There is mixture of Cu(II) chloride and Fe(II) sulphate. The best way to separate the metal ions from the mixture in qualitative analysis is:

Options

A. hydrogen sulphide in acidic medium, where only Cu(II) sulphide will be precipitated
B. ammonium hydroxide buffer, where only Fe(II) hydroxide will be precipitated
C. hydrogen sulphide in acidic medium, where only Fe(II) sulphide will be precipitated
D. ammonium hydroxide buffer, where only Cu(II) hydroxide will be precipitated

Answer

A

Explanation

`Cu^(2+)` is second group radical, gets precipitated first due to having lowerr solubility product `[CuS-K_(sp)=1x10^(-44)]`