Consider titration of NaOH solution versus 1.25 M oxalic acid solution. At the end point following burette readings were obtained.
(i) 4.5 mL
(ii) 4.5 mL
(iii) 4.4 mL
(iv) 4.4 mL
(v) 4.4 mL
If the volume of oxalic acid taken was 10.0 mL then the molarity of the NaOH solution is ________ M. (Rounded off to the nearest integer)

Solution

Chemistry ma’am · 4 steps

Pick the concordant reading first, then remember oxalic acid gives two H⁺ per molecule.

1. Concordant reading. In a titration you use the reading that repeats, not the first one. Here 4.4 mL appears three times in a row, so the volume of NaOH used is 4.4 mL.
2. Balanced equation. Oxalic acid is dibasic: H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O. So one mole of acid needs two moles of NaOH.

   M_(NaOH) × V_(NaOH) = 2 × M_(acid) × V_(acid)

3. Put in numbers. M_(NaOH) × 4.4 = 2 × 1.25 × 10.0 = 25, so M_(NaOH) = 25/4.4 = 5.68 M.
4. So. The molarity of the NaOH solution is 5.68 M, which rounds to 6.

The common slip: 3. Treating oxalic acid as monobasic gives (1.25 × 10)/4.4 = 2.84, which rounds to 3. Each oxalic acid molecule neutralises two NaOH.

The idea to keep. In any titration, take the concordant reading and balance the reaction before equating moles. Oxalic acid has n-factor 2, so it needs twice its moles of NaOH.

Answer from the NTA final answer key. Solution by PrepWiser.