Can a precipitate of barium fluoride form when we mix 100 mL of `1.0xx10^(-3) M Ba(NO_3)_2` (aq) and 200 mL of `1.0xx10^(-3)` M KF (aq) . `K_(sp) (BaF_2)=1.0xx10^(-6)`

Explanation

To determine whether a precipitate of barium fluoride (BaF₂) will form when mixing 100 mL of 1.0 x 10^(-3) M Ba(NO₃)₂ and 200 mL of 1.0 x 10^(-3) M KF, we will follow these steps:

### Step 1: Calculate the total volume of the solution
When we mix the two solutions, the total volume is:
\[ \text{Total Volume} = 100 \, \text{mL} + 200 \, \text{mL} = 300 \, \text{mL} \]

### Step 2: Calculate the concentration of Ba²⁺ ions
Barium nitrate (Ba(NO₃)₂) dissociates into Ba²⁺ and 2 NO₃⁻ ions. The concentration of Ba²⁺ in the final solution can be calculated using the dilution formula:
\[ \text{Concentration of Ba}^{2+} = \frac{C_1 \times V_1}{V_f} \]
Where:
- \(C_1 = 1.0 \times 10^{-3} \, \text{M}\) (initial concentration of Ba(NO₃)₂)
- \(V_1 = 100 \, \text{mL}\) (volume of Ba(NO₃)₂)
- \(V_f = 300 \, \text{mL}\) (final volume)

Calculating:
\[ \text{Concentration of Ba}^{2+} = \frac{(1.0 \times 10^{-3} \, \text{M}) \times (100 \, \text{mL})}{300 \, \text{mL}} = \frac{1.0 \times 10^{-1}}{3} = 3.33 \times 10^{-4} \, \text{M} \]

### Step 3: Calculate the concentration of F⁻ ions
Potassium fluoride (KF) dissociates into K⁺ and F⁻ ions. The concentration of F⁻ in the final solution can also be calculated using the dilution formula:
\[ \text{Concentration of F}^- = \frac{C_2 \times V_2}{V_f} \]
Where:
- \(C_2 = 1.0 \times 10^{-3} \, \text{M}\) (initial concentration of KF)
- \(V_2 = 200 \, \text{mL}\) (volume of KF)

Calculating:
\[ \text{Concentration of F}^- = \frac{(1.0 \times 10^{-3} \, \text{M}) \times (200 \, \text{mL})}{300 \, \text{mL}} = \frac{2.0 \times 10^{-1}}{3} = 6.67 \times 10^{-4} \, \text{M} \]

### Step 4: Calculate the ionic product (IP) for BaF₂
The ionic product (IP) for the precipitation of BaF₂ can be calculated using the formula:
\[ IP = [Ba^{2+}] \times [F^-]^2 \]
Substituting the concentrations we found:
\[ IP = (3.33 \times 10^{-4}) \times (6.67 \times 10^{-4})^2 \]
Calculating:
\[ IP = (3.33 \times 10^{-4}) \times (4.45 \times 10^{-7}) = 1.48 \times 10^{-10} \]

### Step 5: Compare IP with Ksp
The solubility product constant (Ksp) for BaF₂ is given as:
\[ K_{sp} = 1.0 \times 10^{-6} \]
Now, we compare the ionic product with Ksp:
\[ IP (1.48 \times 10^{-10}) < K_{sp} (1.0 \times 10^{-6}) \]

### Conclusion
Since the ionic product is less than the Ksp, a precipitate of barium fluoride (BaF₂) will **not** form when the two solutions are mixed.

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