1. Cadmium(II) ion forms a 4:1 (4 ligand: 1 metal) complex with cyanide ion CN. The equation is Cd + 4 CN <- Cd(CN) Ki for this reaction is 1.29 x 10". What is the concentration of free (not complexed) Cd ion at equilibrium in a solution prepared from 0.0400M CN and 3.7 x 10 M Cd(II) ion? (Hint: ICE tablel)

1. Cadmium(II) ion forms a 4:1 (4 ligand: 1 metal) complex with cyanide ion CN. The equation is Cd + 4 CN <- Cd(CN) Ki for this reaction is 1.29 x 10". What is the concentration of free (not complexed) Cd ion at equilibrium in a solution prepared from 0.0400M CN and 3.7 x 10 M Cd(II) ion? (Hint: ICE tablel)

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter20: Chemistry Of Selected Transition Elements And Coordination Compounds

Section: Chapter Questions

Problem 73QRT

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