   Chapter 17.6, Problem 1.3ACP

Chapter
Section
Textbook Problem

Use the formation constant of [Au(CN)2]− in Appendix K to determine the equilibrium concentration of Au+(aq) in a solution that is 0.0071 M CN‒ and 1.1 × 10−4 M [Au(CN)2]−. Is it reasonable to conclude that 100% of the gold in solution is present as the [Au(CN)2]− complex ion? Explain.

Interpretation Introduction

Interpretation:

The equilibrium concentration of Au+(aq) in solution containing 0.0071 M CN1 ions and 1.1×104[Au(CN)2] has to be calculated. Also has to be predicted if  100 % of gold in solution is present as [Au(CN)2].

Concept introduction:

Some metal ions when present in an aqueous solution containing anions or neutral species called Lewis base or ligands having a tendency to donate electron pairs to metal ions then complex ion formation will take place.

Example of metal ions that form complex ions includes,Cd2+,Fe2+,Zn2+,Ni2+ etc.

Example of Lewis bases includes,NH3,OH etc.

The complex ion remains in equilibrium with the metal ion and the ligand called complex ion formation equilibrium and the equilibrium constant is called as formation constant Kf.

A larger value of Kf implies that the complex ion formed is more stable. Kf is the measure of the strength of the interaction between the metal ions and the Lewis base to form the complex ion.

For example for general complex ion formation reaction,

xM+yL[MxLy]

Kf can be given as,

Kf=[MxLy][M]x[L]y

Here,

• [MxLy] is the equilibrium concentration of complex ion.
• [M] is the equilibrium concentration of metal ion.
• [L] is the equilibrium concentration of the ligand.
• x and y are the coefficients of metal ion and ligand respectively.
Explanation

The equilibrium concentration of Au+(aq) in solution containing 0.0071 M CN1 ions and 1.1×104[Au(CN)2] is calculated below.

Given:

Refer to the Appendix K in the textbook for the value of Kf.

The value of formation constant, Kf for [Au(CN)2] is 2.0×1038.

Concentration of [Au(CN)2] is 1.1×104 M.

Concentration of CN1 is 0.0071 M.

Au+ ions present in solution undergo complex formation with cyanide ions to form [Ag(CN)2] complex. The reaction for complex formation is given as,

Au+(aq)+2CN1(aq)Kf[Au(CN)2](aq)

The expression for Kf is given as,

Kf=[[Au(CN)2]][Au<

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