The value of `K_(sp)` for `CaF_(2)` is `1.7 xx 10^(-10)` . If the concentration of NaF is 0.1 M then new solubility of `CaF_(2)` is

Options

A. `1.7 xx 10^(-9)`
B. `1.7 xx 10^(-5)`
C. `1.7 xx 10^(-8)`
D. `1.7 xx 10^(-7`

Answer

To calculate the new solubility of \( \text{CaF}_2 \) in the presence of \( \text{NaF} \), we will follow these steps:

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

### Step 2: Define the solubility and concentrations
Let the solubility of \( \text{CaF}_2 \) be \( S \). Therefore, at equilibrium:
- The concentration of \( \text{Ca}^{2+} \) will be \( S \).
- The concentration of \( \text{F}^- \) from \( \text{CaF}_2 \) will be \( 2S \).

### Step 3: Account for the common ion effect
Since \( \text{NaF} \) is added to the solution, it dissociates completely to give \( \text{Na}^+ \) and \( \text{F}^- \). The concentration of \( \text{F}^- \) from \( \text{NaF} \) is given as \( 0.1 \, M \). Therefore, the total concentration of \( \text{F}^- \) becomes:
\[ [\text{F}^-] = 2S + 0.1 \]

### Step 4: Write the expression for \( K_{sp} \)
The solubility product \( K_{sp} \) for \( \text{CaF}_2 \) is given by:
\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \]
Substituting the concentrations into the equation, we have:
\[ K_{sp} = S \cdot (2S + 0.1)^2 \]

### Step 5: Substitute the value of \( K_{sp} \)
We know that \( K_{sp} = 1.7 \times 10^{-10} \). Therefore, we can set up the equation:
\[ 1.7 \times 10^{-10} = S \cdot (0.1)^2 \]
Since \( 2S \) is negligible compared to \( 0.1 \), we can approximate:
\[ 1.7 \times 10^{-10} = S \cdot (0.1)^2 \]
This simplifies to:
\[ 1.7 \times 10^{-10} = S \cdot 0.01 \]

### Step 6: Solve for \( S \)
Now, we can solve for \( S \):
\[ S = \frac{1.7 \times 10^{-10}}{0.01} = 1.7 \times 10^{-8} \]

### Conclusion
The new solubility of \( \text{CaF}_2 \) in the presence of \( 0.1 \, M \) \( \text{NaF} \) is:
\[ S = 1.7 \times 10^{-8} \, M \]

---