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 find the new solubility of \( \text{CaF}_2 \) in the presence of \( \text{NaF} \), we can 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: Write the expression for the solubility product constant \( K_{sp} \)
The solubility product \( K_{sp} \) for \( \text{CaF}_2 \) is given by:
\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \]
Given that \( K_{sp} = 1.7 \times 10^{-10} \).

### Step 3: Define the solubility in pure water
Let the solubility of \( \text{CaF}_2 \) in pure water be \( S \). Then, at equilibrium:
- The concentration of \( \text{Ca}^{2+} \) will be \( S \).
- The concentration of \( \text{F}^- \) will be \( 2S \).

Substituting these into the \( K_{sp} \) expression:
\[ K_{sp} = S \cdot (2S)^2 = S \cdot 4S^2 = 4S^3 \]
Setting this equal to the given \( K_{sp} \):
\[ 4S^3 = 1.7 \times 10^{-10} \]

### Step 4: Solve for \( S \)
\[ S^3 = \frac{1.7 \times 10^{-10}}{4} = 4.25 \times 10^{-11} \]
\[ S = \sqrt[3]{4.25 \times 10^{-11}} \approx 3.48 \times 10^{-4} \, \text{M} \]

### Step 5: Consider the effect of \( \text{NaF} \)
When \( \text{NaF} \) is added, it dissociates completely:
\[ \text{NaF} \rightarrow \text{Na}^+ + \text{F}^- \]
Given that the concentration of \( \text{NaF} \) is \( 0.1 \, \text{M} \), the concentration of \( \text{F}^- \) from \( \text{NaF} \) is \( 0.1 \, \text{M} \).

### Step 6: Update the \( K_{sp} \) expression with the new \( \text{F}^- \) concentration
Now, the total concentration of \( \text{F}^- \) becomes \( 0.1 + 2S \). However, since \( S \) is much smaller than \( 0.1 \), we can approximate:
\[ [\text{F}^-] \approx 0.1 \, \text{M} \]
Thus, the new \( K_{sp} \) expression becomes:
\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = S \cdot (0.1)^2 \]
Setting this equal to \( K_{sp} \):
\[ 1.7 \times 10^{-10} = S \cdot (0.1)^2 \]
\[ 1.7 \times 10^{-10} = S \cdot 0.01 \]
\[ S = \frac{1.7 \times 10^{-10}}{0.01} = 1.7 \times 10^{-8} \, \text{M} \]

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