The solubility of `CaF_2` in water at `18^@C" is "2.04 xx 10^(-4)" mole/litre."` Calculate `K_(sp)" of "CaF_2` and its solubility in 0.01 molar NaF solution.

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the dissociation equation for \( CaF_2 \)

The dissociation of calcium fluoride in water can be represented as:
\[ CaF_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2F^{-} (aq) \]

### Step 2: Determine the solubility product constant \( K_{sp} \)

Given that the solubility of \( CaF_2 \) is \( 2.04 \times 10^{-4} \) mol/L, we can denote the solubility as \( s \):
\[ s = 2.04 \times 10^{-4} \, \text{mol/L} \]

From the dissociation equation, we know:
- The concentration of \( Ca^{2+} \) ions will be \( s \).
- The concentration of \( F^{-} \) ions will be \( 2s \).

Thus, we can express \( K_{sp} \) as:
\[ K_{sp} = [Ca^{2+}][F^{-}]^2 = (s)(2s)^2 \]
\[ K_{sp} = s(4s^2) = 4s^3 \]

### Step 3: Substitute the value of \( s \) into the \( K_{sp} \) expression

Now, substituting \( s = 2.04 \times 10^{-4} \):
\[ K_{sp} = 4(2.04 \times 10^{-4})^3 \]

Calculating \( (2.04 \times 10^{-4})^3 \):
\[ (2.04 \times 10^{-4})^3 = 8.48 \times 10^{-12} \]

Now, substituting this back into the \( K_{sp} \) expression:
\[ K_{sp} = 4 \times 8.48 \times 10^{-12} = 3.392 \times 10^{-11} \]

### Step 4: Calculate the solubility of \( CaF_2 \) in a 0.01 M NaF solution

In a 0.01 M NaF solution, the concentration of \( F^{-} \) ions is already provided as \( 0.01 \) M. Let the solubility of \( CaF_2 \) in this solution be \( S \).

Using the \( K_{sp} \) expression:
\[ K_{sp} = [Ca^{2+}][F^{-}]^2 \]
\[ K_{sp} = S(0.01)^2 \]

Substituting the value of \( K_{sp} \):
\[ 3.392 \times 10^{-11} = S(0.01)^2 \]
\[ 3.392 \times 10^{-11} = S(1 \times 10^{-4}) \]

### Step 5: Solve for \( S \)

Now, we can solve for \( S \):
\[ S = \frac{3.392 \times 10^{-11}}{1 \times 10^{-4}} \]
\[ S = 3.392 \times 10^{-7} \, \text{mol/L} \]

### Final Answers
- \( K_{sp} \) of \( CaF_2 \) is \( 3.392 \times 10^{-11} \).
- The solubility of \( CaF_2 \) in a 0.01 M NaF solution is \( 3.392 \times 10^{-7} \, \text{mol/L} \).