20 mL of 0.02 M hypo solution is used for the titration of 10 mL of copper sulphate solution, in the presence of excess of KI using starch as an indicator. The molarity of Cu²⁺ is found to be ____________ × 10⁻² M. [nearest integer]
Given : 2 Cu²⁺+4 I⁻ → Cu₂ I₂+I₂
I₂+2 S₂ O₃²⁻ → 2 I⁻+S₄ O₆²⁻

Solution

Chemistry ma’am · 4 steps

Follow the iodine: each Cu²⁺ ends up matching exactly one thiosulphate ion.

1. Moles of hypo. Hypo is sodium thiosulphate. Millimoles of S₂O₃²⁻ used = 20 × 0.02 = 0.4 mmol.
2. Link through iodine. From the second equation, 2 S₂O₃²⁻ react with 1 I₂, so I₂ formed = 0.2 mmol. From the first equation, 1 I₂ comes from 2 Cu²⁺, so Cu²⁺ = 0.4 mmol.

   2 Cu²⁺ ≡ I₂ ≡ 2 S₂O₃²⁻ ⇒ n_(Cu²⁺) = n_(S₂O₃²⁻)

3. Molarity of copper. Cu²⁺ is in 10 mL, so its molarity = 0.4/10 = 0.04 M.
4. So. 0.04 M = 4 × 10⁻² M, so the answer is 4.

The common slip: 2. Taking moles of Cu²⁺ equal to moles of I₂ (0.2 mmol) gives 0.02 M, that is 2, not 4. Two Cu²⁺ ions give one I₂.

The idea to keep. In iodometry, chain the equations through I₂. For copper, 2 Cu²⁺ → I₂ → 2 S₂O₃²⁻, so moles of Cu²⁺ equal moles of hypo.

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