A reaction `CaF_(2)hArrCa^(2+)+2F^(-)` is at equilibrium. If the concentration of `Ca^(2+)` is increased four times, what will be the change in `F^(-)` concentration as compared to the initial concentration of `F^(-)`?

Options

A. `(1)/(4)` times
B. `(1)/(2)` times
C. `4` times
D. `2` times

Explanation

To solve the problem step by step, we will analyze the equilibrium reaction and how the change in concentration of one ion affects the other ion at equilibrium.

### Step 1: Write the Equilibrium Reaction
The dissociation of calcium fluoride can be represented as:
\[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^- (aq) \]

### Step 2: Write the Expression for the Equilibrium Constant (K)
The equilibrium constant expression for this reaction is given by:
\[ K = \frac{[\text{Ca}^{2+}][\text{F}^-]^2}{1} \]
Since the solid CaF₂ does not appear in the equilibrium expression, we focus on the aqueous ions.

### Step 3: Define Initial Concentrations
Let:
- The initial concentration of \(\text{Ca}^{2+}\) be \(X\).
- The initial concentration of \(\text{F}^-\) be \(Y\).

Thus, the equilibrium expression can be written as:
\[ K = X \cdot Y^2 \]

### Step 4: Change in Concentration of \(\text{Ca}^{2+}\)
If the concentration of \(\text{Ca}^{2+}\) is increased four times, the new concentration becomes:
\[ \text{New } [\text{Ca}^{2+}] = 4X \]

### Step 5: Write the New Equilibrium Expression
The new equilibrium expression will be:
\[ K = (4X) \cdot [\text{F}^-]^2 \]

### Step 6: Set the Two Equilibrium Expressions Equal
Since the equilibrium constant \(K\) remains the same, we can set the two expressions equal to each other:
\[ X \cdot Y^2 = 4X \cdot [\text{F}^-]^2 \]

### Step 7: Cancel Out \(X\)
Assuming \(X \neq 0\), we can cancel \(X\) from both sides:
\[ Y^2 = 4 \cdot [\text{F}^-]^2 \]

### Step 8: Solve for New Concentration of \(\text{F}^-\)
Taking the square root of both sides gives:
\[ Y = 2 \cdot [\text{F}^-] \]

### Step 9: Relate New Concentration to Initial Concentration
This implies that the new concentration of \(\text{F}^-\) is:
\[ [\text{F}^-] = \frac{Y}{2} \]

### Step 10: Conclusion
Thus, the concentration of \(\text{F}^-\) decreases to half of its initial concentration. Therefore, the change in concentration of \(\text{F}^-\) as compared to the initial concentration is:
- The new concentration of \(\text{F}^-\) is \( \frac{1}{2} \) times the initial concentration.

### Final Answer
The concentration of \(\text{F}^-\) decreases to half of its initial concentration.

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