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

###### FREE Expert Solution

Initial:

[Cu2+] = 0.010 M

[Cd2+] = 0.010 M

99.9% of Cu2+ has been precipitated as CuS

Cu2+ left = 100% - 99.9%
Cu2+ left = 0.1%

[Cu2+] = (0.010 M) (0.1%)
[Cu2+] = 1.0x10-5 M

Ksp CuS = 8.5x10-45   → CuS smaller Ksp → will precipitate first
Ksp CdS = 1.0x10-28

Calculate [S2-]:

CuS(s) → Cu2+(aq) + S2-(aq)

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

x = 8.5x10-40

[S2-] = 8.5x10-40 M

Calculate minimum [S2-] to precipitate 0.010 M Cd2+:

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###### Problem Details

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