Q. ${KMnO}_{4}$ reacts with oxalic acid according to the equation ${2MnO}_{4}^{-} + 5C_{2}O_{4}^{2 -} + 16H^{+} \rightarrow 2{{Mn}^{2}}^{+} + 10{CO}_{2} + 8H_{2}O$ Here, 20 mL of 0.1 M ${KMnO}_{4}$ is equivalent to
A) $20\ mL\ of\ 0.5\ M\ H_{2}C_{2}O_{4}$
B) $50\ mL\ of\ 0.1\ M\ H_{2}C_{2}O_{4}$
C) $50\ mL\ of\ 0.1\ M\ H_{2}C_{2}O_{4}$
D) $20\ mL\ of\ 0.1\ M\ H_{2}C_{2}O_{4}$
Explanation
(b)
$2MnO_{4}^{-} + 5C_{2}O_{4}^{2 -} + 16H^{+} \rightarrow 2Mn^{2 +} + 10CO_{2} + 8H_{2}O$
20 mL of 0.1 M $KMnO_{4} = 20 \times 0.1 = 2$m mol
$\because$ 2 mmol of $KMnO_{4} \equiv 5\ mmol\ of\ C_{2}O_{4}^{2 -}$
50 mL of 0.1 M $H_{2}C_{2}O_{4} = 50 \times 0.1 = 5mmol$
Hence, 20 mL of 0.1 M $KMnO_{4}$
$\equiv 50\ mL\ of\ 0.1\ M\ H_{2}C_{2}O_{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.