A mixture of NaHC₂O₄ and KHC₂O₄.H₂C₂O₄ required equal volumes of 0.2 (N) KMnO₄ and 0.12 (N) NaOH separately. What is the molar ratio of NaHC₂O₄ and KHC₂O₄.H₂C₂O₄ in the mixture?

CONCEPT:

Molar Ratio of Two Substances in a Mixture

• Molality calculations determine the quantity of reactants or products in reactions.
• The equivalence between neutralization and oxidation reactions facilitates calculating molar ratios.
• Equivalents represent the molar amount reacting in these processes.

Explanation:-

• Let the volumes of KMnO₄ and NaOH used be V liters.
• Define 'a' as moles of NaHC₂O₄ and 'b' as moles of KHC₂O₄·H₂C₂O₄.
• From the redox reaction with KMnO₄:
  • \(0.2 \times V = 2a + 4b\)
  • ...(Equation 1)
• From the acid-base neutralization with NaOH:
  • \(0.12 \times V = a + 3b\)
  • ...(Equation 2)
• Solving the equations:
  • From Equation 2:
  • \(a + 3b = 0.12V\)
  • From Equation 1:
  • \(2a + 4b = 0.2V\)
  • Using Equation 1 and solving for 'a':
  • \(a = 0.06V\)
  • Substituting 'a' back into Equation 2:
  • \(0.06V + 3b = 0.12V\)
  • \(3b = 0.06V\)
  • \(b = 0.02V\)
• The molar ratio:
  • \(a:b = 0.06V:0.02V = 3:1\)

CONCLUSION:

The correct answer is 3:1