Question

Pre-Lab Assignment

1. A 0.1017 g sample of KBrO3 was dissolved in dilute acid and treated with an excess of KI. BrO3¯ + 9I¯ + 6H+ Br¯ + 3I3¯ + 3H2O The generated I3¯ required 27.39 mL of a Na2S2O3 solution to reduce it to I¯. Use the above equation and those found in the introduction to determine the molarity of the Na2S2O3 solution.

2. The following data was obtained for an iodometric titration of an ascorbic acid tablet.

Mass of tablet: 1.2381 g

Mass of sample analyzed: 0.2483 g

Vol. of 0.0166 mol•L-1 KIO3: 25.00 mL

Vol. of 0.0198 mol•L-1 Na2S2O3 to reach endpoint: 11.01 mL

Using the equations found in the introduction, determine: the mass of ascorbic acid in the sample analyzed and the mass of ascorbic acid contained within the tablet.

Answer 1

1. A 0.1017 g sample of KBrO3 was dissolved in dilute acid and treated with an excess of KI.

BrO3¯ + 9I¯ + 6H+ Br¯ + 3I3¯ + 3H2O

The generated I3¯ required 27.39 mL of a Na2S2O3 solution to reduce it to I¯.

Use the above equation and those found in the introduction to determine the molarity of the Na2S2O3 solution.

First calculate the number of moles of KBrO3 in sample as follows:

Number of moles

= 0.1017 g sample of KBrO3 / molar mass; 167 g/ mole

=6.09*10^-4 moles KBrO3

Or 6.09*10^-4 moles BrO3^-1

Now calculate the moles of I3¯:

6.09*10^-4 moles KBrO3 * 3 moles of I3¯:/6.09*10^-4 moles BrO3^-

= 1.827*10^-3 moles of I3¯

The reaction of I3¯ and Na₂S₂O₃ is as follows:

6 Na₂S₂O₃ + 2 I₃^- = 3 Na₂S₄O₆ + 6 NaI

Moles of Na₂S₂O₃

=1.827*10^-3 moles of I3¯ *6 moles of Na₂S₂O₃ / 2 mole I₃^-

= 5.481*10^-3 moles of Na₂S₂O₃

Given that 27.39 mL of a Na2S2O3

= 0.02739 L

Molarity of Na2S2O3 =Number of moles / volume in L

= 5.481*10^-3 moles of Na₂S₂O₃/0.02739 L

= 0.200 M

Thus the molarity of the Na2S2O3 solution is 0.200 M