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Why is $\ce{Pb^2+}$ present both in Group I and II of the basic radicals? Is it related to the solubity product of its salts?

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Yes, the Ksp for $\ce{PbCl2}$ is ~$10^{-5}$ while for $\ce{PbS}$ is ~$10^{-28}$

So if the concentration of $\ce{Pb^2+}$ is too low to detect by dilute $\ce{HCl}$ precipitation, it is still possible it could be detected as the sulfide precipitate.

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