Question

A sample of $KMnO_4$ solution required 50 ml when titrated against 3 mmol of oxalic acid. The normality of same solution in reaction with alkaline $H_2O_2$ is

• A. 0.120 N
• B. 0.060 N
• C. 0.072 N
• D. 0.036 N

Answer 1

Answer: (C)

0.072 N

Answer 2

To determine the normality of the $KMnO_4$ solution in reaction with alkaline $H_2O_2$, we first need to find the molarity of the $KMnO_4$ solution from its titration with oxalic acid.

Step 1: Determine the Molarity of $KMnO_4$ from titration with Oxalic Acid

1. n-factor for Oxalic Acid ($H_2C_2O_4$):
   Oxalic acid acts as a reducing agent, getting oxidized from $C_2O_4^{2-}$ to $2CO_2$.
   The oxidation state of Carbon in $H_2C_2O_4$ is +3.
   The oxidation state of Carbon in $CO_2$ is +4.
   Change in oxidation state per carbon atom = $4 - 3 = 1$.
   Since there are two carbon atoms in $C_2O_4^{2-}$, the total change in oxidation state is $2 \times 1 = 2$.
   Therefore, the n-factor for oxalic acid is 2.

2. n-factor for $KMnO_4$ in acidic medium:
   The titration of $KMnO_4$ with oxalic acid occurs in an acidic medium. In acidic medium, $MnO_4^-$ is reduced to $Mn^{2+}$.
   The oxidation state of Mn in $MnO_4^-$ is +7.
   The oxidation state of Mn in $Mn^{2+}$ is +2.
   Change in oxidation state = $7 - 2 = 5$.
   Therefore, the n-factor for $KMnO_4$ in acidic medium is 5.

3. Calculate milliequivalents of oxalic acid:
   Given, moles of oxalic acid = 3 mmol.
   Milliequivalents of oxalic acid = moles $\times$ n-factor = $3 \text{ mmol} \times 2 = 6 \text{ meq}$.

4. Calculate normality of $KMnO_4$ in acidic medium:
   At the equivalence point of a titration, milliequivalents of oxidant = milliequivalents of reductant.
   Milliequivalents of $KMnO_4$ = Milliequivalents of oxalic acid = $6 \text{ meq}$.
   Given, volume of $KMnO_4$ solution = 50 ml.
   Normality ($N$) = Milliequivalents / Volume (in ml)
   $N_{KMnO_4 (\text{acidic})}$ = $6 \text{ meq} / 50 \text{ ml} = 0.12 \text{ N}$.

5. Calculate Molarity of $KMnO_4$ solution:
   Molarity ($M$) = Normality / n-factor
   $M_{KMnO_4}$ = $N_{KMnO_4 (\text{acidic})} / \text{n-factor (acidic)}$
   $M_{KMnO_4}$ = $0.12 \text{ N} / 5 = 0.024 \text{ M}$.
   The molarity of the $KMnO_4$ solution is constant regardless of the reaction medium.

Step 2: Determine the Normality of $KMnO_4$ in alkaline $H_2O_2$ reaction

1. n-factor for $KMnO_4$ in alkaline medium:
   In an alkaline medium, $KMnO_4$ ($MnO_4^-$) is typically reduced to $MnO_2$.
   The oxidation state of Mn in $MnO_4^-$ is +7.
   The oxidation state of Mn in $MnO_2$ is +4.
   Change in oxidation state = $7 - 4 = 3$.
   Therefore, the n-factor for $KMnO_4$ in alkaline medium is 3.

2. Calculate Normality of $KMnO_4$ in alkaline medium:
   Normality ($N$) = Molarity $\times$ n-factor (in alkaline medium)
   $N_{KMnO_4 (\text{alkaline})}$ = $0.024 \text{ M} \times 3 = 0.072 \text{ N}$.

The normality of the same $KMnO_4$ solution in reaction with alkaline $H_2O_2$ is 0.072 N.