Project VIII
Polythionates (Oxoanions of Sulphur)
Abstract:
This experiment combined iodine with sodium thiosulfate pentahydrate to form product A,
a polythionate salt called sodium tetrathionate containing a S4O62- anion. A yield percent of 63%
was achieved for product A. The key experimental IR peaks highlighted were 1020-1040cm-1 and
1200-1260cm-1 for the S4O62- anion. There were possible contaminations in the ranges of 3300-
3600cm-1 and 1600-1700cm-1 likely due to ethanol not being completely removed from the product
A precipitate. Finally the nature of product A was investigated by heating up the sodium
tetrathionate with MnO2 and concentrated H2SO4 with a result of no fumes being visible. Thus
indicating no contamination of NaI was present.
Introduction:
The purpose of this lab was to prepare a sodium tetrathionate product by first determining
the amount of Na2S2O3 required to reduce a known amount of I2 from a titration. Secondly 105%
of the calculated iodine was then added to sodium thiosulfate and filtrated to produce the
precipitate of the sodium tetrathionate (Na₂S₄O₆). These were all completed through the reaction
listed below. Sodium Thiosulfate was mixed with iodine in 95% ethanol in part B as it is insoluble
in ethanol allowing the iodine the form a solution and readily react with the sodium thiosulfate.
Sulphur forms oxoanions which can be described as a sulphur atom(s), with the
coordination number of 3 or 4, attached to oxygen(s) with an overall negative 2 charge.
Polythionates are put in the formula of: SnO62- with n being able to reach a value from 1-20.[2]
These are stable only in salt and can only be formed from the reaction of sodium thiosulfate is
combined with As2O3 and sulphur dioxide or with oxidizing agents or by reacting the thionate with
an oxidizing agent, such as the one used in this lab, iodine.
1
Sulfanes are a species contain sulphur and hydrogen together in various quantities but only
containing these two atoms. A possible way to form sulfanes is from aqueous sodium polysulfide
solutions, from crude oils, and a reaction of liquid H2S with halogens.[3] Polythionates can be
viewed as a derivative of sulfanes as they contain the sulphur needed and the only thing left to do
is remove the oxygen and replace it with some hydrogen.
Efflorescence is a salt that grows inside porous materials and can damage the material. The
salt may already be present inside the material or be brought into the material after the facts through
an opening.[5] Once water is added the salt grows inside the permeable open areas and expands.
The following shows the reaction that occurred in this experiment:
[1] 2 Na2S2O3 + I2 Na2S4O6 + 2 NaI
The VSEPR and Lewis structure of Na2S4O6 is as followed:
Lewis Structure VSEPR Structure
Results:
Measurements:
Table 1: Initial Reagent Measurements and Observations for Part A
Chemical Weight: Observations:
KI 1.032 g White Powder Crystals
Iodine 0.113 g Grey Square Chunks
Sodium Thiosulfate
0.673 g White Powder
Pentahydrate
Table 2: Titration of sodium thiosulfatepentahydrate with KI and iodine.
Burette Reading Observations:
Initial 2.12 mL Red/Brown Color
Final 18.10 mL Yellow Color (Apple Juice)
2
Table 3: Titration of sodium thiosulfate pentahydrate with KI and I2 after the addition of a starch solution.
Burette Reading Observations:
Initial 18.10 mL Dark Violet/Grey Color
Final 19.10 mL Colorless
Total Amount Added 16.98 mL -
Table 4: Initial Reagent Measurements and Observations for Part B
Chemical Weight: Observations:
95% Ethanol 20 mL Clear Solution
Iodine 0.119 g Grey Square Chunks
Sodium Thiosulfate
0.228 g White Powder
Pentahydrate
Product A (Na₂S₄O₆) 0.089 g* White Powder
*
is the actual yield
Observations: (see original, page 10)
For part A of the experiment, as the KI and Iodine were first added together the solution
was a red/brown color. After the titration was performed the remaining solution was a
yellow/orange color much like apple juice. Once the starch solution was added the color turned to
a dark violet/grey. After the second titration the solution turned colorless.
For part B of the experiment, as 95% ethanol was added to the iodine the solution
turned a dark red/brown color. As sodium thiosulfate pentahydrate was slowly added the solution
turned to a brown/orange color with a yellow precipitate forming along the bottom of the beaker.
Once filtrated a white powdered precipitate was gathered and measured.
When product A was mixed with MnO2 and H2SO4 and heated the fumes had no color.
Calculations:
[1] 2 Na2S2O3 + I2 Na2S4O6 + 2 NaI
Table 5: Measurements of molecular weight, weight, and moles that are needed to determine the limiting
reagent and yields in reaction [1].
3
Chemical Molecular Weight Moles Theoretical Actual Limiting
Weight Yield Yield Reagent?
