Final answer:
In a 0.15 M solution of potassium sulfide (K2S), each molecule dissociates to form two potassium ions (K+). Thus, multiplying the molarity of K2S by 2 gives us the concentration of K+ ions, which is 0.30 M.
Step-by-step explanation:
The question asks about the concentration of potassium ions (K+) in a 0.15 M solution of potassium sulfide (K2S). Potassium sulfide is an ionic compound that dissociates completely in water to form potassium ions (K+) and sulfide ions (S2-). When K2S dissociates, it produces two K+ ions for every one sulfide ion. Therefore, to find the concentration of K+ ions, we multiply the given concentration of K2S by 2:
K2S → 2K+ + S2-
If the molarity of K2S is 0.15 M, then the molarity of K+ ions will be:
Concentration of K+ = 2 × (0.15 M) = 0.30 M
Therefore, the concentration of K+ ions is 0.30 M.