Find the solubility of `CaF_2` in 0.05 M solution of `CaCl_2` and water. How many times Z is the solubility in the second case greater than in the first? `K_(sp) (CaF_2)= 4 xx 10^(-11)`.

Answer

To solve the problem of finding the solubility of \( \text{CaF}_2 \) in both a 0.05 M solution of \( \text{CaCl}_2 \) and in pure water, we will follow these steps:

### Step 1: Understanding the Dissolution of \( \text{CaF}_2 \)

The dissolution of calcium fluoride can be represented by the following equilibrium equation:

\[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2 \text{F}^- (aq) \]

The solubility product constant \( K_{sp} \) for \( \text{CaF}_2 \) is given by:

\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \]

### Step 2: Solubility in Pure Water

Let the solubility of \( \text{CaF}_2 \) in pure water be \( S \). In pure water, the concentrations at equilibrium will be:

\[ [\text{Ca}^{2+}] = S \quad \text{and} \quad [\text{F}^-] = 2S \]

Substituting these into the \( K_{sp} \) expression:

\[ K_{sp} = S(2S)^2 = 4S^3 \]

Given \( K_{sp} = 4 \times 10^{-11} \):

\[ 4S^3 = 4 \times 10^{-11} \]

Dividing both sides by 4:

\[ S^3 = 10^{-11} \]

Taking the cube root:

\[ S = (10^{-11})^{1/3} = 10^{-11/3} \approx 2.15 \times 10^{-4} \, \text{M} \]

### Step 3: Solubility in 0.05 M \( \text{CaCl}_2 \)

In a 0.05 M solution of \( \text{CaCl}_2 \), the concentration of \( \text{Ca}^{2+} \) ions is already 0.05 M. Let the solubility of \( \text{CaF}_2 \) in this solution be \( S' \). The equilibrium concentrations will be:

\[ [\text{Ca}^{2+}] = 0.05 + S' \approx 0.05 \quad \text{(since \( S' \) is small compared to 0.05)} \]
\[ [\text{F}^-] = 2S' \]

Substituting into the \( K_{sp} \) expression:

\[ K_{sp} = (0.05)(2S')^2 = 4(0.05)(S')^2 \]

Setting this equal to \( K_{sp} \):

\[ 4(0.05)(S')^2 = 4 \times 10^{-11} \]

Dividing both sides by 4:

\[ 0.05(S')^2 = 10^{-11} \]

Solving for \( S' \):

\[ (S')^2 = \frac{10^{-11}}{0.05} = 2 \times 10^{-10} \]

Taking the square root:

\[ S' = \sqrt{2 \times 10^{-10}} = \sqrt{2} \times 10^{-5} \approx 1.414 \times 10^{-5} \, \text{M} \]

### Step 4: Comparing Solubilities

To find how many times the solubility in the 0.05 M \( \text{CaCl}_2 \) solution is greater than in pure water, we take the ratio:

\[ \text{Ratio} = \frac{S'}{S} = \frac{1.414 \times 10^{-5}}{2.15 \times 10^{-4}} \approx \frac{1.414}{2.15} \approx 0.657 \]

To find how many times \( S' \) is greater than \( S \):

\[ \frac{S}{S'} = \frac{2.15 \times 10^{-4}}{1.414 \times 10^{-5}} \approx 15.4 \]

### Final Answers

1. Solubility of \( \text{CaF}_2 \) in water, \( S \approx 2.15 \times 10^{-4} \, \text{M} \)
2. Solubility of \( \text{CaF}_2 \) in 0.05 M \( \text{CaCl}_2 \), \( S' \approx 1.414 \times 10^{-5} \, \text{M} \)
3. The solubility in 0.05 M \( \text{CaCl}_2 \) is approximately 15.4 times greater than in pure water.