The solubility of `CaF_(2)` (`K_(sp)=3.4xx10^(-11)`) in 0.01 M solution of `NaF` would be:

Options

A. `3.4xx10^(-12)` M
B. `3.4xx10^(-9)` M
C. `3.4xx10^(-7)` M
D. `3.4xx10^(-13)` M

Answer

To find the solubility of \( \text{CaF}_2 \) in a 0.01 M solution of \( \text{NaF} \), we need to consider the common ion effect due to the presence of fluoride ions (\( \text{F}^- \)) from \( \text{NaF} \). Here’s a step-by-step solution:

### Step 1: Write the dissociation equation for \( \text{CaF}_2 \)
The dissociation of calcium fluoride in water is represented as:
\[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^- (aq) \]

### Step 2: Define the solubility product (\( K_{sp} \))
The solubility product expression for \( \text{CaF}_2 \) is given by:
\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \]
Given \( K_{sp} = 3.4 \times 10^{-11} \).

### Step 3: Determine concentrations in the presence of \( \text{NaF} \)
In a 0.01 M solution of \( \text{NaF} \), \( \text{NaF} \) dissociates completely:
\[ \text{NaF} \rightarrow \text{Na}^+ + \text{F}^- \]
Thus, the concentration of \( \text{F}^- \) from \( \text{NaF} \) is 0.01 M.

### Step 4: Set up the equilibrium concentrations
Let the solubility of \( \text{CaF}_2 \) be \( S \). At equilibrium:
- The concentration of \( \text{Ca}^{2+} \) will be \( S \).
- The concentration of \( \text{F}^- \) will be \( 0.01 + 2S \). However, since \( S \) is expected to be very small compared to 0.01, we can approximate this as:
\[ [\text{F}^-] \approx 0.01 \, \text{M} \]

### Step 5: Substitute into the \( K_{sp} \) expression
Substituting the equilibrium concentrations into the \( K_{sp} \) expression:
\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = S (0.01)^2 \]
\[ K_{sp} = S \times (0.01)^2 = S \times 1 \times 10^{-4} \]

### Step 6: Solve for \( S \)
Now, substituting the value of \( K_{sp} \):
\[ 3.4 \times 10^{-11} = S \times 1 \times 10^{-4} \]
\[ S = \frac{3.4 \times 10^{-11}}{1 \times 10^{-4}} = 3.4 \times 10^{-7} \]

### Conclusion
The solubility of \( \text{CaF}_2 \) in a 0.01 M solution of \( \text{NaF} \) is:
\[ S = 3.4 \times 10^{-7} \, \text{M} \]

### Final Answer
The solubility of \( \text{CaF}_2 \) in 0.01 M \( \text{NaF} \) is \( 3.4 \times 10^{-7} \, \text{M} \).

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