What is the molarity of `F^(-)` in a saturated solution of In `F_(3)`?`(K_(sp)=7.9xx10^(-10)`

Options

A. `2.3xx10^(-3)`
B. `8.3xx10^(-3)`
C. `1.0xx10^(-3)`
D. `7.0xx10^(-3)`

Answer

To find the molarity of \( F^- \) in a saturated solution of \( InF_3 \) given that \( K_{sp} = 7.9 \times 10^{-10} \), we can follow these steps:

### Step 1: Write the dissociation equation
When \( InF_3 \) dissolves in water, it dissociates as follows:
\[ InF_3 (s) \rightleftharpoons In^{3+} (aq) + 3F^{-} (aq) \]

### Step 2: Define solubility
Let the solubility of \( InF_3 \) be \( s \) mol/L. From the dissociation equation, for every mole of \( InF_3 \) that dissolves, it produces 3 moles of \( F^- \). Therefore, the concentration of \( F^- \) ions in the solution will be:
\[ [F^-] = 3s \]

### Step 3: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for the dissociation can be expressed as:
\[ K_{sp} = [In^{3+}][F^-]^3 \]
Substituting the concentrations in terms of \( s \):
\[ K_{sp} = s \cdot (3s)^3 \]
This simplifies to:
\[ K_{sp} = s \cdot 27s^3 = 27s^4 \]

### Step 4: Substitute the value of \( K_{sp} \)
Now, we can set up the equation:
\[ 27s^4 = 7.9 \times 10^{-10} \]

### Step 5: Solve for \( s \)
Rearranging gives:
\[ s^4 = \frac{7.9 \times 10^{-10}}{27} \]
Calculating the right side:
\[ s^4 = 2.9259 \times 10^{-11} \]

Now, take the fourth root to find \( s \):
\[ s = \sqrt[4]{2.9259 \times 10^{-11}} \approx 1.67 \times 10^{-3} \text{ mol/L} \]

### Step 6: Calculate the molarity of \( F^- \)
Since \( [F^-] = 3s \):
\[ [F^-] = 3 \times 1.67 \times 10^{-3} \approx 5.01 \times 10^{-3} \text{ mol/L} \]

### Final Answer
The molarity of \( F^- \) in a saturated solution of \( InF_3 \) is approximately:
\[ [F^-] \approx 5.01 \times 10^{-3} \text{ mol/L} \]