Question

a) What is the molarity of the solution that was prepared by dissolving 3.25 g of sulfuric acid in water to a total volume of 500.0 mL?

b)What is the molarity of the hydrogen ion in part a if you assume the sulfuric acid ionizes completely? Write a balanced chemical equation.

Answer 1

a) Molar mass of sulfuric acid, H₂SO₄ = (2*1.008 + 1*32.065 + 4*15.9994) g/mol = 98.0786 g/mol.

Mole(s) of H₂SO₄ corresponding to 3.25 g sulfuric acid = (mass of H₂SO₄)/(molar mass of H₂SO₄) = (3.25 g)/(98.0786 g/mol) = 0.03314 mole.

Total volume of solution = 500 mL = (500 mL)*(1 L/1000 mL) = 0.5 L.

Molarity of H₂SO₄ = (moles of H₂SO₄)/(volume of solution in L) = (0.03314 mole)/(0.5 L) = 0.06628 mol/L ≈ 0.0663 M (1 mol/L = 1 M, ans).

b) Write the balanced chemical equation for the dissociation of H₂SO₄ as below.

H₂SO₄ --------> 2 H⁺ + SO₄^2-

As per the balanced stoichiometric equation above,

1 mole H₂SO₄ = 2 moles H⁺.

Therefore, mole(s) of H⁺ = (0.03314 mole H₂SO₄)*(2 mole H⁺/1 mole H₂SO₄) = 0.06628 mole.

Molarity of H⁺ = (0.06628 mole)/(0.5 L) = 0.13256 mol/L ≈ 0.1326 mol/L = 0.1326 M (ans).