Initial:

[Cu^{2+}] = 0.010 M

[Cd^{2+}] = 0.010 M

99.9% of Cu^{2+} has been precipitated as CuS

Cu^{2+} left = 100% - 99.9%

Cu^{2+} left = 0.1%

[Cu^{2+}] = (0.010 M) (0.1%)**[Cu ^{2+}] = 1.0x10^{-5} M**

K_{sp} CuS = 8.5x10^{-45} → CuS smaller K_{sp} → will precipitate first

K_{sp} CdS = 1.0x10^{-28}

**Calculate [S ^{2-}]:**

CuS(s) → Cu^{2+}(aq) + S^{2-}(aq)

${\mathbf{K}}_{\mathbf{sp}}\mathbf{=}\mathbf{\left[}{\mathbf{Cu}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}\mathbf{\left[}{\mathbf{S}}^{\mathbf{2}\mathbf{-}}\mathbf{\right]}\phantom{\rule{0ex}{0ex}}\mathbf{8}\mathbf{.}\mathbf{5}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{45}}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{)}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{8}\mathbf{.}\mathbf{5}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{45}}}{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{)}}\mathbf{=}\frac{\overline{)\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{)}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\overline{)\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{)}}}$

**x = 8.5x10 ^{-40}**

**[S ^{2-}] = 8.5x10^{-40} M**

**Calculate minimum [S ^{2-}] to precipitate 0.010 M Cd^{2+}:**

A solution is 0.010 M in both Cu^{2+} and Cd^{2+}. What percentage of Cd^{2+} remains in the solution when 99.9% of the Cu^{2+} has been precipitated as CuS by adding sulfide?

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