A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 338.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 136.3mL 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.
To calculate the molarity of the sodium hydroxide solution, you need to use the balanced equation for the reaction between sodium hydroxide (NaOH) and oxalic acid (H2C2O4), and take into account the stoichiometry of the reaction.
The balanced equation for the reaction is:
2NaOH + H2C2O4 -> Na2C2O4 + 2H2O
From the equation, you can see that it takes 2 moles of NaOH to react with 1 mole of H2C2O4.
First, calculate the number of moles of oxalic acid used in the reaction. To do this, divide the mass of oxalic acid by its molar mass:
0.338 g / 90.04 g/mol = 0.00376 mol
Next, use the stoichiometry of the balanced equation to calculate the number of moles of NaOH that reacted with the oxalic acid. Since the stoichiometric ratio is 2:1 between NaOH and H2C2O4, the number of moles of NaOH is half the number of moles of H2C2O4:
0.00376 mol / 2 = 0.00188 mol
Now, calculate the molarity of the NaOH solution by dividing the number of moles of NaOH by the volume of the solution used in the titration:
Molarity = number of moles / volume of solution
Molarity = 0.00188 mol / 0.1363 L = 0.0138 M
Therefore, the molarity of the student's sodium hydroxide solution is 0.0138 M, with the correct number of significant digits.
Check the wording of your question, and make sure you use the correct number of Sig figs, which should be three.
No.
338.mg=0.3380g of H2C2O4
0.3380g of H2C2O4*(mole/90.03g)= moles of Acid
Moles of acid/0.25L=M
C=(M*0.25L)/0.1363L