Question on Fuming Sulphuric Acid

What volume of 1M NaOH is required to completely neutralize a solution of oleum that is labeled 109% H₂SO₄?

Answer:

109% oleum means that 9g of H₂O is required to convert all SO₃ present in 100g of it to H₂SO₄.

The following equation shows that, 1 mole (80g) of SO₃ reacts with 1 mole (18g) of H₂O and hence 40g of SO₃ will react with 9g of H₂O.

SO₃ + H₂O  = H₂SO₄

According to the following equations, 80g of NaOH will react with 80g and 98g of SO₃ and H₂SO₄ respectively.

SO₃ + 2NaOH = Na₂SO₄ + H₂O

H₂SO₄ + 2NaOH = Na₂SO₄ + 2H₂O

Weight of SO₃ present in 100g of the oleum sample = 40g

Weight of H₂SO₄ present in 100g of the oleum sample = (100 – 40)g = 60g

Weight of NaOH required to completely neutralize 40g of SO₃ = (80*40)/80 = 40g

Weight of NaOH required to completely neutralize 60g of H₂SO₄ = (80*60)/98 = 48.98g

Total weight of NaOH required for complete neutralization = 40 + 48.98 = 88.98g

Volume of 1M NaOH solution containing 88.98g of NaOH = (1000*88.98)/40 = 2224.5mL