The flocculation value of HCl for arsenic sulphide sol.
is 30 m mol L⁻¹ If H₂SO₄ is used for the flocculation of arsenic sulphide, the amount in mg, of H₂SO₄ in 500 ml required for the above purpose is ______.
(molecular mass of H₂SO₄ = 98 g/mol)

Solution

Chemistry ma’am · 3 steps

The sol is negative, so only H⁺ does the work, and each H₂SO₄ gives you two of them.

1. Which ion coagulates. Arsenic sulphide is a negatively charged sol, so the cation H⁺ coagulates it. The flocculation value of HCl means 30 mmol of H⁺ is needed per litre of sol.
2. Scale to 500 mL. In 0.5 L you need 30 × 0.5 = 15 mmol of H⁺. One H₂SO₄ releases two H⁺, so only 15 / 2 = 7.5 mmol of H₂SO₄ is required.

   m = 7.5 mmol × 98 mg mmol⁻¹ = 735 mg

3. So. The mass of H₂SO₄ needed in 500 mL is 735 mg.

The common slip: 1470. Taking H₂SO₄ one-for-one with HCl uses 15 mmol of acid and gives 1470 mg, forgetting that each H₂SO₄ supplies two H⁺.

The idea to keep. Find the coagulating ion from the sol's charge first. Flocculation value counts that ion, so divide by how many of them each molecule supplies.

Answer from the NTA final answer key. Solution by PrepWiser.