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

The volume of excess NaOH = (Molarity of HCl x Volume of HCl)/ Molarity of NaOH

= (0.125 x 12.62)/0.1028

= 15.3 ml

The volume of NaOH utilised to neutralise H₂C₂O₄ = 50-15.3 = 34.7 ml

Number of moles of NaOH utilised to neutralise H₂C₂O₄ = (Molarity of NaOH x Volume of NaOH)/1000

= (0.1028 x 34.7)/1000

=0.003567 mole

2 mole of NaOH required to neutralise 1 mole of H₂C₂O₄

Number of moles of H₂C₂O₄ present in the sample = 0.003567/2

= 0.001783 mole

weight of H₂C₂O₄ present in the sample = No. of moles of H₂C₂O₄ x MW of H₂C₂O₄

   = 0.001783 x 90.04

= 0.1605 g

weight % of H₂C₂O₄ = (weight of H₂C₂O₄/ weight of the sample)100

= (0.1605/1.762)100

=9.11%