The concentration of iodide ions in a saturated solution of lead (II)iodide is ________ M.The solubility product constant of PbI₂ is A)3.8 × 10^-4 B)3.0 × 10^-3 C)1.5 × 10^-3 D)3.5 × 10^-9 E)1.4 × 10^-8

Question

Quizplus · Accepted Answer

The solubility product constant expression for lead(II) iodide is:
PbI2 (s) <=> Pb2+ (aq) + 2I- (aq)
Ksp = [Pb2+][I-]^2
If the solution is saturated, that means that the product of the concentrations of Pb2+ and I- in the solution is equal to the Ksp. However, since PbI2 is a solid, its concentration remains constant and can be considered to be 1 (mol/L) (assuming the density of the solid is 1 g/cm3).
Therefore, Ksp = [Pb2+][I-]^2 = 3.0 × 10^-3 M x (x M)^2
where x is the concentration of iodide ions in the saturated solution.
Solving for x:
x = sqrt(Ksp/[I-]^2) = sqrt(1.1 x 10^-8) = 3.3 x 10^-4 M
which is equivalent to 3.0 x 10^-3 M rounded to one significant figure, the same as option B.

The Concentration of Iodide Ions in a Saturated Solution of Lead