Write a balanced chemical equation describing the dissolution (dissolving) of strontium fluoride. Determine the molar solubility of SrF2 in a 0.5 mol/L solution of Sr(NO3)2. (Strontium nitrate is a highly soluble salt)
The Correct Answer and Explanation is :
Balanced chemical equation for the dissolution of strontium fluoride:
Strontium fluoride (SrF₂) dissociates in water as follows:
[
\text{SrF}_2 (s) \rightleftharpoons \text{Sr}^{2+} (aq) + 2\text{F}^{-} (aq)
]
This is the chemical equation representing the dissolution of SrF₂ into its ions.
Determining Molar Solubility in 0.5 mol/L Sr(NO₃)₂ Solution
- Ksp Expression: The solubility product constant (Ksp) for SrF₂ is defined by the equation:
[
K_{sp} = [\text{Sr}^{2+}][\text{F}^-]^2
]
Let’s denote the molar solubility of SrF₂ as “s” mol/L. This means that for every mole of SrF₂ that dissolves, “s” moles of Sr²⁺ and “2s” moles of F⁻ are released into solution. However, since Sr(NO₃)₂ is already providing Sr²⁺ ions (0.5 mol/L), we must account for this in the equilibrium calculation.
- Substitute Ion Concentrations:
- [Sr²⁺] in solution is the sum of the strontium from Sr(NO₃)₂ and SrF₂:
[
[\text{Sr}^{2+}] = 0.5 + s
] - [F⁻] comes only from SrF₂, so it is:
[
[\text{F}^-] = 2s
]
Thus, the Ksp expression becomes:
[
K_{sp} = (0.5 + s)(2s)^2
]
For SrF₂, the Ksp is approximately ( 2.6 \times 10^{-9} ) at room temperature. Since “s” is typically very small compared to 0.5 M, we can approximate the equation as:
[
K_{sp} \approx (0.5)(4s^2)
]
Solving for “s”:
[
s^2 \approx \frac{K_{sp}}{2} \times \frac{1}{0.5}
]
[
s^2 = \frac{2.6 \times 10^{-9}}{2}
]
[
s = \sqrt{1.3 \times 10^{-9}}
]
[
s \approx 1.14 \times 10^{-5} \, \text{mol/L}
]
Explanation:
The molar solubility of SrF₂ is the amount of SrF₂ that dissolves in the solution. In the presence of 0.5 M Sr(NO₃)₂, the large concentration of Sr²⁺ from strontium nitrate reduces the solubility of SrF₂ due to the common ion effect. This effect lowers the amount of SrF₂ that can dissolve because adding Sr²⁺ from Sr(NO₃)₂ shifts the dissolution equilibrium of SrF₂ to the left, decreasing its solubility.