Look up the RDA (recommended dietary allowance) for Vitamin C (L-ascorbic acid)

Question

Look up the RDA (recommended dietary allowance) for Vitamin C (L-ascorbic acid) and report the value and your source. Prelab calculations: A Vitamin C Back-Titration Calculations A 2000mg tablet it crushed and brought to final volume with 1.5 M H_2 SO_4 in a 500.00 mL flask. A 25.00 aliquot of the tablet solution is mixed with 2 g of KI (Note: in excess) and 25.00 mL of 0.0130 M KIO_3 into an erlenmeyer flask as the analyte. It is back-titrated with a 0.0311 M sodium thiosulfate solution and requires 31.24 mL of thiosulfate to reach the end point. Calculate the measured mg of L-ascorbic acid in a vitamin C tablet. How many mols of original I_2 were produced from the IO_3^- ? How many mols of I_2 reacted with thiosulfate? How many mols of Vitamin C in the titration? How many mols of Vitamin C in the original 500.00 mL tablet solution? How many mg of Vitamin C in the original tablet?

Explanation / Answer

Question 1:

Molarity of KIO3 solution = 0.0130 M;    Volume of KIO3 solution = 25 mL = 25 *10-3 L = 0.025 L

Number of moles of KIO3 = Molarity * volume = 0.0130 * 0.025 = 3.25 * 10-4 mol

From equation (1) given, it follows that, 1 mole of IO3- forms 3 moles of I2

Hence, 3.25 * 10-4 mol of IO3- (in 3.25 * 10-4 mol of KIO3) forms, 3 * 3.25 * 10-4 = 9.75 * 10-4 mol of I2

Hence, number of moles of original I2 produced from the IO3- is 9.75 * 10-4 mol

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Question 2

Molarity of thiosulphate = 0.0311 M; Volume of thiosulphate = 31.24 mL = 31.24 *10-3 L = 0.03124 L

Number of moles of thiosulphate = Molarity * volume = 0.0311 * 0.03124 = 9.716 * 10-4mol

From equation (3) given, 2 moles of thiosulphate reacts with 1 mole of I2

Hence, 9.716 * 10-4mol of thiosulphate reacts with, 9.716 * 10-4* 1 / 2 = 4.858 * 10-4 mol of I2

Hence, 4.858 * 10-4 mol of I2 reacted with thiosulphate.

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Question 3:

Ascorbic acid reacts with I2 formed from IO3- and after ascorbic acid is completely reacted, the excess I2 is reacted with thiosulphate.

Hence, total amount of I2 formed from IO3-

= amount of I2 that reacted with ascorbic acid + amount of I2 that reacted with thiosulphate

Hence, amount of I2 that reacted with ascorbic acid

= total amount of I2 formed from IO3- - amount of I2 that reacted with thiosulphate

= 9.75 * 10-4 mol - 4.858 * 10-4 mol            ( Using answers of Questions 1 & 2)

= 4.892 * 10-4 mol

Hence, 4.892 * 10-4 mol of I2 reacted with ascorbic acid present in 25.00 aliquot of tablet solution.

From equation (2) given, it follows that, 1 mol I2 reacts with 1 mol ascorbic acid.

Hence, 4.892 * 10-4 mol of I2 reacted with 4.892 * 10-4 mol of ascorbic acid present in 25.00 aliquot of tablet solution.

Hence, 4.892 * 10-4 mol of ascorbic acid is present in 25.00 aliquot of tablet solution

or 4.892 * 10-4 mol of ascorbic acid (Vitamin C) was taken in titration.

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Question 4:

From answer of question 3, it follows that 4.892 * 10-4 mol of vitamin C was involved in the process.

Volume of tablet solution taken = 25 mL

Total volume of tablet solution = 500 mL

In 25 mL of the tablet solution, 4.892 * 10-4 mol of vitamin C is present;

Hence, 500 mL of tablet solution contains 500 * 4.892 * 10-4 / 25 = 97.84 * 10-4 mol of vitamin C is present;

Hence, 97.84 * 10-4 mol  = 9.784 * 10-3 mol of vitamin C is present in original 500 mL tablet solution.

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Question 5:

Since 500 mL tablet solution was made using the original tablet, number of moles of vitamin C in original tablet is same as that present in 500 mL tablet solution and equal to 9.784 * 10-3 mol (from answer of question 4).

Hence, number of moles of vitamin C in original tablet = 9.784 * 10-3 mol

Molar mass of vitamin C = 176.12 g / mol

Mass of vitamin C = number of moles * Molar mass

= 9.784 * 10-3 * 176.12 = 1723.1581 * 10-3 g = 1723.1581 mg   ( since 10-3 g = mg)

Hence, mass of vitamin C in original tablet = 1723.1581 mg  

or 1723.1581 mg of vitamin C is present in the original tablet.

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