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If the molar solubility of lead iodide is 2.4 x 10^-3M. Calculate the Ksp.

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    PbI2 ==> Pb^+2 + 2I^-

    Ksp = (Pb^+2)(I^-)^2
    Therefore, (Pb^+2) = 2.4 x 10^-3M
    and (I^-) = 4.8 x 10^-3 M.
    Plug into Ksp expression and solve for Ksp.

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