what is the concentration of lead ions and sulfide ions in a saturated solution of lead sulfide (PbS) solution at 25 degrees celsius.

PbS ==> Pb^+2 + S^-2

Ksp = (Pb^+2)(S^-2)
Set up an ICE chart, substitute into the Ksp expression and solve. Post your work if you get stuck.

To determine the concentration of lead ions (Pb2+) and sulfide ions (S2-) in a saturated solution of lead sulfide (PbS) at 25 degrees Celsius, we need to refer to the solubility product expression (Ksp) for PbS.

The solubility product expression for PbS is given by:

Ksp = [Pb2+][S2-]

At the saturation point, the concentration of Pb2+ and S2- ions will be equal, as PbS dissociates completely.

Therefore, the concentration of lead ions and sulfide ions in a saturated solution of PbS will be equal, represented by x.

Hence, the concentration of Pb2+ and S2- ions in a saturated solution of PbS at 25 degrees Celsius is x.