Iodimetric Titration of Vitamin C11
Purpose:
The purpose of this experiment is the determination of ascorbic acid. With this determination, it
will be decided whether or not manufacturers’ specifications are met. Titration technique will be
used.
Procedure:
Standardization of Thiosulfate Solution
Prepare starch indicator solution as a group for each lab. This will be done by making a paste of
5g soluble starch in 50mL of water. Pour the paste into 500mL of boiling water, boil until clear.
Prepare about 0.07M Na2S2O3 by dissolving an accurately weighed sample about 8.70g Na2S2O3
5H20 in 500mL freshly boiled water containing 0.05g of Na2CO3. Store this solution in a tightly
capped amber bottle. Prepare about 0.01M KIO3 by accurately weighing about 1g of solid
reagent and dissolving it in a 500mL flask. Standardize the thiosulfate solution using the
following procedure: pipet 50.00mL of KIO3 into an Erlenmeyer flask. Add 2g solid KI and
10mL of 0.5M H2SO4. Immediately titrate with thiosulfate until the solution has lost almost all if
its color (pale yellow). Then add 2mL of starch indicator and complete the titration. Repeat the
titration until you have three good trials.
Analysis of Vitamin C
Weigh out an appropriate amount of ascorbic acid for an approximate 30mL titration and
dissolve in 50mL of 0.3M H2SO4 with stirring. Add 2g of solid KI and 50.00mL of standard
KIO3. Titrate with standard thiosulfate as above, adding 2mL of starch indicator just before the
end point. Repeat this procedure with two weighed vitamin C tablets.
Reactions:
IO3- + 8I + 6H+ ----> 3I3- + 3H2O
I3- + 2S2O32- ----> 3I- + S4O62C6H8O6 + I3- + H2O ----> C6H8O7 + 3I- + 2H+
Data:
Iodimetric Titration Data
Calculations:
Molarity of KIO3:
𝐺𝑟𝑎𝑚𝑠 𝑜𝑓 𝐾𝐼03 ×
1 𝑚𝑜𝑙𝑒 𝐾𝐼0 1
×
214.00277𝑔 𝐿
Example: 0.9992𝑔 𝐾𝐼03 ×
1 𝑚𝑜𝑙𝑒 𝐾𝐼0
214.00277𝑔
×
1
0.500𝐿
= 0.00934𝑀
Molarity of Thiosulfate Solution:
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑜𝑓 𝐾𝐼03 × 𝐿𝑖𝑡𝑒𝑟𝑠 ×
3 𝑚𝑜𝑙 𝐼−
3
1 𝑚𝑜𝑙 𝐼𝑂−
3
×
2 𝑚𝑜𝑙 𝑆2 𝑂2−
3
1
𝑚𝑜𝑙 𝐼−
3
×
1
𝑇ℎ𝑖𝑜𝑠𝑢𝑙𝑓𝑎𝑡𝑒 𝑎𝑑𝑑𝑒𝑑
Example:
0.00934𝑀 𝐾𝐼03 × 0.0500𝐿 ×
3 𝑚𝑜𝑙 𝐼−
3
1 𝑚𝑜𝑙 𝐼𝑂−
3
×
2 𝑚𝑜𝑙 𝑆2 𝑂2−
3
1 𝑚𝑜𝑙 𝐼−
3
×
1
0.03662𝐿
= 0.076𝑀
Grams of Ascorbic Acid:
1 𝑚𝑜𝑙 𝐼−
1 𝑚𝑜𝑙 𝐶6 𝐻8 𝑂6
𝑔
3
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑜𝑓 𝑁𝑎2 𝑆2 𝑂3 × 0.03𝐿 ×
×
×
176.13
𝐴𝑠𝑐𝑜𝑟𝑏𝑖𝑐 𝐴𝑐𝑖𝑑
−
1 𝑚𝑜𝑙 𝐼3
𝑚𝑜𝑙
2 𝑚𝑜𝑙 𝑆2 𝑂2−
3
Example:
1 𝑚𝑜𝑙 𝐼−
1 𝑚𝑜𝑙 𝐶6 𝐻8 𝑂6
𝑔
3
0.075𝑀 𝑁𝑎2 𝑆2 𝑂3 × 0.03𝐿 ×
×
×
176.13
𝐴. 𝐴. = 0.1981𝑔
−
1 𝑚𝑜𝑙 𝐼3
𝑚𝑜𝑙
2 𝑚𝑜𝑙 𝑆2 𝑂2−
3
Weight Percent of Ascorbic Acid (Used for Vitamin C As well):
1 mol S O3
Liters of Thiosulfate added × M of Thiosulfate solution × 1 mol Na 2S
2 2 O3
1 mol S O3
0.00898 L × 0.075 M × 1 mol Na 2S
2 2 O3
1 mol I3
× 2 mol S
2 O3
1 mol I3
× 2 mol S
2 O3
= excess I3
= 0.00033675 mol excess I3
Mol of I3− :
Molarity of KIO3 × 0.05 L = mol KIO3 ×
0.00934 M KIO3 × 0.05 L = 0.000467 mol KIO3 ×
3 mol I3−
= mol I3−
1 mol KIO3
3 mol I3−
= 0.001401mol I3−
1 mol KIO3
mol of I3− − mol of I3− excess = mol of I3−
0.001401 mol − 0.00033675 mol = 0.00106425 mol of I3−
mol of I3− ×
1 mol A. A. molar mass A. A
×
= grams of A. A
1 mol I3−
1 mol
0.00106425 mol of I3− ×
1 mol A. A. 176.1256 g A. A
×
= 0.1874 g of A. A
1 mol I3−
1 mol
Grams calculated
× 100 = % A. A.
grams started with
0.1874 g
× 100 = 94.13 % A. A
0.19908 g
Weight % Claimed by Manufacturer:
0.25 𝑥 0.6255𝑡𝑎𝑏𝑙𝑒𝑡 𝑥
1𝑚𝑜𝑙
𝑥 100 = 74.96%
0.2086𝑔
Standard Deviation of Vitamin C:
𝑠 = √∑(𝑥𝑖 − 𝑥𝑚𝑒𝑎𝑛 ) ÷ 𝑛 − 1
= 80.36% ± 1.577
Conclusion:
The objective of this lab was to compare the determined weight percent of a vitamin C tablet and
compare it to that of the manufacturer’s weight percent. The average weight percent that we
calculated was from the vitamin C was 80.36%. That of the manufacturer was 74.96%. The
percent that we calculated in lab of the vitamin C tablet, 80.36%, exceeded the manufacturer’s
claim. This may be due to the solid not completely dissolving in the solution that was to be
titrated. Also, the amount of thiosulfate added during the titration was not what was expected,
which may have affected the calculations. The undissolved solid from the tablets may have been
impurities, additives, which may be the calcium carbonate.