Question

1a)  From the volume of titrant added, calculate the moles of vitamin C for each titration. Show a sample calculation for trial 1, and then just state the final results for the calculation of trials 2 and 3. Use 5 digits in all intermediate values and report the final value to the appropriate number of significant figures.

Trial 1

Trial 2

Trial 3

b) From the results of question 1, calculate the average value for the number of moles of vitamin C titrated. Report the final value to the appropriate number of significant figures.

c) Calculate the precision for the mole values used to calculate the average in question 2.

Data table:

	Trail1	Trail2	Trail3
Final buret reading	24.50 mL	48.80 mL	42.90 mL
Initial buret reading	0 mL	24.50 mL	18.32 mL
Volume of KlO3	24.50 mL	24.30 mL	24.58 mL

KlO3 concentration= 0.002012 M

Answer 1

Ans:

Volume of titrnt ued KlO3

Trial 1 = 24.50 mL

Trial 2 = 24.30 mL

Trial 3 = 24.58 mL

This redox titration reaction happens in the presence of KI and acid, as below:

The iodate ions are reduced to form iodine and the iodide ions are oxidized to form iodine.

KIO3 + 5 KI + 6H+ → 3I2 + 3H2O

The liberated iodine oxidizes the Vitamin-C (ascorbic acid) as below:

Ascorbic acid + I2 → 2I− + Dehydroascorbic acid

               

Calculations:

1. Note the volume of iodate solution used

2. Calculate the moles of iodate that reacted forming iodine

3. Using the equation of the reaction between the iodate ions and iodide ions as given below calculate the moles of iodine formed

2KIO3 + 10KI + 12H+ → 6I2 + 6H2O

4. From the titration equation (below) determine the moles of ascorbic acid reacting.

Ascorbic acid + I2 → 2I− + dehydroascorbic acid

5. Then calculate the concentration in mol L−1, of ascorbic acid in the solution

Mol. wt. of KIO3 = 214 g/mol

KlO3 concentration= 0.002012 M

The amount of KIO3 in L (1000 mL) = 214 * 0.002012 = 0.430568 g

Quantity of KIO3 in 1 mL = 0.430568 g / 1000 = 0.00043 g

Quantity of KIO3 consumed in each titration trial =

Trial 1 = 24.50 mL* 0.00043 g/mL = 0.010535 g

Trial 2 = 24.30 mL* 0.00043 g/mL = 0.010449 g

Trial 3 = 24.58 mL* 0.00043 g/mL = 0.010569 g

From the step 3 in calculation we know that 2 moles of KIO3 generates 6 moles of iodine or 1 mole of KIO3 will give 3 moles of iodine after redox reaction

2KIO3 + 10KI + 12H+ → 6I2 + 6H2O

KIO3             =   3I2

Mol. Wt.              214 g/mol            = 3 (253.8 g/mol)

214 g of KIO3 = 761.4 g

So, 1 g will react and give = 761.4/214 = 3.5579 g of Iodine

Now, we need to calculate that the amount of Iodine generated from each trial:

Trial 1 = 0.010535 * 3.5579 = 0.03748 g of iodine

Trial 2 = 0.010449 * 3.5579 = 0.03717 g of iodine

Trial 3 = 0.010569 * 3.5579 = 0.03760 g of iodine

Form the step 4 one mole of ascorbic acid reacts with one mole of iodine

Ascorbic acid + I2 → 2I− + dehydroascorbic acid

176.12 g/mol + 253.8 g/mol           

1g of iodine will react with = 176.12/253.8 = 0.69393 g of Vitamin C (ascorbic acid)

Now, quantity of ascorbic acid for each trial can be calculated as below:

Trial 1 = 0.03748 * 0.69393 = 0.02600/176.12 = 0.00015 moles of Vitamin C

Trial 2 = 0.03717 * 0.69393 = 0.02579/176.12 = 0.00015 moles of Vitamin C

Trial 3 = 0.03760 * 0.69393 = 0.02609/176.12 = 0.00015 moles of Vitamin C

Average Value of Vitamin C moles = (0.00015 + 0.00015 + 0.00015) / 3 = 0.00015 Moles

Since all the three trials give same values in 5 significant values, hence the precision (RSD) of this titration is 1.