Question

If 1.830g of soid potassium oxalate ( K2C2O4.H2O) is dissolved in deionised water and diluted to 250mL in a volumetric flask, what is the concentration of the oxalate ions ?

Assume that 15.50mL of potassium permanganate solution is used in titrating 10mL of the standard potassium oxalate solution made in the question above. What is the concentration of the potassium permanganate solution ?

Answer 1

Mass of solid potassium oxalate ( K₂C₂O₄.H₂O) is dissolved in deionised water = 1.830 g

Equivalent mass of K₂C₂O₄.H₂O = 92 g/eq

Number of gram equivalents of K₂C₂O₄.H₂O = 1.830 g/92 g/eq= 0.0199 eq

Hence number of gram equivalents of C₂O₄^2- ions in aqueous solution = 0.0199 eq

Volume of solution = 250 mL = 0.25 L

Hence concentration of C₂O₄^2- ions in aqueous solution

= number of gram equivalents / volume of solution (in L)

= 0.0199 eq / 0.25 L

= 0.0796 gram eq. L^-1 = 0.0796 N

For titration

Volume of the dilute K₂C₂O₄.H₂O solution = V₁ = 10 mL

Strength of the dilute K₂C₂O₄.H₂O solution = S₁ = 0.0796 N

Volume of the dilute KMnO₄ solution = V₂ = 15.5 mL

Strength of the dilute KMnO₄ solution = S₂ = ?

Since chemical compounds react in the ratio of their equivalent weights, therefore

V₁ X S₁ = V₂ X S₂

or S₂ = V₁ X S₁ / V₂  

= 10 mL X 0.0796 N / 15.5 mL​

= 0.0514 N