Final answer:
In a solution that is 0.010 M in both Cu^2+ and Cd^2+, almost all of the original Cd^2+ remains in the solution (99.000%) after 99.9% of Cu^2+ is precipitated as CuS.
Step-by-step explanation:
The common ion effect plays a significant role in the precipitation of ions from a solution. In a solution that is 0.010 M in both Cu2+ and Cd2+, when 99.9% of Cu2+ has been precipitated as CuS by adding sulfide ions (S2-), very little Cd2+ will be precipitated due to the much higher solubility product of CdS (Ksp = 1.0 × 10-28 for CdS) compared to that of CuS.
After the removal of 99.9% of Cu2+, essentially all of the original Cd2+ remains in the solution, or 99.000%, as Cu2+ precipitates out in a much more significant amount due to its lower Ksp value and its reactivity with S2-.