Solutions: Iodine solution: (0.005 mol L^-1) Starch indicator solution: (0.5%) S

Question

Solutions: Iodine solution: (0.005 mol L^-1) Starch indicator solution: (0.5%) Stock “unknown” solution: 250mg of crushed up vitamin c tablet dissolved Solutions: Starch indicator solution: (0.5%) Stock “unknown” solution: 250mg of crushed up vitamin c tablet dissolved
In this experiment, we crushed up half a tablet of vitamin C (250 mg), diluted it in 50 mL of water, then transferred it to a 100 mL volumetric flask and filled it up with water. We pipetted 1mL Of this solution into a 250 flask, added 50 mL of water, and then 1 mL of starch indicator solution. We titrated this sample with the iodine solution until the blue-black color appeared, then repeated the experiment 2 more times.
how would I solve for these calculations?
REPORT SHEET: OXIDATION-REDUCTION TIRTRAT?ONS EXPERIMENT A. Standardization Trial1 Unknown (A, B, or C) Trial 2 Trial 3 volume pipetted Im ImL Titration Final reading gg A 37 - nitial reading 230 40 Volume of lodine soln Calculations Mols of l2 Mols of Vit C Molarity of Unknown Average Molarity

Explanation / Answer

The balanced chemical equation for the reaction of Vitamin C (ascorbic acid) with iodine is given as

ascorbic acid + I2 ---------> 2 I- +dehydroascorbic acid

As per the stoichiometric equation,

1 mole I2 = 1 mole ascorbic acid.

Trial 1

Trial 2

Trial 3

Volume pipetted (mL)

1

1

1

Final reading (mL)

39

37.7

39

Initial reading (mL)

30

30

30

Volume of iodine solution (mL) = (final reading) – (initial reading)

39 – 30 = 9

37.7 – 30 = 7.7

39 – 30 = 9

Calculations

Moles of I2 (mol) = (volume of I2 in L)*(concentration of I2 solution)

(9 mL)*(1 L/1000 mL)*(0.005 mol.L-1) = 4.5*10-5

(7.7 mL)*(1 L/1000 mL)*(0.005 mol.L-1) = 3.85*10-5

(9 mL)*(1 L/1000 mL)*(0.005 mol.L-1) = 4.5*10-5

Moles of Vitamin C (mol) = moles of I2 as per the stoichiometric equation

4.5*10-5

3.85*10-5

4.5*10-5

Molarity of unknown (mol.L-1) = (moles of Vitamin C)/(volume pipetted in L)

(4.5*10-5 mol)/[(1 mL)*(1 L/1000 mL)] = 0.045

(3.85*10-5 mol)/[(1 mL)*(1 L/1000 mL)] = 0.0385

(4.5*10-5 mol)/[(1 mL)*(1 L/1000 mL)] = 0.045

Average Molarity (mol.L-1)

1/3*(0.045 + 0.0385 + 0.045) = 0.0428

Trial 1

Trial 2

Trial 3

Volume pipetted (mL)

1

1

1

Final reading (mL)

39

37.7

39

Initial reading (mL)

30

30

30

Volume of iodine solution (mL) = (final reading) – (initial reading)

39 – 30 = 9

37.7 – 30 = 7.7

39 – 30 = 9

Calculations

Moles of I2 (mol) = (volume of I2 in L)*(concentration of I2 solution)

(9 mL)*(1 L/1000 mL)*(0.005 mol.L-1) = 4.5*10-5

(7.7 mL)*(1 L/1000 mL)*(0.005 mol.L-1) = 3.85*10-5

(9 mL)*(1 L/1000 mL)*(0.005 mol.L-1) = 4.5*10-5

Moles of Vitamin C (mol) = moles of I2 as per the stoichiometric equation

4.5*10-5

3.85*10-5

4.5*10-5

Molarity of unknown (mol.L-1) = (moles of Vitamin C)/(volume pipetted in L)

(4.5*10-5 mol)/[(1 mL)*(1 L/1000 mL)] = 0.045

(3.85*10-5 mol)/[(1 mL)*(1 L/1000 mL)] = 0.0385

(4.5*10-5 mol)/[(1 mL)*(1 L/1000 mL)] = 0.045

Average Molarity (mol.L-1)

1/3*(0.045 + 0.0385 + 0.045) = 0.0428

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