Question

A sample of oxalic acid (a diprotic acid of the formula H2C2O4) is dissolved in enough water to make 1.00 L of solution. A 100.0 mL sample of this solution is titrated with a solution of sodium hydroxide of concentration 0.750 M and requires 20.0 mL of sodium hydroxide to reach the end point. Calculate the mass of the original oxalic acid sample.

Answer 1

Original solution of volume 1.00 L is made by using the sample of the oxalic acid.

Now, 100.0 mL of the oxalic acid solution is titrated with a strong base NaOH.

The reaction of oxalic acid with NaOH can be written as

Hence, 2 moles of NaOH reacts with 1 mol of oxalic acid.

Now, the concentration of NaOH used = 0.750 M = 0.750 mol/L

Volume of NaOH used to reach the end point = 20.0 mL = 0.0200 L

Hence, number of moles of NaOH used can be calculated as

Now, the number of moles of oxalic acid that must have reacted with 0.0150 mol of NaOH is

Hence, 0.0075 mol of oxalic acid must be present in 100.0 mL of solution titrated.

Hence, the number of moles of oxalic acid in the original acid sample is

Hence, the original sample of oxalic acid has 0.075 mol of oxalic acid.

Molar mass of oxalic acid = 90.03 g/mol

Hence, the mass of oxalic acid can be calculated as

Hence, the mass of original oxalic acid sample used to prepare the solution is 6.75 g approximately.