the Ksp of Bi2s3 at room temmperature is 1.60x10-72, what is its solubility in g/100 ml? Please help me with the solution. Our teacher haven't discussed this yet so I have no idea how to solve this.. and if i ever got the answer with your solution. could you please check my answer?

Answer 1

............Bi2S3 --> 2Bi^2+ + 3S^2-
Initial.....solid......0........0
Change......solid......2x.......3x
equilibrium.solid......2x........3x

Ksp = (Bi^3+)^2(S^2-)^3
Substitute the E line into the Ksp equation and solve for x.
x will give you the solubility in mols/L. Convert to g/L by g = mols x molar mass. Then convert to g/100 mL.