What is the concentration of K^+ ions in a 0.15 M solution of K_2S?

• A. 0.30 M
• B. 0.45 M
• C. 0.15 M
• D. 0.075 M

Answer: (A)

Verified step by step guidance

1. Identify the chemical formula of the compound: K_2S, which dissociates in water into potassium ions (K^+) and sulfide ions (S^{2-}).
2. Write the dissociation equation: \(\mathrm{K_2S (aq) \rightarrow 2K^+ (aq) + S^{2-} (aq)}\).
3. Note that for every 1 mole of K_2S that dissolves, 2 moles of K^+ ions are produced.
4. Use the initial concentration of K_2S (0.15 M) and multiply by the stoichiometric coefficient of K^+ (which is 2) to find the concentration of K^+ ions: \([K^+] = 2 \times 0.15\, \mathrm{M}\).
5. Calculate the product from the previous step to get the concentration of K^+ ions in the solution.