NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Chemistry
- Chapter
- Redox Reactions
NTA Abhyas 2020 · Chemistry · Redox Reactions
Hydrogen peroxide solution (20 mL) reacts quantitatively with a...
Hydrogen peroxide solution (20 mL) reacts quantitatively with a solution of KMnO4 (20 mL) acidified with dilute H2SO4. The same volume of the KMnO4 solution is just decolourised by 10 mL of MnSO4 in neutral medium simultaneously forming a dark brown precipitate of hydrated MnO2. The brown precipitate is dissolved in 10 mL of 0.2 M sodium oxalate under boiling condition in the presence of dilute H2SO4. Write the balanced equations involved in the reactions and calculate the molarity of H2O2.
Correct answer
0.10
Step-by-step solution
\left(\text{I}\right) \text{2KMn} {\text{O}_{4} + 3 {\text{H}_{2} \text{S} {\text{O}_{4} + 5 {\text{H}_{2} {\text{O}_{2} \to {\text{K}_{2} \text{S} {\text{O}_{4} + 2 \text{Mn} \text{S} {\text{O}_{4} + 8 {\text{H}_{2} \text{O} + 5 {\text{O}_{2}
\left(\text{I} \text{I}\right) \text{2KMn} {\text{O}_{4} + 3 {\text{Mn} \text{S} \text{O}_{4} + 2 {\text{H}_{2} \text{O} \to 5 \text{Mn} {\text{O}_{2} + \text{K} {}_{2} \text{S} {\text{O}_{4} + {2 \text{H}_{2} \text{S} {\text{O}_{4}<br /><br />\left(\text{I} \text{I} \text{I}\right)\text{Mn} {\text{O}_{2} + {\text{Na}_{2} {\text{C}_{2} {\text{O}_{4} + 2 {\text{H}_{2} {\text{S} \text{O}_{4} \to \text{Mn} {\text{S} \text{O}_{4} + \text{Na} {}_{2} \text{S} {\text{O}_{4} + {2 \text{CO}_{2} + {\text{H}_{2} \text{O}
\text{Reaction (III)} \to {\text{mM of Na}_{2} {\text{C}_{2} {\text{O}_{4} = \left(\text{10 mL} × \text{0.2M}\right) = 2 \text{mM} = \text{mM} \text{of Mn} {\text{O}_{2}
\text{Reaction (II)} \to {\text{mM of} \text{K} \text{MnO}_{4} = \text{2} × \dfrac{\text{2}}{\text{5}} = \dfrac{\text{4}}{\text{5}} \text{mM}
\text{Reaction (I)} \to \text{mM of} {\text{H}_{2} {\text{O}_{2} = \dfrac{4}{\text{5}} × \dfrac{5}{2} = 2 \text{mM}
2 {\text{mM of H}_{2} {\text{O}_{2} = \text{20 mL} × {\text{M}}_{{\text{H}_{2} {\text{O}_{2}}
{\text{M}}_{{\text{H}_{2} {\text{O}_{2}} = \text{0.1 M}
\left(\text{I} \text{I}\right) \text{2KMn} {\text{O}_{4} + 3 {\text{Mn} \text{S} \text{O}_{4} + 2 {\text{H}_{2} \text{O} \to 5 \text{Mn} {\text{O}_{2} + \text{K} {}_{2} \text{S} {\text{O}_{4} + {2 \text{H}_{2} \text{S} {\text{O}_{4}<br /><br />\left(\text{I} \text{I} \text{I}\right)\text{Mn} {\text{O}_{2} + {\text{Na}_{2} {\text{C}_{2} {\text{O}_{4} + 2 {\text{H}_{2} {\text{S} \text{O}_{4} \to \text{Mn} {\text{S} \text{O}_{4} + \text{Na} {}_{2} \text{S} {\text{O}_{4} + {2 \text{CO}_{2} + {\text{H}_{2} \text{O}
\text{Reaction (III)} \to {\text{mM of Na}_{2} {\text{C}_{2} {\text{O}_{4} = \left(\text{10 mL} × \text{0.2M}\right) = 2 \text{mM} = \text{mM} \text{of Mn} {\text{O}_{2}
\text{Reaction (II)} \to {\text{mM of} \text{K} \text{MnO}_{4} = \text{2} × \dfrac{\text{2}}{\text{5}} = \dfrac{\text{4}}{\text{5}} \text{mM}
\text{Reaction (I)} \to \text{mM of} {\text{H}_{2} {\text{O}_{2} = \dfrac{4}{\text{5}} × \dfrac{5}{2} = 2 \text{mM}
2 {\text{mM of H}_{2} {\text{O}_{2} = \text{20 mL} × {\text{M}}_{{\text{H}_{2} {\text{O}_{2}}
{\text{M}}_{{\text{H}_{2} {\text{O}_{2}} = \text{0.1 M}