1.2 g of a salt with their empirical formula KxHy(C2O4)z was dissolved in 50 mL of water and its 10 mL portion required 11.0 mL of a 0.1 M HCl solution to reach the equivalence point. In a separate analysis, 15 mL of the stock solution required 20 mL 0.2475M KOH to reach the equivalence point. Empirical formula of the salt is :

Engineering

Chemistry

Law of Chemical Equivalence

Titration

Question

1.2 g of a salt with their empirical formula K_xH_y(C₂O₄)_z was dissolved in 50 mL of water and its 10 mL portion required 11.0 mL of a 0.1 M HCl solution to reach the equivalence point. In a separate analysis, 15 mL of the stock solution required 20 mL 0.2475M KOH to reach the equivalence point. Empirical formula of the salt is :

x = 2

y = 3

z = 2

x= 3

y = 2

z = 1

x = 1

y = 1

z = 3

x = 1

y = 3

z = 2

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The salt K_xH_y(C₂O₄)_z has acidic protons from H_y and basic oxalate ions. The HCl titration neutralizes the basic oxalate: C₂O₄^2- + 2H⁺ → H₂C₂O₄. The KOH titration neutralizes the acidic protons H_y.

For HCl: Moles in 50mL = (11.0mL * 0.1M * 50/10) / 1000 = 0.0055 mol HCl = moles of 2H⁺ used, so moles of C₂O₄^2- = 0.00275.

For KOH: Moles in 50mL = (20mL * 0.2475M * 50/15) / 1000 = 0.0165 mol KOH = moles of H⁺ from H_y.

Mole ratio H_y : C₂O₄^2- = 0.0165 : 0.00275 = 6:1. Therefore, y=3 and z=1. Charge balance gives x=1.

Final Answer: x=1, y=3, z=1