What is the result of dissolving potassium sulfide in water? Answer choices: 2 K(aq) + S(aq) → 2 K+(aq) + S2–(aq) K2S2(aq) → 2 K+(aq) + S2–(aq) K2+(aq) + S2–(aq)
The Correct Answer and Explanation is:
The correct result of dissolving potassium sulfide (K₂S) in water is:
K₂S (aq) → 2 K⁺(aq) + S²⁻(aq)
Explanation:
When potassium sulfide (K₂S) dissolves in water, it dissociates into its constituent ions. Potassium sulfide is an ionic compound, and like most ionic compounds, it dissolves in water by dissociating into its ions, which are stabilized by water molecules. In this case, potassium sulfide dissociates as follows:
- Potassium sulfide (K₂S) is made of potassium (K⁺) ions and sulfide (S²⁻) ions.
- When K₂S dissolves in water, it breaks apart into two potassium ions (K⁺) and one sulfide ion (S²⁻).
This can be represented by the following equation:K2S(aq)→2K+(aq)+S2−(aq)\text{K}_2\text{S} (aq) \rightarrow 2 \text{K}^+ (aq) + \text{S}^{2-} (aq)K2S(aq)→2K+(aq)+S2−(aq)
Why this occurs:
- Potassium ions (K⁺): Potassium is a Group 1 metal, and its ion (K⁺) is highly soluble in water due to its large size and low charge density. The water molecules surround the K⁺ ions and stabilize them through ion-dipole interactions.
- Sulfide ions (S²⁻): Sulfide is a relatively large and highly charged ion. In water, the S²⁻ ions are surrounded by water molecules, and they are also stabilized by the ion-dipole interactions. The negative charge on S²⁻ makes it highly reactive, and it will typically react with any available hydrogen ions (H⁺) from water to form hydrogen sulfide (H₂S) in certain conditions. However, in pure water, it remains as S²⁻.
In conclusion, when potassium sulfide dissolves in water, it dissociates into 2 potassium ions (K⁺) and 1 sulfide ion (S²⁻), as represented by the equation above.
