Question

In a titration of 48.0 mL of a 0.400 M solution of a diprotic acid H₂C₂O₄･2H₂O (oxalic acid dihydrate) with 0.333 M NaOH, how many mL of base are required to reach the second equivalence point?

Answer 1

To reach the second equivalence point in the titration of 48.0 mL of a 0.400 M solution of oxalic acid dihydrate with 0.333 M NaOH, 115.3 mL of NaOH is required.

Step-by-step explanation:

To determine how many mL of base are required to reach the second equivalence point in a titration of 48.0 mL of a 0.400 M solution of oxalic acid dihydrate (H₂C₂O₄·2H₂O) with 0.333 M NaOH, we must first calculate the total number of moles of the diprotic acid in the initial solution. Since oxalic acid can donate two protons per molecule, the second equivalence point occurs when twice the amount of acid moles have reacted with the base. The moles of oxalic acid are:

Moles of H₂C₂O₄·2H₂O = Volume × Concentration = 0.048 L × 0.400 mol/L = 0.0192 mol

At the second equivalence point, we need 2 moles of NaOH for every mole of H₂C₂O₄·2H₂O. Thus, the moles of NaOH required are 2 × 0.0192 mol = 0.0384 mol. To find the volume of NaOH solution needed, we rearrange the concentration formula:

Volume of NaOH = Moles of NaOH / Concentration of NaOH = 0.0384 mol / 0.333 mol/L

The required volume of NaOH is 115.3 mL (rounded to three significant figures).