In reversible reaction `CaF_(2(s)) hArr Ca^(+2)""_(aq) + 2F_((aq)^(+))^(-)` , the concentration of fluoride ions was made halved, then equilibrium concentration of `Ca^(+2)`

Options

A. increases by 2 times
B. decreases by 2 times
C. increases by 4 times
D. decreases by 4 times

Answer

To solve the problem, we need to analyze the equilibrium reaction and how changes in the concentration of fluoride ions affect the concentration of calcium ions.

### Step-by-Step Solution:

1. **Write the equilibrium expression**:
The given reaction is:
\[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^- (aq) \]
The equilibrium constant \( K_{eq} \) for this reaction can be expressed as:
\[ K_{eq} = [\text{Ca}^{2+}] \cdot [\text{F}^-]^2 \]

2. **Define initial concentrations**:
Let's assume the initial concentrations of \(\text{Ca}^{2+}\) and \(\text{F}^-\) at equilibrium are both 1 M:
\[ [\text{Ca}^{2+}]_1 = 1 \, \text{M}, \quad [\text{F}^-]_1 = 1 \, \text{M} \]

3. **Calculate the initial equilibrium constant**:
Substituting the initial concentrations into the equilibrium expression:
\[ K_{eq} = (1) \cdot (1)^2 = 1 \]

4. **Change in fluoride concentration**:
The problem states that the concentration of fluoride ions is halved:
\[ [\text{F}^-]_2 = \frac{1}{2} \, \text{M} \]

5. **Set up the new equilibrium expression**:
Let \( [\text{Ca}^{2+}]_2 \) be the new concentration of calcium ions at equilibrium. The new equilibrium expression becomes:
\[ K_{eq} = [\text{Ca}^{2+}]_2 \cdot \left(\frac{1}{2}\right)^2 \]

6. **Equate the two equilibrium expressions**:
Since \( K_{eq} \) remains constant, we can set the two expressions equal:
\[ 1 = [\text{Ca}^{2+}]_2 \cdot \left(\frac{1}{2}\right)^2 \]

7. **Solve for the new calcium concentration**:
Rearranging the equation gives:
\[ [\text{Ca}^{2+}]_2 = 1 \cdot \left(\frac{4}{1}\right) = 4 \, \text{M} \]

8. **Conclusion**:
Thus, the equilibrium concentration of calcium ions increases by 4 times when the concentration of fluoride ions is halved.

### Final Answer:
The equilibrium concentration of \( \text{Ca}^{2+} \) increases by 4 times.

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