If the concentration of lead iodide in its saturated solution at `25^(@)C` be `2 xx 10^(-3)` moles per litre, then the solubility product is

Options

A. `4 xx 10^(-6)`
B. `8 xx 10^(-12)`
C. `6 xx 10^(-9)`
D. `32 xx 10^(-9)`

Explanation

To find the solubility product (Ksp) of lead iodide (PbI2) given its concentration in a saturated solution, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Dissociation Equation**:
Lead iodide (PbI2) dissociates in water as follows:
\[ \text{PbI}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{I}^- (aq) \]

2. **Define Solubility**:
Let the solubility of PbI2 be \( S \) moles per liter. According to the dissociation equation:
- The concentration of lead ions, \([\text{Pb}^{2+}]\), will be \( S \).
- The concentration of iodide ions, \([\text{I}^-]\), will be \( 2S \).

3. **Substitute the Given Concentration**:
We are given that the concentration of PbI2 in its saturated solution is \( 2 \times 10^{-3} \) moles per liter. Thus:
\[ S = 2 \times 10^{-3} \, \text{mol/L} \]

4. **Calculate the Concentration of Iodide Ions**:
Since \([\text{I}^-] = 2S\):
\[ [\text{I}^-] = 2 \times (2 \times 10^{-3}) = 4 \times 10^{-3} \, \text{mol/L} \]

5. **Write the Expression for Ksp**:
The solubility product \( K_{sp} \) is given by the formula:
\[ K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2 \]
Substituting the values we found:
\[ K_{sp} = (S)(2S)^2 \]

6. **Substitute the Values of S**:
Substitute \( S = 2 \times 10^{-3} \):
\[ K_{sp} = (2 \times 10^{-3}) \times (4 \times 10^{-3})^2 \]

7. **Calculate Ksp**:
First, calculate \( (4 \times 10^{-3})^2 \):
\[ (4 \times 10^{-3})^2 = 16 \times 10^{-6} = 1.6 \times 10^{-5} \]
Now substitute this back into the Ksp equation:
\[ K_{sp} = (2 \times 10^{-3}) \times (1.6 \times 10^{-5}) = 3.2 \times 10^{-8} \]

8. **Final Result**:
The solubility product \( K_{sp} \) for lead iodide at \( 25^\circ C \) is:
\[ K_{sp} = 3.2 \times 10^{-8} \, \text{mol}^3/\text{L}^3 \]