The concentration of iodide ions in a saturated solution of lead (II)iodide is ________ M. The solubility product constant of PbI₂ is 1.4 × 10^-8. 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

For the equilibrium reaction: 
PbI2 ⇌ Pb2+ + 2I-
The solubility product constant expression is: 
Ksp = [Pb2+][I-]^2
Since PbI2 is a 1:2 electrolyte, at equilibrium, [Pb2+] = [I-]/2. 
Substituting this into the expression for Ksp: 
Ksp = (1/2)[I-]^3
Rearranging the equation: 
[I-] = (2Ksp)1/3
Substituting the given value of Ksp: 
[I-] = (2 × 1.4 × 10^-8)^1/3 = 3.0 × 10^-3 M

Therefore, the concentration of iodide ions in a saturated solution of lead (II)iodide is 3.0 × 10^-3 M which corresponds to choice B.

The Concentration of Iodide Ions in a Saturated Solution of Lead