Achemistry student needs to standardize a fresh solution of sodium hydroxide. she carefully weighs out 385.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 123.7ml of sodium hydroxide solution. calculate the molarity of the student's sodium hydroxide solution.

The molarity of the sodium hydroxide solution is 0.0692 M

Explanation:

Step 1: Data given

Mass of H2C2O4 = 385 mg = 0.385 grams

volume = 250 mL = 0.250 L

Volume of NaOH = 123.7 mL = 0.1237 L

Molar mass H2C2O4 = 90.03 g/mol

Step 2: The balanced equation

2NaOH + H2C2O4 → Na2C2O4 + 2H2O

Step 3: Calculate moles H2C2O4

Moles H2C2O4 = mass H2C2O4 / molar mass H2C2O4

Moles H2C2O4 = 0.385 grams / 90.03 g/mol

Moles H2C2O4 = 0.00428 moles

Step 4: Calculate molarity of H2C2O4

Molarity H2C2O4 = moles / volume

Molarity H2C2O4 = 0.00428 moles / 0.250 L

Molarity H2C2O4 = 0.01712 M

Step 5: Calculate molarity of NaOH

2*Ca*Va = n*Cb*Vb

⇒ with Ca = Molarity of H2C2O4 = 0.01712 M

⇒ Va = volume of H2C2O4 = 0.250 L

⇒Cb = molarity of NaOH = TO BE DETERMINED

⇒ Vb = volume of NaOH = 0.1237 L

Cb = (2*0.01712*0.250)/0.1237

Cb = 0.0691 M

The molarity of the sodium hydroxide solution is 0.0692 M