Hydrogen sulfide gas is bubbled through an excess of potassium hydroxide solution. H2S + 2OH- → S2- + 2H2O Write the balanced chemical equation excluding spectator ions. How do you get there from H2S + KOH -> K2S + H2O?
The correct Answer and Explanation is:
The correct balanced chemical equation, excluding spectator ions, for the reaction of hydrogen sulfide (H₂S) with an excess of potassium hydroxide (KOH) is:
H₂S + 2OH⁻ → S²⁻ + 2H₂O
Explanation:
- Understanding the Reactants:
- Hydrogen sulfide (H₂S) is a weak diprotic acid.
- Potassium hydroxide (KOH) is a strong base that dissociates completely in aqueous solution into potassium ions (K⁺) and hydroxide ions (OH⁻).
- Ionic Perspective of KOH:
- Since KOH is a strong electrolyte, it fully dissociates in solution:
KOH→K++OH−KOH \rightarrow K⁺ + OH⁻
- When KOH is in excess, it provides a surplus of hydroxide ions.
- Acid-Base Neutralization:
- H₂S releases hydrogen ions (H⁺) in two steps due to its diprotic nature:
H2S⇌HS−+H+H₂S \rightleftharpoons HS⁻ + H⁺
HS−⇌S2−+H+HS⁻ \rightleftharpoons S²⁻ + H⁺
- The hydroxide ions (OH⁻) neutralize these hydrogen ions, forming water:
H++OH−→H2OH⁺ + OH⁻ \rightarrow H₂O
- Formation of Sulfide Ion:
- With an excess of OH⁻, both H⁺ ions from H₂S are neutralized, leading to the formation of the sulfide ion (S²⁻):
H2S+2OH−→S2−+2H2OH₂S + 2OH⁻ \rightarrow S²⁻ + 2H₂O
- Comparison with K₂S Formation:
- The reaction H₂S + KOH → K₂S + H₂O suggests the formation of potassium sulfide (K₂S).
- However, when written in ionic form, K₂S dissociates into K⁺ and S²⁻, making K⁺ a spectator ion.
- Removing spectator ions from the equation results in:
H2S+2OH−→S2−+2H2OH₂S + 2OH⁻ \rightarrow S²⁻ + 2H₂O
Thus, the simplified balanced equation represents the core reaction, excluding irrelevant spectator ions.
