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PR3103 Pharmaceutical Analysis II

PR3103 Pharmaceutical Analysis II. Tutorial: “Three Troublesome Neighbors” Question 1 Mah Choon Siong. Question.

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PR3103 Pharmaceutical Analysis II

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  1. PR3103 Pharmaceutical Analysis II Tutorial: “Three Troublesome Neighbors” Question 1 Mah Choon Siong

  2. Question After learning that you are currently doing a course in pharmacy at NUS, your neighbor, Mrs Seow came over with a specific question: “Is it true that I should not cook my vegetables over a copper frying pan. I heard that this will destroy all the vitamin C? What’s the big deal about losing some Vitamin C anyway?” • Is it true that copper destroys Vitamin C? Please provide a reasonable explanation based on your understanding of reduction potential, using the information as provided below. [R] Cu2+ + 2e- Cu Eo = +0.34V [O] DHA + e- + H+ Ascorbyl radical Eo = -0.17V [O] Ascorbyl radical + e- + H+ Ascorbate Eo = +0.28V [O] DHA + 2e- + 2H+ Ascorbate Eo = -0.33V [R] I2 + 2e- 2I- Eo = +0.53V [R] I3- + 2e- 3I- Eo = +0.54V

  3. Eo = +0.28V Eo = -0.17V - 2e-, -2H+ Eo = -0.33V

  4. Answer • Is it true that copper destroys Vitamin C? Please provide a reasonable explanation based on your understanding of reduction potential, using the information as provided below. Yes. Assume standard conditions, First method: E = Eo (reduced) - Eo (oxidised) = +0.34V – (– 0.33V) = +0.67V, feasible [R] Cu2+ + 2e- Cu Eo = +0.34V [O] DHA + 2e- + 2H+ Ascorbate Eo = -0.33V

  5. Answer Assume standard conditions, Second method: E1 = Eo (reduced) - Eo (oxidised) = +0.34V – (+0.28V) = +0.06V, feasible E2 = Eo (reduced) - Eo (oxidised) = +0.06V – (–0.17V) = +0.23V, feasible [R] Cu2+ + 2e- Cu Eo = +0.34V [O] DHA + e- + H+ Ascorbyl radical Eo = -0.17V [O] Ascorbyl radical + e- + H+ Ascorbate Eo = +0.28V

  6. Question b. The BP assay for vitamin C involves a redox titration with iodine. Briefly describe this assay and comment why you think this assay will work?

  7. Assay of ascorbic acid Dissolve 0.150 g in a mixture of 10 ml of dilute sulphuric acid R and 80 ml of carbon dioxide-free water R. Add 1 ml of starch solution R. Titrate with 0.05 M iodine until a persistent violet-blue colour is obtained. 1 ml of 0.05 M iodine is equivalent to 8.81 mg of C6H8O6.

  8. Redox Titration Dissolve 0.150 g in a mixture of 10 ml of dilute sulphuric acid R and 80 ml of carbon dioxide-free water R. • Ascorbate becomes ascorbic acid • Iodine (I2) is solubilized by complexing with iodide in sulphuric acid (I2+ I- ↔ I3-) Add 1 ml of starch solution R. [Indicator] Titrate with 0.05 M iodine until a persistent violet-blue colour is obtained. • Iodine oxidizes ascorbic acid into dehydroascorbic acid (DHA) • Excess iodine subsequently react with starch to form a blue-black complex (endpoint) 1 ml of 0.05 M iodine is equivalent to 8.81 mg of C6H8O6. • 1 mol I2 reacted with 1 mol C6H8O6

  9. Answer Yes. Assume standard conditions, First method: E = Eo (reduced) - Eo (oxidised) = +0.54V – (–0.33V) = +0.87V, feasible [O] DHA + 2e- + 2H+ Ascorbate Eo = -0.33V [R] I2 + 2e- 2I- Eo = +0.53V [R] I3- + 2e- 3I- Eo = +0.54V

  10. Answer Assume standard conditions, Second method: E1 = Eo (reduced) - Eo (oxidised) = +0.54V – (+0.28V) = +0.26V, feasible E2 = Eo (reduced) - Eo (oxidised) = +0.26V – (–0.17V) = +0.42V, feasible [O] DHA + e- + H+ Ascorbyl radical Eo = -0.17V [O] Ascorbyl radical + e- + H+ Ascorbate Eo = +0.28V [R] I2 + 2e- 2I- Eo = +0.53V [R] I3- + 2e- 3I- Eo = +0.54V

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