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

Options

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

Answer

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 equations
The dissociation of \( \text{PbI}_2 \) in water can be represented as:
\[ \text{PbI}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{I}^- (aq) \]
And the dissociation of \( \text{Pb(NO}_3)_2 \) is:
\[ \text{Pb(NO}_3)_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{NO}_3^- (aq) \]

### Step 2: Determine the initial concentration of ions
Since the concentration of \( \text{Pb(NO}_3)_2 \) is \( 0.2 \, \text{M} \), the initial concentration of \( \text{Pb}^{2+} \) ions from this salt is:
\[ [\text{Pb}^{2+}] = 0.2 \, \text{M} \]
The concentration of \( \text{NO}_3^- \) ions will be:
\[ [\text{NO}_3^-] = 2 \times 0.2 = 0.4 \, \text{M} \]

### Step 3: Set up the solubility expression
Let the molar solubility of \( \text{PbI}_2 \) in this solution be \( s \). When \( \text{PbI}_2 \) dissolves, it produces:
- \( [\text{Pb}^{2+}] = s + 0.2 \)
- \( [\text{I}^-] = 2s \)

### Step 4: Write the expression for \( K_{sp} \)
The solubility product \( K_{sp} \) for \( \text{PbI}_2 \) is given by:
\[ K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2 \]
Substituting the concentrations:
\[ K_{sp} = (s + 0.2)(2s)^2 \]
This simplifies to:
\[ K_{sp} = (s + 0.2)(4s^2) \]

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

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

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