The solubility products for a series of iodides are CuI K = 1 x 10-12 AgI Ksp

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The solubility products for a series of iodides are
CuI K = 1 x 10-12
AgI Ksp = 8.3 x 10-17
PbI2 Ksp = 7.1 x 10-9
BiI3 Ksp = 8.1 x 10-19
List these four compounds in order of decreasing molar solubility in
(a) Water.
(b) 0.20 M NaI.

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Fundamentals of Analytical Chemistry

ISBN: 978-0495558286

9th edition

Authors: Douglas A. Skoog, Donald M. West, F. James Holler, Stanley R. Crouch

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Question Posted: Apr 30, 2016 02:12 PM

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The solubility products for a series of iodides are CuI K = 1 x 10-12 AgI Ksp

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CuIs Cu I K sp Cu I 1 10 12 AgIs Ag I K sp Ag I 83 10 17 PbI 2 s Pb 2 ...View the full answer

Fundamentals of Analytical Chemistry

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