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A solution is 0.0100 M in each of the metal ions in the following table:

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Metal sulfide, Kspa FeS, Kspa = 6 x 10^2 NiS, Kspa = 8 x 10^(-1) PbS, Kspa = 3 x 10^(-7) CuS, Kspa = 6 x 10^(-16) HCl is added to the solution so that the H3O concentration ...show more
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In saturated solution the ion concentrations satisfy the equilibrium
Ksp = [X²⁺] ∙[S²⁻]
with X = Fe, Ni, Pb or Cu
The sulfide is set to 0.1M. So the maximum metal ion concentration, for which no sulfide precipitates is:
[X²⁺] = Ksp/[S²⁻]
If this concentration is lower than the 0.01M, which are present in solution, the metal sulfide precipitates.

[Fe²⁺] = 6×10²/0.1 = 6×10³M > 0.01M => no precipitate
[Ni²⁺] = 8×10⁻¹ / 0.1 = 8M > 0.01M => no precipitate
[Pb²⁺] = 6×10⁻⁷/0.1 = 6×10⁻⁶M < PbS preciptates
[Cu²⁺] = 6×10⁻¹⁶/0.1 = 6×10⁻¹⁵M < 0.01M => CuS precipitates

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poster is absolutely correct :)
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A solution is 0.0100 M in each of the metal ions in the following table:
Metal sulfide, Kspa
FeS, Kspa = 6 x 10^2
NiS, Kspa = 8 x 10^(-1)
PbS, Kspa = 3 x 10^(-7)
CuS, Kspa = 6 x 10^(-16)

HCl is added to the solution so that the H3O concentration is 0.300 M. H2S gas is bubbled through the solution to attain a H2S concentration of 0.100 M. Predict which of the sulfides precipitate under the given conditions
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