mL of is reacted with excess in acidic medium. The liberated requires mL of for titration. The value — Redox Reactions and Volumetric Analysis Chemistry Question
Question
$100$ mL of $0.1 M$ $KI$ is reacted with excess $H_2O_2$ in acidic medium. The liberated $I_2$ requires $V$ mL of $0.2 M$ $Na_2S_2O_3$ for titration. The value of $V$ is:
Answer: B
💡 Solution & Explanation
**Step 1:** $2I^- + H_2O_2 + 2H^+ \rightarrow I_2 + 2H_2O$. <br>Meq of $I^-$ ($n=1$) = $100 \times 0.1 = 10$. <br>**Step 2:** $I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-}$. <br>Meq of $I^-$ = Meq $I_2$ liberated = Meq $Na_2S_2O_3$ used. <br>$10 = V \times 0.2 \times 1 \Rightarrow V = 50$ mL.
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