Answer to Question #347422 in Organic Chemistry for Dominic

Question

Calculate the volume of 0.5m potassium hydroxide that will neutralize 45cm³ of 0.1m tetraoxosulphate(VI) acid solution.

(K=39,O=16,H=1)

Answer 1

Solution:

potassium hydroxide = KOH

tetraoxosulphate(VI) acid = sulphuric acid = H₂SO₄

Balanced chemical equation:

2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

According to stoichiometry:

Moles of KOH / 2 = Moles of H₂SO₄

or:

Molarity of KOH × Volume of KOH = 2 × Molarity of H₂SO₄ × Volume of H₂SO₄

Therefore,

Volume of KOH = 2 × Molarity of H₂SO₄ × Volume of H₂SO₄ / Molarity of KOH

Volume of KOH = (2 × 0.1 M × 45 cm³) / (0.5 M) = 18 cm³

Volume of KOH = 18 cm³

Answer: The volume of potassium hydroxide (KOH) solution is 18 cm³