Post a New Question


posted by .

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.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question