What is the molar solubility of `MgF_(2)` in a 0.2 M solution of KF? `(K_(sp)=8xx10^(-8))`

Options

A. `8xx10^(-8)M`
B. `8xx10^(-7)M`
C. `2xx10^(-6)M`
D. `2.7xx10^(-4)M`

Explanation

To find the molar solubility of `MgF2` in a 0.2 M solution of `KF`, we can follow these steps:

### Step 1: Write the Dissociation Reaction
The dissociation of `MgF2` in water can be represented as:
\[ \text{MgF}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2 \text{F}^{-} (aq) \]

### Step 2: Define the Solubility
Let the molar solubility of `MgF2` be \( S \). Therefore, at equilibrium:
- The concentration of `Mg^{2+}` will be \( S \).
- The concentration of `F^{-}` will be \( 2S \).

### Step 3: Consider the Common Ion Effect
Since we are adding `KF`, which dissociates completely into `K^{+}` and `F^{-}`, the concentration of `F^{-}` ions from `KF` is 0.2 M. Thus, the total concentration of `F^{-}` ions in solution will be:
\[ [F^{-}] = 2S + 0.2 \]

### Step 4: Write the Expression for the Solubility Product (Ksp)
The solubility product expression for `MgF2` is given by:
\[ K_{sp} = [\text{Mg}^{2+}][\text{F}^{-}]^2 \]
Substituting the values we have:
\[ K_{sp} = S \cdot (2S + 0.2)^2 \]

### Step 5: Substitute the Known Ksp Value
We know that \( K_{sp} = 8 \times 10^{-8} \). Therefore, we can set up the equation:
\[ 8 \times 10^{-8} = S \cdot (2S + 0.2)^2 \]

### Step 6: Simplify the Equation
Since \( S \) is expected to be much smaller than 0.2 M (due to the common ion effect), we can neglect \( 2S \) in comparison to 0.2. Thus, we simplify to:
\[ 8 \times 10^{-8} \approx S \cdot (0.2)^2 \]
\[ 8 \times 10^{-8} = S \cdot 0.04 \]

### Step 7: Solve for S
Now, we can solve for \( S \):
\[ S = \frac{8 \times 10^{-8}}{0.04} \]
\[ S = 2 \times 10^{-6} \, \text{M} \]

### Conclusion
The molar solubility of `MgF2` in a 0.2 M solution of `KF` is \( 2 \times 10^{-6} \, \text{M} \).

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