Should a precipitate of barium fluoride be obtained when `100mL` of `0.25M NaF` and `100mL` of `0.015M Ba(NO_(3))_(2)` are mixed. `K_(sp)` of `BaF_(2) = 1.7 xx 10^(-6)`

Explanation

To determine whether a precipitate of barium fluoride (BaF₂) will form when mixing 100 mL of 0.25 M NaF and 100 mL of 0.015 M Ba(NO₃)₂, we need to follow these steps:

### Step 1: Calculate the concentration of fluoride ions (F⁻) after mixing
1. **Initial concentration of NaF**: 0.25 M
2. **Volume of NaF solution**: 100 mL
3. **Total volume after mixing**: 100 mL + 100 mL = 200 mL

Using the dilution formula:
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