At ` 25^(@)` C, the solubility product of `Hg_(2)CI_(2)` in water is `3.2xx10^(-17)mol^(3)dm^(-9)` what is the solubility of `Hg_(2)CI_(2)` in water at `25^(@)`C ?

Options

A. `1.2 xx 10^(-12) M`
B. `3.0 xx 10^(-6) M`
C. `2 xx 10^(-6) M`
D. `1.2 xx 10^(-16) M`

Answer

To find the solubility of \( \text{Hg}_2\text{Cl}_2 \) in water at \( 25^\circ C \), we can follow these steps:

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

### Step 2: Define solubility
Let the solubility of \( \text{Hg}_2\text{Cl}_2 \) be \( S \) mol/L. When \( \text{Hg}_2\text{Cl}_2 \) dissolves, it produces:
- \( S \) moles of \( \text{Hg}_2^{2+} \)
- \( 2S \) moles of \( \text{Cl}^- \)

### Step 3: Write the expression for the solubility product (\( K_{sp} \))
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Hg}_2^{2+}][\text{Cl}^-]^2 \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (S)(2S)^2 = S \cdot 4S^2 = 4S^3 \]

### Step 4: Substitute the given \( K_{sp} \) value
We are given that:
\[ K_{sp} = 3.2 \times 10^{-17} \, \text{mol}^3/\text{dm}^9 \]
Thus, we can set up the equation:
\[ 4S^3 = 3.2 \times 10^{-17} \]

### Step 5: Solve for \( S \)
Rearranging the equation gives:
\[ S^3 = \frac{3.2 \times 10^{-17}}{4} \]
Calculating the right-hand side:
\[ S^3 = 0.8 \times 10^{-17} = 8.0 \times 10^{-18} \]
Now, take the cube root:
\[ S = \sqrt[3]{8.0 \times 10^{-18}} \]

### Step 6: Calculate \( S \)
Calculating the cube root:
\[ S = 2 \times 10^{-6} \, \text{mol/L} \]

### Conclusion
The solubility of \( \text{Hg}_2\text{Cl}_2 \) in water at \( 25^\circ C \) is:
\[ \boxed{2 \times 10^{-6} \, \text{mol/L}} \]