Oxalic acid is a diprotic acid. If a solid material contains 53.66 percent of oxalic acid (H 2C 2O 4), by mass, then a 0.6543-g sample of that solid will require mL of 0.3483 M NaOH for neutralization. 11.19 97.78 28.59 1.119 22.39

The volume of NaOH required is 22.39 mL

Explanation:

We are given:

Mass of sample = 0.6543 g

Mass percent of oxalic acid = 53.66 %

This means that 53.66 grams of oxalic acid is present in 100 grams of sample

Mass of oxalic acid in the given amount of sample =

Given mass of oxalic acid = 0.351 g

Molar mass of oxalic acid = 90 g/mol

Putting values in above equation, we get:

The chemical equation for the reaction of oxalic acid and NaOH follows:

By Stoichiometry of the reaction:

1 mole of oxalic acid reacts with 2 moles of NaOH

So, 0.0039 moles of oxalic acid will react with = of NaOH

Moles of NaOH = 0.0078 moles

Molarity of solution = 0.3483 M

Putting values in above equation, we get:

Hence, the volume of NaOH required is 22.39 mL

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explanation: