Answer:
Tin (IV) sulfide and antimony (III) sulfide be separated from copper (II) sulfide and bismuth (III) sulfide by the addition of sodium hydroxide because they become soluble whereas copper (II) sulfide and bismuth (III) sulfide remain insoluble.
Step-by-step explanation:
Sodium hydroxide is a basic solution which is used as a precipitating agent for metallic ions in the laboratory.
When a solution containing a mixture of the sulfides of the Group II cations, antimony (III), copper (II), tin (IV), and bismuth (III), is made basic by the addition of a base such as sodium hydroxide or ammonium hydroxide, the sulfide ion concentration will increase. The sulfides of antimony (III) and tin (IV) will then become soluble because antimony (III) and tin (IV) form stable complexes with sulfide, which are soluble in water, while the sulfides of copper (II) and bismuth (III) do not. The result is the dissolution of the antimony (III) sulfide and tin (IV) sulfide, separating them from the copper (II) sulfide and the bismuth (III) sulfide.
Sb₂S₃(s) + 3S²⁻(aq) ----> 2SbS₃³⁻(aq)
SnS₂(s) + S²⁻(aq) ----> SnS₃³⁻(aq)