If equal volumes of `BaCl_(2)` and `NaF` solutions are mixed, which of these combination will not give a precipitate? `(K_(sp) of BaF_(2) =1.7 xx 10^(-7))`.

Options

A. `10^(-3)BaCl_(2)` and `2xx10^(-2)M NaF`
B. `10^(-3)M BaCl_2` and `1.5 xx 10^(-2)M NaF`
C. `1.5 xx 10^(-2)M BaCl_(2)` and `10^(-2)M NaF`
D. `2xx10^(-2)M BaCl_(2)` and `2xx10^(-2)M NaF`

Explanation

To determine which combination of `BaCl2` and `NaF` solutions will not give a precipitate when mixed, we need to analyze the solubility product (Ksp) of `BaF2` and compare the ionic product (Q) of the solutions.

### Step-by-Step Solution:

1. **Understand the Dissociation of BaF2**:
The dissociation of `BaF2` can be represented as:
\[ BaF2 \rightleftharpoons Ba^{2+} + 2F^{-} \]

2. **Identify the Ksp Expression**:
The solubility product expression for `BaF2` is:
\[ K_{sp} = [Ba^{2+}][F^{-}]^2 \]
Given \( K_{sp} = 1.7 \times 10^{-7} \).

3. **Calculate Ionic Product (Q)**:
When equal volumes of `BaCl2` and `NaF` are mixed, the concentrations of the ions will be halved. Let's denote the initial concentrations of `BaCl2` and `NaF` as \( [Ba^{2+}] \) and \( [F^{-}] \) respectively.

4. **Consider Different Cases**:
We will evaluate different combinations of concentrations for `BaCl2` and `NaF` to find out when \( Q < K_{sp} \).

- **Case 1**: \( [Ba^{2+}] = 10^{-3} \, M \) and \( [F^{-}] = 2 \times 10^{-2} \, M \)
\[ Q = \left( \frac{10^{-3}}{2} \right) \left( \frac{2 \times 10^{-2}}{2} \right)^2 = \left( 5 \times 10^{-4} \right) \left( 1 \times 10^{-2} \right)^2 = 5 \times 10^{-6} \]
Since \( 5 \times 10^{-6} > 1.7 \times 10^{-7} \), a precipitate will form.

- **Case 2**: \( [Ba^{2+}] = 10^{-3} \, M \) and \( [F^{-}] = 1.5 \times 10^{-2} \, M \)
\[ Q = \left( \frac{10^{-3}}{2} \right) \left( \frac{1.5 \times 10^{-2}}{2} \right)^2 = \left( 5 \times 10^{-4} \right) \left( 0.75 \times 10^{-2} \right)^2 = 0.28 \times 10^{-6} \]
Since \( 0.28 \times 10^{-6} > 1.7 \times 10^{-7} \), a precipitate will form.

- **Case 3**: \( [Ba^{2+}] = 1.5 \times 10^{-2} \, M \) and \( [F^{-}] = 10^{-2} \, M \)
\[ Q = \left( \frac{1.5 \times 10^{-2}}{2} \right) \left( \frac{10^{-2}}{2} \right)^2 = \left( 0.75 \times 10^{-2} \right) \left( 0.5 \times 10^{-2} \right)^2 = 0.187 \times 10^{-7} \]
Since \( 0.187 \times 10^{-7} < 1.7 \times 10^{-7} \), no precipitate will form.

- **Case 4**: \( [Ba^{2+}] = 2 \times 10^{-2} \, M \) and \( [F^{-}] = 2 \times 10^{-2} \, M \)
\[ Q = \left( \frac{2 \times 10^{-2}}{2} \right) \left( \frac{2 \times 10^{-2}}{2} \right)^2 = \left( 10^{-2} \right) \left( 10^{-2} \right)^2 = 10^{-6} \]
Since \( 10^{-6} > 1.7 \times 10^{-7} \), a precipitate will form.

5. **Conclusion**:
The combination that will not give a precipitate is when \( [Ba^{2+}] = 1.5 \times 10^{-2} \, M \) and \( [F^{-}] = 10^{-2} \, M \).

### Final Answer:
The combination that will not give a precipitate is the third case: \( [Ba^{2+}] = 1.5 \times 10^{-2} \, M \) and \( [F^{-}] = 10^{-2} \, M \).