Final answer:
The volume of a 0.3300-M solution of sodium hydroxide required to titrate 15.00 mL of 0.1500 M oxalic acid is approximately 13.6 mL.
Step-by-step explanation:
To determine the volume of a 0.3300-M solution of sodium hydroxide (NaOH) required to titrate 15.00 mL of 0.1500 M oxalic acid (C₂O₄H₂), we can use the balanced chemical equation:
C₂O₄H₂(aq) + 2NaOH(aq) ⟶ Na₂C₂O₄(aq) + 2H₂O(l)
From the equation, we can see that the stoichiometric ratio between C₂O₄H₂ and NaOH is 1:2. This means that for every 1 mole of C₂O₄H₂, we would need 2 moles of NaOH. Using the concentration and volume of C₂O₄H₂ and the stoichiometric ratio, we can calculate the volume of NaOH required:
Number of moles of C₂O₄H₂ = concentration × volume = 0.1500 M × 0.01500 L = 0.00225 moles
Number of moles of NaOH required = (2 moles NaOH/1 mole C₂O₄H₂) × 0.00225 moles = 0.00450 moles
Finally, we can use the concentration and volume of NaOH to calculate the required volume:
Volume of NaOH = number of moles/volume = 0.00450 moles/0.3300 M = 0.0136 L or 13.6 mL
Therefore, the volume of a 0.3300-M solution of NaOH required to titrate 15.00 mL of 0.1500 M oxalic acid is approximately 13.6 mL.