Equilibrium question, IMAT 2022

Hydrogen sulfide gas was shaken with a solution of sodium hydroxide in a closed system. An equilibrium was rapidly established.
H2S(g) + 2OH–(aq) ⇌ S2–(aq) + 2H2O(l)
The experiment was repeated at the same temperature and pressure, but one additional
substance was added to the equilibrium mixture.
Adding which of these substances, on its own, will increase the number of molecules of H2S
present at equilibrium?

1 hydrochloric acid
2 solid potassium hydroxide
3 sodium sulfide

A. 1 and 3 only
B. 1 only
C. 2 only
D. 1 and 2 only
E. 2 and 3 only

My reasoning here was that HCl would act as a product if added, and will shift the equilibrium to the reactant’s side, so it is correct. KOH will act as a reactant, so it would do the opposite, hence it is not correct. However I couldn’t figure out if Sodium Sulfide, a salt, would be a reactant or a product. Is my method wrong? Should this question be solved in another way? Answer is A

Sodium sulphide is made of cation from NaOH and anion from H2S which basically means it’s a salt of weak acid but strong base and thus a strong electrolyte which dissociates to give sulphide ions and shift reaction backwards.

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Can you elaborate a bit? A salt made up of a strong base but a weak acid would dissociate to form S2- but not Na+? If Na+ is formed as well, it would react with OH- right? Which would shift eq to the right since there will be more products? Where am I wrong here?

Yes Na+ will also be there as a result of the strong electrolyte salt. And since Na+ if it reacts with OH- the OH- concentration DECREASES in the reactants and thus the reaction moves backward to replenish it.

sodium sulfide is Na2S
when it dissolves it release Na+ and S2-
so the S2- will react with the S2- we all ready have in the products
the c will be increased and so the equilibrium will go on the other side(left) to replenish
thats what i think,if im wrong correct me:)


HCl will dissociate in the water and the H+ ions will react with S2-, favoring the L<–R reaction and the production of H2S.

Na2S will also dissociate in water and produce S2- ions which according to the Le Chatelier’s principle is consumed by favoring the L<–R reaction and thus production of H2S.


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