Oxalic acid (H2C2O4) is present in many plants and vegetables. If 24.0 mL of 0.0100 M KMn04 solution is needed to titrate 1.00 g of a sample of H2C2O4 to the equivalence point, what is the percent by mass of H2C2O4 in the sample? The net ionic equation is
2MNO4- + 16H+ +5C2O4 2- -> 2Mn 2+ + 10CO2 + 8H2O
Answer:
That's 0.024 L x 0.010 mol/L = 0.00024 mol of MnO4(-).
0.00024 mol MnO4 x (5 mol oxalate / 2 mol MnO4) = 0.0006 mol of oxalate that you have (in the form of oxalic acid). Now the molecular weight of oxalic acid = 2 + 24 + 64 = 90 g/mol, so you have titrated 0.0006 mol x 90 g/mol = 0.054 g of oxalic acid.
That's 5.4% of your sample.
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