[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⁺].[S^2-]
As I know the concentration of Hg2+ and the Ksp, I could determine [S^2-].

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

[Borek]

Hint: the final concentration of S^2- 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 S^2-

          Hg²⁺     +         S^2-    ---->       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 S^2-?
And I don't quite understand the part about the 99%, is it something like  this:

KspBi₂S₃=[0,01x0,01]²⁺x[S^2-]³

And I have to add the concentrations of S^2- required to precipitate all mercury with the concentration that precipitates 99% of bismuth?

[Borek]

So I need 0,01 mol of S^2-?

Yes.

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

KspBi₂S₃=[0,01x0,01]²⁺x[S^2-]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 [S^2-] to know the final S^2- concentration.

And I have to add the concentrations of S^2- 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 S^2- 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]

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?

[Borek]

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 S^2- 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⁺.