Calculate the solubility of lead(II) iodide, PbI₂, in 0.025 M KI. K_sp = 7.9 × 10^-9 A) 4.5 × 10^-2 M B) 2.8 × 10^-2 M C) 8.9 × 10^-5 M D) 5.0 × 10^-5 M E) 1.3 × 10^-5 M

Question

Quizplus · Accepted Answer

The solubility of PbI2 in a solution containing a common ion (I-) is calculated using the Ksp expression: Ksp = [Pb2+][I-]^2. In 0.025 M KI, [I-] is 0.025 M. Solving for [Pb2+], we get [Pb2+] = Ksp / [I-]^2 = 7.9 × 10^-9 / (0.025)^2 = 1.3 × 10^-5 M.

Calculate the Solubility of Lead(II) Iodide, PbI2, in 0

Calculate the Solubility of Lead(II) Iodide, PbI₂, in 0