**Passage 12 (continued):** <br> **Q2:** The volume of required for the titration of the liberated i — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

**Step 1:** Moles $I_2$ from limiting $KI = (3/5) \times 0.0025 = 0.0015$. <br>**Step 2:** $I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-}$. <br>**Step 3:** Moles Hypo = $2 \times 0.0015 = 0.003$. <br>**Step 4:** Volume = $0.003 / 0.1 = 0.03$ L = $30$ mL.