Calculation of the Equilibrium Constant from Standard Half-Cell Potentials

Determine the equilibrium constant for the dissolution and dissociation of silver chloride in water, and the silver chloride solubility in water.

Question

Calculation of the Equilibrium Constant from Standard Half-Cell Potentials

Determine the equilibrium constant for the dissolution and dissociation of silver chloride in water, and the silver chloride solubility in water.

Accepted Answer

The reaction is

[latex]AgCl ightarrow Ag ^{+}+ Cl ^{-}[/latex]

which in terms of half-cell reactions we write as

[latex]AgCl +e^{-} ightarrow Ag + Cl ^{-} \quad ext { half-cell standard potential }=+0.22 V[/latex]

[latex]Ag \quad ightarrow Ag ^{+}+e^{-} \quad ext { half-cell standard potential }=-(+0.80 V )[/latex]

[latex] ext { Thus } E^{\circ}=+0.22-(+0.80)=-0.58 V ext {, and }[/latex]

[latex]\ln K_{a}=\frac{n F E^{\circ}}{R T}=\frac{9.6485 imes 10^{4} \frac{ C }{ mol } imes(-0.58) V }{8.314 \frac{ J }{ mol K } imes 298.15 K imes 1 \frac{ C V }{ mol }}=-22.568[/latex]

or

[latex]K_{a}=1.58 imes 10^{-10}[/latex]

which compares well with the value of [latex]1.607 imes 10^{-10}[/latex] computed in Illustration 13.3-2. The equilibrium relation is then

[latex]K_{a}=1.58 imes 10^{-10}=\frac{a_{ Ag ^{+}} a_{ Cl ^{-}}}{a_{ AgCl }}=M_{ Ag +M_{ Cl ^{-}}}=\left(M_{ Ag ^{+}} ight)^{2}[/latex]

so that

[latex]M_{ Ag ^{+}}=M_{ AgCl }=1.257 imes 10^{-5} M =1.257 imes 10^{-5} \frac{ mol }{ kg ext { water }}[/latex]

In writing these last equations we have recognized that the activity of the pure silver chloride solid is unity, assumed that the ion concentrations will be so low that the activity coefficients would be unity, and used the fact that by stoichiometry the concentrations of the silver and chloride ions must be equal.

Question 14.6.2: Calculation of the Equilibrium Constant from Standard Half-C...

Related Answered Questions