The solubility of $$PbI_2$$ at 25°C is 0.7 g L$$^{-1}$$. The solubility product of $$PbI_2$$ at this temperature is (molar mass of $$PbI_2$$ = 461.2 g mol$$^{-1}$$)

The ionic equilibrium involved is the dissolution  $$PbI_2(s) \rightleftharpoons Pb^{2+}(aq)+2\,I^{-}(aq)$$.

Step 1 - Convert the given solubility from g L$$^{-1}$$ to mol L$$^{-1}$$ (molar solubility $$S$$).
Molar mass of $$PbI_2$$ = $$461.2\;\text{g mol}^{-1}$$.
$$S = \frac{0.7\;\text{g L}^{-1}}{461.2\;\text{g mol}^{-1}} = 0.00152\;\text{mol L}^{-1} \approx 1.52 \times 10^{-3}\;\text{mol L}^{-1}$$

Step 2 - Write concentrations in terms of $$S$$ after complete dissolution.
$$[Pb^{2+}] = S$$,
$$[I^{-}] = 2S$$

Step 3 - Write the solubility-product expression.
$$K_{sp} = [Pb^{2+}]\, [I^{-}]^{2} = S \,(2S)^{2} = 4\,S^{3}$$

Step 4 - Insert the numerical value of $$S$$.
$$K_{sp} = 4\,(1.52 \times 10^{-3})^{3}$$
First, cube $$1.52 \times 10^{-3}$$:
$$(1.52)^{3} = 3.51$$ (to three significant figures) and $$(10^{-3})^{3} = 10^{-9}$$,
so $$(1.52 \times 10^{-3})^{3} = 3.51 \times 10^{-9}$$.

Therefore,
$$K_{sp} = 4 \times 3.51 \times 10^{-9} = 14.0 \times 10^{-9}$$.

Hence the solubility product of $$PbI_2$$ at 25 °C is $$1.40 \times 10^{-8}$$, which is written in the provided options as $$14.0 \times 10^{-9}$$.

Option D which is: $$14.0 \times 10^{-9}$$