What are the concentrations of Al^{3+} and Cl^- ions in a 0.12 M solution of AlCl_3, assuming complete dissociation?

• A. Al^{3+}: 0.36 M, Cl^-: 0.12 M
• B. Al^{3+}: 0.04 M, Cl^-: 0.12 M
• C. Al^{3+}: 0.12 M, Cl^-: 0.12 M
• D. Al^{3+}: 0.12 M, Cl^-: 0.36 M

Answer: (D)

Verified step by step guidance

1. Write the dissociation equation for aluminum chloride (AlCl_3) in water: \(\mathrm{AlCl_3 \rightarrow Al^{3+} + 3Cl^-}\).
2. Identify the initial concentration of AlCl_3, which is given as 0.12 M.
3. Since AlCl_3 dissociates completely, the concentration of \(\mathrm{Al^{3+}}\) ions will be equal to the initial concentration of AlCl_3, so \([\mathrm{Al^{3+}}] = 0.12\,M\).
4. For every 1 mole of AlCl_3 that dissociates, 3 moles of \(\mathrm{Cl^-}\) ions are produced, so multiply the initial concentration of AlCl_3 by 3 to find the chloride ion concentration: \([\mathrm{Cl^-}] = 3 \times 0.12\,M\).
5. Summarize the concentrations: \([\mathrm{Al^{3+}}] = 0.12\,M\) and \([\mathrm{Cl^-}] = 0.36\,M\).