PART A 6.217x10-3 - - No
KI 166.0028g/mol 1.032 g mol
Iodine 0.113 g 8.904x10-4
126.9045g/mol - - No
mol
Sodium 0.673 g 2.7117x10-3
Thiosulfate 248.1841g/mol mol - - Yes
Pentahydrate
PART B 9.377x10-4
- - No
Iodine 126.9045g/mol 0.119 g mol
Sodium 0.228 g
9.187x10-4
Thiosulfate 248.1841g/mol - - Yes
mol
Pentahydrate
95% Ethanol 46.0684g/mol 20 mL - - - -
Product A 0.089 g 2.906x10-4 0.14068g 0.089 g
306.2665g/mol mol -
(Na₂S₄O₆)
Sample Calculations:
Total Amount Added = (Final Volume from 2nd Titration - Initial volume added from 1st
titration) = 19.10mL – 2.12mL
Moles of KI = (1.032g) / (37.037g/mol) = 6.217x10-3 mol
Iodine needed for Part B = (0.113g x1.05) = 0.11865 g
Amount of Sodium Thiosulfate Pentahydrate needed for Part B = 0.673g x (1/0.050 L) x
(0.01698 L) = 0.2285508g
Theoretical Value of Na2S4O6 = 4.5935x10-4 mol x 306.2665g/mol = 0.14068g
ICE Tables:
Table 6: An ICE table for reaction [1] Part A
2 Na2S2O3 + I2 Na2S4O6 + 2 NaI
I 2.7117x10-3 mol ~ 6.217x10-3 mol 0 mol 0 mol
C -2.7117x10-3 mol -1.35585x10-3 mol +4.8612x10-3 mol +2.7117x10-3 mol
E 0 mol ~ 4.8612x10-3 mol 4.8612x10-3 mol 2.7117x10-3 mol
Table 7: An ICE table for reaction [1] Part B
2 Na2S2O3 + I2 Na2S4O6 + 2 NaI
I 9.187x10-4 mol 9.377x10-4 mol 0 mol 0 mol
C -9.187x10-4 mol -4.5935x10-4 mol +4.5935x10-4 mol +9.187x10-4 mol
E 0 mol 4.7835x10-4 mol 4.5935x10-4 mol 9.187x10-4 mol
4
% Yield = (Actual Yield / Theoretical Yield) x 100
Product A % Yield = (0.089g / 0.14068g) x 100 = 63.26%
IR Spectra’s:
See attached (Page 8 and 9)
Key IR Stretches:
Product A: [1]
1020-1040cm-1 = S4O62- 3300-3600cm-1 = H+
1200-1260cm-1 = S4O62- 1600-1700cm-1 = O2-
Discussion:
The experiment obtained a 63% yield for sodium tetrathionate. This is likely due to loss
of product from not gathering all precipitate from the filter paper and from over rinsing with 95%
ethanol. The experimental IR spectra (#1) revealed that product A was Na2S4O6 with key peaks at
1020-1040cm-1 and 1200-1260cm-1 highlighting the tetrathionate species. The wide 3300-3600cm-
1
peak indicated the presence of some H+ indicating a possible contaminant. The experimental IR
(#1) matched most closely with the given IR of sodium tetrathionate dehydrate. The only
differences are between the two are a slight change in shape of the 3300-3600cm-1 peak and the
1200-1260cm-1 peak. IR (#1) differed from that of the given IR for sodium thiosulfate pentahydrate
in that the given IR had a much wider and stronger 3300-3600cm-1 peak which is consistent with
the presence of more H+ due to the species of pentahydrate. As well the peak around the range of
1600cm-1-1700cm-1 were much stronger in the given IR of sodium thiosulfate pentahydrate likely
indicating the presence of more O2- ions. The experimental IR of the product obtained has the
peaks pertaining to H+ and O2- in the range of 3300-3600cm-1 and 1600cm-1-1700cm-1,
respectively, likely due to the contamination of ethanol not being completely removed and dried
from the product.
5
Investigating product A for traces of the by-product of NaI came back positive. No violet
fumes were visible when the product was mixed with H2SO4 and MnO2 and then heated. If there
was sodium iodine present, MnO2 would form MnSO4 and I2. Consequently when heated the purple
colored Mn would turn into the violet fume that could have been observed.
The similarities between H2S2O6 and Na2S2O6 is that both can use oxidizing agents to gain
their anion form of S2O62-. They both contain a thionate anion and are both very stable in their
form. Both species can form a sulfane derivative as well.
The differences between H2S2O6 and Na2S2O6 is that H2S2O6 can form sulfanes when added
to water where as Na2S2O6 just decomposes. The H2S2O6 species is an acid and in aqueous form
while Na2S2O6 is a salt and in solid form.
Acid analogs of polythionates are not stable. The following decomposition occurs:
Na2S2O6 + 2 H2O 2 NaOH + 2 H2SO3
The VSEPR and Lewis structure of Na2S4O6 is as followed:
Lewis Structure VSEPR Structure
Conclusion:
The experiment was a success but needs to be reproduced with less errors to achieve higher
yields with less contaminants. The yield achieved was approximately 63% for product A, Na2S4O6.
The products achieved were verified through IR spectra’s identifying the S4O62- anion and
conducting an investigation on product A confirming that there were no contaminants of the by-
product sodium iodine present. The experimental IR spectra of the matched that of sodium
6
thiosulfate dehydrate very closely indicating the presence of some contaminants of H+ and O2-,
likely from ethanol, but were minimal.
References:
[1] Antikainen, P. J. Suomen Kemistilehti B 1958, 31, 223.
[2] Ferraris, G.; Ivaldi, G. X-OH and O-H.O bond lengths in protonated oxoanions. Acta
Crystallographica Section B Structural Science. DOI: 10.1107/s0108768184001671. Published
Online: Feb 1, 1984, 40 (1), 1–6.
[3] Muller, E.; Hyne, J. B. Methods of preparation of sulfanes. Canadian Journal of Chemistry.
DOI: 10.1139/v68-384. Published Online: July 15, 1968, 46 (14), 2341–2346.
[4] R. Roesler. Chem 331 Lab Manual, Project VIII Polythionates (oxoanions of sulphur)
[5] Zehnder, K.; Arnold, A. Crystal growth in salt efflorescence. Journal of Crystal Growth.
DOI: 10.1016/0022-0248(89)90234-0. Published Online: Sept 1989, 97 (2), 513–521.
