Experiment 5: Iodimetric Titration of Vitamin C
Objective:
The purpose of this lab is to determine amounts of Vitamin C in a product compared to their
nutritional claims.
Procedure:
Standardization of Thiosulfate







Prepare Starch Indicator using 5g soluble starch in 50mL water
Add to 500mL boiling water and boil until clear
Dissolve 8.70g Na2S2O3 5H2O in 500mL boiled water with 0.05g Na2CO3
Prepare 0.01M KIO3 by dissolving 1g solid reagent in a 500mL volumetric flask
Standardize Thiosulfate by adding 50mL of KIO3 into an Erlenmeyer Flask with 2g KI and
5mL of 6.0M H2SO4
Titrate to a pale yellow color with thiosulfate
Add 5mL of Starch indicator and complete titration
Analysis of Vitamin C




Weigh approximate amount of ascorbic acid needed for 30mL titration and dissolve it in
50mL of 0.3M H2SO4
Add 2g KI and 50mL of standardized KIO3
Titrate with standardized thiosulfate adding 5mL of starch indicator before completion
Repeat procedure for 2 weighed vitamin C tablets
Equations:
6H+ + IO3- + 8I- → 3I- + 3H2O
I3- + 2S2O32- → 3I- + S4O62C6H8O6 + I3- + H2O →C6H8O7 + 3I- + 2H+
Data:
Standardization of the Thiosulfate
Mass of Soluble Starch (g)
Mass of Na2S2O35H2O (g)
Mass of Na2CO3 (g)
Mass of Solid Reagent (g)
5.0151
8.7012
0.0514
1.0030
Trial
Mass of KI (g)
1
2
3
2.0010
2.0053
2.0117
Average
Volume Thiosulfate Used
(mL)
7.7
7.2
7.6
M Thiosulfate
0.065
0.069
0.066
0.067
Vitamin C Data
Trial
Mass (g)
Mass of KI (g)
Ascorbic Acid
CVS Tablet
Nature’s Bounty
Tablet
0.1383
0.1948
0.1993
2.0057
2.0081
2.0034
Thiofulfate
Volume (mL)
16.8
12.7
10.6
Vitamin C in
Sample (g)
0.1649
0.1892
0.2015
% Vitamin C
in Sample
119.2
97.13
101.1
Calculations:
Title
M Na2S2O3
Formula
Calculation
𝑀 𝑁𝑎2 𝑆2 𝑂3
3𝑚𝑜𝑙 𝐼3−
2𝑚𝑜𝑙 𝑆2 𝑂3
1
= 𝑀 𝐾𝐼𝑂3 (𝐿 𝐾𝐼𝑂3 ) (
)(
)
− )(
1𝑚𝑜𝑙 𝐼𝑂3
1𝑚𝑜𝑙 𝐼3
𝐿 𝑁𝑎2 𝑆2 𝑂3
Moles of I3-
Excess moles
of I3-
