What is the molar solubility of calcium fluoride, CaF2, in a solution of 0.25 M calcium nitrate, Ca(NO3)2 solution? Ksp for calcium fluoride is 3.45 x 10-11. ÂÂ
What is the molar solubility of calcium fluoride, CaF2, in a solution of 0.25 M calcium nitrate, Ca(NO3)2
solution?
The Correct Answer and Explanation is :
To calculate the molar solubility of ( \text{CaF}_2 ) in a solution of ( 0.25 \, \text{M} ) calcium nitrate (( \text{Ca(NO}_3)_2 )), we need to account for the common ion effect caused by the ( \text{Ca}^{2+} ) ions already present in the solution.
Step 1: Write the dissociation equation for ( \text{CaF}_2 )
[
\text{CaF}_2 (s) \leftrightarrow \text{Ca}^{2+} (aq) + 2 \text{F}^- (aq)
]
The solubility product (( K_{sp} )) expression is:
[
K_{sp} = [\text{Ca}^{2+}] [\text{F}^-]^2
]
Step 2: Define the concentrations
Let the molar solubility of ( \text{CaF}_2 ) be ( s ) (in moles per liter).
- The concentration of ( \text{Ca}^{2+} ) from ( \text{CaF}_2 ) is ( s ), but there is already ( 0.25 \, \text{M} ) ( \text{Ca}^{2+} ) from ( \text{Ca(NO}_3)_2 ).
[
[\text{Ca}^{2+}] = 0.25 + s
] - The concentration of ( \text{F}^- ) from ( \text{CaF}_2 ) is ( 2s ).
Step 3: Substitute into ( K_{sp} )
Given ( K_{sp} = 3.45 \times 10^{-11} ),
[
3.45 \times 10^{-11} = (0.25 + s)(2s)^2
]
For small ( s ), ( 0.25 + s \approx 0.25 ). Thus:
[
3.45 \times 10^{-11} = (0.25)(4s^2)
]
[
3.45 \times 10^{-11} = 1.0s^2
]
[
s^2 = 3.45 \times 10^{-11}
]
[
s = \sqrt{3.45 \times 10^{-11}} = 5.88 \times 10^{-6} \, \text{M}
]
Step 4: Interpretation
The molar solubility of ( \text{CaF}_2 ) in a ( 0.25 \, \text{M} ) calcium nitrate solution is approximately ( 5.88 \times 10^{-6} \, \text{M} ). The ( 0.25 \, \text{M} ) ( \text{Ca}^{2+} ) suppresses the solubility significantly compared to pure water due to the common ion effect, which shifts the equilibrium toward the solid ( \text{CaF}_2 ).