The precipitate of `CaF_(2) (K_(sp)=1.7xx10^(-10))` is obtained when equal volumes of the following are mixed

Options

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

Explanation

To determine when the precipitate of \( \text{CaF}_2 \) will form, we need to compare the ionic product (IP) of the solution with the solubility product constant (\( K_{sp} \)). The precipitate will form when the ionic product exceeds the solubility product.

### Given:
- \( K_{sp} \) of \( \text{CaF}_2 = 1.7 \times 10^{-10} \)
- The dissociation of \( \text{CaF}_2 \):
\[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^- (aq) \]

### Steps to Solve:

1. **Identify the Ionic Product Expression**:
The ionic product \( IP \) for \( \text{CaF}_2 \) is given by:
\[ IP = [\text{Ca}^{2+}] \cdot [\text{F}^-]^2 \]

2. **Evaluate Each Option**:
We will evaluate the ionic product for each option provided.

**Option 1**: \( [\text{Ca}^{2+}] = 10^{-2} \, \text{M}, [\text{F}^-] = 10^{-3} \, \text{M} \)
\[ IP = (10^{-2}) \cdot (10^{-3})^2 = 10^{-2} \cdot 10^{-6} = 10^{-8} \]
- Compare with \( K_{sp} \):
\[ 10^{-8} > 1.7 \times 10^{-10} \quad \text{(Precipitate forms)} \]

**Option 2**: \( [\text{Ca}^{2+}] = 10^{-4} \, \text{M}, [\text{F}^-] = 10^{-4} \, \text{M} \)
\[ IP = (10^{-4}) \cdot (10^{-4})^2 = 10^{-4} \cdot 10^{-8} = 10^{-12} \]
- Compare with \( K_{sp} \):
\[ 10^{-12} < 1.7 \times 10^{-10} \quad \text{(No precipitate)} \]

**Option 3**: \( [\text{Ca}^{2+}] = 10^{-3} \, \text{M}, [\text{F}^-] = 10^{-5} \, \text{M} \)
\[ IP = (10^{-3}) \cdot (10^{-5})^2 = 10^{-3} \cdot 10^{-10} = 10^{-13} \]
- Compare with \( K_{sp} \):
\[ 10^{-13} < 1.7 \times 10^{-10} \quad \text{(No precipitate)} \]

**Option 4**: \( [\text{Ca}^{2+}] = 10^{-5} \, \text{M}, [\text{F}^-] = 10^{-3} \, \text{M} \)
\[ IP = (10^{-5}) \cdot (10^{-3})^2 = 10^{-5} \cdot 10^{-6} = 10^{-11} \]
- Compare with \( K_{sp} \):
\[ 10^{-11} < 1.7 \times 10^{-10} \quad \text{(No precipitate)} \]

### Conclusion:
The only case where the ionic product exceeds the solubility product is in **Option 1**. Therefore, the precipitate of \( \text{CaF}_2 \) will form only in this case.

### Final Answer:
**Option 1** is the correct choice for the formation of \( \text{CaF}_2 \) precipitate.

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