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.
The Correct Answer and Explanation is :
To calculate the molar solubility of calcium fluoride (( \text{CaF}_2 )) in a ( 0.25 \, \text{M} ) calcium nitrate (( \text{Ca(NO}_3\text{)}_2 )) solution, we account for the common ion effect caused by ( \text{Ca}^{2+} ) ions from ( \text{Ca(NO}_3\text{)}_2 ).
Step 1: Write the equilibrium expression for ( \text{CaF}_2 ):
[
\text{CaF}_2 (s) \leftrightarrow \text{Ca}^{2+} (aq) + 2 \text{F}^- (aq)
]
The solubility product constant (( K_\text{sp} )) is:
[
K_\text{sp} = [\text{Ca}^{2+}][\text{F}^-]^2
]
Given ( K_\text{sp} = 3.45 \times 10^{-11} ), and the initial concentration of ( \text{Ca}^{2+} ) from ( \text{Ca(NO}_3\text{)}_2 ) is ( 0.25 \, \text{M} ), the additional ( \text{Ca}^{2+} ) from ( \text{CaF}_2 ) is negligible.
Step 2: Define solubility (( s )):
Let ( s ) be the molar solubility of ( \text{CaF}_2 ) in the solution. Upon dissolution:
- ( [\text{Ca}^{2+}] = 0.25 + s ) (approx. ( 0.25 \, \text{M} ), since ( s \ll 0.25 ))
- ( [\text{F}^-] = 2s )
Substituting into the ( K_\text{sp} ) expression:
[
K_\text{sp} = (0.25)(2s)^2
]
[
3.45 \times 10^{-11} = (0.25)(4s^2)
]
[
s^2 = \frac{3.45 \times 10^{-11}}{1.0} = 3.45 \times 10^{-11}
]
[
s = \sqrt{3.45 \times 10^{-11}} = 1.86 \times 10^{-6} \, \text{M}
]
Final Answer:
The molar solubility of ( \text{CaF}_2 ) in ( 0.25 \, \text{M} ) calcium nitrate solution is:
[
\boxed{1.86 \times 10^{-6} \, \text{M}}
]
Explanation:
- The ( \text{Ca(NO}_3\text{)}_2 ) provides a high initial ( \text{Ca}^{2+} ) concentration, suppressing ( \text{CaF}_2 ) dissolution due to the common ion effect.
- The equilibrium shifts to reduce ( \text{F}^- ) concentration, drastically lowering the solubility.
- Using ( K_\text{sp} ), we account for the interaction of ion concentrations in equilibrium.