Question

A 1.766-g sample of impure sodium oxalate (Na2C2O4) is dissolved in water and titrated to methylred end point with 0.250 M HCl. The reaction is: Na2C2O4(s) + 2HCl(aq) o H2C2O4(aq) + 2NaCl(aq) If 44.15 mL of the acid is used for this reaction, what is the percent by mass of sodium oxalate in the sample?

Answer 1

Answer:

41.9(w/w) %

Explanation:

Based on the reaction:

Na₂C₂O₄(s) + 2HCl(aq) → H₂C₂O₄(aq) + 2NaCl(aq)

Where 1 mole of sodium oxalate reacts with 2 moles of HCl

Moles of HCl solution to reach end point are:

44.15mL = 0.04415L ₓ (0.250mol / L) = 0.01104 moles of HCl

As 2 moles of HCl reacts per mole of Na₂C₂O₄:

0.01104mol HCl ₓ (1 mol Na₂C₂O₄ / 2 mol HCl) = 5.519x10⁻³ moles Na₂C₂O₄ are in the sample.

Molar mass of Na₂C₂O₄ is 134g/mol; thus, mass of 5.519x10⁻³ moles Na₂C₂O₄ is:

5.519x10⁻³ moles Na₂C₂O₄ ₓ (134g / mol) = 0.740g of Na₂C₂O₄ in the sample.

Thus, percent by mass of sodium oxalate in the sample is:

0.740g of Na₂C₂O₄ / 1.766g ₓ 100 =

41.9(w/w) %