Question

A mixture of Na2C2O4 and KHC2O4 requires equal volume of 0.1 M KMnO4 in acidic medium and 0.1 M NaOH separately. Find its molar ratio

Answer 1

A mixture of Na₂CO₄ and KHC₂O₄ requires equal volume of 0.1M KMnO₄ in acidic medium and 0.1M NaOH separately.

To find : The molar ratio

solution : let no of moles of Na₂C₂O₄ = x

so, no of moles of C₂O₄²¯ = x

similarly, no of moles of KHC₂O₄.H₂C₂O₄= y (let)

no of moles of H⁺ ions = 3y

and no of moles of C₂O₄²¯ = 2y

here total C₂O₄²¯ = (x + 2y)

Let volume of KMnO₄ or NaOH = V ml

as H⁺ + OH¯ ⇒H₂O

here 1 mole of H⁺ = 1 mole of OH¯

for 3y moles of H⁺ = 3y moles OH¯

but in 0.1M of NaOH, V × 0.1/1000 = V/10000 mol

⇒3y = V/10000

⇒V = 30000y .......(1)

as reaction of oxalate and permanganate is given by,

5C₂O₄²¯ + 2MnO₄²¯ ⇒2Mn²⁺ + 10CO₂

5 moles of C₂O₄²¯ = 2 moles of MnO₄²¯

⇒(x + 2y) moles of C₂O₄²¯ = 2(x + 2y)/5 moles of MnO₄²¯

⇒2(x + 2y)/5 = V × 0.1/1000

⇒2(x + 2y) = 5 × 30000y/10000

⇒2x + 4y = 15y

⇒2x = 11y

⇒x/y = 11/2 = 5.5/1

Therefore the ratio of moles of Na₂C₂O₄ to KHC₂O₄.H₂C₂O₄ is 5.5 : 1.