Question

17.11 g of Al2(SO4)3 is dissolved in water to make 500 mL solution. What is the concentration of aluminum ion and sulfate ion in the resulting solution

Answer 1

Mass of Al₂(SO₄)₃ = 17.11 g

Molar mass of Al₂(SO₄)₃ = 342.15 g/mol

Moles of Al₂(SO₄)₃ = mass of Al₂(SO₄)₃ / molar mass of Al₂(SO₄)₃

                               = 17.11 g / 342.15 g/mol

                               = 0.05001 mol

Now, Al₂(SO₄)₃ dissolves in water to give 2 Al³⁺ and 3 SO₄^2- ions.

Al₂(SO₄)₃2Al³⁺ + 3SO₄^2-

Thus, moles of Al³⁺ in the solution = 2 x 0.05001 mol = 0.1000 mol

And the moles of SO₄^2- in the solution = 3 x 0.05001 mol = 0.1500 mol

Volume of solution = 500 mL = 0.500 L

Molarity = moles of solute/liter of solution

Concentration of Al³⁺ = moles of Al³⁺/liter of solution

                                   = 0.1000 mol/0.500 L

                                   = 0.200 M

Concentration of SO₄^2- = moles of SO₄²⁺/liter of solution

                                       = 0.1500 mol/0.500 L

                                       = 0.300 M