What is the molarity of `F^(-)` ions in a saturated solution of `BaF_(2)`? `(K_(sp)=1.0xx10^(-6))`

Options

A. `1.0xx10^(-2)`
B. `1.0xx10^(-3)`
C. `1.26xx10^(-2)`
D. `6.3xx10^(-3)`

Answer

To find the molarity of \( F^- \) ions in a saturated solution of \( BaF_2 \) given that the solubility product constant \( K_{sp} = 1.0 \times 10^{-6} \), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of \( BaF_2 \) in water can be represented as:
\[ BaF_2 (s) \rightleftharpoons Ba^{2+} (aq) + 2F^{-} (aq) \]

### Step 2: Define the solubility
Let the solubility of \( BaF_2 \) be \( S \) mol/L. This means that in a saturated solution:
- The concentration of \( Ba^{2+} \) ions will be \( S \) mol/L.
- The concentration of \( F^- \) ions will be \( 2S \) mol/L (since 2 moles of \( F^- \) are produced for every mole of \( BaF_2 \) that dissolves).

### Step 3: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for the dissociation can be expressed as:
\[ K_{sp} = [Ba^{2+}][F^-]^2 \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (S)(2S)^2 = S \cdot 4S^2 = 4S^3 \]

### Step 4: Set up the equation with \( K_{sp} \)
Now we can set this equal to the given \( K_{sp} \):
\[ 4S^3 = 1.0 \times 10^{-6} \]

### Step 5: Solve for \( S \)
To find \( S \), we rearrange the equation:
\[ S^3 = \frac{1.0 \times 10^{-6}}{4} = 2.5 \times 10^{-7} \]
Now, take the cube root:
\[ S = (2.5 \times 10^{-7})^{1/3} \]

### Step 6: Calculate \( S \)
Calculating the cube root:
\[ S \approx 6.30 \times 10^{-3} \, \text{mol/L} \]

### Step 7: Find the molarity of \( F^- \) ions
Since the concentration of \( F^- \) ions is \( 2S \):
\[ [F^-] = 2S = 2 \times 6.30 \times 10^{-3} \approx 1.26 \times 10^{-2} \, \text{mol/L} \]

Thus, the molarity of \( F^- \) ions in a saturated solution of \( BaF_2 \) is approximately \( 1.26 \times 10^{-2} \, \text{mol/L} \).

### Summary
The final answer is:
\[ \text{Molarity of } F^- \text{ ions} \approx 1.26 \times 10^{-2} \, \text{mol/L} \]