Common-Ion Effect

AP Chemistry· difficulty 2/5

A student measures the solubility of PbI2 in pure water and in 0.10 M KI by saturating each solvent at 25 C and titrating the I- in the saturated supernatant. In pure water, [I-] = 2.6e-3 M at saturation. In 0.10 M KI, the additional I- from PbI2 dissolution is negligible, and [Pb^2+] is found to be 7.9e-7 M.

What value of Ksp is calculated from the 0.10 M KI trial, and how does it compare to the pure-water value?

  • A

    1.6e-7; common ion increases Ksp

  • B

    0.10; Ksp depends on [KI]

  • C

    7.9e-9; values agree within experimental error

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  • D

    7.9e-7; KI dissolved Pb^2+

Explanation

Ksp = [Pb^2+][I-]^2 = (7.9e-7)(0.10)^2 = 7.9e-9. This is consistent with 8.8e-9 from the first trial.

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