Question

A student pipets 24 ml of a sulphuric acid solution and titrates it with .09928 M NaOH solution. A total of 45.99ml of base are required to exactly neutralize the acid sample. Calculate the molarity of the acid. If the total volume of the acid samplewas 350.0ml, how many grams of sulphuric acid were in the sample?

Answer 1

At neutralization, M₁V₁ = M₂V₂
M₁ = ? ; V₁ = 24 mL = 0.024 L ; M₂ = 0.09928 M or mol/L ; V₂ = 45.99 mL = 0.04599 L
M₁ * 0.024 L = 0.09928 mol/L * 0.04599 L
M₁ = 0.1902453 M or mol/L = molarity of sulphuric acid
moles of sulphuric acid =  0.1902453 mol/L * (350 / 1000) L = 0.066585 mol
Molar mass of sulphuric acid, H₂SO₄ = 98.079 g/mol
(grams of H₂SO₄ in the sample / Molar mass of H₂SO₄ ) = mol of H₂SO₄
grams of sulphuric acid in sample =  0.066585 mol * 98.079 gm/mol = 6.53067 gms