Given the two reactions PbCl2⇌Pb2++2Cl−, K3 = 1.85×10−10, and AgCl⇌Ag++Cl−, K4 = 1.13×10−4, what is the...

Given the two reactions PbCl2⇌Pb2++2Cl−, K3 = 1.85×10−10, and AgCl⇌Ag++Cl−, K4 = 1.13×10−4, what is the equilibrium constant Kfinal for the following reaction? PbCl2+2Ag+⇌2AgCl+Pb2+

Solutions

Expert Solution

PbCl2(s)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.85×10−10
        k3 = [Pb^+2][Cl-]^2/[pbcl2]
    AgCl(s)⇌Ag+(aq)+Cl−(aq) K4 = 1.13×10−4
       k4 = [Ag^+][Cl]/[AgCl]
       K4^2 = [Ag^+]^2[Cl]^2/[AgCl]^2
   PbCl2+2Ag+⇌2AgCl+Pb2+
       k5   = [AgCl]^2[pb^+2]/[pbcl2][Ag^+]^2

K5 = K3/K4^2

          = [Pb^+2][Cl-]^2/[pbcl2]/[Ag^+]^2[Cl]^2/[AgCl]^2
          = [AgCl]^2[pb^+2]/[pbcl2][Ag^+]^2
          = 1.85×10−10/(1.13×10−4)^2
           = 1.5*10-2

queen_honey_blossom answered 1 year ago

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