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SYNTHESIS OF A SOLID ACID 12-TUNGSTOSILICIC ACID,
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SYNTHESIS OF A SOLID ACID 12-TUNGSTOSILICIC ACID, H4SiW12O40*7H2O
Minh Phong Nguyen
Inorganic chemistry Lab
Dr. Vivian Fernand Abstract:
In this experiment, we need to made H4SuW12O40*7H2O as a solid by using the sodium
tungstate dihydrate and sodium silicate solution. Also adding concentrated hydrochloric acid
dropwise during the reaction. Furthermore, we also used and apply technique of separator funnel
in order to observe our product. At the result, we would collect the white crystalline solid after
heat it at 70 Celsius about two hours. We also determined the acidity of our product by using
titration. At the result, the clear liquid solution turns in to orange then reddish during titration.
The last thing that we also need to do is testing our product as a solid acid. We would observe two solutions A and B, which solution A will appear clear liquid and solution B will have
brownish color of bromine.
Tungstate is the inorganic compound with the formula Na2WO4. Tungstate is a white solid and
soluble in water. The octahedron MoO6 is a basic building block of these isopolyanions, these
unites can be connected by sharing corners, edges, but not faces. Tungsten exhibits very similar
in this regard.
Sodium silicate is the member of this series is sodium metasilicate Na2SiO3, as known as water
glass or liquid glass. The pure compositions are colorless or white, but commercial samples are
often greenish or blue owing to the presence of iron-containing impurities.
Tungstosilicic acid is the most commonly encountered heteropoly acid. Tungstosilicic acid is an
acid with yellow color, it is not usually used in our daily life but tungstosilicic have been used a
lot in chemical industry.
Prepare 14.996 g of sodium tungstate dehydrate, Na2WO4*H2O, together added with 30ml of
water and 1.16 g of sodium silicate solution. Stir and heat solution with warm temperature, it
could make the reaction reacted faster. Prepare 10 ml of concentrated hydrochloric acid, started
dropwise the solvent into the solution while the solution was still stirring about 30 minutes. After
30 minutes, cooled the solution by ice bath until the precipitates appeared inside the solution.
Applying suction filtration to collected all the impurities out of the solution. Transfer all the
liquid that have been collected into the separator funnel. Using 12 ml of diethyl ether, together
with our solution inside the separator funnel. Shake and allow the solution inside the separator
funnel separated. There would be three layer appeared inside the separator funnel. The product would be the layer at the bottom of the separator funnel. Withdraw the bottom layer and repeated
the same techniques above until the yellow product in the middle layer completely removed.
Returned the ether extracted together with product of 4 ml concentrated hydrochloric acid in 12
ml of water. At the result, run off the lower layer into an evaporating dish and allow the solvent
to evaporate. Dry the white crystalline product at 70 Celsius for the next lab.
12Na2WO4*2H2O (s) + Na2SiO3 (s)+ 26HCl (aq) H4SiW12O40*7H2O (s) + 5 H2O (l) +26
NaCl Apply the titration techniques for testing the acidic of the product. Using 40ml aliquots of the
solution with 0.1 M NaOH solution from the instructor. Chlorophenol red has been given by the
instructor. The solution has been tested two times in order to get exact number on the burette
when the color has been changed.
2.94ml*1L/1000ml*X mol/L NaOH= 0.6045 g KHP* 1mol/ 204.22 g * 25ml/250ml
X= 0.1007 mol/L= 0.1007M NaOH
Last experiments, testing of H4SiW12O40*7H2O as a solid acid. Prepare two test tube for this
test. For the solution A, added 6 drops of cyclohexanol to 3ml of cyclohexane, shake to dissolve
and added 0.2 g of H4SiW12O40 and shake for few minutes. For the solution B, same for A
without added H4SiW12O40. Then respectively add 10 drops of saturated Br2 water in 20mls of
distilled water to each solution. Solution A should decolor, and white precipitates would appear
at the bottom at the test tube. Solution B would appear the brownish color from Br2.
Result and discussion:
About 8.1548 g of white solid H4SiW12O40*7H2O has been collected, there is no way that
percentage yield would be calculate because there is no amount number hydration of the solid
product. Titration is very important in this experiment, it helped to determine the acidic of the product.
22.35ml of the standardized NaOH solution has been for the second titration. Because the first
titrations have been passed to equivalence point which it is very hard to tell when the color have
changed. There are 4 H+ available per mol of product based on these calculation:
22.35 ml*1 L/1000ml *0.1007 mol/L= 2.2506*10^-3 mol NaOH
4.0225g *1 mol/3004.3g* 40ml/100ml= 5.360*10^-4 H4SiW12O40*7H2O
2.2506*10^-3 mol/5.360^106-4 mol= 4.200 ration of H+ to H4SiW12O40*7H2O
The ration 4.2 is very closed to the prediction have been made which was 4.0 hydrogens released
per unit of product. However, there was still error occurs during this experiment based on
inaccurate measurement and also inaccuracy of the equivalence point base on the color of the
For the last testing experiment with bromine, solution A containing acid, so it has been decolored
and there are plenty of white precipitate at the bottom of the test tube. Solution B appeared
brownish color from the bromine reaction as expected. At a result, the acid was successful for
this for testing the product H4SiW12O40*7H2O as a solid acid.
This experiment has showed us how to make the product H4SiW12O40*7H2O by using
separator funnel in order to observed the product layer. We also learn and understand the acidic
of the product by using titration for testing the color change at the equivalence point of the
product, and also the bromine testing to determine to product was a solid acid.
1. A Keggin unit is the naturally occurring form of various oxo-metallic acids. It is a
complex chemical acids that contain a transition metal like molybdinum or vanadium, oxygens, acidic hydrogens, and a heteroatom like P or Si. These types of acids will selfassemble in aqueous solution to form Keggin structures. The Keggin structure consists of
a central heteroatom bonded to four oxygen atoms. This central chemical species is
surrounded by a cage-like structure consisting of twelve repeating units of the transition
metal bonded to six separate oxygen atoms.
2. First example would be H3PMo12O40.
Second example would be H3PMo12O40
Both of them can be used as acid catalysts very efficient.
1. Ncchemist. "What Is a Keggin Unit?" ENotes. N.p., 13 Feb. 2016. Web. 7 Oct. 2016.
http://www.enotes.com/homework-help/what-keggin-unit-621685 2. https://en.wikipedia.org/wiki/Heteropoly_acid
This question was answered on: Feb 21, 2020
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