A student investigates the common-ion effect on the dissolution of lead(II) iodide by preparing saturated solutions under two conditions at \(25^\circ\text{C}\):

Condition 1: \(\text{PbI}_2(s)\) dissolved in pure distilled water

Condition 2: \(\text{PbI}_2(s)\) dissolved in \(0.10\text{ M KI}(aq)\)

The dissolution equilibrium and its solubility product constant at \(25^\circ\text{C}\) are given below.
\[\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\,\text{I}^-(aq) \quad K_{sp} = 4.0 \times 10^{-9}\]

Assuming that the added \(0.10\text{ M KI}(aq)\) provides the predominant source of \(\text{I}^-(aq)\) in Condition 2, what is the value of the ratio of the molar solubility of \(\text{PbI}_2\) in pure water (\(s_1\)) to its molar solubility in \(0.10\text{ M KI}(aq)\) (\(s_2\)), expressed as \(\dfrac{s_1}{s_2}\)?