Calculate the concentration of chloride ions in a saturated lead(II)chloride (K_sp = 2.4 × 10^-4)solution.

Question

Quizplus · Accepted Answer

The solubility product expression for lead(II) chloride is:

PbCl2 ⇋ Pb2+ + 2 Cl-

Ksp = [Pb2+][Cl-]2

At saturation, the concentrations of Pb2+ and Cl- are equal to the solubility product. Therefore:

Ksp = [Pb2+] [Cl-]2

Solving for [Cl-], we have:

[Cl-] = √(Ksp/[Pb2+])

Substituting the given values, we get:

[Cl-] = √(2.4 × 10-4/1) = 7.8 × 10-2 M = 7.8 × 10-2 mol/L

Therefore, the concentration of chloride ions in the saturated solution of lead(II) chloride is 7.8 × 10-2 M or 7.8 × 10-2 mol/L, which is answer choice E.

Calculate the Concentration of Chloride Ions in a Saturated Lead(II)chloride