[Anagkai]

I got an exercise where I need to calculate the pH of H2S in an alkaline solution.
The pKa values given for the two deprotonations are 7 and 12.9. So I made two equations for those, then divided by the equation for the autoprotolysis to get the reactions of h2s and hs- with oh.
But it gave me a nonlinear equation system I couldn't solve and that isn't probably solvable analytically at all.
Any tips.
The exercise is to calculate the ph of a mixture of 100mL of 10^-2 mol/L H2S (I suppose this means that it's entirely dissolved - no need to calculate how many gas goes into the water) and 10 mL 10^-1mol/L NaOH.

[Borek]

How many moles of H₂S? How many moles of NaOH? Assuming the neutralization went to completion - what is present in the solution? This is a pretty specific case.

Assuming NaOH is a strong base you should get 4^th degree polynomial, these are solvable analytically. Problems start with 5^th degree polynomials. See https://en.wikipedia.org/wiki/Abel–Ruffini_theorem

[Anagkai]

There's the same concentration of OH- and H2S at the beginning, so I can make it all HS- and start from there. There must be a way with at most second degree. I know about Cardano, but they tell us all exercises we get are solvable with second degree equations.

I think I might have found a solution: I start from HS- and use the reactions in both directions (H3O+ and OH-). This gives m two constants that are 1/pka1 and pka2/kap (kap meaning autoprotolysis constant)
Then I use c1/c2 and c1*c2 which gives me two equations. I multiply the first with kap again which makes it second degree and the other equation can easily give one variable in dependence of the other, which I can use to solve the second degree equation.

[Borek]

Actually the answer doesn't require even the quadratic.

http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt

[Anagkai]

Actually I don't have a salt and neither do I have only HS- at the beginning which would be equivalent to a salt.
When I calculate the pH without adding OH-, I can neglect the second protolysis because there's a factor of more than ten to the four between the pka values, but still I have H2S and HS- at the same time.

[Borek]

Your solution is not different from the one prepared by dissolving a salt, which is why you can treat it as being HS^- only initially. Doesn't matter how you prepared a solution of a given composition - every solution with identical composition has the same pH.

In the derivation both hydrolysis and dissociation are taken care of.

[Anagkai]

I did know that identical solutions had the same pH, but I didn't know that this meant, I could use SH- to start.
What do I make of the  "source of the ions" part? Can I leave it out because I don't have a salt?

[Borek]

I did know that identical solutions had the same pH, but I didn't know that this meant, I could use SH- to start.

What if you mix equimolar amounts of NaOH and H₂S, dry the solution out, and dissolve the NaHS back?

What do I make of the  "source of the ions" part? Can I leave it out because I don't have a salt?

Please elaborate, no idea what you refer to.