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

The dissolution equilibrium for $$PbI_2$$ is:

$$PbI_2 \rightleftharpoons Pb^{2+} + 2I^-$$

The solubility product expression is $$K_{sp} = [Pb^{2+}][I^-]^2 = 8.0 \times 10^{-9}$$.

In a 0.1 M lead nitrate solution, the lead ion concentration from lead nitrate is 0.1 M. Let the solubility of $$PbI_2$$ in this solution be $$s$$ mol/L.

Then $$[Pb^{2+}] = 0.1 + s \approx 0.1$$ M (since $$s$$ is very small compared to 0.1), and $$[I^-] = 2s$$.

Substituting into the $$K_{sp}$$ expression:

$$8.0 \times 10^{-9} = (0.1)(2s)^2 = 0.1 \times 4s^2 = 0.4s^2$$

$$s^2 = \frac{8.0 \times 10^{-9}}{0.4} = 2.0 \times 10^{-8}$$

$$s = \sqrt{2.0 \times 10^{-8}} = \sqrt{2} \times 10^{-4} = 1.41 \times 10^{-4} \text{ mol/L}$$

This equals $$141 \times 10^{-6}$$ mol/L, so the value of $$x$$ is $$\textbf{141}$$.