Equal volume of the following `Ca^(+2)` and `F^(-)` solutions are mixed. In which case will percipitation occur? (`K_(sp)` of `CaF_(2) = 7 xx 10^(-10)`).

Options

A. `10^(-2) M Ca^(+2) + 10^(-5)MF^(-)`
B. `10^(-3) M Ca^(+2) + 10^(-5)MF^(-)`
C. `10^(-4) M Ca^(+2) + 10^(-2)MF^(-)`
D. `10^(-2) M Ca^(+2) + 10^(-6)MF^(-)`

Explanation

To determine whether precipitation occurs when equal volumes of `Ca^(+2)` and `F^(-)` solutions are mixed, we need to compare the ion product (K_ip) to the solubility product constant (K_sp) of calcium fluoride (CaF₂). The precipitation will occur if K_ip exceeds K_sp.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The dissolution of calcium fluoride can be represented as:
\[ CaF_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2F^{-} (aq) \]
The solubility product (K_sp) expression is given by:
\[ K_{sp} = [Ca^{2+}][F^{-}]^2 \]

2. **Given Data**:
- \( K_{sp} \) of \( CaF_2 = 7 \times 10^{-10} \)

3. **Calculating Ion Product (K_ip)**:
For each case, we will calculate K_ip using the formula:
\[ K_{ip} = [Ca^{2+}][F^{-}]^2 \]

4. **Case 1**:
- Concentration of \( Ca^{2+} = 10^{-2} \, M \)
- Concentration of \( F^{-} = 10^{-5} \, M \)
\[ K_{ip} = (10^{-2})(10^{-5})^2 = (10^{-2})(10^{-10}) = 10^{-12} \]
- Since \( K_{ip} (10^{-12}) < K_{sp} (7 \times 10^{-10}) \), no precipitation occurs.

5. **Case 2**:
- Concentration of \( Ca^{2+} = 10^{-3} \, M \)
- Concentration of \( F^{-} = 10^{-5} \, M \)
\[ K_{ip} = (10^{-3})(10^{-5})^2 = (10^{-3})(10^{-10}) = 10^{-13} \]
- Since \( K_{ip} (10^{-13}) < K_{sp} (7 \times 10^{-10}) \), no precipitation occurs.

6. **Case 3**:
- Concentration of \( Ca^{2+} = 10^{-4} \, M \)
- Concentration of \( F^{-} = 10^{-2} \, M \)
\[ K_{ip} = (10^{-4})(10^{-2})^2 = (10^{-4})(10^{-4}) = 10^{-8} \]
- Since \( K_{ip} (10^{-8}) > K_{sp} (7 \times 10^{-10}) \), precipitation occurs.

7. **Case 4**:
- Concentration of \( Ca^{2+} = 10^{-2} \, M \)
- Concentration of \( F^{-} = 10^{-12} \, M \)
\[ K_{ip} = (10^{-2})(10^{-12})^2 = (10^{-2})(10^{-24}) = 10^{-26} \]
- Since \( K_{ip} (10^{-26}) < K_{sp} (7 \times 10^{-10}) \), no precipitation occurs.

### Conclusion:
Precipitation occurs in **Case 3** where \( K_{ip} > K_{sp} \).

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