The value of `K_(sp)` is `HgCl_(2)` at room temperature is `4.0 xx 10^(-15)`. The concentration of `Cl^(-)` ion in its aqueous solution at saturation point is

Options

A. `1 xx 10^(-5)`
B. `2xx10^(-5)`
C. `2xx10^(-15)`
D. `8xx10^(-15)`

Explanation

To find the concentration of Cl⁻ ions in the saturated solution of HgCl₂, we can follow these steps:

### Step 1: Write the dissociation equation for HgCl₂
The dissociation of mercury(II) chloride in water can be represented as:
\[ \text{HgCl}_2 (s) \rightleftharpoons \text{Hg}^{2+} (aq) + 2 \text{Cl}^- (aq) \]

### Step 2: Define the solubility (S)
Let the solubility of HgCl₂ in water be \( S \) mol/L. According to the dissociation equation:
- The concentration of Hg²⁺ ions at equilibrium will be \( S \) mol/L.
- The concentration of Cl⁻ ions at equilibrium will be \( 2S \) mol/L (since 2 moles of Cl⁻ are produced for every mole of HgCl₂ that dissolves).

### Step 3: Write the expression for Ksp
The solubility product constant \( K_{sp} \) for the dissociation of HgCl₂ can be expressed as:
\[ K_{sp} = [\text{Hg}^{2+}][\text{Cl}^-]^2 \]

Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (S)(2S)^2 \]
\[ K_{sp} = S \cdot 4S^2 = 4S^3 \]

### Step 4: Substitute the given Ksp value
We know that \( K_{sp} \) for HgCl₂ is given as \( 4.0 \times 10^{-15} \):
\[ 4S^3 = 4.0 \times 10^{-15} \]

### Step 5: Solve for S
Dividing both sides by 4:
\[ S^3 = 1.0 \times 10^{-15} \]

Now, take the cube root of both sides to find \( S \):
\[ S = (1.0 \times 10^{-15})^{1/3} \]

Calculating the cube root:
\[ S = 1.0 \times 10^{-5} \, \text{mol/L} \]

### Step 6: Calculate the concentration of Cl⁻ ions
Since the concentration of Cl⁻ ions is \( 2S \):
\[ [\text{Cl}^-] = 2S = 2 \times (1.0 \times 10^{-5}) \]
\[ [\text{Cl}^-] = 2.0 \times 10^{-5} \, \text{mol/L} \]

### Final Answer
The concentration of Cl⁻ ions in the saturated solution of HgCl₂ is:
\[ \boxed{2.0 \times 10^{-5} \, \text{mol/L}} \]

---