A solution is 0.010 M in both Cu²⁺and Cd²⁺. What percentage ofCd²⁺remains in the solution when 99.9% of the Cu²⁺ has been precipitated as CuS by adding sulfide?

Sulfide is an inorganic anion of sulfur with the chemical formula S^2- or a compound containing one or more S²⁻ ions. Solutions of sulfide salts are corrosive. Sulfide also refers to chemical compounds large families of inorganic and organic compounds, e.g. lead sulfide and dimethyl sulfide.

If99.9% of the Cu²⁺is precipitated, then 0.1% remains in the solution

\begin{aligned}{\left[{{\rm{C}}{{\rm{u}}^{2+}}}\right]=\frac{{0.1}}{{100}}\times 0.01{\rm{mol}}/L=1\times{{10}^{-5}}M}\\{{K_{sp}}=\left[{C{u^{2+}}}\right]\left[{{S^{2-}}}\right]=6.7\times{{10}^{-42}}}\\{\left[{{S^{2-}}}\right]=7\times{{10}^{-37}}M}\end{aligned}

\begin{aligned}{{K_{sp}}=\left[{C{d^{2+}}}\right]\left[{{S^{2-}}}\right]=\left[{C{d^{2+}}}\right]\times7\times{{10}^{-37}}=2.8\times{{10}^{-35}}}\\{\left[{C{d^{2+}}}\right]=40M}\end{aligned}

If the concentration ofCd²⁺ before precipitation is 40Mand after equals to 0.01M 100 ofCd²⁺ is dissolved