Answer:
2 M Al³⁺(aq) and 6 M NO₃⁻(aq)
Step-by-step explanation:
Al(NO₃)₃ is a strong electrolyte that ionizes according to the following equation.
Al(NO₃)₃(aq) → Al³⁺(aq) + 3 NO₃⁻(aq)
The solution 2.0 M contains 2.0 moles of Al(NO₃)₃ per liter of solution.
The molar ratio of Al(NO₃)₃ to Al³⁺ is 1:1. The concentration of Al³⁺ is:
2.0 mol Al(NO₃)₃/L × (1 mol Al³⁺/1 mol Al(NO₃)₃) = 2.0 mol Al³⁺/L = 2.0 M Al³⁺
The molar ratio of Al(NO₃)₃ to NO₃⁻ is 1:3. The concentration of NO₃⁻ is:
2.0 mol Al(NO₃)₃/L × (3 mol NO₃⁻/1 mol Al(NO₃)₃) = 6.0 mol NO₃⁻/L = 6.0 M NO₃⁻