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Question

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equation 3 is as follows......

C6H8O6+I2---> C6H6O6 + (2H^+) + 2I-

question 1. Identify the oxidizing agent and the reducing agent

question 2. In (equation 3) two moles of H+ are produced. Explain why H+ ions are formed.

question 3. A stock solution of I2 is standardized using a 0.1430 g sample of pure ascorbic acid. The titration to the starch end point required 25.45 mL of the I2 solution. What is the molarity of the I2 solution?

question 4. A vitamin C tablet was divided into three equal mass portions. One portion was crushed into a fine powder, dissolved in water and titrated with the standardized I2 solution from question 3, above. The titration to end point required 27.6 mL of the I2 solution. How many mg of Vitamin C are in the original tablet?

Explanation / Answer

Q.1: An oxidizing agent is the one that receives electron (itself reduced) and helps in oxidation. Here I2 receives electron. Hence I2 is oxidizing agent.

A reducing agent is the one that donates electron (itself oxidized) and helps in reduction. Here C6H8O6 donates electron. Hence C6H8O6 is reducing agent.

Q.2: H+ are prduced as a result of donation of electron by C6H8O6.

Q.3:Given the mass of C6H8O6 = 0.1430 g

molar mass of C6H8O6 = 176.12 g/mol

Hence moles of ascorbic acid taken, C6H8O6 taken = 0.1430 g / 176.12 g/mol = 8.12x10-4 mol

suppose the molarity of I2 solution be 'M' mol/L

Volume of I2 solution required to reach end point = 25.45 mL = 0.02545 L

Hence moles of I2 solution required to reach end point = MxV = 0.02545M mol

From equation (3) it is clear that equal moles of ascorbic acid and I2 react each other. Hence

moles of ascorbic acid = moles of I2

=> 8.12x10-4 mol = 0.02545M mol

=> M = 8.12x10-4 mol / 0.02545 = 0.0319 mol/L (answer)

Q.4: moles of I2 required = MxV = 0.0319 mol/L x 0.0276 L = 8.8044x10-4 mol

Since the tablet is divided into 3 parts, moles of vitamin C (ascorbic acid) in the tablet

= 3 x 8.8044x10-4 mol = 2.64132x10-3 mol

Hence mass of vitamin C in the tablet = 2.64132x10-3 mol x 176.12 g/mol = 0.4652 g =465.2 mg (answer)

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