The molar solubility of `PbI_(2)` in `0.2 M Pb(NO_(3))_(2)` solution in terms of solubility product, `K_(sp)`

Options

A. `(K_(sp)//0.2)^(1//2)`
B. `(K_(sp)//0.8)^(1//3)`
C. `(K_(sp)//0.4)^(1//2)`
D. `(K_(sp)//0.8)^(1//2)`

Explanation

To find the molar solubility of \( \text{PbI}_2 \) in a \( 0.2 \, \text{M} \, \text{Pb(NO}_3)_2 \) solution in terms of the solubility product \( K_{sp} \), we can follow these steps:

### Step 1: Write the Dissociation Equation
The dissociation of lead(II) iodide (\( \text{PbI}_2 \)) in water can be represented as:
\[ \text{PbI}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{I}^- (aq) \]

### Step 2: Define Variables
Let \( S \) be the molar solubility of \( \text{PbI}_2 \) in the solution. Therefore, at equilibrium:
- The concentration of \( \text{Pb}^{2+} \) ions will be \( S \).
- The concentration of \( \text{I}^- \) ions will be \( 2S \).

### Step 3: Consider the Common Ion Effect
Since we have \( 0.2 \, \text{M} \) of \( \text{Pb(NO}_3)_2 \) in the solution, it will dissociate to provide additional \( \text{Pb}^{2+} \) ions:
\[ \text{Pb(NO}_3)_2 \rightarrow \text{Pb}^{2+} + 2 \text{NO}_3^- \]
Thus, the total concentration of \( \text{Pb}^{2+} \) ions in the solution will be:
\[ \text{Total } [\text{Pb}^{2+}] = S + 0.2 \]

### Step 4: Write the Expression for \( K_{sp} \)
The solubility product \( K_{sp} \) for \( \text{PbI}_2 \) can be expressed as:
\[ K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2 \]
Substituting the concentrations:
\[ K_{sp} = (S + 0.2)(2S)^2 \]
This simplifies to:
\[ K_{sp} = (S + 0.2)(4S^2) \]

### Step 5: Neglect \( S \) Compared to \( 0.2 \)
Since \( S \) is expected to be very small compared to \( 0.2 \), we can approximate:
\[ K_{sp} \approx (0.2)(4S^2) \]
This gives:
\[ K_{sp} = 0.8S^2 \]

### Step 6: Solve for \( S \)
Rearranging the equation to find \( S \):
\[ S^2 = \frac{K_{sp}}{0.8} \]
Taking the square root:
\[ S = \sqrt{\frac{K_{sp}}{0.8}} \]

### Final Answer
Thus, the molar solubility of \( \text{PbI}_2 \) in \( 0.2 \, \text{M} \, \text{Pb(NO}_3)_2 \) solution in terms of \( K_{sp} \) is:
\[ S = \sqrt{\frac{K_{sp}}{0.8}} \]