[kdbmvp]

Hello,

calculate pH if [HS^-] = 10^-3 M and [S^2-] = 10^-6 M. So when HS^- dissociates, we get:

HS^- -> H⁺ + S^2-. I calculated the [H⁺] concentration to be ~1.2*10^-11. Which results in a pH of approximately 10, if you use that pH = -log[H⁺]. But does this make any sense? I mean, how can some (although very little) [H⁺] yield a basic solution? Given the very small amount of [H⁺] I'd rather think the solution is neutral..?

[Borek]

You are told what are the equilibrium concentrations, and you have to find out at which pH ratio of concentrations [HS^-]/[S^2-] is 10^-3/10^-6. HS^- is not a source of H⁺ here, pH is kept at whatever value it has by some other means (like some other buffer).

I mean, how can some (although very little) [H⁺] yield a basic solution? Given the very small amount of [H⁺] I'd rather think the solution is neutral..?

In general it just means you ignored water autodissociation. Not that it matters here, see above.

[kdbmvp]

Thanks! But didnt I find the correct H+ concentration, and correspondingly the right pH?

[AWK]

What value of pKa2 you used. Different sources give the values ​​from 12.2 to 15

[kdbmvp]

I used a pKa,2 value of 13.9.

[AWK]

And you got pH 10.9?

[kdbmvp]

Yeah. I don't quite understand how the pH can be higher than 7?

[AWK]

A concentration of HS^- in H₂S solution is only slightly greater than 10^-7 with pH slightly below 7. Your concentration indicated very diluted hydrogen sulfide salt (eg KHS with c=10^-3 M) and pH will be ~10 to 11.

[kdbmvp]

So we have a basic solution, even though we have [H+] in solution? How is that possible?

[Borek]

Yeah. I don't quite understand how the pH can be higher than 7?

I told you: pH is forced on the system by other things present in the solution.

The simplest approach here is to start with the dissociation constant and solve it for [H⁺]. You will have three knowns on the RHS and unknown on the LHS. Just plug and chug.

[kdbmvp]

Alright, thanks for the *delete me* much appreciated.

[Corribus]

Sulfide and bisulfide solutions are known to be strongly alkaline because they both react with water to produce hydroxide ions.

https://en.wikipedia.org/wiki/Sulfide
https://en.wikipedia.org/wiki/Bisulfide

Your treatment of HS- as an acid is probably why you are getting confused.