Question

1. A 0.516 g portion of a sample that contains sodium oxalate is dissolved in water to which sulfuric acid has been added. The end point of the titration of the solution with 0.02074 M potassium permanganate is 9.75 mL.

A. Write the balanced chemical reaction for the titration assuming that the reaction is performed in highly acid solution.

B. Use the concentration and endpoint volume of permanganate to calculate the moles of permanganate used in the titration.

C. Use the moles of permanganate calculated in 2b with the balanced titration reaction to calculate the moles of oxalate that reacted with the added permanganate.

D. Use the moles of oxalate that reacted during the titration and the molar mass of sodium oxalate (133.998 g) to calculate the mass of sodium oxalate in the sample.

E. Use the mass of sodium oxalate calculates in 2D and the total mass of the sample (0.516 g) to calculate the percentage of sodium oxalate in the sample.

Answer 1

A.

The balanced chemical reaction for the titration is,

5Na₂C₂O₄(aq) + 2KMnO₄(aq) + 8H₂SO₄(aq) ---> 2MnSO₄(aq) + K₂SO₄(aq) + 5Na₂SO₄(aq) + 10 CO₂(g) + 8 H₂O(l)

B. moles of permanganate used in the titration.:

Moles of KMnO4 = 0.02074 moles/litre * 0.00975 L) = 0.000202215 moles

= 2.02 * 10^-4  moles

C.moles of oxalate that reacted:

0.000202215 moles KMnO4 * 5 Na2C2O4 = 0.0005055375 moles Na2C2O4

1 2 moles KMnO4

5.06 * 10^-4 moles NaC2O4

D. 5.06 * 10^-4 moles NaC2O4 * 134 g Na2C2O4 = 0.0677420 g NaC2O4

1 1 mole Na2C2O4

6.77 * 10^-2 g of Na2C2O4

E. percentage of sodium oxalate

0.0677420 g NaC2O4 * 100 = 13.1 %

0.516 g of sample