A chemistry student needs to standardize a fresh solution of sodium hydroxide. H

Question

A chemistry student needs to standardize a fresh solution of sodium hydroxide. He carefully weighs out 131.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 31.8mL 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.

Explanation / Answer

The reaction equation will be

2NaOH + H2C2O4 ---> Na2C2O4 + 2H2O

So with each mole of oxalic acid two moles of NaOH will be required to react to reach the equivalence point

Moles of Oxalic acid used = mass / Molar mass = 131mg / 90g/mole = 1.46 mmol

moles of NaOH used = 2 X moles of oxalic acid = 2 X 1.46 mmol = 2.92 mmol

Molarity of NaOH = millimoles of NaOH / volume in mL = 2.92 /31.8 = 0.0918 M = 9.18 X 10-2 M

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