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Luisa2901:
If a solution contains 0.01 mol/L of Hg2+ and 0.01 mol/L of Bi3+ what mass of S2+ ions will be necessary to precipitate at least 99% of the bismuth present in one liter of this solution? And if the solution had lead instead of mercury, would the sulfate's mass be the same?

KspPbS=3,4x10-28
KspHgS=2×10-53
KspBi2S3=1,6X10-72

So, for letter a, I initially calculated which one of the cations would precipitate first(HgS). After that, I considered the following equilibrium:

KspHgS=[Hg2+].[S2-]
As I know the concentration of Hg2+ and the Ksp, I could determine [S2-].

I don't know what I have to do after that, could someone help me?

Borek:
Hint: the final concentration of S2- must be that required to keep the Bi3+ at 1% of 0.01 M.

And to get there you need to first precipitate all mercury and 99% of bismuth.

Luisa2901:
 I have to consider that all Hg2+ reacted with S2-

          Hg2+     +         S2-    ---->       H2S
B            0,01                  X                          0
R            -0,01               -0,01                   +0,01
E              0                    X-0,01                   0,01

So I need 0,01 mol of S2-?
And I don't quite understand the part about the 99%, is it something like  this:

KspBi2S3=[0,01x0,01]2+x[S2-]3

And I have to add the concentrations of S2- required to precipitate all mercury with the concentration that precipitates 99% of bismuth?

Borek:

--- Quote from: Luisa2901 on April 16, 2024, 05:16:26 PM ---So I need 0,01 mol of S2-?
--- End quote ---

Yes.


--- Quote ---And I don't quite understand the part about the 99%, is it something like  this:

KspBi2S3=[0,01x0,01]2+x[S2-]3
--- End quote ---

Yes - this is actually trivial when you think about it, you were told 99% was precipitated, so the final concentration of Bi3+ is 1% of the initial, just plug it into Ksp and solve for [S2-] to know the final S2- concentration.


--- Quote ---And I have to add the concentrations of S2- required to precipitate all mercury with the concentration that precipitates 99% of bismuth?

--- End quote ---

More like amounts than concentrations. That is just a simple stoichiometry, amount of S2- required for the precipitation of mercury and 99% of bismuth, assuming reactions went to completion.

But you actually need an excess, to get the concentration calculated above. Otherwise some Bi2S3 will dissolve.

Luisa2901:

--- Quote from: Borek on April 16, 2024, 05:46:16 PM ---
But you actually need an excess, to get the concentration calculated above. Otherwise some Bi2S3 will dissolve.

--- End quote ---

Could you please explain this? Do I have to calculate the excess? If yes, how do I do it?

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