In the presence of CN", Fe3+ forms the complex ion Fe(CN)63- Theequilibrium concentrations of Fe3+ and Fe(CN)63- are 8.5 × 10¬40 M and1.4 x 10-3 M, respectively, in a 0.12 M KCN solution. Calculate the valuefor the overall formation constant of Fe(CN)63-.Fe"(aq) + 6CN (aq) → Fe(CN)63-K =

Given :- [Fe3+] = 8.5 × 10-40 M [Fe(CN)6]3- = 1.4 × 10-3 M [CN]- = 0.12M

In the presence of CN", Fe3+ forms the complex ion Fe(CN)63-. The equilibrium concentrations of Fe3+ and Fe(CN)63- are 8.5 x 10-40 M and 1.4 x 10-3 M, respectively, in a 0.12 M KCN solution. Calculate the value for the overall formation constant of Fe(CN)63-. Fe3*(aq) + 6CN (aq) → Fe(CN)63- K=

In the presence of CN", Fe3+ forms the complex ion Fe(CN)63-. The equilibrium concentrations of Fe3+ and Fe(CN)63- are 8.5 x 10-40 M and 1.4 x 10-3 M, respectively, in a 0.12 M KCN solution. Calculate the value for the overall formation constant of Fe(CN)63-. Fe3*(aq) + 6CN (aq) → Fe(CN)63- K=

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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In the presence of CN", Fe3+ forms the complex ion Fe(CN)63- The
equilibrium concentrations of Fe3+ and Fe(CN)63- are 8.5 × 10¬40 M and
1.4 x 10-3 M, respectively, in a 0.12 M KCN solution. Calculate the value
for the overall formation constant of Fe(CN)63-.
Fe"(aq) + 6CN (aq) → Fe(CN)63-
K =

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