10.0 mL of 0.05 M KMnO₄ solution was consumed in a titration with 10.0 mL of given oxalic acid dihydrate solution. The strength of given oxalic acid solution is _________ × 10⁻² g/L.
(Round off to the nearest integer)

Solution

Chemistry ma’am · 4 steps

Get the mole ratio from the balanced equation, then use the dihydrate's molar mass, 126, not 90.

1. Balanced reaction. In acid, permanganate takes 5 electrons per ion and each oxalate gives 2. So 2 permanganate ions react with 5 oxalate ions.

   2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

2. Moles of oxalic acid. Permanganate used is 10.0 × 0.05 = 0.5 mmol. Oxalic acid is 0.5 × 5/2 = 1.25 mmol, and this is in 10.0 mL, so its molarity is 0.125 M.
3. Convert to g/L. The solution was made from H₂C₂O₄.2H₂O, so its molar mass is 90 + 36 = 126 g mol⁻¹. Strength is 0.125 × 126 = 15.75 g L⁻¹.
4. So. 15.75 g L⁻¹ = 1575 × 10⁻² g L⁻¹, so the answer is 1575.

The common slip: 1125. Using 90, the molar mass of anhydrous oxalic acid, gives 11.25 g L⁻¹, which is 1125. The question asks about the dihydrate, which is 126.

The idea to keep. In a KMnO₄ and oxalate titration, the electron count gives the ratio: 5 for MnO₄⁻ and 2 for C₂O₄²⁻. For strength in g/L, use the molar mass of the compound that was actually weighed out.

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