Calculate Molar Solubility of Caf in 0.25 Cano32
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Reviewed by Calculator Editorial Team
This calculator determines the molar solubility of calcium fluoride (CaF₂) in a 0.25 M calcium nitrate (Ca(NO₃)₂) solution. The calculation accounts for the common ion effect and equilibrium constants.
Introduction
The molar solubility of a compound is the maximum amount of that compound that can dissolve in a given volume of solvent at a specific temperature. For calcium fluoride (CaF₂) in an aqueous solution, the solubility is influenced by the presence of other ions, particularly the common ion effect.
When CaF₂ dissolves, it dissociates into Ca²⁺ and F⁻ ions. If Ca(NO₃)₂ is also present in solution, the Ca²⁺ ions from the salt will compete with the CaF₂ for the F⁻ ions, reducing the overall solubility of CaF₂.
Formula
The molar solubility (s) of CaF₂ in the presence of Ca(NO₃)₂ can be calculated using the following equation:
Where:
- Ksp is the solubility product constant of CaF₂ (1.5 × 10⁻¹⁰ at 25°C)
- α is the common ion factor, calculated as (initial concentration of Ca²⁺) / (solubility of CaF₂ in pure water)
The solubility of CaF₂ in pure water is approximately 3.5 × 10⁻⁴ M.
Calculation
The calculation involves these steps:
- Determine the initial concentration of Ca²⁺ from the Ca(NO₃)₂ solution
- Calculate the common ion factor α
- Use the formula to find the molar solubility of CaF₂
The calculator performs these calculations automatically when you input the concentration of Ca(NO₃)₂.
Example
Let's calculate the molar solubility of CaF₂ in a 0.25 M Ca(NO₃)₂ solution:
- Initial concentration of Ca²⁺ = 0.25 M
- Solubility of CaF₂ in pure water = 3.5 × 10⁻⁴ M
- Common ion factor α = 0.25 / 3.5 × 10⁻⁴ ≈ 7142.86
- Solubility product constant Ksp = 1.5 × 10⁻¹⁰
- Molar solubility s = √(1.5 × 10⁻¹⁰ / (1 + 7142.86)) ≈ √(2.1 × 10⁻¹⁴) ≈ 1.45 × 10⁻⁷ M
The molar solubility of CaF₂ in this solution is approximately 1.45 × 10⁻⁷ M.
Interpretation
The result shows that the presence of Ca(NO₃)₂ significantly reduces the solubility of CaF₂. This is due to the common ion effect, where the added Ca²⁺ ions compete with the CaF₂ for the F⁻ ions, effectively lowering the overall solubility.
This calculation is important in chemical engineering, environmental science, and analytical chemistry where precise solubility information is required.