Questions: A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 210 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 102.4 mL 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.

Solution Steps

First, we need to calculate the number of moles of oxalic acid (\(\text{H}_2\text{C}_2\text{O}_4\)) used. The molar mass of oxalic acid is calculated as follows:

\[ \text{Molar mass of } \text{H}_2\text{C}_2\text{O}_4 = (2 \times 1.008) + (2 \times 12.01) + (4 \times 16.00) = 90.034 \, \text{g/mol} \]

The mass of oxalic acid used is 210 mg, which is 0.210 g. The number of moles of oxalic acid is:

\[ \text{Moles of } \text{H}_2\text{C}_2\text{O}_4 = \frac{0.210 \, \text{g}}{90.034 \, \text{g/mol}} = 0.002332 \, \text{mol} \]

Oxalic acid is a diprotic acid, meaning it can donate two protons. The balanced chemical equation for the reaction with sodium hydroxide (\(\text{NaOH}\)) is:

\[ \text{H}_2\text{C}_2\text{O}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2\text{H}_2\text{O} \]

From the equation, 1 mole of \(\text{H}_2\text{C}_2\text{O}_4\) reacts with 2 moles of \(\text{NaOH}\).

Using the stoichiometry of the reaction, the moles of \(\text{NaOH}\) are twice the moles of \(\text{H}_2\text{C}_2\text{O}_4\):

\[ \text{Moles of NaOH} = 2 \times 0.002332 \, \text{mol} = 0.004664 \, \text{mol} \]

The volume of the sodium hydroxide solution used is 102.4 mL, which is 0.1024 L. The molarity (\(M\)) of the sodium hydroxide solution is:

\[ M = \frac{\text{Moles of NaOH}}{\text{Volume in liters}} = \frac{0.004664 \, \text{mol}}{0.1024 \, \text{L}} = 0.04554 \, \text{M} \]

Final Answer