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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.


Experiment section:


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.


Answer question:


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.


3. Reference:


1. Ncchemist. "What Is a Keggin Unit?" ENotes. N.p., 13 Feb. 2016. Web. 7 Oct. 2016. 2.




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