Question

Ascorbic acid or Vitamin C is a simple compound with the following chemical formula C6H8O6. Besides being an acid, it is also a reducing agent. One method for determining the amount of vitamin C in a sample is to titrate it with a solution of bromine, Br2, a good oxidizing agent based on the following chemical equation:

Br2 (aq) + C6H8O6 (aq) -->  2HBr (aq) + C6H6O6 (aq)

Suppose a 1.00-g “chewable” vitamin C tablet requires 27.85 mL of 0.102 M Br2 for titration to the equivalence point. The tablet bottle is labelled 85% active vitamin C ingredient. Is the bottle mislabelled? Back up your answer (whether yes or no) with all required calculations.

Answer 1

The given reaction is Brz (aq) + C6H8O6 (ar) --→ 2 Hbr (aq) + C64606 (ar) Accueling to balanced Cheriaal Machion, it is evident that 1 mole of ascorbic acid (vitamin c) is neutralized by 2 more of bromine Experimentally 2,007 vitamin C in neutralized by 27.85m2 of 0.102 m Br2 to the equivalence point Now 27.85 mL of 0.102 M Br2 = 0.027852 of 0.102M Brz - 002785 LX 0:102 mol 0.102 mol -3 = 2.8407810 mol vitamin c. 1 According to balanced reaction 1 mole of ascorbic and is neutralized by 1 mole Br2. .: Moth of Br2 reacted 2.8407 x 103 mol Now molar mass of Br2 = 159.81 g/mol .:: 2.8407X163 mol brez 2.8407x16%mo*159.81 g/mol = 0.4540 g Brz 1 g of vitamin C tablet contains 0.454 og of active vitamin c ingredients: so percentage of active vitamin C gradients in 1.00g of chewable vitamin C tablet = 0;145409 x100 % = 45:40% Therefore the label on tablet bottle (85% active vitamin e) is mislabelled. To be accurately labelled, it should have been 45.40% active vitamin C ingradient. lo g