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The Ksp of PbI2 is 8.7x10^-9. Find the solubility of lead(II) iodide on mol/L.

a. 1.3x10^-3
b. 9.3x10^-9
c. 6.6x10^-5
d. 1.6x10^-3

  • chem - ,

    PbI2 ==>Pb^+2 + 2I^-
    Ksp = (Pb^+2)(I^-)^2
    Let S = solubility, then (Pb^+2) = S and I^-) = 2S.
    Substitute into the Ksp expression and solve for S. Post your work if you get stuck.

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