Solubility of `PbI_(2)` is `0.005 M`. Then, the solubility product of `PbI_(2)` is

Options

A. `6.8 xx 10^(-6)`
B. `6.8 xx 10^(6)`
C. `2.2 xx 10^(-9)`
D. None of these

Explanation

To find the solubility product (Ksp) of lead(II) iodide (PbI₂) given its solubility, we can follow these steps:

### Step 1: Write the Dissolution Equation
When lead(II) iodide (PbI₂) dissolves in water, it dissociates into lead ions (Pb²⁺) and iodide ions (I⁻):
\[ \text{PbI}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{I}^- (aq) \]

### Step 2: Define Solubility
Let the solubility of PbI₂ be \( S \). According to the problem, the solubility \( S \) is given as:
\[ S = 0.005 \, M \]

### Step 3: Determine Ion Concentrations
From the dissolution equation, for every 1 mole of PbI₂ that dissolves, 1 mole of Pb²⁺ and 2 moles of I⁻ are produced. Therefore, the concentrations of the ions at equilibrium will be:
- \([ \text{Pb}^{2+} ] = S = 0.005 \, M\)
- \([ \text{I}^- ] = 2S = 2 \times 0.005 \, M = 0.010 \, M\)

### Step 4: Write the Expression for Ksp
The solubility product (Ksp) is given by the expression:
\[ K_{sp} = [ \text{Pb}^{2+} ][ \text{I}^- ]^2 \]

### Step 5: Substitute the Ion Concentrations into the Ksp Expression
Substituting the values of the concentrations into the Ksp expression:
\[ K_{sp} = (0.005)(0.010)^2 \]

### Step 6: Calculate Ksp
Calculating the value:
\[ K_{sp} = (0.005)(0.0001) = 0.0000005 = 5 \times 10^{-7} \]

### Final Answer
Thus, the solubility product \( K_{sp} \) of PbI₂ is:
\[ K_{sp} = 5 \times 10^{-7} \]

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