Chapter 16, Problem 63E

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

Chapter
Section

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

# 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 1.5 × 10−3 M, respectively, in a 0.11-M KCN solution. Calculate the value for the overall formation constant of Fe(CN)63−. Fe 3+ ( a q ) ​   +   6 CN − ( a q )   ⇌   F e ( C N ) 6 3 − ( a q )   K overall   =   ?

Interpretation Introduction

Interpretation: The equilibrium concentration of Fe3+ and Fe(CN)63 in a solution of 0.11M KCN is given. By using these values, the overall formation constant of Fe(CN)63 is to be determined.

Concept introduction: The equilibrium constant for the formation of complex is known as formation constant.

The expression for the equilibrium constant is,

K=ConcentrationofproductsConcentrationofreactants

Explanation

Explanation

To determine: The overall formation constant of Fe(CN)63 for the given reaction.

The reaction for the formation of Fe(CN)63 is,

Fe3+(aq)+6CN(aq)Fe(CN)63(aq)

The overall formation constant of Fe(CN)63 for the given reaction is 1.0×1042_ .Given

The equilibrium concentration of Fe3+ is 8.5×1040M .

The equilibrium concentration of Fe(CN)63 is 1.5×103M .

Concentration of KCN is 0.11M .

The reaction that takes place is,

Fe3+(aq)+6CN(aq)Fe(CN)63(aq)

It is assumed that the concentration of CN is same as the concentration of KCN . Hence, concentration of CN is,

[CN]=0

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