Answer to Question #118861 in Chemistry for queen

Question

From the following dissociation question:
Al2(SO4)3 ----> 2Al^3 + 3SO4^2-

What will be the concentration of the CATION present if 25.5g aluminum sulfate is dissolved in 1.00L of water?

Answer 1

Solution:

The dissociation equation:

Al₂(SO₄)₃ ---> 2Al³⁺ + 3SO₄^2- 

Al³⁺ is the cation.

According to the equation: n(Al₂(SO₄)₃) = n(Al³⁺)/2

Moles of Al₂(SO₄)₃ = n(Al₂(SO₄)₃) = Mass of Al₂(SO₄)₃ / Molar mass of Al₂(SO₄)₃

The molar mass of Al₂(SO₄)₃ is 342.15 g/mol

n(Al₂(SO₄)₃) = (25.5 g) / (342.15 g/mol) = 0.07453 mol

n(Al³⁺) = 2 × n(Al₂(SO₄)₃) = 2 × 0.07453 mol = 0.14906 mol

Concentration of Al³⁺ = C_M(Al³⁺) = Moles of Al³⁺ / Volume of solution

C_M(Al³⁺) = (0.14906 mol) / (1.00 L) = 0.14906 mol/L = 0.149 mol/L

The concentration of Al³⁺ is 0.149 mol/L

Answer: The concentraion of Al³⁺ is 0.149 mol/L.