Lab Determination of Vitamin C 1) C6H8O6------ C6H6O6+ 2H9Plus+ 2I minus identif

Question

Lab Determination of Vitamin C 1) C6H8O6------ C6H6O6+ 2H9Plus+ 2I minus identify the oxidizing and reducing agent. 2) C6H8O6------ C6H6O6+ 2H9Plus+ 2I minus In eq. two mole of Hplus produced. Eaplain why H+ ions are formed. 3) A stock solution of I2 is standardizing using 0.1430g sample of pure ascorbic acid. The titration to the strach endpoint required 25ml of I2 solution. What is the molarity of I2 solution? 4) A vitamin C tablet was divide into three equal mass portion. One portion was crushed into fine powder, dissolved in water and titrated with standardized I2 solution from question 3 above. The titration to endpoint required 27.6ml of I2 solution. How many mg of Vitamin C are in the original tablet?

Explanation / Answer

1) C6H8O6 --------> C6H6O6 + 2H+ + 2I-

The reaction must have been done in iodine.

In iodine, oxidation state of I = 0

In I-, oxidation state of I = -1

So, iodine is being reduced, thus it is acting as oxidising agent.

In ascorbic acid, oxidation state of H is 0

In H+, oxidation state of H is +1.

So, hydrogen is oxidised, thus it acts as reducing agent.

Ascorbic acid is reducing agent and I2 is oxidising agent.

2) In the given equation, two H+ are formed. In ascorbic acid, there are two hydroxyl groups attached to the alkene of 5-membered ring. During redox reaction with iodine, these hydrogens are removed as H+ and the carbons get carbonyl groups. That is why two H+ are formed.

3) Mass of ascorbic acid = 0.1430 g

molar mass of ascorbic acid = 176 g/mol

Number of moles of ascorbic acid = mass/molar mass = 0.1430/176 = 8.1*10-4 mol

Moles of I2 required = moles of ascorbic acid = 8.1*10-4 mol

volume = 25 ml = 0.025 L

Molarity of I2 solution = number of moles/volume = 8.1*10-4 mol/0.025 L = 0.0325 M

4) Molarity of I2 solution = 0.0325 M

Volume of I2 solution required = 27.6 ml = 0.0276 L

Number of moles = Molarity*volume = 0.0325 M*0.0276 L = 8.97*10-4 mol

Moles of vitamin C = moles of I2 = 8.97*10-4 mol

Mass of vitamin C = molar mass*moles = 176*8.97*10-4 = 0.1579 g = 157.9 mg

The tablet was divided into 3 portions.

So, total Vitamin C = 157.9 mg*3 = 473.6 mg

Vitamin C in original tablet = 473.6 mg

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.