JEE MainChemistryIonic Equilibrium
The solubility product of Bismuth(III) sulfide is K . If its molar mass is M , its solubility in g/L is:
Options
- AM ( K 72 )^ 1/5
- BM ( K 108 )^ 1/5
- C1 M ( K 108 )^ 1/5
- DM K^ 1/5
Correct answer
B. M ( K 108 )^ 1/5
Step-by-step solution
The chemical formula of Bismuth(III) sulfide is Bi ₂ S ₃ . The dissociation equilibrium is: Bi ₂ S ₃ 2 Bi ³⁺ + 3 S ²⁻ Let the molar solubility of Bi ₂ S ₃ be S mol/L . At equilibrium, [ Bi ³⁺] = 2S and [ S ²⁻] = 3S . The solubility product K is given by: K = [ Bi ³⁺]^2 [ S ²⁻]^3 K = (2S)^2 (3S)^3 = 4S^2 27S^3 = 108S^5 Solving for molar solubility S : S = ( K 108 )^ 1/5 mol/L To find the solubility in g/L , we multiply the molar solubility by the molar mass M : Solubility in g/L = S M = M ( K 108 )^ 1/5 Answer: M (