In basic medium CrO₄²⁻ oxidises S₂O₃²⁻ to form SO₄²⁻ and itself changes into Cr(OH)₄ -. The volume of 0.154 M CrO₄²⁻ required to react with 40 mL of 0.25 M S₂O₃²⁻ is __________ mL. (Rounded off to the nearest integer)

Solution

Chemistry ma’am · 4 steps

Find how many electrons each ion moves, then match electrons given to electrons taken.

1. Chromate gains. In CrO₄²⁻, Cr is +6. In Cr(OH)₄⁻, Cr is +3. So each chromate ion gains 3 electrons.
2. Thiosulphate loses. In S₂O₃²⁻, each S averages +2. In SO₄²⁻, S is +6. Each S loses 4 electrons, and there are 2 S atoms, so each thiosulphate ion loses 8 electrons.
3. Electrons to supply. Thiosulphate present is 40 × 0.25 = 10 mmol. It releases 10 × 8 = 80 mmol of electrons, and chromate must take up all 80.

   V × 0.154 × 3 = 40 × 0.25 × 8 = 80

4. So. V = 80/0.462 = 173.2 mL, which rounds to 173.

The common slip: 87. Counting only one S in thiosulphate (n-factor 4 instead of 8) gives 87 mL, half the right volume.

The idea to keep. In redox titrations, equate electrons, not moles: molarity × volume × n-factor on both sides. Count every atom that changes, so thiosulphate has two S atoms.

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