July 17, 2024, 06:04:08 AM
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Topic: solubility equilibrium  (Read 2916 times)

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Offline Luisa2901

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solubility equilibrium
« on: April 16, 2024, 04:08:20 PM »
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?

Offline Borek

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Re: solubility equilibrium
« Reply #1 on: April 16, 2024, 04:49:45 PM »
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.
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Offline Luisa2901

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Re: solubility equilibrium
« Reply #2 on: April 16, 2024, 05:16:26 PM »
 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?

Offline Borek

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Re: solubility equilibrium
« Reply #3 on: April 16, 2024, 05:46:16 PM »
So I need 0,01 mol of S2-?

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

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?

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.
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Offline Luisa2901

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Re: solubility equilibrium
« Reply #4 on: April 16, 2024, 06:18:00 PM »

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

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

Offline Borek

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Re: solubility equilibrium
« Reply #5 on: April 17, 2024, 03:22:03 AM »
It is just to prepare solution of the calculated concentration.

In the end you have a solution with a precipitate. It contains sulfide anion both in solids and in the solution itself. All you need to do now is to calculate how much S2- is in the solids, how much is in the liquid, and sum these amounts.

This excess (as compared with the amount in solids) reflects the concentration in the solution, and can be calculated from Ksp, as you know the final concentration if Bi3+.
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