In qualitative analysis Cu^(2+), Cd^(2+) and Pb^(2+) ions are precipate out as sulphide of 2^(nd) group an Ni^(2+) , Zn^(2+) , Mn^(2+) are precipate out as sulphide group and of 3^(rd) B group. In this analysis the reactant HCI + H_2S and NH_4Cl + NH_4OH + H_2O successively added. What is the reason for that ?

In qualitative analysis Cu^(2+), Cd^(2+) and Pb^(2+) ions are precipat

1.	In qualitative analysis Cu^(2+), Cd^(2+) and Pb^(2+) ions are precipate out as sulphide of 2^(nd) group an Ni^(2+) , Zn^(2+) , Mn^(2+) are precipate out as sulphide group and of 3^(rd) B group. In this analysis the reactant HCI + H_2S and NH_4Cl + NH_4OH + H_2O successively added. What is the reason for that ?
Answer» Solution :In SECOND group `K_(SP)` is very LESS so `CU^(2+), Cd^2, Pb^(2+)` form `PP^+` with `S^(-2)`. Because of common ion `[H^+]` to `[S^(2-)]`concentration decreases so ppt of 3rd group will not form and only 2nd group precipitant.