We appreciate your visit to A chemistry student needs to standardize a fresh solution of sodium hydroxide He carefully weighs out 86 mg of oxalic acid tex H 2C 2O. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
The molarity of the student's sodium hydroxide solution is approximately **0.00976 M**.
The balanced chemical equation for the reaction between sodium hydroxide (NaOH) and oxalic acid (H₂C₂O₄) is:
[tex]\(2 \, \text{NaOH} + \text{H₂C₂O₄} \rightarrow \text{Na₂C₂O₄} + 2 \, \text{H₂O}\)[/tex]
From the equation, you can see that 2 moles of sodium hydroxide react with 1 mole of oxalic acid.
Given the volume of sodium hydroxide solution used (97.6 mL) and the mass of oxalic acid (86 mg), we can first calculate the number of moles of oxalic acid:
Number of moles of oxalic acid = [tex]\(\frac{\text{mass}}{\text{molar mass}}\)Molar mass of oxalic acid (H₂C₂O₄) = \(2 \times \text{atomic mass of H} + 2 \times \text{atomic mass of C} + 4 \times \text{atomic mass of O} = 2 \times 1.01 + 2 \times 12.01 + 4 \times 16.00 = 90.04 \, \text{g/mol}\)Number of moles of oxalic acid = \(\frac{0.086 \, \text{g}}{90.04 \, \text{g/mol}} = 0.000955 \, \text{mol}\)Since 2 moles of NaOH react with 1 mole of H₂C₂O₄, the number of moles of NaOH used is also \(0.000955 \, \text{mol}\).[/tex]
Now we can calculate the molarity of the sodium hydroxide solution:
Molarity (M) = [tex]\(\frac{\text{moles of solute}}{\text{volume of solution (L)}}\)Volume of solution (L) = \(97.6 \, \text{mL} = 0.0976 \, \text{L}\)[/tex]
Molarity = [tex]\(\frac{0.000955 \, \text{mol}}{0.0976 \, \text{L}} = 0.00976 \, \text{M}\)[/tex]
**Summary:** The molarity of the student's sodium hydroxide solution is approximately **0.00976 M**.
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