Calculate the molarity of the student's sodium hydroxide solution. Be sure your answer has the correct number of significant digits.

Question

A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 68.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 her sodium hydroxide solution. When the titration reaches the equivalence point, the student finds she has used 64.7mL

 of sodium hydroxide solution.

Calculate the molarity of the student's sodium hydroxide solution. Be sure your answer has the correct number of significant digits.

Answer 1

Molar mass oxalic acid = 90.0 g / mole (assuming it is anhydrous)

Molar mass sodium hydroxide (NaOH) = 40.0 g / mole

Write the balanced equation for the neutralization reaction:

2NaOH + H₂C₂O₄ ==> 2H₂O + Na₂C₂O₄

Moles H₂C₂O₄ present = 68. mg x 1 g / 1000 mg x 1 mol / 90.0 g = 7.56x10^-4 moles

Moles NaOH needed to reach equivalence = 7.56x10^-4 moles H₂C₂O₄ x 2 mols NaOH / mole H₂C₂O₄ = 1.51x10^-3 moles NaOH because from the balanced equation it takes 2 moles NaOH for each mole of H₂C₂O₄

Molarity (moles/liter) of NaOH = 1.51x10^-3 mols / 64.7 ml x 1000 ml / L = 0.023 M (to 2 sig.figs.)