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A chemistry student needs to standardize a fresh solution of sodium hydroxide. He carefully weighs out 32.mg of oxalic acid H2C2O4, a diprotic acid that can be purchased inexpensively in high purity, and dissolves it in 250.mL of distilled water. The student then titrates the oxalic acid solution with his sodium hydroxide solution. When the titration reaches the equivalence point, the student finds he has used 60.9mL of sodium hydroxide solution. Calculate the molarity of the student's sodium hydroxide solution. Round your answer to 2 significant digits. M

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Answer:

0.012 M

Explanation:

The reaction that takes place is:

  • H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O

First we convert 32 mg of H₂C₂O₄ to moles, using its molar mass:

  • 32 mg ÷ 90 mg/mmol = 0.36 mmol

Then we calculate how many NaOH moles would react with that amount, using the stoichiometric coefficients:

  • 0.36 mmol H₂C₂O₄ * [tex]\frac{2mmolNaOH}{1mmolH_2C_2O_4}[/tex] = 0.72 mmol NaOH

Finally we divide NaOH moles by the used volume, to calculate the molarity:

  • 0.72 mmol NaOH / 60.9 mL = 0.012 M

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