[Borek]

charge balance:
[Cl^-] + [S^2-] + [HS^-] = [H⁺] + [Pb²⁺]

I am not checking anything else - but think it over... Especially S^2- and Pb²⁺ part.

Also note that in general you have omitted one ion (and one equilibrium). As this ion is in minute concentrations at pH 1 and is not taking part in any of the equilbria important for your calculations, that's OK. This ion (plus one other) has been omitted in the previous question for the similar reasons.

[zephyrblows]

charge balance:
[Cl^-] + [S^2-] + [HS^-] = [H⁺] + [Pb²⁺]

I am not checking anything else - but think it over... Especially S^2- and Pb²⁺ part.

Ow...then I must have had some wrong conceptions...
If I put all the cations in one side of charge-balance equation and anions in the other, will something go wrong?

[sdekivit]

well the method of Borek has the same maths as the method i use, only i substitue other things to become the same equation. Both ways are correct as stated by borek and it's up to you which method you prefer. In any case, the solubility of PbS in 1 M HCl is indeed 8,9 x 10^-4 M. I'll write my calculation here so you can see how it can be calculated:

8 x 10^-28 = [Pb²⁺] [S^2-]

Then for the other equilibria where S^2- is involved:

10¹⁴ = [HS^-] / ( [S^2-][H⁺] )

10⁷ = [H₂S] / ( [HS^-] [H⁺] )

Because in the first equation we want to express [S^2-] in terms of [H₂S], we use the above 2 equations:

[S^2-] = [HS^-] / ( [H⁺]*10¹⁴ )

And also:

[HS^-] = [H₂S] / ( [H⁺] * 10⁷ )

Now substitue [HS^-] in the expression of [S^2-] to become:

[S^2-] = [H₂S] / ( [H⁺]² * 1 x 10²¹ )

This can be plugged in into the solubility product of PbS in water to become:

8 x 10^-28 = [Pb²⁺] * [H₂S] / ( [H⁺]² * 1 x 10²¹ )

Now we have our equation and an equilibrium:

PbS + 2 H⁺ <--> Pb²⁺ + H₂S

When dissolving x mol in 1 L 1 M HCl, we gain x mol/L Pb²⁺ and H₂S. The concentration H⁺ will be 1 - 2x M

Plug these into our equation we get exactly the same as with the other question:

8 x 10^-7 = x² / (1-2x)²

Solve for x to find the solubility of 8,9 x 10^-4 M

[Borek]

If I put all the cations in one side of charge-balance equation and anions in the other, will something go wrong?

That's OK, but you have to account for the fact that some ions have different charges. Think in terms of CaCl₂ solution. Is charge balance

[Ca²⁺] = [Cl^-]

or

2[Ca²⁺] = [Cl^-] ?

[zephyrblows]

Aw, right! I've totally missed that!!

But I don't understand what you said:

Also note that in general you have omitted one ion (and one equilibrium). As this ion is in minute concentrations at pH 1 and is not taking part in any of the equilbria important for your calculations, that's OK. This ion (plus one other) has been omitted in the previous question for the similar reasons.

You mean, I don't need to put [Cl-] in the equations? Without [Cl^-], how should I derive the charge equation?

[Borek]

You mean, I don't need to put [Cl-] in the equations?

No. OH^- and K_w.

[zephyrblows]

Borek, sorry but I can't understand it intuitively ...
If the charge of Cl^- is neglected, this solution no more holds a neutral charge...
So, I shouldn't use "putting all the cations in one side of charge-balance equation and anions in the other"...

And I don't know what to do now...

All I can write down is:
PbS <-> Pb²⁺ + S^2- (8*10^-28)
H⁺ + S^2- <-> HS^-
H⁺ + HS^- <-> H₂S
H2O <->H⁺ + OH^-

mass: [OH^-] + [H⁺]₀ = [H⁺] + [HS^-] + [H₂S]
charge:() [H⁺] + 2[Pb²⁺] = [OH^-] + [HS^-] + 2[H₂S]

I know there must be something wrong...Is there another method to build charge equation without considering the net charge of the system?

[Borek]

Wipe everything about Cl^-, I have not said a word about it.

I was referring JUST to the fact, that every water solution contains both H⁺ and OH^- and their concentrations are connected through K_w=[H⁺][OH^-]. No idea where did you get this Cl^- thing - it must be some your misconception that I am not able to understand.

mass: [OH^-] + [H⁺]₀ = [H⁺] + [HS^-] + [H₂S]

These should be separate mass balances for different substances. Like:

[Pb²⁺] = [S^2-] + [HS^-] + [H₂S]

which reflects the fact that there is exactly the same amount of sulfide as lead in the solution.

charge:() [H⁺] + 2[Pb²⁺] = [OH^-] + [HS^-] + 2[H₂S]

In the charge balance you should use ONLY charged particles. Or was 2[H₂S] just a typo? And Cl^- must be included:

[H⁺] + 2[Pb²⁺] = [OH^-] + [HS^-] + 2[S^2-] + [Cl^-]

Now, what you can safely omit here is OH^-. Its concentration is much lower that that of the other things and it doesn't play any role in the whole system that we are interested in.

[zephyrblows]

No wonder...finally I get what you mean.
Sorry, my English is not good enough, so I misunderstood some of your words.
It's not so difficult as I thought.
Thanks a lot !!