Reacted
moles of I3Mass of
Vitamin C
% Vitamin C
in Sample
3𝑚𝑜𝑙𝐼3−
𝑀𝑜𝑙𝑒𝑠 𝐼3− = 𝑀𝐾𝐼𝑂3 (𝐿 𝐾𝐼𝑂3 ) (
)
1𝑚𝑜𝑙 𝐾𝐼𝑂3
𝐸𝑥𝑐𝑒𝑠𝑠 𝐼3−
1𝑚𝑜𝑙 𝑆2 𝑂32−
1𝑚𝑜𝑙 𝐼3−
= 𝑀 𝑁𝑎2 𝑆2 𝑂3 (𝐿 𝑁𝑎2 𝑆2 𝑂3 ) (
)(
)
1𝑚𝑜𝑙 𝑁𝑎2 𝑆2 𝑂3 2𝑚𝑜𝑙 𝑆2 𝑂32−
𝑅𝑒𝑎𝑐𝑡𝑒𝑑 𝑚𝑜𝑙 𝐼3− = 𝑡𝑜𝑡𝑎𝑙 − 𝑒𝑥𝑐𝑒𝑠𝑠
𝑀𝑎𝑠𝑠 𝑉𝑖𝑡𝑎𝑚𝑖𝑛 𝐶
1𝑚𝑜𝑙 𝑉𝑖𝑡𝐶 𝑚𝑎𝑠𝑠 𝐴𝑠𝑐𝑜𝑟𝑏𝑖𝑐 𝑎𝑐𝑖𝑑
= 𝑚𝑜𝑙𝑒𝑠 𝑟𝑒𝑎𝑐𝑡𝑒𝑑 𝐼3− (
)(
)
1𝑚𝑜𝑙 𝐼3−
1𝑚𝑜𝑙 𝐴𝐴
% 𝑉𝑖𝑡 𝐶 = (
𝑚𝑎𝑠𝑠 𝑣𝑖𝑡 𝐶
) ∗ 100
𝑚𝑎𝑠𝑠 𝑆𝑎𝑚𝑝𝑙𝑒
𝑀 𝑁𝑎2 𝑆2 𝑂3 = 0.01 𝐾𝐼𝑂3 (0.050𝐿 𝐾𝐼𝑂3 )
3𝑚𝑜𝑙 𝐼3−
2𝑚𝑜𝑙 𝑆2 𝑂3
1
(
)(
)
− )(
1𝑚𝑜𝑙 𝐼𝑂3
1𝑚𝑜𝑙 𝐼3
0.0077𝐿 𝑁𝑎2 𝑆2 𝑂3
=0.065M
3𝑚𝑜𝑙𝐼3−
𝑚𝑜𝑙𝑒𝑠 𝐼3− = 0.01𝑀 𝐾𝐼𝑂3 (0.050𝐿 𝐾𝐼𝑂3 ) (
)
1𝑚𝑜𝑙 𝐾𝐼𝑂3
−
= 0.0015𝑚𝑜𝑙 𝐼3
𝐸𝑥𝑐𝑒𝑠𝑠 𝐼3− = 0.067𝑀 𝑁𝑎2 𝑆2 𝑂3 (0.0168𝐿 𝑁𝑎2 𝑆2 𝑂3 )
1𝑚𝑜𝑙 𝑆2 𝑂32−
1𝑚𝑜𝑙 𝐼3−
(
)(
)
1𝑚𝑜𝑙 𝑁𝑎2 𝑆2 𝑂3 2𝑚𝑜𝑙 𝑆2 𝑂32−
= 0.0005628𝑚𝑜𝑙 𝐼3−
𝑅𝑒𝑎𝑐𝑡𝑒𝑑 𝑚𝑜𝑙 𝐼3− = 0.0015𝑚𝑜𝑙 − 0.0005628𝑚𝑜𝑙
= 0.0009372𝑚𝑜𝑙
𝑀𝑎𝑠𝑠 𝑉𝑖𝑡𝑎𝑚𝑖𝑛 𝐶
1𝑚𝑜𝑙 𝑉𝑖𝑡𝐶 176.0𝑔 𝐴𝐴
= 0.0009372𝑚𝑜𝑙 𝐼3− (
)(
)
1𝑚𝑜𝑙 𝐼3−
1𝑚𝑜𝑙 𝐴𝐴
= 0.1649𝑔
0.1649𝑔 𝑣𝑖𝑡 𝐶
% 𝑉𝑖𝑡 𝐶 = (
) ∗ 100 = 119.2%
0.1383𝑔 𝑆𝑎𝑚𝑝𝑙𝑒
Conclusion:
The percent vitamin C obtained is way too high meaning there is great error. For the CVS tablet
there was 97.13% and Nature’s Bounty had 101.1%. Both of these values are much higher than the
claims on the nutritional label. The main problem with this lab occurred before the actual experiment
started. The lab preps are incompetent and did not know how to create the correct molarity solutions to
be used. The actual value had to be estimated because the value was unknown and off from its claimed
molarity. This caused problems in the calculations with the rest of the moles used because the values
are based off the molarity of this original solution. Also in the procedure half of a tablet was used and
this may not have been perfectly half of the tablet. Finally, in titrating the sample the tablet had to be
dissolved first. If it was not fully dissolved the equivalence point may have shown up earlier than
expected.