In qualitative analysis, the metals of Group-I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag^+ and Pb^(2+) at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl^- concentration is 0.10 M. What will the concentrations of Ag^+ and Pb^(2+) be at equilibrium ? (K_(sp) for AgCl = 1.8xx10^(-10), K_(sp) for PbCl_2=1.7xx10^(-5) )

In qualitative analysis, the metals of Group-I can be separated from o

1.	In qualitative analysis, the metals of Group-I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag^+ and Pb^(2+) at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl^- concentration is 0.10 M. What will the concentrations of Ag^+ and Pb^(2+) be at equilibrium ? (K_(sp) for AgCl = 1.8xx10^(-10), K_(sp) for PbCl_2=1.7xx10^(-5) )
Answer» `[Ag^+]=1.8xx10^(-7) M , [Pb^(+2)]=1.7xx10^(-6)` M `[Ag^+]=1.8xx10^(-11) M , [Pb^(+2)]=8.5xx10^(-5)` M `[Ag^+] = 1.8xx10^(-9)M , [Pb^(+2)]=1.7xx10^(-3)` M `[Ag^+]=1.8xx10^(-11) M, [Pb^(+2)]=8.5xx10^(-4)` M SOLUTION :`K_(sp)=[Ag^+][Cl^-]` `1.8xx10^(-10)= [Ag^+][0.1]` `[Ag^+]= 1.8xx10^(-9)` M `K_(sp)=[Pb^(+2)][Cl^-]^2` `1.7xx10^(-5) = [Pb^(+2)][0.1]^2` `[Pb^(+2)]=1.7xx10^(-3)` M