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The complete reaction between NaOH and oxalic acid is
2NaOH + (COOH)2 = (COONa)2 + 2H2O
2 mol 1 mol
15.00 mL of a 0.46 M oxalic acid solution = 0.46 mol L-1x 0.015 L = 0.0069 mol of oxalic acid
From the above reaction
1 mol of oxalic acid requires 2 moles of NaOH for complete reaction.
So 0.0069 mol of oxalic acid will require 2 x 0.0069 i.e. 0.0138 mol of NaOH
Hence the concentration of NaOH = 0.0138 mol/0.01583 L = 0.87 M