[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
CuS(s) → Cu2+(aq) + S2-(aq)
x = 8.5x10-40
[S2-] = 8.5x10-40 M
Calculate minimum [S2-] to precipitate 0.010 M Cd2+:
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?
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