If solubility product for `CaF_(2)` is `1.7xx10^(-10)` at 298 K, calculate the solubility in mol `L^(-1)`.

Answer

To calculate the solubility of calcium fluoride (CaF₂) given its solubility product (Ksp), we can follow these steps:

### Step 1: Write the dissociation equation
Calcium fluoride dissociates in water as follows:
\[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2 \text{F}^{-} (aq) \]

### Step 2: Define solubility
Let the solubility of CaF₂ be \( S \) mol/L. This means:
- The concentration of \(\text{Ca}^{2+}\) ions at equilibrium will be \( S \).
- The concentration of \(\text{F}^{-}\) ions at equilibrium will be \( 2S \) (since two fluoride ions are produced for each formula unit of CaF₂).

### Step 3: Write the expression for the solubility product (Ksp)
The solubility product expression for CaF₂ is given by:
\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^{-}]^2 \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (S)(2S)^2 \]
\[ K_{sp} = S \cdot 4S^2 = 4S^3 \]

### Step 4: Substitute the given Ksp value
We know that \( K_{sp} = 1.7 \times 10^{-10} \). Therefore:
\[ 4S^3 = 1.7 \times 10^{-10} \]

### Step 5: Solve for S
To find \( S \), we rearrange the equation:
\[ S^3 = \frac{1.7 \times 10^{-10}}{4} \]
\[ S^3 = 4.25 \times 10^{-11} \]

Next, we take the cube root:
\[ S = \sqrt[3]{4.25 \times 10^{-11}} \]

### Step 6: Calculate the value of S
Using a calculator:
\[ S \approx 3.5 \times 10^{-4} \, \text{mol/L} \]

### Final Answer
The solubility of CaF₂ at 298 K is approximately:
\[ S \approx 3.5 \times 10^{-4} \, \text{mol/L} \]

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