A suitable method for the determination of vitamin C (ascorbic acid, C6H3O6) is a titration with potassium iodate (KIO3). In this reaction, potassium iodate is used as a titrant and when added to an ascorbic acid solution that contains strong acid and potassium iodide (KI), the potassium iodate reacts with potassium iodide, liberating molecular iodine (I2) as shown in the reaction below: [1] KIO3 + 5KI + 6H* → 312 + 6K* + 3H20 During the titration, as long as the solution contains ascorbic acid, the l2 produced in equation I is used up in a rapid reaction with ascorbic acid (equation 2), during which dehydroascorbic acid (C6H6O6) and iodide ion (1) are formed: [2] C6H8O6 + 2 → C6H6O6 + 21 + 2H* Once all the ascorbic acid has been consumed, any excess iodine (1) will remain in solution. This excess iodine reacts with starch, to form an intensely blue colored complex, indicating that the endpoint is reached.
A suitable method for the determination of vitamin C (ascorbic acid, C6H3O6) is a titration with potassium iodate (KIO3). In this reaction, potassium iodate is used as a titrant and when added to an ascorbic acid solution that contains strong acid and potassium iodide (KI), the potassium iodate reacts with potassium iodide, liberating molecular iodine (I2) as shown in the reaction below: [1] KIO3 + 5KI + 6H* → 312 + 6K* + 3H20 During the titration, as long as the solution contains ascorbic acid, the l2 produced in equation I is used up in a rapid reaction with ascorbic acid (equation 2), during which dehydroascorbic acid (C6H6O6) and iodide ion (1) are formed: [2] C6H8O6 + 2 → C6H6O6 + 21 + 2H* Once all the ascorbic acid has been consumed, any excess iodine (1) will remain in solution. This excess iodine reacts with starch, to form an intensely blue colored complex, indicating that the endpoint is reached.
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Redox Titration: Explain what happened in the experiment through logical explanations and arguments.
![A suitable method for the determination of vitamin C (ascorbic acid, C6H8O6) is a titration with potassium iodate (KIO3). In this reaction, potassium
iodate is used as a titrant and when added to an ascorbic acid solution that contains strong acid and potassium iodide (KI), the potassium iodate
reacts with potassium iodide, liberating molecular iodine (I2) as shown in the reaction below:
[1] KIO3 + 5KI + 6H* → 312 + 6K* + 3H20
During the titration, as long as the solution contains ascorbic acid, the l produced in equation I is used up in a rapid reaction with ascorbic acid
(equation 2), during which dehydroascorbic acid (C6H606) and iodide ion (1) are formed:
[2] C6H8O6 + 12 → C6H6O6 + 21 + 2H*
Once all the ascorbic acid has been consumed, any excess iodine (1) will remain in solution. This excess iodine reacts with starch, to form an
intensely blue colored complex, indicating that the endpoint is reached.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F18ee930a-7d51-45e0-998f-136654caf94f%2F124229c1-680e-4fe8-a812-2520caa31422%2F2mf08yf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A suitable method for the determination of vitamin C (ascorbic acid, C6H8O6) is a titration with potassium iodate (KIO3). In this reaction, potassium
iodate is used as a titrant and when added to an ascorbic acid solution that contains strong acid and potassium iodide (KI), the potassium iodate
reacts with potassium iodide, liberating molecular iodine (I2) as shown in the reaction below:
[1] KIO3 + 5KI + 6H* → 312 + 6K* + 3H20
During the titration, as long as the solution contains ascorbic acid, the l produced in equation I is used up in a rapid reaction with ascorbic acid
(equation 2), during which dehydroascorbic acid (C6H606) and iodide ion (1) are formed:
[2] C6H8O6 + 12 → C6H6O6 + 21 + 2H*
Once all the ascorbic acid has been consumed, any excess iodine (1) will remain in solution. This excess iodine reacts with starch, to form an
intensely blue colored complex, indicating that the endpoint is reached.
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