Question

A 1.762 g sample of impure H2C2O4 (MW =90.04) was dissolved in water and the impurities removed by filtration. 50.00 mL of .01028 M NaOH was added to the filtrate, this being in excess of the amount needed to neutralize both hydrogens of the H2C2O4. The excess NaOH required 12.62 mL of 0.1251M HCl to reach the equivalence point. What was the weight % of H2C2O4 in the sample.

Answer 1

2MnO4 ^1- + 16H+ + 5C2O4^2- ------------> 2Mn^(2+) + 10CO2 + 8H2O

48.0 mL of 0.100 M KMnO4 required to titrate the sample including H2C2O4.

2 Moles of MnO4^1- required to titrate 5 moles of C2O4^2-

moles of MnO4 was used = 0.0480 L x 0.100 M = 0.00480 mol

C2O4^2- titrated. 5/2 x 0.00480 mol = 0.0120 mol

Since oxalic acid is 1:1 with C2O4^2-;

Molar mass oxalic acid (H2C2O4) = 90.03 g/ mol

mass of oxalic acid = 0.0120 mol x 90.03 g/ mol = 1.08036 g

find the % weight/weight of H2C2O4 in sample of 2.00 g = (1.08 g / 1.762g ) x 100% = 61.3 %

% weight/weight of H2C2O4 in sample = 61.3%