Calculate the solubility of `CaF_(2)` in water at 298 K which is 70% dissociated. `K_(sp)` of `CaF_(2)` is `1.7 xx 10^(-10)`. If answer is x `xx 10^(-4)` mol/ltr then x = ____?

Answer

To calculate the solubility of \( \text{CaF}_2 \) in water at 298 K, given that it is 70% dissociated and the \( K_{sp} \) of \( \text{CaF}_2 \) is \( 1.7 \times 10^{-10} \), we can follow these steps:

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

### Step 2: Define solubility
Let the solubility of \( \text{CaF}_2 \) be \( S \) mol/L. Since it is 70% dissociated, the concentrations of the ions will be:
- Concentration of \( \text{Ca}^{2+} \): \( S \times \frac{70}{100} = S \times 0.7 \)
- Concentration of \( \text{F}^{-} \): \( 2S \times \frac{70}{100} = 2S \times 0.7 = 1.4S \)

### Step 3: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^{-}]^2 \]
Substituting the concentrations we found:
\[ K_{sp} = (0.7S)(1.4S)^2 \]

### Step 4: Substitute \( K_{sp} \) value
Substituting the value of \( K_{sp} \):
\[ 1.7 \times 10^{-10} = (0.7S)(1.4^2 S^2) \]
Calculating \( 1.4^2 \):
\[ 1.4^2 = 1.96 \]
Thus, we have:
\[ 1.7 \times 10^{-10} = 0.7S \cdot 1.96S^2 \]
\[ 1.7 \times 10^{-10} = 1.372S^3 \]

### Step 5: Solve for \( S^3 \)
Rearranging the equation gives:
\[ S^3 = \frac{1.7 \times 10^{-10}}{1.372} \]
Calculating the right side:
\[ S^3 \approx 1.239 \times 10^{-10} \]

### Step 6: Calculate \( S \)
Taking the cube root of both sides:
\[ S = \sqrt[3]{1.239 \times 10^{-10}} \approx 4.98 \times 10^{-4} \text{ mol/L} \]

### Step 7: Find the value of \( x \)
Since the answer is in the form \( x \times 10^{-4} \) mol/L, we find that:
\[ x \approx 4.98 \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{4.98} \]