A solution which is 10^(-3) M each in Mn^(2+),Fe^(2+),Zn^(2+)andHg^(2+) is treated with 10^(-16)M sulphide ion. If K_(sp) od MnS, ZnS and HgS are 10^(-15),10^(-25),10^(-20)and10^(-54) respectively, which one will precipitate first ?

A solution which is 10^(-3) M each in Mn^(2+),Fe^(2+),Zn^(2+)andHg^(2+

1.	A solution which is 10^(-3) M each in Mn^(2+),Fe^(2+),Zn^(2+)andHg^(2+) is treated with 10^(-16)M sulphide ion. If K_(sp) od MnS, ZnS and HgS are 10^(-15),10^(-25),10^(-20)and10^(-54) respectively, which one will precipitate first ?
Answer» FeS MnS HgS ZnS Solution :Ionic product in the solution `=10^(-3)xx10^(16)=10^(-19)` The metal sulphide having the lowest solubility will precipitate FIRST provided the ionic product is higher than the `K_(SP)`. Here, all salts are of the same valence-type. So the sulphide having the lowest `K_(sp)` value will precipitate first provided `K_(sp)lt10^(-19)`, HgS has lowest `K_(sp)` value `(10^(-54))`, so will precipitate first.