Calculate the solubility of AgBr in water at 298 K.

Question

Accepted Answer

From Table 14.1, [latex]\varepsilon ^{o,Ag}=0.7991[/latex] volts and [latex]\varepsilon ^{o,Br}=1.0652[/latex] volts. Therefore, for the reaction

[latex]Ag_{(s)}+\frac{1}{2}Br_{2(l)}=Ag^{+}_{(m)}+Br^- _{(m)} [/latex]

[latex]\Delta G^\circ _{298 K}=-f(-0.7991+1.0652)=25,775[/latex] J

For

[latex]Ag_{(s)}+\frac{1}{2}Br_{2(l)}=AgBr_{(s)} [/latex]

[latex]\Delta G^\circ _{298 K}=-95,670 [/latex] J

and hence for

[latex]AgBr_{(s)} = Ag^{+}_{(m)} + Br^{-}_{(m)}[/latex]

[latex]\Delta G^\circ _{298K}=25,675+95,670[/latex]

= 121,345 J

[latex]=-8.3144 imes 298\ln K_{sp}[/latex]

Thus, [latex] K_{sp}=(\gamma_{\pm} m_{AgBr})^2=5.4 imes 10^{-22}[/latex] or [latex] m_{AgBr}=2.3 imes 10^{-11}[/latex] , which indicates that AgBr is virtually insoluble in water.

Question 14.13: Calculate the solubility of AgBr in water at 298 K.

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