Which of the following is(are) correct when `0.1L` of `0.0015M MgCl_(2)` and `0.1L` of `0.025M NaF` are mixed togther? `(K_(sp) of MgF_(2) = 3.7 xx 10^(-8))`.

Options

A. `MgF_(2)` remains in solution
B. `MgF_(2)` precipitates out
C. `MgCl_(2)` precipitates out
D. `Cl^(Theta)` ions remains in solution

Explanation

To solve the problem, we need to analyze the mixing of the two solutions and determine whether a precipitate forms based on the ionic product and the solubility product (Ksp) of MgF2.

### Step-by-Step Solution:

1. **Calculate the new concentrations after mixing**:
- When 0.1 L of 0.0015 M MgCl2 is mixed with 0.1 L of 0.025 M NaF, the total volume becomes 0.2 L.
- The concentration of MgCl2 after mixing:
\[ \text{Concentration of } MgCl_2 = \frac{0.0015 \, \text{mol/L} \times 0.1 \, \text{L}}{0.2 \, \text{L}} = \frac{0.00015 \, \text{mol}}{0.2 \, \text{L}} = 0.00075 \, \text{M} = 7.5 \times 10^{-4} \, \text{M} \]
- The concentration of NaF after mixing:
\[ \text{Concentration of } NaF = \frac{0.025 \, \text{mol/L} \times 0.1 \, \text{L}}{0.2 \, \text{L}} = \frac{0.0025 \, \text{mol}}{0.2 \, \text{L}} = 0.0125 \, \text{M} = 1.25 \times 10^{-2} \, \text{M} \]

2. **Determine the concentrations of ions in solution**:
- From MgCl2, it dissociates into Mg²⁺ and Cl⁻ ions:
\[ [Mg^{2+}] = 7.5 \times 10^{-4} \, \text{M} \]
- From NaF, it dissociates into Na⁺ and F⁻ ions:
\[ [F^{-}] = 1.25 \times 10^{-2} \, \text{M} \]

3. **Calculate the ionic product (Qsp) for MgF2**:
- The ionic product for MgF2 is given by:
\[ Q_{sp} = [Mg^{2+}][F^{-}]^2 \]
- Substituting the values:
\[ Q_{sp} = (7.5 \times 10^{-4}) \times (1.25 \times 10^{-2})^2 \]
- Calculating \( (1.25 \times 10^{-2})^2 \):
\[ (1.25 \times 10^{-2})^2 = 1.5625 \times 10^{-4} \]
- Now calculating \( Q_{sp} \):
\[ Q_{sp} = (7.5 \times 10^{-4}) \times (1.5625 \times 10^{-4}) = 1.171875 \times 10^{-7} \, \text{(approximately } 1.17 \times 10^{-7}) \]

4. **Compare Qsp with Ksp**:
- Given \( K_{sp} \) of MgF2 is \( 3.7 \times 10^{-8} \).
- Since \( Q_{sp} (1.17 \times 10^{-7}) > K_{sp} (3.7 \times 10^{-8}) \), a precipitate of MgF2 will form.

5. **Conclusion**:
- MgF2 will precipitate because the ionic product exceeds the solubility product.
- MgCl2 remains in solution as it is a strong electrolyte and completely dissociates.

### Final Answer:
- MgF2 precipitates.
- MgCl2 remains in solution.
- NaF remains in solution.