Redox Reactions and Volumetric AnalysishardMATCH FOLLOWING

**Match Column I (Mixture) with Column II (Endpoint volume relationship for HCl titration):** <br> *Redox Reactions and Volumetric Analysis Chemistry Question

Question

**Match Column I (Mixture) with Column II (Endpoint volume relationship for HCl titration):** <br> **Column I:** <br> (A) $NaOH + Na_2CO_3$ <br> (B) $Na_2CO_3 + NaHCO_3$ <br> (C) $NaOH + NaHCO_3$ <br> (D) $NaOH + Na_2CO_3 + NaHCO_3$ <br> **Column II (where $V_{ph}$ is Phenolphthalein and $V_{mo}$ is Methyl Orange):** <br> (p) $V_{ph} > 1/2 V_{mo}$ <br> (q) $V_{ph} < 1/2 V_{mo}$ <br> (r) $V_{ph} = V_{mo}$ <br> (s) $V_{ph} > 1/2 V_{mo}$ (Wait, logic check)

Answer: A

💡 Solution & Explanation

**Logic:** <br> (A) $V_{ph} = V_{NaOH} + 1/2 V_{Na_2CO_3}$; $V_{mo} = V_{NaOH} + V_{Na_2CO_3}$. Since $V_{ph}$ is more than half of $V_{mo}$, (p) fits. <br> (B) $V_{ph} = 1/2 V_{Na_2CO_3}$; $V_{mo} = V_{Na_2CO_3} + V_{NaHCO_3}$. Here $V_{ph} < 1/2 V_{mo}$ because of the extra $NaHCO_3$. <br> (C) Note: $NaOH$ and $NaHCO_3$ react to form $Na_2CO_3$. <br> (D) Complex mixture where $V_{ph} > 1/2 V_{mo}$ typically implies excess $NaOH$.

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