For a solution in which μ = 7.0 x 10-2, calculate K'sp fora. AgSCN(Ksp (AgSCN) = 1.1 × 10-¹2, Ag+ = 0.25, ascN = 0.35)K's =spb. Pbl₂(Ksp (PbI2) = 7.9 x 10-9, apb²+ = 0.45, ar =0.3)K'sp =c. La(103) 3(Ksp (La(103) 3) = 1.0 x 10-¹¹, L²+ = 0.9, 10,- = 0.35)K'sp=

The solubility product constant (Ksp) is related to the chemical potential (μ) and the activity…

(Ksp (AgSCN) = 1.1 x 107 K'sp = " = Ag+ = 0.25, ascN = 0.35 b. PbI2 (Ksp (Pbl₂) = 7.9 x 10-9, apb²+ = ( K'sp = c. La(103)3 (Ksp (La(103) 3) = 1.0 x 10-¹1, a¹ = 0.9, α10,- = 0.35 K'sp , apb²+ = 0.45, ar = 0.3)

(Ksp (AgSCN) = 1.1 x 107 K'sp = " = Ag+ = 0.25, ascN = 0.35 b. PbI2 (Ksp (Pbl₂) = 7.9 x 10-9, apb²+ = ( K'sp = c. La(103)3 (Ksp (La(103) 3) = 1.0 x 10-¹1, a¹ = 0.9, α10,- = 0.35 K'sp , apb²+ = 0.45, ar = 0.3)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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