Respuesta :
Answer:
The equilibrium constant of the given reaction is 0.01351.
Explanation:
[tex]PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)[/tex]
The equilibrium constant of the reaction = [tex]K_3=1.85\times 10^{-10}[/tex]
[tex]K_3=\frac{[Pb^{2+}][Cl^-]^2}{[PbCl_2]}[/tex]...[1]
[tex]AgCl(aq)\rightleftharpoons Ag^{+}(aq)+Cl^-(aq)[/tex]
The equilibrium constant of the reaction = [tex]K_4=1.17\times 10^{-4}[/tex]
[tex]K_4=\frac{[Ag^+][Cl^-]}{[AgCl]}[/tex]..[2]
[tex][Cl^-]=\frac{K_4\times [AgCl]}{[Ag^+]}[/tex]
[tex]PbCl_2(aq)+2Ag^+(aq)\rightleftharpoons 2AgCl(aq)+Pb^{2+}(aq)[/tex]
The expression of equilibrium constant of the creation is ;
[tex]K=\frac{[AgCl]^2[Pb^{2}]}{[PbCl_2]][Ag^+]^2}[/tex]
Dividing [1] by [2]
[tex]\frac{K_3}{K_4}=\frac{\frac{[Pb^{2+}][Cl^-]^2}{[PbCl_2]}}{\frac{[Ag^+][Cl^-]}{[AgCl]}}[/tex]
[tex]\frac{K_3}{K_4}=\frac{[Pb^{2+}][Cl^-][AgCl]}{[PbCl_2][Ag^+]}[/tex]
Substituting the value of [tex][Cl^-][/tex] from [2] :
[tex]\frac{K_3}{K_4}=\frac{[Pb^{2+}][AgCl]}{[PbCl_2][Ag^+]}\times \frac{K_4\times [AgCl]}{[Ag^+]}[/tex]
[tex]\frac{K_3}{K_4}=K_4\times K[/tex]
[tex]K=\frac{K_3}{(K_4)^2}=\frac{1.85\times 10^{-10}}{(1.17\times 10^{-4})^2}[/tex]
[tex]K=0.01351[/tex]
The equilibrium constant of the given reaction is 0.01351.
