Why can tin (IV) sulfide and antimony (III) sulfide be separated from copper (II) sulfide and bismuth (III) sulfide by the addition of sodium hydroxide

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.

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)

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm