Question

An alternative titration method for determining the concentration of vitamin C in a sample is to use an iodine solution. In this reaction iodine oxidizes the ascorbic acid to C₆H₆O₆.

A 20.00 mL sample of a 1.00 mg/mL vitamin C solution is placed in a flask along with 1 mL of a 1% starch solution to serve as an indicator. A dilute solution of iodine is placed in a buret. A titration is performed using the intense blue color produced by the reaction of excess I₂ with the starch as an indicator that the endpoint has been achieved. The titration was repeated using a 20.00-mL sample of juice. The following data were obtained in this experiment.

	Initial Buret Reading	Final Buret Reading
Standard Vitamin C sample	3.23 mL	8.54 mL
Juice Sample	8.54 mL	11.39 mL

a. Write the two balanced half reactions for this titration and the balanced net ionic equation.

b. What was the concentration (molarity) of iodine in the titrant?

c.   What was the concentration (mg/mL) of Vitamin C in the juice sample?

Answer 1

ascorbic acid + I2 → 2 I− + dehydroascorbic acid

since meq of ascorbic acid = meq of I2

hence concentraion of vitaminc C in the standard sample = 20 mg in in 20 ml

hence conc inmolarity = (20 X 10^-3 / 176.12 ) 1000 / 20

= 5.68 X10^-3 M

hence

20 X 5.68 X10^-3 = (8.54-3.23) X C _I2

C _I2 = 0.0214 M

hence now we can calculate teh concentraionof the vitaminc C in juice sample =

as 20 X C _{vitamin C} = 0.0214 X (11.39-8.54)

C _{vitamin C} =3.05 X 10^-3 M

or conc in g per ml = 3.05 X 10^-3 X 176.12 = 0.537 mg in 20 ml