Question

The following results were obtained when a sample of pineapple juice was analyzed to determine its vitamin C content. A 19.8 mL sample of pineapple juice was placed in a flask with 20 ml of 0.5 M H2SO4, 1.0 g of potassium iodide, KI, and 25.00 mL of 0.01988 M potassium iodate, KIO3. Titration with sodium thiosulfate required the use of 23.85 mL of 0.08014 M Na2S2O3. Calculate the concentration of vitamin C in the sample in mg vitamin C per 100 mL juice.

Answer 1

This is an example of a back titration; i.e., the KI/KIO3 produces I2, the I2 reacts with some of the vit C, the excess I2 is titrated with thiosulfate.
6H^+ + I^- + IO3^- ==> 3I2 + 3H2O
Then vitC + I2 ==> 2I^- + vitCx
mols IO3 added initially = M x L = 0.025 x 0.01988 = approx and you must do this more carefully = 0.0005
mols I2 formed initially = 0.0005 x (3 mol I2/1 mol IO3^-) = 0.0015.

How much I2 was left; i.e., the amount left is the amount vit C did NOT react with.
2S2O3^2- + I2 ==> 2I^- + S4O6^2-
mols thiosulate = M x L = 0.08014 x 0.02385 = approx 0.002
Convert 0.002 mols S2O3^2- to mols I2. From the equation that is 0.002 x 1/2 = about 0.001 mols I2 not used by the vit C. So the difference between initial and end is amount used by vit C. That is 0.0015-0.001 = approx 0.0005 mols vit C analyzed.
g vit C = mols x molar mass in 19.8 mL sample. Convert to mg/100 mL.