The solubility of `CaF_(2)` in a solution of `0.1M Ca(NO_(3))_(2)` is

Options

A. `[Ca^(2+)]`
B. `2[F^(Theta)]`
C. `([F^(Theta)])/(2)`
D. `2[NO_(3)^(Theta)]`

Explanation

To find the solubility of \( \text{CaF}_2 \) in a \( 0.1 \, \text{M} \, \text{Ca(NO}_3)_2 \) solution, we can follow these steps:

### Step 1: Write the dissociation equation for \( \text{CaF}_2 \)
The dissociation of calcium fluoride in water can be represented as:
\[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2 \text{F}^- (aq) \]
This means that for every mole of \( \text{CaF}_2 \) that dissolves, one mole of \( \text{Ca}^{2+} \) and two moles of \( \text{F}^- \) are produced.

### Step 2: Define solubility in pure water
Let the solubility of \( \text{CaF}_2 \) in pure water be \( S \). Then:
- The concentration of \( \text{Ca}^{2+} \) ions will be \( S \).
- The concentration of \( \text{F}^- \) ions will be \( 2S \).

### Step 3: Consider the effect of the \( 0.1 \, \text{M} \, \text{Ca(NO}_3)_2 \) solution
When \( \text{CaF}_2 \) is dissolved in a \( 0.1 \, \text{M} \, \text{Ca(NO}_3)_2 \) solution, the concentration of \( \text{Ca}^{2+} \) ions from the \( \text{Ca(NO}_3)_2 \) will contribute to the total concentration of \( \text{Ca}^{2+} \) ions in the solution. Therefore, the total concentration of \( \text{Ca}^{2+} \) ions becomes:
\[ \text{Total } [\text{Ca}^{2+}] = S + 0.1 \]

### Step 4: Concentration of \( \text{F}^- \) ions
The concentration of \( \text{F}^- \) ions remains \( 2S \) since \( \text{CaF}_2 \) does not contribute to the \( \text{F}^- \) ions from \( \text{Ca(NO}_3)_2 \).

### Step 5: Set up the equilibrium expression
At equilibrium, the solubility product \( K_{sp} \) for \( \text{CaF}_2 \) can be expressed as:
\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \]
Substituting the concentrations:
\[ K_{sp} = (S + 0.1)(2S)^2 \]
This can be simplified to:
\[ K_{sp} = (S + 0.1)(4S^2) \]

### Step 6: Solve for \( S \)
Since we are interested in the solubility \( S \) in the presence of \( 0.1 \, \text{M} \, \text{Ca(NO}_3)_2 \), we can assume that \( S \) will be very small compared to \( 0.1 \) and thus can be neglected in the \( S + 0.1 \) term:
\[ K_{sp} \approx (0.1)(4S^2) \]
From this, we can isolate \( S \):
\[ S^2 = \frac{K_{sp}}{0.4} \]
Thus,
\[ S = \sqrt{\frac{K_{sp}}{0.4}} \]

### Step 7: Determine fluoride concentration
The concentration of \( \text{F}^- \) ions will be:
\[ [\text{F}^-] = 2S \]
Substituting \( S \):
\[ [\text{F}^-] = 2 \sqrt{\frac{K_{sp}}{0.4}} \]

### Final Answer
The solubility of \( \text{CaF}_2 \) in a \( 0.1 \, \text{M} \, \text{Ca(NO}_3)_2 \) solution is \( \frac{[\text{F}^-]}{2} \).