operatorname{SrF}_{2} (s) \rightleftharpoons \operatorname{Sr}^{2+} (aq) + 2\mathrm{F}^{-} (aq) \]. The solubility product constant \( K_{sp} \) is defined as \( K_{sp} = [\operatorname{Sr}^{2+}][\mathrm{F}^{-}]^{2} \).

Determine Ksp value

Look up the \( K_{sp} \) value for \( \operatorname{SrF}_{2} \). Assume \( K_{sp} = 2.5 \times 10^{-9} \) for this problem.

Calculate molar solubility in Sr(NO₃)₂

In a solution containing \( 0.010 \mathrm{M} \operatorname{Sr}(\mathrm{NO}_{3})_{2} \), the concentration of \( \operatorname{Sr}^{2+} \) ions is already \( 0.010 \) M. Let the molar solubility of \( \operatorname{SrF}_{2} \) be \( s \) M. The equilibrium concentrations will then be \( [\operatorname{Sr}^{2+}] = 0.010 + s \) and \( [\mathrm{F}^{-}] = 2s \). Approximate by assuming \( s \ll 0.010 \), hence \( [\operatorname{Sr}^{2+}] \approx 0.010 \). Substitute into the \( K_{sp} \) expression: \[ K_{sp} = (0.010)(2s)^{2} = 4s^{2} \times 0.010 \]. Solve for \( s \): \[ s = \sqrt{\frac{2.5 \times 10^{-9}}{4 \times 0.010}}. \]

Solve for s in Sr(NO₃)₂

Calculate the value of \( s \) from the equation: \[ s = \sqrt{\frac{2.5 \times 10^{-9}}{0.04}} = \sqrt{6.25 \times 10^{-8}} \]. This gives \( s = 7.9 \times 10^{-4} \) M.

Calculate molar solubility in NaF

In a solution of \( 0.010 \mathrm{M} \mathrm{NaF} \), the concentration of \( \mathrm{F}^{-} \) ions is \( 0.010 \) M. Let the molar solubility of \( \operatorname{SrF}_{2} \) be \( s \) M. The equilibrium concentrations are \( [\operatorname{Sr}^{2+}] = s \) and \( [\mathrm{F}^{-}] = 0.010 + 2s \). Again, assume \( s \ll 0.010 \) so \( [\mathrm{F}^{-}] \approx 0.010 \). Substitute into the \( K_{sp} \) expression: \[ K_{sp} = s(0.010)^{2} \]. Solve for \( s \): \[ s = \frac{2.5 \times 10^{-9}}{0.0001} \].

Solve for s in NaF

Calculate the value of \( s \): \[ s = \frac{2.5 \times 10^{-9}}{0.0001} = 2.5 \times 10^{-5} \] M.

Key concepts

Solubility Equilibrium

In chemistry, solubility equilibrium describes the state of a system where a solute (solid substance) dissolves in a solvent until the rate of dissolution equals the rate of precipitation. This balance between dissolution and precipitation results in a stable, saturated solution. The solubility equilibrium for a slightly soluble salt, like strontium fluoride (\( \operatorname{SrF}_{2} \)), can be expressed through a chemical equation. In this case, it is:\[ \operatorname{SrF}_{2} (s) \rightleftharpoons \operatorname{Sr}^{2+} (aq) + 2\mathrm{F}^{-} (aq) \]This means that for every molecule of \( \operatorname{SrF}_{2} \) that dissolves, one \( \operatorname{Sr}^{2+} \) ion and two \( \mathrm{F}^{-} \) ions are formed in the solution. Understanding solubility equilibrium helps in predicting how much of a substance will dissolve in a given amount of solvent. This balance is critical, especially when dealing with substances with low solubility, such as many metal fluoride salts.

Solubility Product Constant

The solubility product constant (\( K_{sp} \)) is a special type of equilibrium constant used to describe the solubility of a sparingly soluble compound. For a general compound \( \text{AB}_2 \), which dissociates into \( \text{A}^{2+} \) and \( 2\text{B}^{-} \), the \( K_{sp} \) is defined as:\[ K_{sp} = [\text{A}^{2+}][\text{B}^{-}]^2 \]For strontium fluoride, \( \operatorname{SrF}_{2} \), it's given as\[ K_{sp} = [\operatorname{Sr}^{2+}][\mathrm{F}^{-}]^2 \]The value of the \( K_{sp} \) for a compound reflects how soluble the compound is - the smaller the \( K_{sp} \), the less soluble the compound. In our example, the \( K_{sp} \) for \( \operatorname{SrF}_{2} \) is \( 2.5 \times 10^{-9} \), indicating it is not very soluble. By using this concept, chemists can calculate how much of the compound will dissolve under specific conditions, which is crucial in various chemical applications and processes.

Chemical Equilibrium Calculations

Chemical equilibrium calculations involve determining the concentrations of reactants and products in a reaction at equilibrium. When you know the solubility product constant (\( K_{sp} \)), you can compute the molar solubility of a substance. This involves a series of steps:

• Identify initial concentrations and set up expressions for concentration changes ("s") under equilibrium.
• Apply the assumption that if a compound is sparingly soluble, the change in its concentration ("s") is much smaller than that of the other ions.
• Substitute these values into the \( K_{sp} \) expression to set up an algebraic equation.
• Solve for "s," which represents the molar solubility in the given conditions.

These calculations are greatly simplified by making justified approximations, like assuming that the change in concentration is negligible compared to the initial concentrations. Practicing these calculations helps reinforce understanding of both equilibrium concepts and algebraic manipulation.

Strontium Fluoride

Strontium fluoride (\( \operatorname{SrF}_{2} \)) is an example of a sparingly soluble salt. It consists of strontium and fluoride ions, arranged in a lattice structure. In water, only a small fraction of \( \operatorname{SrF}_{2} \) will dissolve, releasing strontium ions and fluoride ions into the solution. Despite its limited solubility, \( \operatorname{SrF}_{2} \) is important in various applications. It can be used in materials science, specifically in optics and lenses due to its transparency in a wide range of wavelengths. Additionally, understanding its solubility behavior is essential because it can indicate how the compound would behave in natural waters, soil, or even in biological systems. By learning about the properties and behaviors of substances like strontium fluoride, students can grasp key concepts of solubility and apply them in real-world scenarios.