Lead(II) iodide (PbI2) has a solubility of 1.52 × 10–3 mol/L.?
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A. Write the dissolution reaction of PbI2, including all states. B. Write the expression for Ksp for PbI2. C. What is the concentration of Pb2+ in the equilibrium solution? D. ...show more
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Answer:
A. PbI2(s) ===> Pb2+(aq) + 2I-(aq) B. Ksp = [Pb2+][I-]^2 C. 1.52 x 10^-3 M. It is equal to the moles/L of PbI2 that go into solution. D. 2 x 1.52 x 10^-3 = 3.04 x 10^-3 M E. Ksp = (1.52x10^-3)(2.31x10^-6) = 3.51 x 10^-9
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