Question

Question: Using the information and equations below determine the concentration of Vitamin C in an unknown soft drink.

Hints: In this experiment, the amount of potassium iodate determines the amount of iodine that is produced. In addition, the first part of the experiment requires the determination of the concentration of the thiosulphate solution. This is done by generating a known amount of iodine via the reaction between potassium iodide and a known amount of potassium iodate in an acidic medium. In the second part of the titration, the same amount of iodine is generated as in the first part, and to this solution is added the vitamin C containing solution. Thus some of the iodine generated reacts with the vitamin C. The amount of unreacted iodine is determined by titration with the standardized thiosulfate solution. Once the amount of iodine that reacts with the vitamin C is known, the amount and then the concentration of vitamin C in the unknown solution

can be calculated.

Standardization of Sodium Thiosulphate

In Erlenmeyer flask: 10ml of 0.3M sulphuric acid + 20ml 1.0002x10^-2 M potassium iodate stock solution

Starch indicator added (1ml) and titration was carried out

Titration 1:

V_i= 0.00ml

V_f=23.00ml

Iodine=2.000g

Titration 2:

V_i= 0.00ml

V_f=20.10ml

Iodine=2.000g

Titration 3:

V_i= 0.00ml

V_f=22.60ml

Iodine=2.000g

Quantification of Ascorbic Acid in Drink Samples

In Erlenmeyer flask: 10ml of 0.3M sulphuric acid + 20ml 1.0002x10^-2 M potassium iodate stock solution + 20ml of an unknown citric acid solution

Starch indicator added (1ml) and titration was carried out

Titration 1:

V_i= 0.00ml

V_f=15.70ml

Iodine=2.000g

Titration 2:

V_i= 0.00ml

V_f=28.90ml

Iodine=2.000g

Titration 3:

V_i= 0.00ml

V_f=40.90ml

Iodine=2.000g

Equations:

NaOH (aq) + HCl (aq) _ H2O (l) + NaCl (aq)

Amount of NaOH present before reaction with the unknown acid

= (0.1000 M)(2.500x10^-2L)

=2.500x10^-3 moles

Amount of NaOH present after reaction with the unknown acid

=(0.1000 M)(1.550x10^-2L)

=1.550x10^-3 moles

NaOH (aq) + HA (aq) _ H2O (l) + NaA (aq)

Amount of NaOH that reacted with the unknown acid

=2.500x10^-3 moles - 1.550x10^-3 moles

=9.5x10^-4

concentration of the unknown acid = 3.80 x 10^-3

IO3^– + 5 I^– + 6 H⁺ à 3 I2 + 3 H2O

I2 + 2S2O3^2– à 2 I– + S4O6^2–

Answer 1

Standardization of Thiosulphate:

If an excess of iodide is used to quantitatively reduce a chemical species while simultaneously forming iodine, and if the iodine is subsequently titrated with thiosulfate, the technique is iodometry.

As iodine is volatile and insoluble it cannot be used as a titrant. So iodide is added to iodine solution which converts iodine to triiodide which is soluble in water.

                                                                   I₂ + I^-   I₃

This triiodate reacts with thiosulphate to give iodine

                                                          2S₂O₃^2–+ I₃^– S₄O₆^2–+ 3I^–

Here 10ml of 0.3M sulphuric acid is added to 20ml 1.0002x10^-2 M potassium iodate then excess of iodine is added.

So the all iodate is reacts with excess of iodine to give triiodate

                                               IO₃^- + 8I^- + 6H⁺ 3I₃^- + 3 H₂O

So one mole of IO₃^- gives 3 moles of triiodate and one mole of triiodate reacts with 2 moles of thiosulphate.

That is one mole IO₃- reduces 6 moles of thiosulphate.

                                                                V₁M₁ = 6V₂M₂

Here 20ml of 1.0002x10^-2 M potassium iodate contains ( 20 ml x 1.0002x10^-2 mmol/ml ) 0.2004 mmol of potassium iodate.

Titration:1

                                                          0.2004 = 6 x 23 x M₂

                                                                              M₂   = 0.2004/6 x 23

                                               M₂   = 0.00144M

Titration:2

                                                          0.2004 = 6 x 20.10 x M₂

                                                                              M₂   = 0.2004/6 x 20.10

                                               M₂   = 0.00166M

Titration:3

                                                          0.2004 = 6 x 22.60 x M₂

                                                                              M₂   = 0.2004/6 x 22.60

                                               M₂   = 0.00147M

The concentration of thiosulphate is 0.00152M

Quantification of Ascorbic Acid in Drink Samples:

Here as in the previous experiment the same amount of triiodate is generated by adding 20ml 1.0002x10^-2 M potassium iodate and excess of iodine.

The ascorbic acid is added.

Ascorbic acid is oxidized to dehydroascorbic acid by iodine

So the formed triiodate reacts with ascorbic acid to form dehydroascorbic acid.

The concentration of unreacted triiodate is determined by titrating with thiosulphate.

From the initial concentration and unreacted concentration, the unknown concentration of ascorbic acid can be determined.

Titration 1:

                                                           V₁M₁ = 6V₂M₂

                                  6 x 15.70 x 0.00152 = 50 x M₁

Concentation of unreacted Iodate M₁ = 0.00286 M

Initial concentration of Iodate               = 0.2004 mmol / 50 mL

                                                               = 0.004008M

The concentration of ascorbic acid ( 0.004008 - 0.00286 ) is 0.001148M

Titration 2:

                                                           V₁M₁ = 6V₂M₂

                                  6 x 28.90 x 0.00152 = 50 x M₁

Concentation of unreacted Iodate M₁ = 0.00527 M

Initial concentration of Iodate               = 0.004008 M

Here the unreacted iodate concentration is (0.00527) greater than initial concentration (0.004008) This is not possibel so the either V_i or V_f might be wrong.

[Assuming that you have titrated using the same burette without refilling.

So

Vi = 15.70ml(previous titration end point)

Vf = 28.90 ml

V = Vf - Vi = 28.90 - 15.70 = 13.20ml

Using this

                                                         V₁M₁ = 6V₂M₂

                                  6 x 13.20 x 0.00152 = 50 x M₁

Concentation of unreacted Iodate M₁ = 0.00240 M

Initial concentration of Iodate               = 0.004008 M

The concentration of ascorbic acid ( 0.004008 - 0.00240 ) is 0.001600M ]

Titration 3:

                                                           V₁M₁ = 6V₂M₂

                                  6 x 40.90 x 0.00152 = 50 x M₁

Concentation of unreacted Iodate M₁ = 0.00746 M

Initial concentration of Iodate               = 0.004008 M

Here the unreacted iodate concentration is (0.00746) greater than initial concentration (0.004008).

[Again assuming that you have titrated using the same burette without refilling.

So

Vi = 28.90ml(previous titration end point)

Vf = 40.90 ml

V = Vf - Vi = 40.90 - 28.90= 12.00ml

Using this

                                                         V₁M₁ = 6V₂M₂

                                  6 x 12.00 x 0.00152 = 50 x M₁

Concentation of unreacted Iodate M₁ = 0.00218 M

Initial concentration of Iodate               = 0.004008 M

The concentration of ascorbic acid ( 0.004008 - 0.00218 ) is 0.001819M ]