100ml of 0.1M of Al₂(SO₄)₃ is mixed with 100ml of 0.1M AlCl₃. Molarity of Al⁺³ ion in the resulting solution is ____?

Explanation

From Al₂(SO₄)₃:

• Moles of Al₂(SO₄)₃:  Moles = Molarity × Volume = 0.1 M × 0.1 L = 0.01 moles
• Each Al₂(SO₄)₃ dissociates into 2 Al³⁺ ions:  Moles of Al³⁺ = 0.01 moles × 2 = 0.02 moles

From AlCl₃:

Moles of AlCl₃:

Moles = Molarity × Volume = 0.1 M × 0.1 L = 0.01 moles

Each AlCl₃ dissociates into 1 Al³⁺ ion:

Moles of Al³⁺ = 0.01 moles × 1 = 0.01 moles

Total moles of Al³⁺ = 0.02 moles + 0.01 moles = 0.03 moles

Total volume = 100 mL + 100 mL = 200 mL = 0.2 L

Molarity = Total moles of Al³⁺ / Total volume (L)

Molarity = 0.03 moles / 0.2 L = 0.15 M

Final Answer:

Molarity of Al³⁺ ions = 0.15 M