If the solubility product of lead iodide `(Pbl_(2))` is `3.2 xx 10^(-8)`, then its solubility in moles/litre will be

Options

A. `2 xx 10^(-3)`
B. `4 xx 10^(-4)`
C. `1.6 xx 10^(-5)`
D. `1.8 xx 10^(-5)`

Explanation

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

### Step 1: Write the dissociation equation
Lead iodide dissociates in water according to the following equation:
\[ \text{PbI}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{I}^- (aq) \]

### Step 2: Define solubility
Let the solubility of PbI₂ be \( S \) moles per liter. When PbI₂ dissolves:
- The concentration of \(\text{Pb}^{2+}\) ions will be \( S \).
- The concentration of \(\text{I}^-\) ions will be \( 2S \) (since two iodide ions are produced for each formula unit of PbI₂).

### Step 3: Write the expression for Ksp
The solubility product (Ksp) expression for PbI₂ is given by:
\[ K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2 \]
Substituting the concentrations from Step 2:
\[ K_{sp} = (S)(2S)^2 \]
\[ K_{sp} = S \cdot 4S^2 = 4S^3 \]

### Step 4: Substitute the given Ksp value
We know that \( K_{sp} = 3.2 \times 10^{-8} \):
\[ 4S^3 = 3.2 \times 10^{-8} \]

### Step 5: Solve for S
Now, we can solve for \( S \):
\[ S^3 = \frac{3.2 \times 10^{-8}}{4} \]
\[ S^3 = 0.8 \times 10^{-8} \]
\[ S^3 = 8.0 \times 10^{-9} \]

Now, take the cube root of both sides:
\[ S = \sqrt[3]{8.0 \times 10^{-9}} \]
\[ S = 2.0 \times 10^{-3} \]

### Conclusion
The solubility of lead iodide (PbI₂) in moles per liter is:
\[ S = 2.0 \times 10^{-3} \, \text{mol/L} \]

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