Example The solubility of AgCl is found experimentally to be 1.92X10-3 gL at 25C. Calculate the value of K for AgCl. Answer:We know M(AgCl)=143.3 S=1.92x103 mol.L=1.34X10mol.L 143.3 AgCl(s)=Ag"(aq)+Cl (ag) Equilibrium mol.L-1 S S K9(AgC)={c(Ag)}{c(CI)}=S2=1.80X10-10Example ：The solubility of AgCl is found experimentally to be 1.92×10-3 g·L-1 at 25oC. Calculate the value of for AgCl. Equilibrium / mol.L-1 SS M 3.143 r Answer：We know (AgCl)= 1 5 1 3 mol .1L 1034 mol L 3.143 .1 1092 − − − − S = × =⋅ × ⋅ AgCl(s) Ag (aq) Cl (aq) + − + 2 10 (AgCl ({) Ag )}{ (Cl )} .1 1080 + − − Ksp = Scc == × Ksp