Another approach
First u must know that oxalic acid is a diprotic acid which means it gives 2 hydrogen ions, therefore, 1 mole of oxalic acid will react with 2 moles of oxalic acid.
My method is:
Calculate the mole of NaOH used: (37.98ml)(0.2283M) = 8.6708mmol
mmol of NaOH = 2 X mmol of oxalic acid
Therefore,
mmol of oxalic acid = 0.5 X 8.6708 =4.3354 mmol =0.0043354 mole
Mass of oxalic acid present = (0.0043354)(2X1.00794+2X12.011+4X15.9994)
=0.3910g#
Percent of Oxalic acid in sample = (0.3910/0.7984) X 100% =48.97%##
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