Question

A mixture of acetic acid (CH3 CO2H monoprotic) and oxalic acid (H2C2O4; diprotic) requires 27.15 mL of 0.100 M NaOH to neutralize it. When an identical amount of the mixture is titrated, 15.05 mL of 2.68x10-2 M KMnO, is needed for complete reaction.

Answer 1

# moles of KMno_4 reacted = 15.05 times 2.68 times 10^-2 = 0.40334 mnwl 5 mol of oxalic acid react with 2 mol opf KMw_4 Therefore # mol of Oxalic acid = 5/2 times 0.40334 times 10^-3 = 1.008 mmol Man of oxalic acid = 1.008 times 10^-3 times 90 = 0.0907 g \r\n # mol of oxalic acid = 1.008 mmol # mol of NaoH = 27.15 times 0.1 = 2.715 times 10^-3 mol Therefore # ol of Acetic acid = 2.715 - (2 times 1.008) = 0.699 mmol 1mol of oxalic acid react with 2 mol d Naoll. Mass of Acetic acid = 0.699 times 60 times 10^3 = 0.0419 g Total mass = 0.0907f 0.0419 = 0.1326g % of oxalic acid = 0.0907/0.1326 times 100 = 68.40% %of Acetic acid = 0.0419/0.1326 times 100 = 31.6%