# Show the acid nature of hydrogen sulfide

We are studying the chalcogens at school. I should prove (show) the acid nature of $$\ce{H2S}$$ by writing two chemical interactions.

I am really sorry if the terms I am using aren't the right ones. It's the first time I've ever used this part of the language. Which reactions should I write, and why? I think the first one must be neutralization:

$$\ce{H2S + 2 NaOH -> Na2S + 2 H2O}$$

• How about the reaction with sodium carbonate? – Waylander Nov 10 '19 at 12:58
• @Waylander, thank you for the response! I haven't seen this reaction in school, but it might be the reaction I am searching for. What are the reasons you wrote it? I mean what does it show? – LYI Nov 10 '19 at 13:04
• Do you know how an (Arrhenius) acid is defined? That may also suggest an equation which can illustrate the acidic behaviour of H2S. – orthocresol Nov 10 '19 at 13:10
• It will show that H2S will give a proton to carbonate, a reaction that is consistent with an acidic nature – Waylander Nov 10 '19 at 13:18
• I suggest you do. It is difficult to describe something if you don't know what it is. – orthocresol Nov 10 '19 at 13:37

As orthocresol suggested, to illustrate "classic" acidity as it's understood on a school level you unquestionably want to start with Arrhenius definition, which attributes acidic properties to any substance which increase concentration of hydronium $$\ce{H3O+}$$ ions in water.

That's said, the very first reaction you probably want to write is dissociation of $$\ce{H2S}$$ in water and demonstrating that it is a weak acid by showing the corresponding acid dissociation constants $$K_\mathrm{a}$$ (the larger the $$\mathrm{p}K_\mathrm{a},$$ the weaker the acid; data from [1, p. 5-87]):

\begin{align} \ce{H2S + H2O &<=> HS- + H3O+} &\quad \mathrm{p}K_\mathrm{a1} &= 7.05\\ \ce{HS- + H2O &<=> S^2- + H3O+} &\quad \mathrm{p}K_\mathrm{a2} &= 19 \end{align}

Some reactions illustrating acidic properties of $$\ce{H2S}$$:

### Reaction with sodium hydroxide

Note that the reaction

$$\ce{H2S(aq) + 2 NaOH(aq) -> Na2S(aq) + 2 H2O(l)}$$

is for complete neutralization (concentrated $$\ce{NaOH}$$ solution). If the reactants are taken in 1:1 ratio (diluted $$\ce{NaOH}$$ solution), sodium hydrosulfide is formed:

$$\ce{H2S(aq) + NaOH(aq) -> NaHS(aq) + H2O(l)}$$

### Reaction with sodium carbonate

Sodium hydrosulfide is also formed when saturated $$\ce{H2S}$$ solution reacts with $$\ce{Na2CO3}$$:

$$\ce{H2S(aq) + Na2CO3(aq) -> NaHS(aq) + NaHCO3(aq)}$$

### Reaction with ammonia

Reaction with aqueous ammonia solution yields exclusively in ammonium hydrogen sulfide $$\ce{NH4HS},$$ even when concentrated ammonia solution is used:

$$\ce{H2S(aq) + NH4OH(aq) -> NH4HS(aq) + H2O(l)}$$

Ammonium sulfide $$\ce{(NH4)2S}$$ doesn't form as ammonium $$\ce{NH4+}$$ $$(\mathrm{p}K_\mathrm{a} = 9.25)$$ is a much stronger acid than hydrosulfide $$\ce{HS-}$$ by a factor of about $$10^{19 - 9.25}\approx\pu{6E9}.$$

So far the reactions between well-soluble compounds that were carried in aqueous solutions where Arrhenius definition works fine. However, it cannot be used to explain acidity in non-aqueous media, where a Brønsted–Lowry definition of an acid comes in handy: the acid is a proton donor. This can be illustrated by the reaction with liquid ammonia:

$$\ce{H2S(g) + 2 NH3(l) ->[\pu{-40 °C}] (NH4)2S(s)}$$

### References

1. Haynes, W. M.; Lide, D. R.; Bruno, T. J., Eds, CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data; 97th Edn.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016-2017. ISBN 978-1-4987-5429-3.