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

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Presentation on theme: "PR3103 Pharmaceutical Analysis II Tutorial: “Three Troublesome Neighbors” Question 1 Mah Choon Siong."— Presentation transcript:

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?” a.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 Eo = -0.33V - 2e-, -2H+

4 Answer a.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 = E o (reduced) - E o (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: E 1 = E o (reduced) - E o (oxidised) = +0.34V – (+0.28V) = +0.06V, feasible E 2 = E o (reduced) - E o (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 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 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 = E o (reduced) - E o (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: E 1 = E o (reduced) - E o (oxidised) = +0.54V – (+0.28V) = +0.26V, feasible E 2 = E o (reduced) - E o (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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