A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 300.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 123.3mL of sodium hydroxide solution.Calculate the molarity of the student's sodium hydroxide solution. Round your answer to 3 significant digits.

To calculate the molarity of the sodium hydroxide solution, we need to follow these steps:

1. First, we need to convert the mass of oxalic acid to moles. The molar mass of oxalic acid (H2C2O4) is approximately 90.03 g/mol. So, 300 mg is equal to 0.300 g. Therefore, the number of moles of oxalic acid is 0.300 g / 90.03 g/mol = 0.00333 mol.

2. Oxalic acid is a diprotic acid, which means it can donate two protons. Therefore, the mole ratio of oxalic acid to sodium hydroxide is 1:2. So, the moles of sodium hydroxide is 2 * 0.00333 mol = 0.00667 mol.

3. The volume of the sodium hydroxide solution used is 123.3 mL, which is equal to 0.1233 L.

4. Finally, we can calculate the molarity of the sodium hydroxide solution by dividing the moles of sodium hydroxide by the volume of the solution in liters. So, the molarity is 0.00667 mol / 0.1233 L = 0.0541 M.

So, the molarity of the student's sodium hydroxide solution is approximately 0.054 M.

To calculate the molarity of the sodium hydroxide solution, we first need to calculate the moles of oxalic acid used.

1. Convert the mass of oxalic acid to moles. The molar mass of oxalic acid (H2C2O4) is approximately 90.03 g/mol.

   Moles of oxalic acid = mass/molar mass = 300 mg / 90.03 g/mol = 0.003 kg / 90.03 g/mol = 0.000033 mol

2. Since oxalic acid is a diprotic acid, it can donate two protons. Therefore, the moles of NaOH would be twice the moles of oxalic acid.

   Moles of NaOH = 2 * moles of oxalic acid = 2 * 0.000033 mol = 0.000066 mol

3. The volume of NaOH used is 123.3 mL, which is 0.1233 L.

4. Molarity is defined as moles of solute per liter of solution. So, the molarity of the NaOH solution is:

   Molarity = moles of NaOH / volume of NaOH in L = 0.000066 mol / 0.1233 L = 0.000535 M

So, the molarity of the student's sodium hydroxide solution is approximately 0.00054 M when rounded to three significant digits.