The solubility product of `PbI_2` is `8.0 xx 10^(-9)`. The solubility of lead iodide in 0.1 molar solution of lead nitrate is ` x xx 10^(-6)` mol /L. The value of x is __________. (Rounded off to the nearest integer ).
[Given : ` sqrt(2)` = 1.41]

Answer

To solve the problem, we need to find the solubility of lead iodide (PbI₂) in a 0.1 M solution of lead nitrate (Pb(NO₃)₂) given that the solubility product (Ksp) of PbI₂ is \(8.0 \times 10^{-9}\).

### Step-by-Step Solution:

1. **Write the Dissociation Equation:**
The dissociation of lead iodide in water can be represented as:
\[ PbI_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2I^{-} (aq) \]

2. **Define the Solubility:**
Let the solubility of PbI₂ in pure water be \(S\). Therefore, at equilibrium:
- The concentration of \(Pb^{2+}\) ions will be \(S\).
- The concentration of \(I^{-}\) ions will be \(2S\).

3. **Write the Expression for Ksp:**
The solubility product \(K_{sp}\) is given by:
\[ K_{sp} = [Pb^{2+}][I^{-}]^2 = S \cdot (2S)^2 = 4S^3 \]
Given \(K_{sp} = 8.0 \times 10^{-9}\), we can set up the equation:
\[ 4S^3 = 8.0 \times 10^{-9} \]

4. **Solve for S:**
Rearranging the equation gives:
\[ S^3 = \frac{8.0 \times 10^{-9}}{4} = 2.0 \times 10^{-9} \]
Taking the cube root:
\[ S = \sqrt[3]{2.0 \times 10^{-9}} = \sqrt[3]{2} \times 10^{-3} \]

5. **Calculate the Value of S:**
We know that \(\sqrt[3]{2} \approx 1.26\), hence:
\[ S \approx 1.26 \times 10^{-3} \text{ mol/L} \]

6. **Consider the Effect of Lead Nitrate:**
In a 0.1 M solution of lead nitrate, the concentration of \(Pb^{2+}\) ions is already 0.1 M. Therefore, the total concentration of \(Pb^{2+}\) will be:
\[ [Pb^{2+}] = 0.1 + S \approx 0.1 \text{ (since } S \text{ is very small compared to 0.1)} \]

7. **Set Up the New Ksp Expression:**
Now, the Ksp expression becomes:
\[ K_{sp} = [0.1][2S]^2 = 0.1 \cdot 4S^2 \]
Setting this equal to the Ksp:
\[ 0.1 \cdot 4S^2 = 8.0 \times 10^{-9} \]
Simplifying gives:
\[ 4S^2 = \frac{8.0 \times 10^{-9}}{0.1} = 8.0 \times 10^{-8} \]
Thus:
\[ S^2 = \frac{8.0 \times 10^{-8}}{4} = 2.0 \times 10^{-8} \]

8. **Calculate S Again:**
Taking the square root:
\[ S = \sqrt{2.0 \times 10^{-8}} = \sqrt{2} \times 10^{-4} \]
Given \(\sqrt{2} \approx 1.41\):
\[ S \approx 1.41 \times 10^{-4} \text{ mol/L} \]

9. **Convert to the Required Format:**
We need to express \(S\) in terms of \(x \times 10^{-6}\):
\[ S = 1.41 \times 10^{-4} = 141 \times 10^{-6} \text{ mol/L} \]
Thus, \(x = 141\).

### Final Answer:
The value of \(x\) is **141**.