\(\ce{PbI2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)}\)

• Let the solubility of \(\ce{PbI2} = x\ \text{mol L}^{-1}\)

\begin{array} {|c|c|c|c|}
\hline  \text{Concentration (mol/L)} & \ce{PbI2} & \ce{Pb^{2+}} & \ce{2I^-} \\
\hline \text{Initial} & – & 0 & 0.1 \\
\hline \text{Change} & – & +x & +2x \\
\hline \text{Equilibrium} & – & x & 0.1 +2x \\
\hline \end{array}

• As \(x\) is very small, assume  \(0.1 +2x \approx 0.1\)

\(K_{sp} \ce{=[Pb^{2+}][I^-]^2} =9.8 \times 10^{-9}\)

\(K_{sp} = 0.1^2 \times x =9.8 \times 10^{-9}\)

\(\therefore x=\dfrac{9.8 \times 10^{-9}}{0.1^2}\)

\(\Rightarrow D\)