What is the concentration of `Ba^(2+)` when `BaF_(2)` `(K_(sp)=1.0xx10^(-6))` begins to precipitate from a solution that is 0.30 M `F^(-)` ?

Options

A. `9.0xx10^(-7)`
B. `3.3xx10^(-5)`
C. `1.1xx10^(-5)`
D. `3.0xx10^(-5)`

Answer

To find the concentration of \( \text{Ba}^{2+} \) when \( \text{BaF}_2 \) begins to precipitate from a solution that is 0.30 M \( \text{F}^- \), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of \( \text{BaF}_2 \) in water can be represented as:
\[ \text{BaF}_2 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + 2\text{F}^- (aq) \]

### Step 2: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for the dissociation of \( \text{BaF}_2 \) is given by:
\[ K_{sp} = [\text{Ba}^{2+}][\text{F}^-]^2 \]
Given \( K_{sp} = 1.0 \times 10^{-6} \).

### Step 3: Define the variables
Let the concentration of \( \text{Ba}^{2+} \) at the point of precipitation be \( S \). The concentration of \( \text{F}^- \) is given as 0.30 M.

### Step 4: Substitute the known values into the \( K_{sp} \) expression
Substituting the values into the \( K_{sp} \) expression gives:
\[ 1.0 \times 10^{-6} = S \cdot (0.30)^2 \]

### Step 5: Calculate \( (0.30)^2 \)
Calculating \( (0.30)^2 \):
\[ (0.30)^2 = 0.09 \]

### Step 6: Substitute and solve for \( S \)
Now substituting \( 0.09 \) into the equation:
\[ 1.0 \times 10^{-6} = S \cdot 0.09 \]
To find \( S \):
\[ S = \frac{1.0 \times 10^{-6}}{0.09} \]

### Step 7: Perform the division
Calculating \( S \):
\[ S = \frac{1.0 \times 10^{-6}}{0.09} \approx 1.11 \times 10^{-5} \, \text{M} \]

### Conclusion
Thus, the concentration of \( \text{Ba}^{2+} \) when \( \text{BaF}_2 \) begins to precipitate is approximately \( 1.11 \times 10^{-5} \, \text{M} \).

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