Question

Determination of Vitamin C Concentration by Redox Titration 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 C&H&O 2. 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 l2 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 3.23 mL 8.54 mL Final Buret Reading 8.54 mL 11.39 mL Standard Vitamin C sample Juice Sample 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

Part a.

Oxidation-half reaction:

C₆H₈O₆ C₆H₆O₆ + 2H⁺ + 2e^-

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(Where C₆H₈O₆ = ascorbic acid, and C₆H₆O₆ = dehydroascorbic acid)

Reduction-half reaction:

I₂ + 2e^- $\rightarrow$ 2I^-

Balanced net ionic equation:

C₆H₈O₆ + I₂ $\rightarrow$ C₆H₆O₆ +  2H⁺ +?? 2I^-

Part b.

The concention of vitamin C in the standard solution (M1) = {(20 mL * 1 mg/mL)/(176 g/mol)} * 1/20 mL = 0.00568 M

Volume of vitamin C (V1) = 20 mL

The volume of iodine (V2) = 8.54-3.23 = 5.31 mL

The concentration of iodine (M2) = ?

Formula: M1V1 = M2V2

i.e. 0.00568 M * 20 mL = M2 * 5.31 mL

i.e. M2 = 0.02139 M

Part c.

The concentration of iodine (M1) = 0.02139 M

The volume of iodine (V1) = 11.39-8.54 = 2.85 mL

The volume of juice sample (V2) = 20 mL

The concentration of vitamin C in juice sample (M2) = ?

Now, M1V1 = M2V2

i.e. 0.02139 M * 2.85 mL = M2 * 20 mL

i.e. M2 = 0.00305 M