There is mixture of Cu(II) chloride and Fe(II) sulphate. The best way to separate the metal ions fro — Qualitative and Quantitative Analysis Chemistry Question

💡 Solution & Explanation

Step 1: Cu2+ belongs to analytical Group II, while Fe2+ belongs to Group III/IV. Group II cations are selectively precipitated as sulfides in an acidic medium. Step 2: Passing H2S gas through a solution acidified with dil. HCl suppresses the sulfide ion concentration (common ion effect of H+), keeping it high enough to precipitate only the highly insoluble CuS (Ksp ~ 10^-36) while leaving Fe2+ in solution. Step 3: Thus, passing H2S in an acidic medium selectively precipitates copper(II) sulfide, separating it from Fe(II), corresponding to option (a).