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 ?

Explanation

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.