Question

in the experiment, a vitamin C tablet was dissolved in water and diluted to 100 ml. A student took 50 ml of solution to titrate. The titration volume of 0.025 M KIO3 (aq) was 15.8 ml. Calculate the ascorbic acid content per tablet. The molar mass of vitamin C is 176 g/mol

IO3- + 5I- + 6H+ = 3I2 + H20 C6H8O6 + I2 = C6H8O6 + 2I- + 2H+

According to the experiment, whihc of the following is correct

a) vitamin c is also called ascorbic acid

b) Iodide ion I- react with stgarch indicator to form the complex

c) Vitamin C is an ionic compound and soluble in water

d) vitamin C is used as a reducing agent

Answer 1

part 1:

IO3- + 5I- + 6H+ = 3I2 + H20

here KIO3 (IO3-) reacts to form I2

number of moles of KIO3 = Molarity of KIO3 * Volume of KIO3

                                        = 0.025 M * 15.8 mL

                                         = 0.395 mmol

From above reaction:

number of moles of I2 formed = 3* number of moles of KIO3 = 3 * 0.395 = 1.185 mmol

Now look at 2nd reaction:

C6H8O6 + I2 = C6H6O6 + 2I- + 2H+

Here 1 mol ascorbic acid reacts with 1 mol of I2

number of moles of ascorbic acid reacted = number of moles of I2 = 1.185 mmol = 0.001185 mol

mass of ascorbic acid = number of moles of ascorbic acid * molar mass of ascorbic acid

                                  = 0.001185 mol* 176 g/mol

                                  =0.20856 gm

                                 =0.21 gm (approx)

Answer: ascorbic acid content per tablet = 0.21 gm

Part 2:

In 2nd reaction iodine is reduced; it gains two electrons and we have the REDUCTION of iodine.

So, Vitamin C is oxidized by losing two electrons. So we have the OXIDATION of vitamin C and hence it will act as reducing agent.

Answer : d) vitamin C is used as a reducing agent

This question was very lengthy. I have tried my best to explain it to you.